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(1)Complex Functions Examples c7 Applications of the Calculus of Residues Leif Mejlbro. Download free books at.

(2) Leif Mejlbro. Complex Functions Examples c-7 Applications of the Calculus of Residues. Download free eBooks at bookboon.com.

(3) Complex Functions Examples c-7 – Applications of the Calculus of Residues © 2008 Leif Mejlbro & Ventus Publishing ApS ISBN 978-87-7681-390-1. Download free eBooks at bookboon.com.

(4) Complex Funktions Examples c-7. Contents. Contents Introduction. 6 7 7 8 8 8. 1.5 1.6. Some practical formulæ in the applications of the calculation of residues Trigonometric integrals Improper integrals in general Improper integrals, where the integrand is a rational function Improper integrals, where the integrand is a rational function time a trigonometric function Cauchy’s principal value Sum of some series. 2.. Trigonometric integrals. 13. 3.. Improper integrals in general. 25. 4.. Improper integral, where the integrand is a rational function. 44. 5.. Improper integrals, where the integrand is a rational function times a trigonometric function. 72. 1. 1.1 1.2 1.3 1.4. 10 12. www.sylvania.com. We do not reinvent the wheel we reinvent light. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. 4 Download free eBooks at bookboon.com. Click on the ad to read more.

(5) Complex Funktions Examples c-7. Kapiteloverskrift NL. 6.. Improper integrals, where the integrand is a rational function times an exponential function. 98. 7.. Cauchy’s principal value. 114. 8.. Sum of special types of series. 130. 360° thinking. .. 360° thinking. .. 360° thinking. .. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. © Deloitte & Touche LLP and affiliated entities.. Discover the truth 5at www.deloitte.ca/careers Click on the ad to read more Download free eBooks at bookboon.com. © Deloitte & Touche LLP and affiliated entities.. D.

(6) Complex Funktions Examples c-7. Introduction. Introduction This is the seventh book containing examples from the Theory of Complex Functions. In this volume we shall apply the calculations or residues in computing special types of trigonometric integrals, some types of improper integrals, including the computation of Cauchy’s principal value of an integral, and the sum of some types of series. We shall of course assume some knowledge of the previous books and the corresponding theory. Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the ﬁrst edition. It is my hope that the reader will show some understanding of my situation. Leif Mejlbro 19th June 2008. 6 Download free eBooks at bookboon.com.

(7) Complex Funktions Examples c-7. 1. Some practical formulæ in the applications of the calculation of residues. Some practical formulæ in the applications of the calculation of residues. 1.1. Trigonometric integrals. We have the following theorem: Theorem 1.1 Given a function R(x, y) in two real variables in a domain of R 2 . If the formal extension, given by 2 z − 1 z2 + 1 , , R 2z 2iz is an analytic function in a domain Ω ⊆ C, which contains the unit circle |z| = 1, then . . . 2π. R(sin θ, cos θ) dθ = 0. R |z|=1. z1 − 1 z2 + 1 , 2iz 2z. . dz . iz. In most applications, R(sin θ, cos θ) is typically given as a “trigonometric rational function”, on which the theorem can be applied, unless the denominator of the integrand is zero somewhere in the interval [0, 2π].. We will turn your CV into an opportunity of a lifetime. Do you like cars? Would you like to be a part of a successful brand? We will appreciate and reward both your enthusiasm and talent. Send us your CV. You will be surprised where it can take you.. 7 Download free eBooks at bookboon.com. Send us your CV on www.employerforlife.com. Click on the ad to read more.

(8) Complex Funktions Examples c-7. 1.2. Some practical formulæ in the applications of the calculation of residues. Improper integrals in general. We shall now turn to the improper integrals over the real axis. The general result is the following extension of Cauchy’s residue theorem: Theorem 1.2 Given an analytic function f : Ω → C on an open domain Ω which, apart from a ﬁnite number of points z1 , . . . , zn , all satisfying Im zj > 0, j = 1, . . . , n, contains the closed upper half plane, i.e. Ω ∪ {z1 , . . . , zn } ⊃ {z ∈ C | Im z ≥ 0}. If there exist constants R > 0, c > 0 and a > 1, such that we have the estimate, |f (z)| ≤. c , |z|a. when both |z| ≥ R and Im z ≥ 0,. then the improper integral of f (x) along the X-axis is convergent, and the value is given by the following residuum formula, +∞ n f (x) dx = 2πi res (f ; zj ) = 2πi res (f ; zj ) . −∞ j=1 Im zj >0. 1.3. Improper integrals, where the integrand is a rational function. We have the following important special case, where f (z) is a rational function with no poles on the real axis. When this is the case, the theorem above is reduced to the following: P (z) without poles on the real axis. If the degree of Q(z) the denominator polynomial is at least 2 bigger than the degree of the numerator polynomial, then the improper integral of f (x) along the real axis exists, and its value is given by a residuum formula, +∞ f (x) dx = 2πi res (f ; zj ) . −∞ Im zj >0 Theorem 1.3 Given a rational function f (z) =. 1.4. Improper integrals, where the integrand is a rational function time a trigonometric function. If the integrand is a rational function time a trigonometric function, we even obtain a better result, because the exponent of the denominator in the estimate can be chosen smaller: Theorem 1.4 Assume that f : Ω → C is an analytic function on an open domain Ω, which, apart from a ﬁnite number of points z1 , . . . , zn , where all Im zj > 0, j = 1, . . . , n, contains all of the closed upper half plane, i.e. Ω ∪ {z1 , . . . , zn } ⊃ {z ∈ C | Im z ≥ 0}. If there exist constants R > 0, c > 0 and a > 0, such that we have the estimate |f (z)| ≤. c , |z|a. if both |z| ≥ R and Im z ≥ 0,. 8 Download free eBooks at bookboon.com.

(9) Complex Funktions Examples c-7. Some practical formulæ in the applications of the calculation of residues. then the improper integral of f (x) eimx along the X-axis exists for every m > 0, and its value is given by the following residuum formula, . +∞. −∞. n res f (z) eimz ; zj = 2πi res f eimz ; zj .. . f (x) eimx dx = 2πi. Im. j=1. zj >0. In the special case, where f (z) is a rational function, we of course get a simpler result: Theorem 1.5 Given f (z) =. P (z) imz · e , where P (z) and Q(z) are polynomials. Assume that Q(z). 1) the denominator Q(z) does not have zeros on the real axis, 2) the degree of the denominator is at least 1 bigger than the degree of the numerator, 3) the constant m is a real positive number. Then the corresponding improper integral along the real axis is convergent and its value is given by a residuum formula, +∞ P (z) imz P (x) imx · e ; zj . ·e dx = 2πi res Q(z) −∞ Q(x) Im zj >0 The ungraceful assumption m > 0 above can be repared by the following: Theorem 1.6 Assume that f (z) is analytic in C\{z1 , . . . , zn }, where none of the isolated singularities zj lies on the real axis. If there exist positive constants R, a, c > 0, such that |f (z)| <. c , |z|a. for |z| ≥ R,. then . +∞ −∞. f (x) eixy dx =.

(10) ⎧ ⎨ 2πi Im ⎩. −2πi. zj >0.

(11) Im. zj <0. res f (z) eizy ; zj. for y > 0,. res f (z) eizy ; zj. for y < 0.. In the ﬁnal theorem of this section we give some formulæ for improper integrals, containing either cos mx or sin mx as a factor of the integrand. We may of course derive them from the theorem above, but it would be more helpful, if they are stated separately:. 9 Download free eBooks at bookboon.com.

(12) Complex Funktions Examples c-7. Some practical formulæ in the applications of the calculation of residues. Theorem 1.7 Given a function f (z) which is analytic in an open domain Ω which – apart from a ﬁnite number of points z1 , . . . , zn , where Im zj > 0 – contains the closed upper half plane Im zj > 0. Assume that f (x) ∈ R is real, if x ∈ R is real, and that there exist positive constants R, a, c > 0, such that we have the estimate, c for Im z ≥ 0 and |z| ≥ R. |f (z)| ≤ a , |z| Then the improper integrals given by . +∞ −∞. +∞. ⎧ ⎨ f (x) cos(mx) dx = Re 2πi ⎩. and . ⎧ ⎨. +∞. f (x) sin(mx) dx = Im −∞. cos(mx) dx are convergent for every m > 0 with the values sin(mx). f (x). −∞. 2πi. ⎩. Im. zj >0. Im. ⎫ ⎬ res f (z) eimz ; zj , ⎭ . res f (z) eimz ; zj. zj >0. ⎫ ⎬ ⎭. ,. respectively.. 1.5. Cauchy’s principal value. If the integrand has a real singularity x0 ∈ R, it is still possible in some cases with the right interpretation of the integral as a principal value, i.e. x0 −ε +∞ +∞ f (x) dx, f (x) dx := lim + v.p. −∞. ε→0+. −∞. x0 +ε. to ﬁnd the value of this integral by some residuum formula. Here v.p. (= “valeur principal”) indicates that the integral is deﬁned in the sense given above where one removes a symmetric interval around the singular point, and then go to the limit. Using the deﬁnition above of the principal value of the integral we get Theorem 1.8 Let f : Ω → C be an analytic function on an open domain Ω, where Ω ⊇ {z ∈ C | Im z ≥ 0} \ {z1 , . . . , zn } . Assume that the singularities zj , which also lie on the real axis, all are simple poles. If there exist constants R > 0, c > 0 and a > 1, such that we have the estimate c |f (z)| ≤ a for Im z ≥ 0 and |z| ≥ R, |z| +∞ then Cauchy’s principal value v.p. −∞ f (x) dx exists, and its value is given by the following residuum formula, +∞ f (x) dx = 2πi res (f ; zj ) + πi res (f ; zj ) . v.p. −∞ Im zj >0 Im zj =0. 10 Download free eBooks at bookboon.com.

(13) Complex Funktions Examples c-7. Some practical formulæ in the applications of the calculation of residues. This formula is easily remembered if one think of the real path of integration as “splitting” the residuum into two equal halves, of which one half is attached to the upper half plane, and the other half is attached to the lower half plane. It is easy to extend the residuum formula for Cauchy’s principal value to the previous cases, in which we also include a trigonometric factor in the integrand.. I joined MITAS because I wanted real responsibili� I joined MITAS because I wanted real responsibili�. Real work International Internationa al opportunities �ree wo work or placements. �e Graduate Programme for Engineers and Geoscientists. Maersk.com/Mitas www.discovermitas.com. � for Engin. M. Month 16 I was a construction M supervisor ina cons I was the North Sea supe advising and the N he helping foremen advi ssolve problems Real work he helping International Internationa al opportunities �ree wo work or placements ssolve p. 11 Download free eBooks at bookboon.com. Click on the ad to read more.

(14) Complex Funktions Examples c-7. 1.6. Some practical formulæ in the applications of the calculation of residues. Sum of some series. Finally, we mention a theorem with some residuum formulæ, which can be used to determine the sum of special types of series, Theorem 1.9 Let f : Ω → C be an analytic function in a domain of the type Ω = C \ {z 1 , . . . , zn }, / Z. where every zj ∈ If there exist constants R, c > 0 and a > 1, such that |f (z)| ≤. c |z|a |. then the series +∞ n=−∞. for |z| ≥ R,.

(15) +∞. n=−∞. f (n) = −π. n . f (n) is convergent with the sum res (cot(πz) · f (z); zj ) .. j=1. Furthermore, the alternating series +∞ n=−∞. (−1)n f (n) = −π. n j=1. res.

(16) +∞. n=−∞ (−1). n. f (n) is also convergent. Its sum is given by. f (z) ; zj . sin(πz). 12 Download free eBooks at bookboon.com.

(17) Complex Funktions Examples c-7. 2. Trigonometric integrals. Trigonometric integrals. Example 2.1 Compute. 2π 0. e2 cos θ dθ.. Here, the auxiliary function is given R(ξ, η) = e2η , in which ξ does not enter. The function 2 2 z +1 z − 1 z2 + 1 = exp , R 2z z 2iz is analytic in C \ {0}, so 2π 1 1 2πi 1 dz 2 cos θ ;0 . exp z + res = e dθ = exp z + z z i z iz 0 |z|=1 We note that both z = 0 and z = ∞ are essential singularities, so we are forced to determine the Laurent series of the integrand in 0 < |z|. However, there is a shortcut here, because we shall only be interested in the coeﬃcient a−1 . We see from +∞ +∞ 1 1 m 1 1 1 1 1 1 = exp z · exp = z exp z + , z m=0 m! z z z n! z n z n=0. z = 0,. that a−1 is obtained by a Cauchy multiplication as the coeﬃcient, which corresponds to m = n, thus . 2π 0. e2 cos θ dθ = 2π. +∞ . 1 , (n!)2 n=0. which can be shown to be equal to J0 (2i), where J0 (z) is the zeroth Bessel function. Example 2.2 Compute. 2π 0. dθ . 2 + cos θ. θ This integral can of course be computed in the traditional real way (change to tan , where one of 2 course must be careful with the singularity at θ = π). We have in fact, √ 2π 2π π2 2π 4 3 π δθ dθ dt du 4 +∞ · =√ . = = 2·2 = = 2 t + sin2 t 1 θ θ 2 3 2 + cos θ 3 3 cos 3 2 0 0 0 0 1 + u2 3 cos2 + sin 3 2 2 If we instead apply the Complex Function Theory, then we have the following computation 2π dz dθ 1 −2i = = dz 2 2 + 4z + 1 iz 2 + cos θ z z + 1 0 |z|=1 |z|=1 2+ 2z √ 2π 1 1 √ =√ , = (−2i) · 2πi res ; −2 + 3 = 4π lim √ z 2 + 4z + 1 3 3 z→−2+ 3 z + 2 + √ √ where we have applied that z 2 + 4z + 1 has the roots −2 ± 3, of which only −2 + 3 lies inside |z| = 1.. 13 Download free eBooks at bookboon.com.

(18) Complex Funktions Examples c-7. Trigonometric integrals. Example 2.3 Prove that 2π π cos 2θ , dθ = (a) 18 5 − 3 cos θ 0. . 2π. (b) 0. π cos 3θ . dθ = 54 5 − 3 cos θ. (a) We shall use the substitution z = eiθ , where in particular, 1 1 1 2iθ z2 + 2 . e + e−2iθ = cos 2θ = 2 z 2 Then . 2π. 0. cos 2θ dθ 5 − 3 cos θ. 1 1 2 z + 2 z2 + 2 1 dz z z = = dz 1 iz 3 i |z|=1 −3z 2 + 10z − 3 |z|=1 z+ 5− z 2 1 1 z4 + 1 z4 + 1 dz dz = − = − 1 3i |z|=1 2 3i |z|=1 2 2 10 (z − 3) z+1 z z − z z− 3 3 2πi z4 + 1 1 z4 + 1 ; 0 + res res = . ; −3i z 2 z 2 − 10 z 2 z − 13 (z − 3) 3 3 z+1 1 2. We obtain by Rule I, res. z4 + 1 1 ; 1 2 z z − 3 (z − 3) 3. . 1 +1 41 82 34 = =− , = 1 1 12 3 · (−8) −3 32 3. and res. z4 + 1 ;0 = z 2 z 2 − 10 3 z+1 =. z4 + 1 z 2 − 10 3 z+1 4 z + 1 2z − 10 10 4z 3 3 . = − 2 10 2 10 3 z − 3 z+1 z2 − z + 1. d 1 lim 1! z→0 dz lim. z→0. . 3. Finally, by insertion, 2π π 2π −41 + 40 41 10 2πi cos 2θ . = =− − + dθ = 18 12 3 3 12 −3i 5 − 3 cos θ 0 (b) For the substitution z = eiθ , where we see that in particular, 1 1 1 3iθ −3iθ 3 z + 3 , e +e = cos 3θ = 2 z 2. 14 Download free eBooks at bookboon.com.

(19) Complex Funktions Examples c-7. Trigonometric integrals. we get 1 1 1 3 2π z + 3 1 z3 + 3 cos 3θ dz z 2 z 2 = − 3i dθ = dz 10 3 1 2 iz 5 − 3 cos θ 0 |z|=1 5 − 2 z + z |z|=1 z − 3 z + 1 2 1 2π z6 + 1 1 z6 + 1 . res ; = − · 3 ; 0 + res 3 z z 2 − 10 z 3 z − 13 (z − 3) 3 3 z+1 Here, ⎛. ⎞ 1 1 z3 + 3 + 33 3 365 1 730 ⎜ ⎟ 3 z , res ⎝ = ; ⎠= =− 1 36 9·8 z − 13 (z − 3) 3 −3 3. 15 Download free eBooks at bookboon.com. Click on the ad to read more.

(20) Complex Funktions Examples c-7. Trigonometric integrals. and ⎞ ⎞ ⎛ 1 1 3 ⎟ ⎟ ⎜ ⎜ z3 ; 0⎠ = res ⎝ 2 10 ; 0⎠ res ⎝ 2 10 z z − 3 z+1 z − 3 z+1 ⎛. =. =. z3 +. 2z − 10 1 1 1 d2 d 3 lim = lim − 2 2 z→0 dz 2! z→0 dz 2 z 2 − 10 z 2 − 10 3 z+1 3 z+1 2 2 2z − 10 91 2 100 1 3 , = lim − 2 + 2 = −1 + 10 10 2 2 9 9 2 z→0 z − z+1 z − z+1 3. 3. hence by insertion, 2π π π 2π 364 − 364 365 91 2π cos 3θ . = = · = + − dθ = − 54 3 · 18 36 3 9 36 3 5 − 3 cos θ 0. Example 2.4 Prove that 2π 2π dθ = , 2 1 − a2 1 + a − 2a cos θ 0. for 0 < a < 1.. Find also the value, when a > 1. We get by the substitution z = eiθ that 1 1 dz , z+ and cos θ = dθ = z 2 iz thus . 2π. 0. dθ 2 1 + a − 2a cos θ. . 1 . = |z|=1. =. i a. 1 + a2 − z +. |z|=1. 2π 0. 1+. a2. dθ − 2a cos θ. =. |z|=1. az 2. dz − (1 + a2 ) z + a. 1 . Of these, only z = a lies inside the circle |z| = 1, hence a ⎛ ⎞. ⎜ i · 2πi · res ⎜ ⎝ a. = −. a. 1 dz =− i iz. dz . 1 2 z+1 z − a+ a. The integrand has the poles z = a and z = . 1 z. . 2π lim a z→1. . 1. z2 − a +. 1 1 z− a. =−. 1 a. 2π · a. z+1 1. 1 a− a. ⎟ ; a⎟ ⎠. =−. 2π 2π = . 1 − a2 −1. a2. 16 Download free eBooks at bookboon.com.

(21) Complex Funktions Examples c-7. Trigonometric integrals. 1 < 1, and it follows from the above that a 2π 1 2π 2π dθ dθ 1 = 2 . = 2· = 2 2 1 a 0 a 1 + a2 − 2a cos θ −1 a 1 1 1 − 1+ − 2 · cos θ a2 a a. If a > 1, then 0 < . 2π. 0. Remark 2.1 We note that the case a < 0 gives the same values, only dependent on if |a| < 1 or |a| > 1. Finally, the case a = 0 is trivial. ♦ Summing up, 2π 1+. 0. a2. 2π dθ = , |1 − a2 | − 2a cos θ. for a ∈ R \ {−1, 1}.. The integral is divergent, if a = ±1. Example 2.5 Prove that if a > 1, then . 2<pi 0. 2π dt =√ . a + sin t a2 − 1. It follows from 2π R(sin θ, cos θ) dθ = 0. that 0. R |z|=1. 2π. dt = a + sin t. |z|=1. z2 − 1 z2 + 1 , 2iz 2z. dz 1 =2 z 2 − 1 iz a+ 2iz. |z|=1. . dz , iz. 1 dz. z 2 + 2i a z − 1. 1 has the two simple poles z 2 + 2i a z − 1 z = −i a ± −a2 − 1. √ Of these only z = i a2 − 1 − 1 lies inside the unit circle. Hence 2π ! 1 dt 2 = 2 · 2πi · res ;i a −1−1 z 2 + 2i a z − 1 a + sin t 0 1 1 2π √ = 2 · 2πi · √ =√ . = 2 · 2πi √lim 2i a2 − 1 a2 − 1 z→i{ a2 −1−a} z + i a2 − 1 + i a. The function. 17 Download free eBooks at bookboon.com.

(22) Complex Funktions Examples c-7. Trigonometric integrals. Example 2.6 Compute 2π cos(2 cos θ) dθ, 0. expressed as a sum.

(23) +∞. n=0. an .. Applying the substitution z = eiθ we get 2π 1 1 2πi 1 dz ;0 cos z + res = cos(2 cos θ) dθ = cos z + z z i z iz 0 |z|=1 1 1 ;0 . cos z + = 2π · res z z It follows from 1 1 1 1 1 1 + exp −i z + exp i z + = · cos z + z z z 2 z z i i 1 1 + exp(−i z) · exp − exp(i z) · exp · = z z z 2 +∞ +∞ +∞ +∞ 1 1 1 m m 1 in 1 1 1 · = i z · (−i)m z m · (−i)n · n , + z 2 m=0 m! z n! z n m=0 m! n! n=0 n=0 that the coeﬃcient a−1 in the Laurent series expansion for 1 1 cos z + z z is determined by m = n, i.e. a−1 =. +∞ +∞ +∞ 1 (−i)n (−i)n 1 in in (−1)n · + · = 2 , 2 n=0 n! n! 2 n=0 n! n! n=0 (n!). which can be shown to be equal to J0 (2), where J0 (z) is the zeroth Bessel function. Hence we conclude that at 2π +∞ 1 1 (−1)n ; 0 = 2πa−1 = 2π cos z + cos(2 cos θ) dθ = 2π res = 2π J0 (2). z z (n!)2 0 n=0. 18 Download free eBooks at bookboon.com.

(24) Complex Funktions Examples c-7. Trigonometric integrals. Example 2.7 (a) Determine the Taylor series from z = 0 of 1 , 1 2 z+1 z − a+ a. where 0 < a < 1,.

(25) +∞ in the form p=0 ap z p . Find the radius of convergence r of the series. (b) Find the Laurent series of f (z) =. z n + z −n , 1 z+1 zn − a + a. n ∈ N,. 0 < a < 1,. in the domain given by 0 < |z| < r, by using the result of (a), and then ﬁnd the residuum of f at the point z = 0. (c) Compute . 2π 0. cos(n v) dv, 1 + a2 − 2a cos v. n ∈ N,. 0 < a < 1,. by transforming the integral into a line integral in the complex plane.. (a) First note that we have the factor expansion 1 1 . z + 1 = (z − a) z − z2 − a + a a 1 , it follows by a decomposition and an application of the geometric series, If |z| < a < a . 1. z2 − a +. 1 a. . 1 1 1 · + · 1 1 1 1 z−a −a z− a− (z − a) z − a a a a 1 1 1 1 −1 1 · · + · · = 1 1 a 1− z 1 1 − az − a a− − a a a a +∞ +∞ +∞ +∞ 1 1 " z #p a2 p p 1 p p+2 p = − 2 − a z = z − a z a − 1 p=0 a 1 − a2 p=0 ap 1 − a2 p=0 p=0 =. z+1. =. 1 . =. 1. +∞ 1 1 − a2p+2 p · z . ap 1 − a2 p=0. The radius of convergence is of course r = a, which e.g. follows from the fact that z = a is the pole, which is closest to 0. We may also easily obtain this result by the criterion of roots.. 19 Download free eBooks at bookboon.com.

(26) Complex Funktions Examples c-7. Trigonometric integrals. (b) If 0 < |z| < a and n ∈ N, then it follows from (a) that f (z). +∞ 1 1 − a2p+2 p z n + z −n = z n + z −n · z 1 ap 1 − a2 p=0 z+1 z2 − a + a +∞ +∞ 1 1 − a2p+2 1 1 − a2p+2 p−n p+n = · z + · z , ap ap 1 − a2 1 − a2 p=0 p=0. =. which we may reduce to the Laurent series f (z) =. n+1 p=−n. 1 ap+n. 2n +∞ a + 1 1 − a2p+2 p 1 − a2p+2+2n p 1 · z + · z , ap+n 1 − a2 1 − a2 p=n. although this result is not much nicer. We know that the residuum is given by p = −1, so res(f ; 0) = a−1 =. no.1. Sw. ed. en. nine years in a row. 1 an−1. ·. 1 − a2n . 1 − a2. STUDY AT A TOP RANKED INTERNATIONAL BUSINESS SCHOOL Reach your full potential at the Stockholm School of Economics, in one of the most innovative cities in the world. The School is ranked by the Financial Times as the number one business school in the Nordic and Baltic countries.. Stockholm. Visit us at www.hhs.se. 20 Download free eBooks at bookboon.com. Click on the ad to read more.

(27) Complex Funktions Examples c-7. Trigonometric integrals. (c) If we put z = eiv , then cos nv =. 1 inv e + e−inv 2. and. dv =. dz . iz. Then we get by insertion, reduction and an application of the residuum theorem (with the two poles z = 0 and z = a inside the unit circle |z| = 1), . 1 dz cos(nv) z n + z −n dv = · 2 − 2a cos v 2 − a (z + z −1 ) 2 iz 1 + a 1 + a 0 |z|=1 1 1 z n + z −1 1 · 2πi {res(f ; 0) + res(f ; a)} dz = − = · 1 2ia 2i |z|=1 −a 2 z+1 z − a+ a ⎧ ⎫ ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎨ 2n n −n ⎬ 2n n −n ⎬ π 1 1−a 1 a +a z +z 1 1−a = − · + lim · + · = −π n−1 n 1 ⎪ 1 ⎪ z→a ⎪ a⎪ a 1 − a2 1 − a2 ⎩a ⎭ ⎭ ⎩a z− a− a a π 1 − a2n − a2n − 1 1 a2n + 1 an 1 1 − a2n =− n · · − · = 2π · . = −π n 2 n 2 2 a a 1−a 1−a a 1−a 1 − a2 2π. 21 Download free eBooks at bookboon.com.

(28) Complex Funktions Examples c-7. Trigonometric integrals. Example 2.8 Given the function f (z) =. 1 − ei2z . z2. (1) Prove that f has a simple pole at z = 0, so we have for z = 0, f (z) =. b1 + g(z), z. where g is an entire function, i.e. analytic in C. Find the residuum b1 . Consider for r > 0 the half circle γr , given by the parametric description γr (t) = r eit ,. 0 ≤ t ≤ π.. (2) Prove that f (z) dz → b1 π i γr. (3) Prove that f (z) dz → 0 γr. for r → 0.. for r → +∞.. 3. 2. 1. –3. –2. –1. 0. 1. 2. 3. –1. Figure 1: The curve Γε,R . Let Γε,R = I + II + III + IV denote the simple, closed curve on the ﬁgure, where II = γ R and IV = −γε . (4) Compute f (z) dz, Γε,R. and prove the formula 2 +∞ sin x π dx = . x 2 0. 22 Download free eBooks at bookboon.com.

(29) Complex Funktions Examples c-7. Trigonometric integrals. 1) It follows from the Laurent series expansion +∞ n 1 1 i n n i2z 1−e = 2 1− ·2 z f (z) = z z2 n! n=0 +∞ 1 in n+2 n+2 ·2 1 − 1 − 2iz + = z z2 (n + 2)! n=0 = −. +∞ 2i in +4 · 2n · z n , z (n + 2)! n=0. |z| > 0,. that f has a simple pole at 0 and that g(z) = 4. +∞ . in · 2n z n , (n + 2)! n=0. z ∈ C,. is an entire function, and that res(f ; 0) = −2i = b1 . 2) When we use the parametric description of the half circle, we get π 2i 2i − + g(z) dz = − it · i r eit dt + f (z) dz = g(z) dz z re 0 γr γr γr π dt + g(z) dz = 2π + g(z) dz, = 2 0. γr. γr. where 2π = −2i · iπ = b1 · iπ. In particular, g(z) is continuous, so |g(z)| ≤ c for |z| ≤ 1. Therefore, if 0 < r < 1, then we get the estimate % % % % % ≤ c·πr → 0 % g(z) dz for r → 0 + . % % γr. It follows that f (z) dz = b1 πi = 2π. lim r→0+. γr. 3) Assume that r > 0 is large. It follows from ei2z = exp(2ir · (cost + i sin t)) = exp(−2r sin t) exp(2ir cos t), and r > 0 and 0 < t < π that −2r sin t < 0, hence % i2z % %e % ≤ for z ∈ Γr . This implies the estimate % % % π % % 2π % % % 1+1 %= %≤% % →0 · r dt f (z) dz % % % % 2 r r 0 γr. for r → +∞.. 23 Download free eBooks at bookboon.com.

(30) Complex Funktions Examples c-7. Trigonometric integrals. 4) Now, f (z) is analytic everywhere inside Γε,R , so it follows from Cauchy’s integral theorem that . . 0= Γε,R. f (z) dz = −. . R. f (z) dz + ε. γε. 1 − e2ix dx + x2. . . −ε. f (z) dz + γR. −R. 1 − e2ix dx. x2. Since cosine is an even function and sine is an odd function, it follows by the symmetry that . −ε 1 − e2ix 1 − e2ix dx + dx x2 x2 ε −R R −ε −ε R 1 − cos 2x sin 2x 1 − cos 2x sin 2x dx − dx + dx − dx = 2 2 2 x x x x2 ε ε −R −R 2 R R sin x 1 − cos 2x dx = 2 dx. = 2 x x2 ε ε R. Then by insertion and taking the limits ε → 0+ and R → +∞, 0. −. =. . . f (z) dz + γε. γR. . → −π + 0 + 2. +∞ 0. R. f (z) dz + 2 . sin x x. ε. 2. . sin x x. 2 dx. dx.. This limit is of course also equal to 0, so by a rearrangement, 0. +∞. . sin x x. 2 dx =. π . 2. 24 Download free eBooks at bookboon.com. Click on the ad to read more.

(31) Complex Funktions Examples c-7. 3. Improper integrals in general. Improper integrals in general. Example 3.1 Compute the improper integrals +∞ x 1 1 exp cos dx, 2 x2 + 1 x2 + 1 −∞ x + 1. . +∞. and −∞. 1 exp 2 x +1. . x 2 x +1. . sin. 1 2 x +1. dx.. When we split into the real and the imaginary part, we get x 1 1 = 2 +i 2 , x +1 x +1 x−i so it is quite natural to consider the analytic function 1 1 , for z ∈ C \ {−i, i}. exp f (z) = 2 z−i z +1 Since. 1 → 0 for z → ∞, there clearly exists an R > 1, such that we have the estimate z−i. |f (z)| ≤. 2 |z|2. for |z| ≥ R.. Then the assumptions of an application of the residuum formula are satisﬁes, so we conclude by the linear transform w = z − i that +∞ x+i 1 1 1 ;i exp dx = 2πi · res exp 2 x2 + 1 z2 + 1 z−i −∞ x + 1 1 1 ;0 . = 2πi · res exp w2 + 2iw w Now, w0 = 0 is an essential singularity of the function 1 1 , exp w w2 + 2iw so in order to ﬁnd the residuum we shall expand into a Laurent series from w0 = 0, then perform 1 a Cauchy multiplication and ﬁnally determine a−1 by collecting all the coeﬃcients of . When w 0 < |w| < 2, we get m +∞ +∞ 1 1 −w 1 1 1 1 1 1 1 1 1 · = · exp · · · exp = · . w 2i w m=0 2i w 2i w 1 + w w 2i + w n! wn n=0 2i 1 Since we have separated the factor , it follows that a−1 is equal to the constant term in the product w of the two series, i.e. m = n. Thus a−1. n +∞ 1 1 1 i 1 1 i 1 , cos + i sin = exp = = 2 2 2i 2 2i n=0 n! 2 2i. 25 Download free eBooks at bookboon.com.

(32) Complex Funktions Examples c-7. Improper integrals in general. and we conclude that +∞ 1 1 x 1 cos + i sin dx exp 2 x2 + 1 x2 + 1 x2 + 1 −∞ 1 + x +∞ 1 1 1 x+i + i sin . = exp dx = 2πia = π cos −1 2 2 2 x2 + 1 −∞ x + 1 When we separate the real and the imaginary parts, we get +∞ x 1 1 1 exp cos dx = π · cos , 2+1 2+1 2+1 x x 2 x −∞ and . +∞ −∞. 1 exp 2 x +1. . x 2 x +1. . sin. 1 2 x +1. . 1 dx = π · sin . 2. Alternatively we may use that the function 1 1 1 1 exp = g(w) = 2 exp w w(w + 2i) w + 2iw w is analytic in C \ {0, −2i}, so if we include the residuum at ∞, then the sum of the residues is zero. Hence +∞ x+i 1 exp dx = 2πi · res(g(w) ; 0) = −2πi{res(g(w) ; −2i) + res(g(w) ; ∞)}. 2 x2 + 1 −∞ x + 1 Here, −2i is a simple pole, so by Rule Ia, i 1 1 1 . exp = · exp − res(g(w) ; −2i) = − lim z→−2i w 2 2i w 1 Furthermore, limw→∞ exp = exp 0 = 1, so w = ∞ is a zero of order 2 of w 1 1 1 , g(w) = 2 · · exp 2i w w 1+ w and it follows from Rule IV that res(g(w) ; ∞) = 0. Then by insertion, +∞ 1 1 i x+i 2πi 1 , + i sin = π cos · exp · exp dx = 2πi · res(g(w) ; , ∞) = 2 2 2 2 x2 + 1 2i −∞ x + 1 and the results follow as above by separating the real and the imaginary parts.. 26 Download free eBooks at bookboon.com.

(33) Complex Funktions Examples c-7. Improper integrals in general. Example 3.2 Assume that x > 0. Find the limit value A 1 1 dt. − lim A→+∞ −A t + ix t − ix Here we get without using Complex Function Theory, A A 1 t − ix − t − ix 1 dt = lim lim − dt A→+∞ −A A→+∞ −A t 2 + x2 t + ix t − ix & 'A A t dt Arctan −2ix = −2i lim = −2iπ. = lim 2 2 A→+∞ A→+∞ x −A −A t + x. 27 Download free eBooks at bookboon.com. Click on the ad to read more.

(34) Complex Funktions Examples c-7. Improper integrals in general. Example 3.3 Let a ∈ R be a constant. Prove that the integral +∞ 2 e−(x+ia) dx I(a) = −∞. is independent of a. Hint: We may assume that a ∈ R+ . Denote by C the rectangle of the corners −b, b, b + ia and −b + ia. Show that exp −z 2 dz = 0. C. Then prove that % a % % % −(b+iy)2 % % ≤ e−b2 e dy % % 0. a. 2. ey dy.. 0. By letting b → +∞, prove that I(a) = I(0).. 1.5. 1. 0.5. –2. 0. –1. 1. 2. –0.5. Figure 2: Example of one of the curves C. Here, a = 1 and b = 2. Clearly, we may assume that a > 0, because we otherwise might consider an analogous curve in the lower half plane. Now exp −z 2 is analytic in C, so exp −z 2 dz = 0. C. We estimate the line integrals along the vertical lines by % a % % a % a % % % % 2 −(±b+iy)2 −b2 ∓2ibu+y 2 −b2 % % % % e dy % = % e dy % ≤ e ey dy → 0 % 0. Since . 0. −z 2. e C. . b. dz =. −x2. e −b. for b → +∞.. 0. . −(b+iy)2. dx + i 0. dy −. b. −(x+ia)2. e −b. dx − i. a 0. 28 Download free eBooks at bookboon.com. 2. e−(−b+iy) dy = 0,.

(35) Complex Funktions Examples c-7. and since. +∞. 2. e−x dx is an improper convergent integral, it follows by taking the limit b → +∞ that. −∞. . Improper integrals in general. +∞. 2. e−(x+ia) dx =. I(a) =. . −∞. +∞. 2. e−x dx = I(0).. −∞. Note that we also have I(−a) = I(a) = I(0) = I(0).. Example 3.4 Compute +∞ 2 I(0) = e−x dx. +∞. Hint: Use that {I(0)}2 =. +∞. 2. e−x dx. . −∞. +∞. 2. e−y dy =. . 2. e−(x. +y 2 ). dx dy.. R2. −∞. Then use polar coordinates. 2 2 Since e−(x +y ) > 0 for every (x, y) ∈ R2 , and since the function is continuous, all the transforms below are legal, if only the improper plane integral exists. (The only thing which may go wrong is that the value could be +∞). Hence, +∞ +∞ 2 2 2 2 e−x dx · e−y dy = e−(x +y ) dx dy I(0)2 =. −∞ 2π +∞. . −∞. −r 2. e. = 0. 0. &. R2. 1 −r2 e r dr dθ = 2π · 2. '+∞ = π, 0. and thus I(0) =. √. π.. ♦. 2. Example 3.5 Integrate the function eiz by using Cauchy’s theorem along a triangle of corners 0, a and a(1 + i), where a > 0. Prove that the integral along the path from a to a(1 + i) tends to 0 for a → +∞, and then prove that +∞ +∞ +∞ 2 1 + i +∞ −x2 eix dx = cos x2 dx + i sin x2 dx = √ e dx. 2 0 0 0 0 The integrand is analytic, so it follows from Cauchy’s theorem that a a a 2 2 ix2 i(a+it)2 e dx + e i dt − ei(1+i) t (1 + i) dt. 0= 0. 0. 0. 29 Download free eBooks at bookboon.com.

(36) Complex Funktions Examples c-7. Improper integrals in general. 1. 0.8. 0.6. 0.4. 0.2. 0. –0.5. 0.5. 1. 1.5. Figure 3: The curve C when a = 1.. We ﬁrst consider a 2 I2 = ei(a+it) i dt. 0. Here we get the estimate % % |I2 | = %%. a. 2. ei(a. −t2 ) −2at. e. 0. % % i dt%% ≤. a. 2. e−2at dt =. 0. 1 1 − e−2a . < 2a 2a. It follows immediately that . a(1+i) 0. 2. eiz dz → 0. for a → +∞.. √ Then we introduce the substitution u = t 2 into the latter integral, a 2 2 I3 = ei(1+i) t (1 + i) dt. 0. We get here I3. a. i(1+i)2 t2. =. e. →. 1+i √ 2. 0. 0. +∞. . a. (1 + i) dt = (1 + i). −2t2. e 0. 2. e−x dx. 1+i dt = √ 2. √ a 2. . 2. e−u du. 0. for a → +∞.. We ﬁnally conclude that the ﬁrst integral I1 is also convergent for a → +∞, and +∞ +∞ +∞ 2 1 + i +∞ −x2 eix dx = cos x2 dx + i sin x2 dx = √ e dx. 2 0 0 0 0. 30 Download free eBooks at bookboon.com.

(37) Complex Funktions Examples c-7. Improper integrals in general. Remark 3.1 If we use the result of Example 3.4, it follows by the symmetry that √ +∞ 1+i π ix2 , e dx = √ · 2 2 0 hence +∞. . 2. cos x 0. . . +∞. dx =. . 2. sin x 0. . 1 dx = 2. (. π . 2. ♦. Excellent Economics and Business programmes at:. “The perfect start of a successful, international career.” CLICK HERE. www.rug.nl/feb/education. 31 Download free eBooks at bookboon.com. to discover why both socially and academically the University of Groningen is one of the best places for a student to be. Click on the ad to read more.

(38) Complex Funktions Examples c-7. Improper integrals in general. Example 3.6 1) Find the domain of analyticity of the function f (z) =. Log z . z2 − 1. Explain why f has a removable singularity at z = 1. 2) Let Cr,R denote the simple, closed curve on the ﬁgure, where 0 < r < R < +∞.. 1.2. 1. 0.8. 0.6. 0.4. 0.2. –0.2. 0.2. 0.4. 0.6. 0.8. 1. 1.2. –0.2. Compute the line integral f (z) dz. (1) Cr,R. 3) Show that the improper integral . +∞ 0. ln x dx −1. x2. is convergent, and then ﬁnd its value, e.g. by letting r → 0+ and R → +∞ in (1).. 1) Clearly, f is deﬁned and analytic, when z ∈ C \ (R− ∪ {0, 1}) , and the singularity at z = 1 is at most a simple pole, +∞. f (z) =. Log z res(f ; 1) + = an (z − 1)n , 2 z −1 z−1 n=0. 0 < |z − 1| < 1.. But since Log 1 Log z = 0, = z→1 z + 1 2. res(f (f ; 1) = lim. 32 Download free eBooks at bookboon.com.

(39) Complex Funktions Examples c-7. Improper integrals in general. it follows that the Laurent series of f from z = 1 is a power series, so the singularity at z = 1 is removable. Alternatively, both the numerator and the denominator are 0 for z = 1, so we get by l’Hospital’s rule that 1 1 Log z z = , = lim lim f (z) = lim 2 z→1 2z z→1 z→1 z − 1 2 so the singularity is removable, and we may consider ⎧ Log z ⎪ ⎪ x ∈ C \ (R− ∪ {0, 1}) , ⎪ ⎨ z2 − 1 , f (z) = ⎪ ⎪ 1 ⎪ ⎩ , z = 1, 2 as an analytic function in C \ (R− ∪ {0}). Alternatively it follows by a series expansion of Log z = Log(1 + (z − 1)). for 0 < |z − 1| < 1,. that f (z) =. +∞ +∞ 1 (−1)n+1 1 1 (−1)n Log z n · = · (z − 1) (z − 1)n . = z + 1 z − 1 n=1 z + 1 n=0 n + 1 z2 − 1 n. 1 is continuous in all of the disc |z − 1| < 1, so we conclude again that z = 1 is a z+1 removable singularity and that f can be analytically extended to z = 1 by putting 1 1 1 +0 = . f (1) := 2 1+1 1+0. Here. 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2. –0.2. 0.2. 0.4. 0.6. 0.8. 1. 1.2. 1.4. 1.6. –0.2. Figure 4: The path of integration Cr,R with the removable singularity at z = 1.. 33 Download free eBooks at bookboon.com.

(40) Complex Funktions Examples c-7. Improper integrals in general. 2) Since we may consider f as an analytic function in C \ (R− ∪ {0}), we conclude from Cauchy’s integral theorem that f (z) dz = 0. (2) Cr,R. 3) When we restrict the analytic function to R+ , we get a continuous function. ln x , supplied by −1. x2. 1 at x = 1. Since we only have ln x = 0 for x = 1, we see that x = 1 is the only possiple 2 ln x 1 zero. However, the value is here > 0, so we conclude by the continuity that 2 is positive x −1 2 (and continuous) for x ∈ R+ . Then we have the splitting the value. . +∞. 0. ln x dx = x2 − 1. . 1 2. 0. ln x dx + x2 − 1. . 2 1 2. ln x dx + x2 − 1. . +∞. 2. ln x dx. x2 − 1. The estimate 12 1 1 4 2 ln x % % dx ≤ | ln x| dx = (− ln x) dx 2 %1 % 3 0 0 x −1 % − 1% 0 %4 % 1 2 4 4 1 2 [−x ln x + x]0+ = · (1 + ln 2) = (1 + ln 2) < +∞, 3 3 3 2. 0 <. =. 1 2. implies that the ﬁrst integral exists. ln x It was mentioned above that we could consider 2 as a continuous function in the closed x −1 )1 * bounded interval 2 , 2 , from which we conclude that the second integral also is convergent. Finally, it follows from the magnitudes of the functions, when x → +∞ that there exists a constant C > 0, such that +∞ +∞ √ ln x 1 dx < C 2 < +∞, 0< 3 dx = C 2 x −1 x2 2 2 and we conclude that the last integral also is convergent. Summing up we have proved that the improper integral. +∞ ln x dx is convergent. 0 x2 − 1. When we expand (2), then . R. 0 = r. = r. R. π2 R Log R eiθ Log(it) iθ · R i e dθ − · i dt − 2 R2 e2iθ − 1 r (it) − 1 0 0 R ln t + i π π2 π2 ln x ln R + iθ 2 dt + i dx + i · R eiθ dθ + i 2 2 e2iθ − 1 1 + t x2 − 1 R r 0 0 ln x dx + 2 x −1. . π 2. 34 Download free eBooks at bookboon.com. Log r eiθ r i eiθ dθ r2 e2iθ − 1 ln r + iθ · r eiθ dθ, 1 − r 2 e2iθ.

(41) Complex Funktions Examples c-7. Improper integrals in general. hence by a rearrangement, . R ln t ln x dx + i dt 2−1 1 + t2 x r r π2 π2 π R dt ln R + iθ ln r + iθ iθ · R e dθ − i −i · r eiθ dθ. = 2 e2iθ − 1 2 r 1 + t2 R 1 − r 2 e2iθ 0 0 R. By taking the limits r → 0+ and R → +∞ on each of the terms on the right hand side we get π r→0+ R→+∞ 2 lim. . R. lim. r. π dt = 2 2 1+t. . +∞ 0. π2 π π dt = · , = 2 2 1 + t2 4. and % % π π2 % % 2 ln R + π2 ln R + iθ % % iθ · R e · R dθ dθ ≤ % %i % % 0 R2 e2iθ − 1 R2 − 1 0 π R ln R + π2 · = →0 2 R2 − 1. for R → +∞,. In the past four years we have drilled. 89,000 km That’s more than twice around the world.. Who are we?. We are the world’s largest oilfield services company1. Working globally—often in remote and challenging locations— we invent, design, engineer, and apply technology to help our customers find and produce oil and gas safely.. Who are we looking for?. Every year, we need thousands of graduates to begin dynamic careers in the following domains: n Engineering, Research and Operations n Geoscience and Petrotechnical n Commercial and Business. What will you be?. careers.slb.com Based on Fortune 500 ranking 2011. Copyright © 2015 Schlumberger. All rights reserved.. 1. 35 Download free eBooks at bookboon.com. Click on the ad to read more.

(42) Complex Funktions Examples c-7. Improper integrals in general. and % π % % % π r | ln r| + π 2 ln r + iθ % % 2iθ 2 e dθ% ≤ · →0 %i % 0 % 1 − r2 1 − r2 2. for r → 0+,. respectively. Hence, by summing up, +∞ +∞ π2 ln x ln t dx + i dt = . 2 2 x −1 t +1 4 0 0 Finally, by separating the real and the imaginary parts, +∞ +∞ x2 ln x ln x dx = og dx = 0. x2 − 1 x2 + 1 4 0 0. 36 Download free eBooks at bookboon.com.

(43) Complex Funktions Examples c-7. Improper integrals in general. Example 3.7 Given the function f (z) =. ez . 1 + e4z. (1) Find all the isolated singularities of f in C. Determine the type of each of them and their residuum. Given for each r1 > 0 and r2 > 0 the closed curve γr1 ,r2 = Ir1 ,r2 + IIr2 + IIIr1 ,r2 + IVr1 (cf. the ﬁgure), which form the boundary of the domain Ar1 ,r2 = {z ∈ C | −r1 < Re(z) < r2 and 0 < Im(z) < π}.. 6. 4. 2. –10. –5. 5. 10. –2. Figure 5: The curve γr1 ,r2 with the direction given on Ir1 ,r2 = [−r1 , r2 ] and IIIr1 ,r2 .. (2) Prove that √. f (z) dz = γr1 ,r2. 2 π. 2. (3) Prove that the line integrals along the vertical curves IIr2 and IVr1 tend to 0 for r2 and r1 tending to +∞. (4) Find . +∞. −∞. ex dx. 1 + e4x. 1) Since ez = 0 for every z ∈ C, the singularities are determined by e4z = −1 = ei(π+2pπ) ,. p ∈ Z,. 37 Download free eBooks at bookboon.com.

(44) Complex Funktions Examples c-7. Improper integrals in general. so the isolated singularities are π! π , p ∈ Z. +p· zp = i 2 4 We see from d 4z e + 1 |z=z = 4 e4zp = −4 = 0, p dz that these singularities are all simple poles with the residues √ " π # " π π !# 2 1 =− (1 + i) · exp i · p , exp i +p res (f ; zp ) = 2 2 8 −4 4. p ∈ Z.. 2) We have inside the curve γr1 ,r2 only the two poles z0 and z1 , hence by Cauchy’s residuum theorem, √ ez 2 dz = 2πi {res (f ; z0 ) + res (f ; z1 )} = 2πi − (1 + i)(1 + i) 4z 8 γr1 ,r2 1 + e √ √ π 2 2 π=√ . · 2i = = 2πi · − 2 8 2 3) We may choose the parametric descriptions of the vertical paths of integration in the form z(t) = r + it, t ∈ [0, π], where either r = r2 or r = −r1 . If r = r2 > 0, then we get the estimate % % % % % % π% r2 +it r2 % % % % e % dt ≤ π · e →0 for r2 → +∞. f (z) dz % ≤ % % % 4r +4it r % % IIr 1+e 2 e 2 −1 0 2 If r = −r1 < 0, then we get instead % % % % % % π% e−r1 +it %% e−r1 % % % f (z) dz →0 ≤ dt ≤ π · % % % % −4r +4it % % IVr 1+e 1 1 − e−4r1 0 1 4) Finally, . . r2. f (z) dz = Ir1 ,r2. −r1. and . . −r1. f (z) dz = IIIr1 ,r2. ex dx, 1 + e4x. r2. ex eiπ dx = 1 + e4x ei4π. . r2 −r1. ex dx. 1 + e4x. It follows from (2) and (3) that √ +∞ 2 ex π= lim f (z) dz = 2 dx, 4x r1 , r2 →+∞ γ 2 −∞ 1 + e r1 ,r2 hence +∞ −∞. √ 2 ex π. dx = 4 1 + e4x. 38 Download free eBooks at bookboon.com. for r1 → +∞..

(45) Complex Funktions Examples c-7. Improper integrals in general. Remark 3.2 It is possible to ﬁnd the value of the improper integral (which clearly is convergent) without using the calculus of residues. First we get by the substitution t = e x , +∞ +∞ ex dt dx = . 4x 1 + t4 1 + e −∞ 0 Then we decompose the integrand in the following way, 1 1 + t4. = =. 1 1 1 √ = √ 2 = 2 √ 2 (t4 + 2t2 + 1) − 2t2 2 t + 2 t + 1 t2 − 2 t + 1 2t (t + 1) − ct + d at + b √ √ + , t 2 + 2 t + 1 t2 − 2 t + 1. hence. " # " # √ √ 1 = (at+b) t2 − 2t+1 + (ct+d) t2 + 2t+1 √ √ √ √ = (a+c)t3 +(− 2a+b+ 2c+d)t2 +(a− 2b+c+ 2d)t+(b+d).. We get a + c = 0, i.e. c = −a, and b + d = 1, so √ √ √ −2 2 a + 1 = 0 and − 2 b + 2 d = 0. It follows that 1 a = √ = −c 2 2 thus 1 1 + t4. =. and. b=d=. 1 , 2. √ √ 1 2t − 2 1 2t + 2 1 1 1 1 √ · √ √ √ √ + · − √ · + · . 2 2 2 2 4 2 t + 2t + 1 4 t + 2t + 1 4 2 t − 2t + 1 4 t − 2t + 1. Finally, we get the primitive √ √ √ 2t + 2 2t + 2 t2 + 2 t + 1 √ √ √ − dt = ln → 0, t2 + 2 t + 1 t2 − 2 t + 1 t2 − 2 t + 1 for t → 0+, and for t → +∞. We therefore conclude that +∞ +∞ 1 1 1 +∞ ex dt √ √ + dt dx = = 4x 4 0 1 + t4 t 2 + 2 t + 1 t2 − 2 t + 1 −∞ 1 + e 0 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ 1 +∞ ⎨ 1 1 = + dt 2 2 ⎪ 4 0 1 1 1 1⎪ ⎪ ⎪ ⎪ ⎪ √ √ t− + + ⎭ ⎩ t+ 2 2 2 2 √ + " # " #, ∞ √ √ 2 Arctan 2 t + 1 + Arctan 2t − 1 = 0 √ √4 2π 2 π π π π! . ♦ = + − + = 4 4 4 2 4 2. 39 Download free eBooks at bookboon.com.

(46) Complex Funktions Examples c-7. Improper integrals in general. Example 3.8 Denote by A the domain A = C \ {z ∈ C | Re(z) = 0 and Im(z) ≤ 0}, √ and denote by z the branch of the square root which is analytic in A, and which is equal to the usual real square root on the positive real axis R+ . Furthermore, let Γr,R = Ir,R + IIR + IIIr,R + IVr. for 0 < r < 1 < R,. denote the simple closed curve on the ﬁgure.. 2.5. 2. 1.5. 1. 0.5. –2. –1. 0. 1. 2. –0.5. Figure 6: The closed curve Γr,R med Ir,R = [r, R] and the circular arc IIR with a direction, (and IIIr,R and IVr follow in a natural way). The pole i of f (z) is indicated inside Γr,R . Put 1 f (z) = √ . z (z 2 + 1) 1) Prove that π f (z) dz = √ (1 − i). 2 Γr,R 2) Prove that the integrals of f along the half circles IIR and IVr tend to 0 when R tends to ∞, and r tends to 0. 3) Prove that the integral 0. +∞. 1 √ dx x (x2 + 1). is convergent and ﬁnd its value.. 40 Download free eBooks at bookboon.com.

(47) Complex Funktions Examples c-7. Improper integrals in general. 1) The only singularity of f (z) inside Γr,R is the simple pole z = i, so it follows by Cauchy’s residuum theorem that 1 1 1 f (z) dz = 2πi res √ , i = 2πi lim √ = 2πi · 2 + 1) 1 + i z→i z (z z · 2z Γr,R √ · 2i 2 1−i π = π · √ = √ (1 − i). 2 2 2) We estimate the integral along the curve IVr of the parametric description z(t) = r ei(π−t) , t ∈ [0, π] and 0 < r < 1, by % π % √ % % π r 1 r dt % % √ √ →0 for r → +∞. dz ≤ = % 1 − r2 z (z 2 + 1) % r · (1 − r 2 ) 0 IVr Along med IIR we choose the parametric description z(t) = R · eit , t ∈ [0, π], R > 1, and then get the estimate √ % π % % % π R 1 R % % √ √ dt = 2 →0 for R → +∞. dz ≤ % R −1 z (z 2 + 1) % R · (R2 − 1) 0 IIR. American online LIGS University is currently enrolling in the Interactive Online BBA, MBA, MSc, DBA and PhD programs:. ▶▶ enroll by September 30th, 2014 and ▶▶ save up to 16% on the tuition! ▶▶ pay in 10 installments / 2 years ▶▶ Interactive Online education ▶▶ visit www.ligsuniversity.com to find out more!. Note: LIGS University is not accredited by any nationally recognized accrediting agency listed by the US Secretary of Education. More info here.. 41 Download free eBooks at bookboon.com. Click on the ad to read more.

(48) Complex Funktions Examples c-7. Improper integrals in general. 3) We obtain along IIIr,R , √. IIIr,R. 1 dz = 2 i z (z + 1). . −r −R. 1 dx = 2 i |x| (x + 1). . R r. dx √ = −i x (x2 + 1). . R r. √. dx . x (x2 + 1). Taking the limits r → 0+ and R → +∞ and applying the results of (1) and (2) we get (1 − i) hence +∞ 0. +∞ 0. π dx √ = √ (1 − i), x (x2 + 1) 2. π dx √ =√ . 2 x (x + 1) 2. Alternatively we may change the variable to t = . +∞. 0. √. dx =2 x (x2 + 1). . +∞ 0. √. x, x = t2 , t ∈ R+ ,. dt . t4 + 1. Then we decompose in the following way, 1 1 + t4. = =. 1 1 1 √ = √ 2 = 2 √ 2 (t4 + 2t2 + 1) − 2t2 2 t + 2 t + 1 t2 − 2 t + 1 2t (t + 1) − ct + d at + b √ √ + , 2 2 t + 2t + 1 t − 2t + 1. hence. " # " # √ √ 1 = (at+b) t2 − 2t+1 + (ct+d) t2 + 2t+1 √ √ √ √ = (a+c)t3 +(− 2a+b+ 2c+d)t2 +(a− 2b+c+ 2d)t+(b+d).. We get a + c = 0, thus c = −a, and b + d = 1, so √ √ √ −2 2 a + 1 = 0 and − 2 b + 2 d = 0. Then 1 a = √ = −c 2 2. and. b=d=. 1 , 2. and √ √ 1 2t − 2 1 1 2t + 2 1 1 1 1 √ √ √ √ + · − √ · + · . = √ · 2 2 2 2 1 + t4 4 4 t + 2t + 1 4 2 t − 2t + 1 t − 2t + 1 4 2 t + 2t + 1 Finally, we see that the primitive is given by √ √ √ 2t + 2 t2 + 2 t + 1 2t + 2 √ √ √ − dt = ln → 0, t2 − 2 t + 1 t2 + 2 t + 1 t2 − 2 t + 1. 42 Download free eBooks at bookboon.com.

(49) Complex Funktions Examples c-7. Improper integrals in general. for t → 0+, and for t → +∞. We therefore conclude that +∞ +∞ 1 1 1 +∞ 1 dt √ √ √ + dt = dx = 2 2 0 1 + t4 x (x2 + 1) t 2 + 2 t + 1 t2 − 2 t + 1 0 0 ⎫ ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎬ 1 +∞ ⎨ 1 1 = + dt 2 2 ⎪ 2 0 1 1 1 1⎪ ⎪ ⎪ ⎪ ⎪ t− √ + + ⎭ ⎩ t+ √ 2 2 2 2 √ + " # " #, ∞ √ √ 2 Arctan 2 t + 1 + Arctan 2t − 1 = 0 √ √2 2π 2 π π π π! . = + − + = 2 4 4 2 2 2. .. 43 Download free eBooks at bookboon.com. Click on the ad to read more.

(50) Complex Funktions Examples c-7. 4. Improper integral, where the integrand is a rational function. Improper integral, where the integrand is a rational function. Example 4.1 Find the value of the improper integral +∞ dx . 4 −∞ x + 1 1) It is possible to compute the integral by a real decomposition; but this is not an easy method. We shall here shortly sketch it in order to demonstrate the diﬃculties connected with it: By “adding something and then subtracting it again, followed by factorizing the diﬀerence of two squares” we get ! ! √ !2 √ √ 2 2 x = x2 + 2 x + 1 x2 − 2 x + 1 . x4 = x4 + 2x2 + 1 − 2x2 = x2 + 1 − We conclude that there exist real constants A, B, C and D ∈ R, such that 1 x4 + 1. = =. 1. √ √ + 1} − 2x = + 2 x + 1 x2 − 2 x + 1 Cx + D Ax + B √ √ + . x2 + 2 x + 1 x 2 − 2 x + 1. {x2. √. 2. 2. . x2. Then by the usual decomposition, 1 A= √ , 2 2. B=. 1 C=− √ , 2 2. 1 , 2. and we ﬁnd a primitive of. x4. D=. 1 , 2. 1 in the usual way. +1. 2) A variant of the method of decomposition above is to note that all four poles z j are simple, so res (f ; z1 ) res (f ; z2 ) res (f ; z3 ) res (f ; z4 ) 1 = + + + , z − z1 z − z2 z − z3 z − z4 z4 + 1 where res (f ; zj ) =. 1 z0 1 = 4 = − zj , 4z03 4 4z0. by Rule II. We see that the zj are complex (they occur in complex conjugated pairs), so the terms shall afterwards be paired together in the same way, before we ﬁnd the real primitives. 3) Finally, we show that it is much easier to use the residuum formula. We shall ﬁrst check the assumptions. The integrand is a rational function with a zero of order 4 at ∞ and no pole on the real axis. This implies that the improper integral is convergent and we can ﬁnd it value by ipπ , computing the residues of the poles in the upper half plane. The four simple poles are exp 4 p = 1, 3, 5, 7, of which only −1 + i 3iπ 1+i iπ = √ og exp = √ exp 4 4 2 2. 44 Download free eBooks at bookboon.com.

(51) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. lies in the upper half plane. If we as above use Rule II with A(z) = 1 and B(z) = z 4 + 1, where B (z) = 4z 3 , then we get for any of the poles z0 that res (f ; z0 ) =. 1 A (z0 ) z0 1 = 3 = 4 = − z0 , 4z0 B (z0 ) 4z0 4. because z04 = −1 for all of them. Then by the residuum formula, +∞ 1 1 1+i −1 + i dx √ √ + res = 2πi res ; ; 4 z4 + 1 z4 + 1 2 2 −∞ x + 1 π 1 + i −1 + i 2πi 2i 1 √ + √ ·√ =√ . =− = 2πi · − 4 4 2 2 2 2. Example 4.2 The improper integral. +∞ −∞. nevertheless use the residuum formula.. x2. x dx is not convergent. Discuss what happens if one +1. The only singularity of the analytic extension of the integrand in the upper half plane is the simple pole at z0 = i. Here we have 1 z z = , ; i = lim res z→i 2z 2 z2 + 1 so if we unconsciously put this into the residuum formula, then +∞ z x “ dx = 2πi · res ; i = πi ”. 2 z2 + 1 −∞ x + 1 This is of course not true, because if we could attach the improper integral a value (it is not convergent, so one should at least use “Cauchy’s principal value” in order just to get a little sense into this expression), and then it is obvious that a possible value should be reel and not at all imaginary. The example shows that residuum formulæ formally often can be applied in cases, in which their assumptions are not fulﬁlled. If so, they will usually give a wrong result. Example 4.3 Compute +∞ dx (a) 2, (1 + x2 ) −∞. . +∞. (b) −∞. . x2 (1 + x2 ). 2. dx,. +∞. (c) −∞. dx. 3.. (1 + x2 ). 2 (a) The integrand has a zero of fourth order at ∞, and since 1 + x2 = 0 for every x ∈ R, the integral is convergent. The integrand has the two double poles ±i, of which only +i lies in the upper half plane, so +∞ 1 dx d 1 1 lim = 2πi · res ; i = 2πi · 2 2 2 1! z→i dz (z + i)2 (1 + z 2 ) −∞ (1 + x ) −4πi π −2 = = . = 2πi lim z→i (z + i)3 (2i)3 2. 45 Download free eBooks at bookboon.com.

(52) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. (b) The diﬀerence of degrees is 2 where the denominator is dominating, and the integrand has only the singularities ±i (double poles, which do not lie on R). Hence, the integral exists and +∞ z2 1 x2 d z2 lim dx = 2πi · res ; i = 2πi · 2 2 2 1! z→i dz (z + i)2 (z 2 + 1) −∞ (1 + x ) 2z 2i 2z 2 2i2 = 2πi lim = 2πi − − z→i (z + i)2 (2i)3 (z + i)3 (2i)3 2 3 (2i) π 1 (2i) = −π = . − 2 3 2 2 (2i) (2i) Alternatively, we of course also have x2 (1 +. 2 x2 ). =. x2 + 1 − 1 (1 +. 2 x2 ). =. 1 1 − 2, 2 1+x (1 + x2 ). and then it follows from (a) that +∞ +∞ +∞ π π x2 1 1 dx = dx − dx = π − = . 2 2 )2 2 )2 2 2 1 + x (1 + x (1 + x −∞ −∞ −∞ (c) The integrand has a zero of order 6 at ∞ and no singularity on the x-axis, and poles of order 3 at z = ±i. Hence, +∞ 1 1 dx d2 1 lim = 2πi · res ; i = 2πi · 3 2 2 2! z→i dz 2 (z + i)3 (z 2 + 1) −∞ (1 + x ) 3π −3 12πi 12πi d (−3)(−4) . = = πi lim = = = πi lim 4 5 5 z→i dz z→i 8 (z + i) (2i) 32i (z + i) Example 4.4 Prove that √ +∞ π 2 x2 (a) , dx = 4 2 −∞ x + 1. . +∞. (b) −∞. 2π 4π x−1 . sin dx = 5 5 x5 − 1. x2 has a zero of second order at ∞, and no singularity on the x-axis. The +1 poles in the upper half plane,. (a) The integrand 1+i z1 = √ 2. x4. and. z2 =. −1 + i √ , 2. are both simple with z1 · z2 = −1, so 2 2 +∞ z z x2 dx = 2πi res ; z1 + res ; z2 4 z4 + 1 z4 + 1 −∞ x + 1 2πi 1 πi 1 z2 z2 (z1 + z2 ) = =− + lim + = 2πi lim 3 3 z→z1 4z z→z2 4z z1 4 2 z2 √ π 2 π πi 2i π 1 + i −1 + i √ + √ . =− ·√ = √ = = − 2 2 2 2 2 2 2. 46 Download free eBooks at bookboon.com.

(53) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. (b) First we note that we have a removable singularity at x = 1, because lim. x→1. 1 x−1 1 = lim = 5 x5 − 1 x→1 5x4. either by l’Hospital’s rule, or by a simple division, 1 1 x−1 = 4 → 5 3 2 x +x +x +x+1 5 x −1. for x → 1.. There is no other singularity on R than the removable singularity at z = 1, and the integrand has a zero of order 4 at ∞. We therefore conclude that the improper integral is convergent. The poles in the upper half plane, 2iπ 4iπ and z2 = exp , z1 = exp 5 5. Join the best at the Maastricht University School of Business and Economics!. Top master’s programmes • 3 3rd place Financial Times worldwide ranking: MSc International Business • 1st place: MSc International Business • 1st place: MSc Financial Economics • 2nd place: MSc Management of Learning • 2nd place: MSc Economics • 2nd place: MSc Econometrics and Operations Research • 2nd place: MSc Global Supply Chain Management and Change Sources: Keuzegids Master ranking 2013; Elsevier ‘Beste Studies’ ranking 2012; Financial Times Global Masters in Management ranking 2012. Visit us and find out why we are the best! Master’s Open Day: 22 February 2014. Maastricht University is the best specialist university in the Netherlands (Elsevier). www.mastersopenday.nl. 47 Download free eBooks at bookboon.com. Click on the ad to read more.

(54) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. are both simple, so the value of the improper integral is . +∞. −∞. = = = =. z−1 z−1 x−1 dx = 2πi res ; z ; z + res 1 2 z5 − 1 z5 − 1 x5 − 1 z−1 z−1 + lim 2πi lim 4 z→z1 5z z→z2 5z 4 2πi z2 − z z2 − z 2πi 2 lim = z1 − z1 + z22 − z2 + lim 5 5 z→z1 z→z2 5 z z 5 4iπ 8iπ 2iπ 4iπ 2πi − exp + exp − exp exp 5 5 5 5 5 2π 4π 2π 2πi 2πi 2πi 2πi . sin = · −2i sin = − exp exp − 5 5 5 5 5 5 5. Remark 4.1 Since √ 1+ 5 π cos = 4 5. π sin = 5. and. √ 10 − 2 5 , 4. it follows by insertion that . +∞ −∞. x−1 dx = x5 − 1 =. Example 4.5 Compute +∞ dx (a) , 2 + 1)n (x −∞. √ # √ π " π π 4π · 1 + 5 · 10 − 2 5 · 2 sin · cos = 10 5 5 5 √ √ π π 10 + 2 5. 10 + 8 5 = 5 10. n ∈ N,. (b) 0. +∞. x2 + 1 dx. x2 + 1. In both cases we see that the improper integrals are convergent (the denominator is dominating, and the diﬀerence of the degrees is at least 2, and we have no singularity on the real axis R), so the values can be found by means of the residues in the upper half plane. (a) Since z = i is an n-tuple pole, we ﬁnd . +∞ −∞. dx n (x2 + 1). . . 1 dn−1 lim n−1 (z + i)−n (n − 1)! z→i dz 2πi 1 (−n)(−n − 1) · · · (−2n + 2) lim = z→i (z + i)2n−1 (n − 1)! 1 (−1)n−1 · (2n − 2)! 1 π 2n − 2 · 2n−1 · 2n = 2n−2 . = 2πi · n−1 2 i 2 (n − 1)!(n − 1)! = 2πi · res. 1 n ; i (z 2 + 1). = 2πi ·. (b) If z0 is one of the simple poles, then z04 = −1, so the residuum is given by 2 z +1 z02 + 1 z0 2 res ; z z0 + 1 . = =− 0 z4 + 1 4z03 4. 48 Download free eBooks at bookboon.com.

(55) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Then by using the symmetry, (the integrand is an even function), 1 +∞ x2 + 1 x2 dx = dx 2 −∞ x4 + 1 x4 + 1 0 2 2 " π # z +1 3π z +1 + res ; exp i ; exp i = πi res 4 4 z4 + 1 z4 + 1 # " " # # " 3π π π 3π πi +1 exp i + 1 + exp i exp i exp i = − 2 4 2 4 4 πi πi 4i · πi π = − √ {(1 + i)(1 + i) + (−1 + i)(1 − i)} = − √ {2i − (−2i)} = − √ = √ . 4 2 4 2 4 2 2. . +∞. Remark 4.2 It is possible, though far from easy to compute the value of the integral by only using the known real methods of integration from Calculus, i.e. by a decomposition. We shall here only sketch the method. If we only can factorize the denominator, the rest is standard, though still diﬃcult. The trick is here the usual one: Add something and then subtract it again, " #" # √ √ √ 2 x4 + 1 = x4 + 2x2 + 1 − 2x2 = x2 + 1 − ( 2 x)2 = x2 + 2 x + 1 x2 − 2 x + 1 . Now we have written the denominator as a product of polynomials of degree 2, so we can in principle decompose and then compute the integral. However, the factorization of the denominator shows that this will be fairly diﬃcult to carry through in practice. ♦. Example 4.6 Compute +∞ dx , (a) 6 −∞ 1 + x. . +∞. (b) −∞. x2 dx. 1 + x6. The denominator is dominating with at least 4 degrees in the exponent, and there are no poles on the x-axis. Therefore, both improper integrals are convergent, and we can compute them by a residuum formula. 1 (a) The integrand has in the upper half plane the three simple poles 1 + z6 " π# " π# 5π . = i, exp i , exp i exp i 6 2 6 Let z0 be anyone of these poles. Then in particular z06 = −1, and it follows that 1 z0 1 1 ; z 0 = 5 = 6 = − z0 . res 6z0 6 1 + z6 6z0 By insertion; . +∞. −∞. dx 1 + x6. " π # 1 1 1 5π + res = 2πi res ; exp i ; i + res exp i 1 + z6 1 + z6 6 1 + z6 6 ! # " 2π π πi 5π π 2πi . 2i sin + i = =− + i + exp i exp i = − 3 6 3 6 6 6. 49 Download free eBooks at bookboon.com.

(56) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Remark 4.3 The denominator can be factorized in the following way 1 + x6. 1 + x2 x4 − x2 + 1 = 1 + x2 x4 + 2x3 + 1 − 3x2 # #" # √ √ √ " 2 2 " = x2 + 1 x + 1 − ( 3 x)2 = x2 + 1 x2 + 3 x + 1 x2 − 3 x + 1 , =. so we can in principle decompose the integrand and then integrate in the usual way known from Calculus. However, the coeﬃcients clearly show that this will be very diﬃcult to carry through. ♦ (b) Here we must not forget what we learned in the “kindergarten”: +∞ *+∞ 1 +∞ d x3 1 ) π x2 Arctan x3 −∞ = . dx = = 2 6 3 −∞ 1 + (x3 ) 3 3 −∞ 1 + x Alternatively (and this time far more diﬃcult) we see that we have the same simple poles as in (a), and then we get by the general expression of the residuum at the pole z 0 , where z06 = −1, 2 z z02 z03 1 res = ; z = = − z03 , 0 5 6 1 + z6 6z0 6z06. > Apply now redefine your future. - © Photononstop. AxA globAl grAduAte progrAm 2015. axa_ad_grad_prog_170x115.indd 1. 19/12/13 16:36. 50 Download free eBooks at bookboon.com. Click on the ad to read more.

(57) Complex Funktions Examples c-7. hence +∞ −∞. Improper integral, where the integrand is a rational function. 2 2 2 " π # z z z 5π x2 +res dx = 2πi res ; exp i ; i +res ; exp i 6 6 1 + x6 1 + z6 1 + z6 1 + z6 # " π π πi 2πi 5π + i3 + exp i exp i = − {i − i + i} = . = − 3 2 3 6 2. Example 4.7 Compute. +∞ −∞. dx . 1 + x8. 1 has a zero of order 8 at ∞ and no singularity on the real axis. Hence the improper 1 + z8 integral is convergent, and its value can be found by the residues at the poles in the upper half plane. All poles are simple, and we have in the upper plane the four poles " π# 3π 5π 7π , z2 = exp i , z3 = exp i , z4 = exp i . z1 = exp i 8 8 8 8 The function. We have for every pole zk that zk9 = −1, so 1 1 1 = 7 = − z7 . ; z res k 8 8zk 8 1+z Then by the residuum formula, +∞ π π π 3π π 1 2πi dx = · 2 sin · cos · {z1 + z2 + z3 + z4 } = − · 2i sin + sin = 2πi − 8 8 4 2 8 8 8 8 1 + x −∞ . ( 1 + cos π4 π√ 1 π 2· = 1+ √ . = 2 2 2 2 Alternatively, it is possible to decompose. Here, we shall only show how one factorizes the denominator 1 + x8 : #2 " #" # √ √ 2 "√ 1 + x8 = x8 + 2x4 + 1 − 2x4 = x4 + 1 − 2 · x2 = x4 + 2 x2 + 1 x4 − 2 x2 + 1 ! √ √ # ! " 4 x4 + 2x2 + 1 − 2 + 2 x2 = x + 2x2 + 1 − (2 − 2)x2 2 2 √ √ 2 2 2 2 x +1 − 2− 2·x 2 + 2x = x +1 − =. √ √ √ √ 2 2 2 2 x + 2− 2 x+1 x − 2− 2 x+1 x + 2+ 2 x+1 x − 2+ 2 x+1 .. Obviously, the following decomposition becomes very diﬃcult, although it can in principle be carried through.. 51 Download free eBooks at bookboon.com.

(58) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Example 4.8 Compute +∞ x dx (a) 2, 2 −∞ (x + 4x + 13). . +∞. (b). x2 dx (x2 + a2 ). 0. 2,. a ∈ R+ .. (a) It follows from x2 + 4x + 13 = (x + 2)2 + 32 , that the integrand has the double poles −2 ± 3i, which do not lie on the real axis. The diﬀerence of the degrees is 3 with the denominator dominating, so the integral exists, and the value can be expressed by the residuum at z = −2 + 3i: . +∞. . x dx. z. 2 ; −2 + 3i (z 2 + 4z + 13) z 2π 1 2z d lim = = 2πi lim − z→−2+3i (z + 2 + 2i)2 1! z→−2+3i dz (z + 2 + 3i)2 (z + 2 + 3i)3 2πi 2(−2 + 2i) π 1 = − {6i + 4 − 6i} = − . = 2πi 2 3 3 (6i) (6i) 27 (6i). −∞. (x2 + 4x + 13). 2. = 2πi · res. . (b) Here we have the double poles ±i a ∈ / R, and since the diﬀerence in degrees is 2 with the denominator dominating, it follows by the symmetry that 1 +∞ x2 z2 dx = πi · res 2 dx = 2 2 ; ia 2 2 2 (x2 + a2 ) (z 2 + a2 ) 0 −∞ (x + a ) z2 2z 2z 2 d = πi lim = πi lim − z→i a dz z→i a (z + i a)2 (z + i a)2 (z + i a)3 2i a π (2i a)2 π 2(i a)2 1 (2i a)3 . = πi = = − − 2 3 2 (2i a) 2a (2i a) 4a 2 (2i a)3 (2i a). . +∞. x2. Example 4.9 Compute +∞ x2 − x + 2 (a) dx, 4 2 −∞ x + 10x + 9. (a) The integrand z 2 = −5 ±. (b) 0. +∞. x2 − 1 dx. x4 + 5x2 + 4. z2 − z + 2 has a zero of second order at ∞ and its poles are given by + 10x2 + 9. x4. √. 25 − 9 = −5 ± 4,. i.e. the simple poles are 3i,. −3i,. i and. − i.. 52 Download free eBooks at bookboon.com.

(59) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. None of these lies on the x-axis. Since f (z) is analytic outside the poles and since we have the estimate |f (z)| ≤. C |z|2. for |z| ≥ 4 og Im(z) ≥ 0,. we conclude that +∞ x2 − x + 2 dx = 2πi {res(f ; i) + res(f ; 3i)} 4 2 −∞ x + 10x + 9 −1 − i + 2 −9 − 3i + 2 z2 − z + 2 z2 − z + 2 + = 2π lim + lim 2 = 2πi z→i (z + i) (z 2 + 9) z→3i (z + 1) (z + 3i) −8 · 6i 2i · 8 5π π . {3 − 3i + 9 + 31 − 2} = = 12 24 Alternatively, one may apply the traditional real method of integration, by using that we have proved that the integral exists. In particular, +∞ −x dx = 0, 4 2 −∞ x + 10x + 9 because the integrand is an odd function. Then . +∞. −∞. x2 − x + 2 dx = x4 + 10x2 + 9. . +∞. 1 x2 + 2 dx = 2 + 9) (x2 + 1) 8 (x −∞ 5π 10π 7 1 . = = + = π 12 24 8 3·8. . +∞. −∞. 7 dx + x2 + 1 8. . +∞. −∞. dx x2 + 9. (b) The diﬀerence of the degrees is 2 where the denominator is dominating, and the denominator is furthermore positive for every real x. Hence, the improper integral is convergent. Since the integrand is an even function, it follows by the symmetry, followed by an application of the residuum formula that +∞ 1 +∞ x2 − 1 x2 − 1 dx = dx = πi {res(f ; i) + res(f ; 2i)} 4 2 2 2 −∞ (x + 4) (x2 + 1) x + 5x + 4 0 −5 −2 z2 − 1 z2 − 1 + = πi lim 2 + lim 2 = πi z→i (z + 4) (z + i) z→2i (z + 1) (z + 2i) 3 · 2i (−3) · 4i π 1 5 . = − = π 12 12 3 Alternatively we decompose: . +∞ 2 +∞ dx 5 +∞ dx x2 − 1 x2 − 1 dx = dx = − + 3 0 x4 + 5x2 + 4 (x2 + 4) (x2 + 1) x2 + 1 3 0 x2 + 4 0 0 ' & " x # +∞ π 1 5 2 5 2 π . = − = − Arctan x + · Arctan = 12 2 0 2 3 3 6 3 2 +∞. 53 Download free eBooks at bookboon.com.

(60) Complex Funktions Examples c-7. Example 4.10 Compute +∞ dx , (a) 2+x+1 x −∞. Improper integral, where the integrand is a rational function. . +∞. (b) −∞. dx . (x2 + 1) (x2 + 4). 1 has a zero of second order at ∞ and the poles z2 + z + 1 √ 3 1 ∈ / R. z =− ±i 2 2. (a) The integrand. Hence, √ the improper integral exists, and we may ﬁnd its value by means of the residuum at 3 1 , i.e. at the pole in the upper half plane: − +i 2 2 √ +∞ 3 1 dx 1 1 = 2πi lim √ = 2πi · res ; − +i 2 2 2 z2 + z + 1 −∞ x + x + 1 z→− 12 +i 23 2z + 1 = 2πi ·. 2π 2πi 1 √ = √ =√ . 3 i 3 −1 + i 3 + 1. 54 Download free eBooks at bookboon.com. Click on the ad to read more.

(61) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Alternatively, the traditional computation gives . +∞ −∞. dx = 2 x +x+1. . ⎞⎤+∞ 1 ⎜ x + 2 ⎟⎥ 2π 1 ⎢ π dx ⎜ ( ⎟⎥ Arctan =( ⎢ =( =√ . 2 ⎠ ⎣ ⎦ ⎝ 3 3 3 3 1 3 x+ + 4 4 4 2 4 −∞ ⎡. +∞. −∞. ⎛. 1 has a zero of order 4 at ∞ and the simple poles ±i and ±2i ∈ / R. + 1) (z 2 + 4) Hence, the improper integral exists, and its value can be found by means of the residues at the poles in the upper half plane. We get. (b) The integrand. (z 2. 1 1 dx = 2πi res ; i +res ; 2i 2 2 (z 2 +1) (z 2 +4) (z 2 +1) (z 2 +4) −∞ (x + 1) (x + 4) π 1 1 1 1 = . − =π + = 2πi 6 3 6 2i · 3 (−3)4i. . +∞. Alternatively, . +∞ −∞. dx (x2 + 1) (x2 + 4). = =. Example 4.11 Compute +∞ dx (a) , 2 + 2x + 2 x −∞. '+∞ & x 1 1 +∞ dx 1 +∞ dx 1 Arctan Arctan x − − = 2 −∞ 2 3 −∞ x2 + 1 3 −∞ x2 + 4 3 π! π 1 = . π− 6 2 3. . +∞. (b) −∞. dx . (x2 + 1) (x2 + 2x + 2). 1 has a zero of second order at ∞ and simple poles at z = −1 ± i ∈ / R. Hence, z 2 + 2z + 2 the improper integral is convergent, and its value can be found by a residuum formula. However, the easiest method here is actually the traditional one, +∞ +∞ dx dx = = [Arctan(x + 1)]+∞ −∞ = π. 2 + 2x + 2 2+1 x (x + 1) −∞ −∞. (a) Here. For comparison we get by the calculus of residues, . +∞. −∞. dx x2 + 2x + 2. . 1 = 2πi · res ; −1 + i z 2 + 2z + 2 2πi 2πi = π. = = 2i −1 + i + 1 + i. = 2πi. 55 Download free eBooks at bookboon.com. lim. z→−1+i. 1 z+1+i.

(62) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. (b) The integrand has a zero of order 4 at ∞, and the simple poles ±i, −1 ± i ∈ / R, so we conclude that the improper integral is convergent, and its value is given by . +∞. dx + 1) (x2 + 2x + 2) −∞ 1 1 ; i + res ; −1i = 2πi res (z 2 + 1) (z 2 + 2z + 2) (z 2 + 1) (z 2 + 2z + 2) 1 1 1 1 1 + + · = π = 2πi 1 + 2i 1 − 2i 2i −1 + 2i + 2 {(−1 + i)2 + 1} · 2i 2π 1 − 2i + 1 + 2i . = π· = 5 1+4 (x2. Example 4.12 1) Explain why the improper integral +∞ x2 dx 2 2 −∞ (x + 1) (x + 4) is convergent, and ﬁnd its value. 2) Compute the complex line integral z 2 dz . 2 2 2 |z|=3 (z + 1) (z + 4) 1) The integrand is a rational function with a zero of order 4 at ∞ and with no poles on the real axis. The poles are z = ±i (double poles) and z = ±2i (simple poles), so the integral is convergent, and its value can be found by a residuum formula, +∞ x2 dx z2 z2 = 2πi res ; i + res ; 2i . 2 2 2 2 (z 2 + 1) (z 2 + 4) (z 2 + 1) (z 2 + 4) −∞ (x + 1) (x + 4) Here we get straight away, z2 1 d z2 lim ;i = res 2 1! z→i dz (z + i)2 (z 2 + 4) (z 2 + 1) (z 2 + 4) 2z 2 z 2 · 2z 2z lim − − z→i (z+i)2 (z 2 +4) (z+i)3 (z 2 +4) (z+i)2 (z 2 +4)2 2i 2i −2 (−1) · 2i 2i 2i − + − − =− 3 3 2 2 (2i) 3 (2i) · 3 (2i) · 3 4·3 8·3 9·4 5i i (−6 + 3 − 2) = − , = 36 36 =. and res. . . z2 (z 2. + 1). 2. (z 2. + 4). ; 2i. = lim. z→2i (z 2. z2 2. + 1) (z + 2i). =. 4i −4 , = 2 36 (−3) · 4i. 56 Download free eBooks at bookboon.com.

(63) Complex Funktions Examples c-7. so by insertion, +∞. Improper integral, where the integrand is a rational function. π 1 4i 5i . = = 2π · + = 2πi − 2 18 36 36 36 (x2 + 1) (x2 + 4). −∞. x2 dx. Alternatively, we may ﬁrst decompose to get u C B A + + = . (u + 1)2 (u + 4) u + 4 n + 1 (u + 1)2 Here we immediately get A=. −4 4 =− (−3)2 9. and. C=. 1 −1 =− . 3 3. Then by insertion, rearrangement and reduction, B u+1. = = = =. u 1 1 4 1 + + (u + 1)2 (u + 4) 9 u + 4 3 (u + 1)2 1 1 · 9u + 4(u + 1)2 + 3(u + 4) 2 9 (u + 1) (u + 4) 1 1 · 4(u + 1)2 + 12(u + 1) 2 9 (u + 1) (u + 4) 1 4 u+1+3 4 . = · · 9 u+1 9 (u + 1)(u + 4). Then put u = x2 to get . +∞. −∞. x2. dx 2 (x2 + 1) (x2 + 4) 4 +∞ dx 1 +∞ 4 +∞ dx dx + − =− 9 −∞ x2 + 4 9 −∞ x2 + 1 3 −∞ (x2 + 1)2 " x #,+∞ 4 4 1 + 1 +∞ dx Arctan =− · + [Arctan x]+∞ − −∞ 2 2 −∞ 9 9 2 3 −∞ (x + 1)2 +∞ 2π 1 dx − = . 3 −∞ (x2 + 1)2 9. We can now compute the integral +∞ dx −∞. (x2 + 1). 2. in a number of ways: a) We get by the calculus of residues, +∞ 1 1 dx d 1 lim = 2πi · res = 2πi · 2 2 ; i 2 1! z→i dz (z + i)2 (z 2 + 1) −∞ (x + 1) π 2 2 2πi · (−2i) = . = 2πi · − = = 2πi lim − z→i 2 (z + i)3 (2i)3 8. 57 Download free eBooks at bookboon.com.

(64) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. b) Alternatively, we get by a partial integration & '+∞ +∞ +∞ +∞ 2 2 x +1 −2 x dx 2x · x = + dx 2 dx = 2 2 2 x2 + 1 −∞ (x2 + 1) −∞ x + 1 −∞ (x + 1) −∞ +∞ +∞ dx dx −2 = 2 2, 2+1 2 x −∞ −∞ (x + 1) and then by a rearrangement, +∞ dx π 1 +∞ dx = . = 2 2 2 2 2 −∞ x + 1 −∞ (x + 1) Finally, by insertion, +∞ x2 −∞. (x2. + 1). 2. (x2. + 4). dx =. π π 2π π . (4 − 3) = − = 18 18 6 9. 2) Since all pole lie inside |z| = 3, and since we have a zero of order 4 at ∞, we get by changing the direction on the path of integration, z2 z 2 dz z 2 dz ; ∞ = 0. =− = −2π i · res 2 2 2 2 2 2 (z 2 + 1) (z 2 + 4) |z|=3 (z + 1) (z + 4) |z|=3 (z + 1) (z + 4). Need help with your dissertation? Get in-depth feedback & advice from experts in your topic area. Find out what you can do to improve the quality of your dissertation!. Get Help Now. Go to www.helpmyassignment.co.uk for more info. 58 Download free eBooks at bookboon.com. Click on the ad to read more.

(65) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Example 4.13 1) Find all complex solutions of the equation z 4 + 5z 2 + 4 = 0. 2) Prove that the improper integral . +∞. 0. 2x2 − 1 dx x4 + 5x2 + 4. is convergent, and ﬁnd its value. 1) We get by the factorization 0 = z 4 + 5z 2 + 4 = z 2 + 1 z 2 + 4 , the four roots −i,. i,. 2i,. −2i.. 2) The integrand is a rational function with no pole on the x-axis and with a zero of second order at ∞. Hence, the improper integral is convergent. Since the integrand is even it follows by a reﬂection and the residues at the singularities i and 2i in the upper half plane that . 1 +∞ 2x2 − 1 2x2 − 1 dx = dx 2 −∞ x4 + 5x2 + 4 x4 + 5x2 + 4 0 2z 2 − 1 2πi 2z 2 − 1 res = ; i + res ; 2i 2 z 4 + 5z 2 + 4 z 4 + 5z 2 + 4 2z 2 − 1 2z 2 − 1 = πi lim 3 + lim z→i 4z + 10z z→2i 4z 3 + 10z π 9 1 −8 − 1 1 1 −2 − 1 = . =π − + · + · = πi 4 2 16 i −4 + 10 2i −16 + 10 +∞. Alternatively, the traditional method of decomposition gives that 2x2 − 1 2x2 − 1 1 3 = =− 2 + , x + 1 x2 + 4 x4 + 5x2 + 4 (x2 + 1) (x2 + 4) hence +∞ 0. 2z 2 − 1 dx = − x4 + 5x2 + 4. . +∞ 0. 1 dx + 2 x +1. 0. +∞. x2. π π 3 π 3 dx = − + · = . 4 2 2 2 +4. 59 Download free eBooks at bookboon.com.

(66) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Example 4.14 Prove that the improper integral +∞ 1 + x2 dx 4 −∞ 1 + x is convergent, and ﬁnd its value. We estimate the integrand for |x| > 1 in the following way, 2 1 1+ 1 2 1 + x2 x = 2 0 < g(x) := 4 < 2 . 4 x x 1+x 1 1+ x Since the improper integral. +∞ 1 dx is convergent, the given integral is also convergent. 1 x2. 1. 0.5. –2. –1. 1. 2. –0.5. –1. i 1 Figure 7: The curve CR for R > 1 and the singularities ± √ ± √ . 2 2 We can now ﬁnd the value of the improper integral as a Cauchy principal value via the residuum theorem. The denominator has the simple poles at the points " π π# , p ∈ {0, 1, 2, 3}, zp = exp i + p 2 4 where the former two lie inside the circle of integration CR . We get by a small computation ⎧ 1 1+i ⎪ p = 0, − · √ (1 + i), ⎪ ⎪ ⎨ 4 2 2 1 + zp 1 2 res (g(z); zp ) = = − zp 1 + z p = ⎪ 4 4zp3 ⎪ +i ⎪ − 1 · −1 ⎩ √ (1 − i), p = 1. 4 2 i Both residues are − √ , so 2 2 R R −i √ = 2πi · g(z) dz = g(z) dz + g R eit · i R eiθ dθ, 2 −R 0 CR. 60 Download free eBooks at bookboon.com. R > 1..

(67) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. If R > 2, then we have the following estimate on CR , % % % % 1 %1 + 1 e−2it % 1+ 2 % % 2 % iθ % 1 1 R R < 1 · %g R e % = %≤ ·% · 1 1 −4it %% R2 R2 %% R2 1− 4 % %1 + R4 e R. 5 4 15 16. .. It follows easily that the line integral along the circular arc tends to zero, when R → +∞, so we ﬁnally get by taking this limit, +∞ √ 1 + x2 dx = π 2. 4 −∞ 1 + x. Brain power. By 2020, wind could provide one-tenth of our planet’s electricity needs. Already today, SKF’s innovative knowhow is crucial to running a large proportion of the world’s wind turbines. Up to 25 % of the generating costs relate to maintenance. These can be reduced dramatically thanks to our systems for on-line condition monitoring and automatic lubrication. We help make it more economical to create cleaner, cheaper energy out of thin air. By sharing our experience, expertise, and creativity, industries can boost performance beyond expectations. Therefore we need the best employees who can meet this challenge!. The Power of Knowledge Engineering. Plug into The Power of Knowledge Engineering. Visit us at www.skf.com/knowledge. 61 Download free eBooks at bookboon.com. Click on the ad to read more.

(68) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Example 4.15 Given the function f (z) =. 1 , z3 + 1. and for every R > 1 UR = z = r eit. the closed curve γR = I + II + III (see the ﬁgure), enclosing the domain % % % 0 < r < R and 0 < t < 2π . % 3. Figure 8: The curve γR = I + II + III, enclosing UR . Here, I = [0, R] is an interval on the x-axis, II is the circular arc, and III is the oblique line.. 1) Find f dz. γR. 2) Prove that the line integral along the circular arc II tends towards 0, when R tends towards +∞. 3) Prove that +∞ 0. x3. 2π 1 dx = √ . +1 3 3. 1 has the simple poles +1 " π# " π# , z3 = exp i . z1 = −1, z2 = exp −i 3 3 " π# If R > 1, then only z3 = exp i lies inside γR , so it follows by the residuum theorem that 3 1 2πi z3 · f (z) dz = 2πi · res (f ; z3 ) = 2πi · 2 = 3z 3 z33 γR 3 √ ! " π# π √ 2πi 1 3 2πi 3−i . +i = =− exp i = − 3 3 2 2 3 3. 1) The function f (z) =. z3. 62 Download free eBooks at bookboon.com.

(69) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. 2) We get along II the estimate % % % 2π dz %% 1 % R→0 · ≤ 3 % 3 + 1% 3 z R − 1 II. for R → +∞.. 3) Along III we choose the parametric description 2π , r ∈ [0, R], (R − r) exp i 3 so . . 2π R − exp i 2π dx 3 dr = − exp i . 3+1 1 + (R − r)3 exp(2πi) 3 x 0. R. f (z) dz = 0. III. Then by insertion and the limit R → +∞, π √ ( 3 − i) 3. +∞ 2π dx = lim f (z) dz = 0 + 1 − exp i 3 R→+∞ γ 3 x +1 0 R √ +∞ √ +∞ 3 3 √ dx dx 3 ( 3 − i) = −i = , 3 3 2 2 x +1 x +1 2 0 0 . . . so by a rearrangement, +∞ 2π dx = √ . 3+1 x 3 3 0 Remark 4.4 The integral can in fact also be computed by more elementary methods. We get by a decomposition, 1 x3 + 1. =. =. hence +∞ 0. dx 3 x +1. 3 − x2 + x − 1 1 1 1 1 · + = (x + 1) (x2 − x + 1) 3 x + 1 3 (x2 − x + 1) (x + 1) 3 1 − x− 1 x−2 1 1 1 1 1 2 2 − · − · 2 = · , 2 3 x+1 3 3 x+1 3 x −x+1 1 3 x− + 2 4. =. 1 3. 0. +∞. 1 dx − x+1 6. 0. +∞. 1 2x − 1 dx + 2 2 x −x+1. . +∞ 0. dx 2 1 3 x− + 2 4 6+∞. . 5 & 2 '+∞ x − 12 x + 2x + 1 1 1 2 √ √ ln · Arctan = + 3 6 2 x2 − x + 1 3 0 2 0 2π 1 π π! 1 π 1 = √ . + =√ + Arctan √ = 0+ √ 6 3 3 3 2 3 3 2. 63 Download free eBooks at bookboon.com. ♦..

(70) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Example 4.16 Given the function f (z) =. z2 . z4 + z2 + 1. 1) Find all isolated singularities of f in C, and specify their types. 2) Prove by using the calculus of residues that the improper integral 0. +∞. x4. x2 dx + x2 + 1. π is convergent of the value √ . 2 3 One may use that f (x) is an even function.. 1) First note that 2 z − 1 z4 + z2 + 1 = z6 − 1 = 0 for. " pπ # , z = exp i 3. p ∈ Z.. When we again remove the roots z = ±1 of the auxiliary factor z 2 − 1, we see that the simple poles are √ √ " π# 1 3 1 2π 3 , =− +i , exp i = +i exp i 2 2 3 2 2 3 √ 3 π# 1 , = −i exp −i 2 2 3 ". √ 3 1 2π . =− −i exp −i 2 2 3. 2) Since we have a zero of second degree at ∞, and since we do not have any pole on the x-axis, we conclude that the improper integral is convergent. The integrand is even, so we get by an extended residuum theorem that +∞ 1 +∞ x2 x2 dx = dx 2 −∞ x4 + x2 + 1 x4 + x2 + 1 0 " " π ## 2π , + res f (z) ; exp i = πi res f (z) ; exp i 3 3 " π# 2π are the only singularities in the upper half plane. and exp i because exp i 3 3 Using the rearrangement z2 z2 − 1 z2 f (z) = 4 = , z + z2 + 1 z6 − 1. 64 Download free eBooks at bookboon.com.

(71) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. we get " π ## res f (z) ; exp i 3 ". 5. 6 z2 z2 − 1 2iπ 1 −1 = = · exp(iπ) · exp 3 6 6z 5 z=exp(i π ) 3 √ 1 3 1 1 {3 − i 3}, − +i −1 = = − 12 2 2 6. and 2π res f (z) ; exp i 3. 5. 6 z2 z2 − 1 4iπ 1 = − 1 · 1 · exp = 3 6 6z 5 z=exp(i 2π 3 ) √ 1 3 1 1 {−3 − i 3}, − −i −1 = = − 12 2 2 6. hence by insertion, . +∞ 0. √ √ π π 3 πi x2 = √ . · (−2i 3) = dx = 6 12 x4 + x2 + 1 2 3. 65 Download free eBooks at bookboon.com. Click on the ad to read more.

(72) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Remark 4.5 Since 2 z 4 + z 2 + 1 = z 4 + 2z 2 + 1 − z 2 = z 2 + 1 − z 2 = z 2 + z + 1 z 2 − z + 1 , it is of course also possible – though not quite easy – to use the method of decomposition. This variant is left to the reader as an exercise. ♦. Example 4.17 Given the function f (z) =. z2 , z4 + 1. and for every R > 1 a positively oriented curve ΓR = IR + IIR + IIIR , (cf. the ﬁgure), which surrounds the domain % π! % . UR = z = r eit % 0 < r < R og 0 < t < 2. 2. 1.5. 1. 0.5. –0.5. 0. 0.5. 1. 1.5. 2. –0.5. " π# Figure 9: The curve ΓR , starting with IR = [0, R] on the x-axis and with the singularity exp i 4 inside the curve.. 1) Prove that π f (z) dz = √ (1 + i). 2 2 ΓR 2) Show that the line integral along the circular arc IIR tends towards 0 for R tending towards +∞, and ﬁnd the value of +∞ x2 dx. x4 + 1 0. 66 Download free eBooks at bookboon.com.

(73) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. 1) The function f (z) has the four simple poles " π π !# , p ∈ {0, 1, 2, 3}. +p zp = exp i 2 4 Of these only " π# 1 = √ (1 + i) z0 = exp i 4 2 lies inside ΓR , when R > 1. Then by Cauchy’s residuum theorem, 2 z2 z πi 1 · ; z = 2πi · 03 = f (z) dz = 2πi res 0 4 2 z0 z +1 4z0 ΓR √ π π i(1 − i) π 2 = √ (1 + i). =√ · = i· · 2 2 1+i 2 2 2 + π, 2) We use along the circular arc IIR the parametric description z(t) = R eit , t ∈ 0 , , so we get 2 the estimate for R > 1, % % π 2 % % 1 π z2 R2 % %≤ dz · R dt = · →0 % % 4+1 4−1 1 2 z R IIR 0 R− 3 R when R → +∞. Finally, we use along IIIR on the imaginary axis the parametric description z(t) = (R − t)i, t ∈ [0, R], giving R R z2 (R − t)2 i2 t2 dz = · (−i)dt = i dt. 4 4 4 4 0 (R − t) i + 1 0 t +1 IIIR z + 1 Then by (1) we get by insertion and taking the limit R → +∞, +∞ π x2 √ (1 + i) = (1 + i) dx, x4 + 1 2 2 0 hence +∞ 0. π x2 dx = √ . 4 x +1 2 2. Alternatively, we get by a decomposition, x2 x4 + 1. = =. thus x2. x2 x2 x2 √ √ = = √ 2 2 x4 + 2x2 + 1 − 2x2 x2 + 2 x + 1 x2 − 2 x + 1 (x2 + 1) − 2x ax + b cx + d √ √ + , 2 2 x + 2x + 1 x − 2x + 1. " # " # √ √ = (ax + b) x2 − 2 x + 1 + (cx + d) x2 + 2 x + 1 √ √ √ √ = (a+c)x3 +(b− 2a+d+ 2c)x2 +(a− 2b+c+ 2d)x+b+d.. 67 Download free eBooks at bookboon.com.

(74) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. By identifying the coeﬃcients we clearly obtain that a+c=0. and. b + d = 0,. so √. 2 (−a + c) = 1,. 1 thus c = −a = √ , 2. and b = d = 0. Hence x 1 x x2 √ √ √ = − x4 + 1 2 2 x2 − 2 x + 1 x2 + 2 x + 1 √ √ √ √ 2x + 2 1 2x − 2 2 2 √ √ √ √ √ + − + = 4 2 x2 − 2 x + 1 x2 − 2 x + 1 x2 + 2 x + 1 x2 + 2 x + 1 ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ √ √ ⎪ ⎪ ⎨ ⎬ 1 2x − 2 2x + 2 1 1 1 √ √ √ = − + + . 2 2 4⎪ 4 2 x2 − 2 x + 1 x 2 + 2 x + 1 1 1 1⎪ ⎪ ⎪ ⎪ x − √1 ⎪ x+ √ + + ⎭ ⎩ 2 2 2 2 Clearly, the improper integral . +∞. 0. =. =. 1 x2 dx = 2 x4 + 1. . +∞ 0. x2 dx is convergent, and x4 + 1. +∞. x2 dx 4 −∞ x + 1 √ √ R 2x + 2 2x − 2 1 1 √ √ √ lim − dx 2 R→+∞ 4 2 −R x2 − 2 x + 1 x2 + 2 x + 1 "√ # √ "√ #,+∞ 1 1 +√ 2 Arctan 2 x − 1 + 2 Arctan 2x + 1 + · 2 4 −∞ 6R 5 √ 2 π π 1 1 x − 2x + 1 √ √ lim ln + √ · (π + π) = 0 + √ = √ . 2 R→+∞ 2 2 2 8 2 4 2 x + 2x + 1 −R. 68 Download free eBooks at bookboon.com.

(75) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Example 4.18 Given the rational function f (z) =. z2 + z + 1 . z4 + z2 + 1. 1) Find all the isolated singularities of f in C, and specify their types. 2) Prove by calculus of residues that . +∞. p.v. −∞. 2π x2 + x + 1 dx = √ . x4 + x2 + 1 3. 1) First variant. If z = ±1, then (z + 1)(z − 1) z 2 + z + 1 z2 + z + 1 z3 − 1 z+1 1 = = (z + 1) · 6 = 3 = 2 , 4 2 2 4 2 z +1 z −z+1 z +z +1 z −1 (z − 1) (z + z + 1) and the simple poles are √ 3 1 and z1 = + i 2 2. z2 =. √ 3 1 . −i 2 2. Challenge the way we run. EXPERIENCE THE POWER OF FULL ENGAGEMENT… RUN FASTER. RUN LONGER.. RUN EASIER…. READ MORE & PRE-ORDER TODAY WWW.GAITEYE.COM. 1349906_A6_4+0.indd 1. 22-08-2014 12:56:57. 69 Download free eBooks at bookboon.com. Click on the ad to read more.

(76) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Second variant. Obviously, z = ±1 are not poles. Now, 2 z − 1 z4 + z2 + 1 = z6 − 1 = 0 for. " pπ # z = exp i , 3. for p ∈ {0, 1, 2, 3, 4, 5},. and since we shall remove p = 0 and p = 3, because they stem from the auxiliary factor z 2 − 1, the singularities are √ " π# 1 3 , = +i z˜1 = exp i 2 2 √ 3 3 1 2π , z˜2 = exp i =− +i 2 2 3 √ 3 1 4π z˜4 = exp i , =− −i 2 2 3 √ 3 1 5π z˜5 = exp i . = −i 2 2 3 Each one of these is at most a simple pole, and they could even be removable singularities. Analogously, (z − 1) z 2 + z + 1 = z 3 − 1, so since z = 1 is a “false” singularity coming from the auxiliary factor z − 1, the numerator has the roots √ √ 1 3 1 3 , and z˜4 = − + i , z˜2 = − + i 2 2 2 2 which will cancel the same zeros in the denominator. Thus √ 3 sqrt3 1 1 , and z˜4 = − + i , z˜2 = − + i 2 2 2 2 are removable singularities, while √ √ 3 3 1 1 z˜1 = + i , and z˜5 = − i , 2 2 2 2 are simple poles. 2) The integrand is deﬁned on R, and since the integrand has a zero of order 2 at ∞, the improper integral is convergent, and we do not need the notation “p.v.” (= “principal value”). The improper integral can be computed in a number of ways. First method. By a simple integration (without using the calculus of residues) it follows from the ﬁrst solution above that +∞ 2 +∞ +∞ x +x+1 dx dx dx = = 2 4 2 2 1 2 −∞ x + x + 1 −∞ x − x + 1 −∞ x− + 2 3 '+∞ & 1 2 2 2π x− Arctan √ = √ =√ . 2 3 3] 3 −∞. 70 Download free eBooks at bookboon.com.

(77) Complex Funktions Examples c-7. Improper integral, where the integrand is a rational function. Second method. We shall in the calculus of residues use that only z1 =. √ " π# 1 3 +i = exp i 3 2 2. lies in the upper half plane and that z13 = −1. Then +∞ 2 z+1 z1 + 1 2πi z12 + z1 x +x+1 · dx = 2πi res ; z = 2πi = 1 2 4 2 z3 + 1 3 3z1 z13 −∞ x + x + 1 √ 2π 2πi 2πi 3 {˜ z2 + z˜1 } = − · 2·i =√ . = − 3 3 2 3. Third method. Calculus of residues without a reformulation gives the following diﬃcult computations, 2 +∞ 2 z +z+1 x +x+1 dx = 2πi res ; z1 4 2 z4 + z2 + 1 −∞ x + x + 1 2 z1 + z1 + 1 z˜2 + z˜1 + 1 = 2πi = 2πi 4z13 + 2z1 4z˜3 + 2˜ z1 ⎧ ⎫ √ √ ⎪ ⎪ 3 1 3 1 ⎪ ⎪ ⎪ + +i + 1⎪ ⎪ ⎪ ⎨ −2 + i 2 ⎬ 2 2 = 2πi √ ⎪ ⎪ ⎪ ⎪ 1 3 ⎪ ⎪ ⎪ ⎪ 4(−1) + 2 +i ⎩ ⎭ 2 2 √ √ 2πi 1 + i 3 1+i 3 √ = √ · √ = 2πi · −4 + 1 + i 3 3 − 3+i √ 2π i− 3 2π =√ . = √ · √ 3 3 − 3+i. This e-book is made with. SetaPDF. SETA SIGN. PDF components for PHP developers. www.setasign.com 71 Download free eBooks at bookboon.com. Click on the ad to read more.

(78) Complex Funktions Examples c-7. 5. Improper integrals, where the integrand is a rational function times .... Improper integrals, where the integrand is a rational function times a trigonometric function. Example 5.1 The transfer function of a RC-ﬁlter is given by f (z) =. 1 . 1 + 2π i RC z. Find the corresponding answer. The corresponding answer is given by the improper integral +∞ +∞ 1 1 h(t) = ei2πxt dx. f (x) e2πixt dt = i 2πiRC −∞ −∞ x− 2πRC i is the only pole of the corresponding analytic function 2πRC. Here, z1 = f (z) =. 1 · 2πiRC. 1 z−. i 2πRC. ,. and it is obvious that there exist constants k, and r > |f (z)| <. k |z|. 1 , such that we have the estimate 2πRC. for |z| > r.. Since f (z) does not have any singularity in the lower half plane, we conclude from the corresponding residuum formula, which here is empty that +∞ +∞ 1 1 h(t) = f (x) e2πixt dx = for t < 0. ei2πxt dx = 0 i 2πiRC −∞ −∞ x − 2πRC If instead t > 0, then, since we have already checked the assumptions of the validity of the residuum formula, ⎞ ⎛ +∞ i2πzt t i ⎟ 1 2πi ⎜ e 2πixt . · exp − res ⎝ ; f (x) e = h(t) = ⎠= i RC 2πRC RC 2πiRC −∞ z− 2πRC. 72 Download free eBooks at bookboon.com.

(79) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.2 Compute the improper integrals +∞ +∞ x cos x x sin x dx og dx. 2+1 2+1 x x −∞ −∞ Here we must consider the analytic function z eiz , z2 + 1. for z = ±i,. z cos z z sin z z and 2 . Clearly, the rational function 2 has a zero of ﬁrst order at ∞ and z2 + 1 z +1 z +1 no pole on the X-axis, so the assumptions of the residuum formula are fulﬁlled. Since m = 1 > 0, the pole z = i in the upper half plane is the only relevant singularity. Hence by the residuum formula, +∞ z eiz i ei·i πi x eix dx = 2πi · res ; i = 2πi · = . 2+1 2+1 e x z i + i −∞. instead of. Then by separating into the real and the imaginary parts, +∞ +∞ π x cos x x sin x dx = 0 og dx = . 2+1 2+1 e x x −∞ −∞. Example 5.3 Compute +∞ x sin x (a) dx, 2 −∞ x + 9. (b) 0. +∞. cos π x dx. x4 + 4. x is a rational function of real coeﬃcients and with a zero of ﬁrst order at ∞. The +9 denominator does not have real zeros and % % % z % C % % for |z| ≥ 4, % z 2 + 9 % ≤ |z|. (a) Here. x2. so we conclude that the improper integral is convergent. Using that sin x = Im ei·1·x , where 1 > 0, it follows by the residuum formula that . +∞. −∞. z ei z 3i e−3 x sin x z eiz dx = Im 2πi · res ; 3i = 2π Re lim = 2π Re z→3i z + 3i x2 + 9 z2 + 9 3i + 3i = 2π ·. π 3 e−3 = 3. e 6. 73 Download free eBooks at bookboon.com.

(80) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... 1 has a zero of fourth order at ∞ and no poles on the x-axis. Hence, the integral is z4 + 1 cos πx is an even function, it follows by the symmetry and a residuum formula convergent. Since 4 x +4 that +∞ iπx +∞ 1 +∞ cos πx 1 cos πx e dx = dx = (Re) dx 4 2 −∞ x4 + 4 2 x4 + 4 0 −∞ x + 4 iπz iπz e e = πi res ; 1 + i + res ; −1 + i z4 + 4 z4 + 4 & & ' ' & iπz ' & iπ z ' z eiπz e ze eiπz = πi + + = πi 4z 3 z=1+i 4z 3 z=−1+i 4z 4 z=1+i 4z 4 z?−1+i ! πi 1 πi e−π · 2i = − π e−π . · (1 + i)eπ(−1+i) + (−1 + i)eπ(−1−i) = = 16 8 4 · (−4). (b) Here. Example 5.4 Compute +∞ cos x (a) 3 dx, (1 + x2 ) 0. . +∞. (b) −∞. cos x dx. (1 + x2 ) (4 + x2 ). 1 3 has a zero or order 6 at ∞ and no real pole. Hence the improper integral (1 + x2 ) exists. The integrand is an even function, so by the symmetry, followed by an application of a residuum formula, +∞ +∞ 1 cos x eix eiz dx = πi · res 3 dx = 2 (Re) 3 ; i 2 3 (1 + x2 ) (1 + z 2 ) 0 −∞ (1 + x ) i eiz eiz πi 3 eiz 1 d2 d lim lim 2 = − = πi · 2 z→i dz (z + i)3 2! z→i dz (z + i)3 (z + i)4 e−1 πi πi 6i eiz 12 eiz 6i e−1 e−1 −eiz − lim = = − + − + 12 · 2 2 z→i (z + i)3 (z + i)4 (z + i)5 (2i)3 (2i)4 (2i)5 7π 1 π 2πi . −(2i)2 − 6i · (2i) + 12 = {4 + 12 + 12} = · = 5 4 16 e 4·2 e 4(2i) e. (a) We see that. (b) We get by a decomposition that 1 1 1 1 1 = · 2 − · , (1 + x2 ) (4 + x2 ) 3 x + 1 3 x2 + 4 so it follows immediately that the integral is convergent. Then by the residuum formula, +∞ 1 +∞ cos x 1 +∞ cos x cos x dx = dx − dx 2 2 3 −∞ x2 + 1 3 −∞ x2 + 4 −∞ (1 + x ) (4 + x ) iz iz e e 1 1 Re 2πi · res Re 2πi · res = ; i − ; 2i 3 3 z2 + 1 z2 + 4 e−1 e−2 1 1 1 π 1 1 π Re 2πi · = − Re 2πi · = · − · 2 = 2 (2e − 1). 3 2e 3 3 3 e 6e 2i 4i. 74 Download free eBooks at bookboon.com.

(81) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.5 Prove that +∞ +∞ π cos x cos x dx = 2 dx = e . 2 2 1 + x −∞ −∞ (1 + x ) Clearly, both integrals are convergent, and we can apply the residuum formula. Thus iz +∞ +∞ e e−1 π cos x eix dx = Re 2πi · res ;i = 2π Re i · = , dx = Re 2 2+1 2+1 e 1 + x x z 2i −∞ −∞ and +∞ −∞. . cos x (1 + x2 ). 2. +∞. dx = Re. . eix. . = Re 2πi · res. eiz. . 2 ; i (z 2 + 1) & ' eiz i eiz 2 eiz 1 d lim = 2π Re i − = 2π Re i · 1! z→i dz (z + i)2 (z + i)2 (z + i)3 z=i −1 π 1 1 2 e−1 e−1 2π 2e−1 ie = . + · − = 2π Re = − = 2π Re i e 4 4 e (2i)2 (2i)3 4 8i2 −∞. (x2 + 1). 2. dx. . www.sylvania.com. We do not reinvent the wheel we reinvent light. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. 75 Download free eBooks at bookboon.com. Click on the ad to read more.

(82) Complex Funktions Examples c-7. Example 5.6 Compute +∞ x cos x (a) dx, 2 − 2x + 10 x −∞. Improper integrals, where the integrand is a rational function times .... . +∞. (b) −∞. x sin x dx. x2 − 2x + 10. It follows from +∞ +∞ +∞ x eox x cos x x sin x dx = dx + i dx, 2 − 2x + 10 2 − 2x + 10 2 − 2x + 10 x x x −∞ −∞ −∞ that it suﬃces to prove that +∞ x eix dx 2 −∞ x − 2x + 10 exists and to ﬁnd the value of this integral. z has a ﬁrst order zero at ∞ and simple poles at z = 1 ± 3i ∈ / R, hence the z 2 − 2z + 10 improper integral exists. Since m = 1 > 0, we can compute the integral by a residuum formula, +∞ z eiz x eix z eiz dx = 2πi · res ; 1 + 3i = 2πi lim 2 z→1+3i z − 1 + 3i z 2 − 2z + 10 −∞ x − 2x + 10 We see that. π (1 + 3i)ei(1+3i) = (1 + 3i)e−3 {cos 1 + i sin 1}. 3 6i Then by a separation into the real and the imaginary part, = 2πi ·. (a). . +∞ −∞. (b). . +∞ −∞. π x cos x dx = 3 (cos 1 − 3 sin 1), 3e x2 − 2x + 10. π x sin x dx = 3 (3 cos 1 + sin 1). 3e x2 − 2x + 10. Example 5.7 Compute +∞ x sin x dx, (a) 2 + 4x + 20 x −∞ (a) The function. . +∞. (b) −∞. dx . 1 + x2. z has a zero of ﬁrst order at ∞. The poles are z 2 + 4z + 20. −2 ± 4i ∈ / R, so by a residuum formula, +∞ +∞ x sin x x eix dx = Im dx 2 2 −∞ x + 4x + 20 −∞ x + 4x + 20 z eiz z eiz = Im 2πi · res ; −2 + 4i = 2π · Im i · lim z→−2+4i z + 2 + 4i z 2 + 4z + 20 i(−2+4i) (−2 + 4i)e π = 2π · Im i · = (2 cos 2 + sin 2). 2 e4 8i. 76 Download free eBooks at bookboon.com.

(83) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... (b) We have of course, . +∞. −∞. dx = [Arctan ]+∞ −∞ = π. 1 + x2. Alternatively, it follows by a residuum formula that +∞ 2πi 1 dx 1 = π. = = 2πi · res ; i = 2πi lim 2 2 z→i 2i 1+z z+i −∞ 1 + x. Example 5.8 Prove that +∞ π cos x dx = π. cosh x −∞ cosh 2 cos z along a rectangle with the corners −R, R, R + πi and −R + πi, Hint: Integrate the function cosh z and let R → +∞.. 4. 3. 2. 1. –4. –2. 0. 2. 4. Figure 10: The curve Cπ with the singularity z0 = i. π inside Cπ . 2. We shall use the hint, so we call the curve CR . It follows from cosh z = 0. for z = i. π + i pπ, 2. p ∈ Z,. π that z = i is the only singularity (a simple pole) lying inside CR for every R > 0. Hence by Cauchy’s 2 integral formula " π# π # " cosh cos i π cos z cos z 2 = 2π cosh π . 2 " π # = 2πi · = 2πi · dz = 2πi · res ;i π 2 2 cosh z CR cosh z i sin sinh i 2 2. 77 Download free eBooks at bookboon.com.

(84) Complex Funktions Examples c-7. On the other hand, cos z dz = CR cosh z. . Improper integrals, where the integrand is a rational function times .... R cos x cos(x + iπ) dx − dx −R cosh x −R cosh(x + iπ) π π cos(R + iy) cos(−R + iy) dy − i dy. +i 0 cosh(R + iy) 0 cosh(−R + iy) R. We ﬁrst note that R R cos(x + iπ) cos x · cosh π − i sin x · sinh π − dx = − dx −R cosh(x + iπ) −R cosh x · cos π + i sinh x · sin π R R cos x sin x dx − i sinh π dx = cosh π cosh x cosh x −R −R R cos x dx + 0, = cosh π · cosh x −R because the latter integral has an odd integrand. Summing up we get for the ﬁrst two terms, R R R cos x cos(x + iπ) cos x dx − dx = (1 + cosh π) dx. cosh x cosh(x + iπ) cosh x −R −R −R Clearly, this integral is convergent for R → +∞, because the numerator of the integrand is bounded, and its denominator tends exponentially towards 0 by the limits x → ±∞. We only have to show that the contributions from the vertical axes tend to zero for R → +∞. It follows from cos R · cosh y − i sin R · sinh y cos(R + iy) , = cosh R · cos y + i sinh R · sin y cosh(R + iy) when 0 ≤ y ≤ π that % % % cos(R + iy) %2 % % % cosh(R + iy) % =. cos2 R + sinh2 y cos2 R · cosh2 y + sin2 R · sinh2 y = cosh2 R · cos2 y + sinh2 R · sin2 y sinh2 R + cos2 y cosh2 π 1 + sinh2 π = . ≤ 2 sinh R sinh2 R The length of the path of integration is π, so we conclude that % % π % % cos(R + iy) % ≤ π · cosh π → 0 % for R → +∞. dy % % sinh R cosh(R + iy) 0. Since also % % % cos(−R + iy) % cosh π % % % cosh(−R + iy) % ≤ sinh R , it follows in the same way that the latter integral tends to 0 for R → +∞. Summing up we get by this limit, +∞ π cos x (1 + cosh π) dx = 2π cosh , 2 cosh x −∞ and since 1 + cosh π = 1 + 2 cosh2. ! π π − 1 = 2 cosh2 , 2 2. 78 Download free eBooks at bookboon.com.

(85) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... we ﬁnally get that . +∞ −∞. π 2π cosh π cos x 2 = dx = π. 2 π cosh x 2 cosh cosh 2 2. Example 5.9 Compute +∞ cos x (a) , 2 −∞ x + 4. . +∞. (b) −∞. x2. sin 2x dx. +x+1. The denominator is in both cases a polynomial of degree grad 2 without zeros on the x-axis. The numerators are purely trigonometric, so we get by a residuum formula, (a) . +∞. −∞. cos x dx = (Re) x2 + 4. . +∞ −∞. 360° thinking. .. eix dx = 2πi · res x2 + 4. . eiz ; 2i z2 + 4. . ei·2i π · 2πi = 2 . 2e 4i. 360° thinking =. .. 360° thinking. .. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. © Deloitte & Touche LLP and affiliated entities.. Discover the truth79at www.deloitte.ca/careers Click on the ad to read more Download free eBooks at bookboon.com. © Deloitte & Touche LLP and affiliated entities.. D.

(86) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... (b) . +∞. −∞. e2ix dx 2 −∞ x + x + 1 √ 3 1 e2iz ; − +i = Im 2πi · res 2 2 z2 + z + 1 √ ei(−1+i 3) e2iz √ = Re 2π lim √ = Re 2π · −1 + i 3 + 1 z→− 12 +i 23 2z + 1 √ √ 2π 2π − 3−i = Re √ (−i)e = − √ e− 3 sin 1. 3 3. sin 2x dx = Im 2 x +x+1. Example 5.10 Compute +∞ 3 x sin x dx, (a) x4 + 1 0. . (b) 0. +∞. +∞. x2 cos 3x (x2 + 1). 2. dx.. The integrands are in both cases even functions, so they may be extended by symmetry to all of R. Furthermore, the diﬀerence of degrees of the numerator and the denominator of the rational function of the integrands is at least 1, where the denominators are dominating, so the integrals are convergent, and we can ﬁnd their values by a residuum formula. (a) The zeros of the denominator are determined by z 4 + 1 = 0, so 1 1 z = ±√ ± i √ , 2 2 and we get +∞ 3 ix 1 +∞ x3 sin x 1 x3 sin x x e dx = dx = Im dx 4 2 −∞ x4 + 1 2 x4 + 1 0 −∞ x + 1 & 3 iz 3 iz ' z e 1 1 i z e 1 i √ √ √ √ Im 2πi res = ; + + res ; − + . 2 z4 + 1 z4 + 1 2 2 2 2. . +∞. Let z0 be any pole. Then z04 = −1, and 3 iz z03 eiz0 z e 1 ; z = res = eiz0 , 0 3 4 4 z +1 4z0 hence by insertion, i 1 1 i 1 1 x3 sin x √ √ √ √ + exp i − + i · + exp i dx = π Im i · 4 4 x4 + 1 2 2 2 2 0 1 1 π i i 1 1 π . cos √ = exp − √ exp − √ · · exp √ Im i · exp − √ = 2 2 2 2 2 2 2 2. . +∞. 80 Download free eBooks at bookboon.com.

(87) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... (b) Here z = i is a double pole. It lies in the upper half plane, so we start by computing its residuum: 2 3iz z e 2ze3iz 1 3iz 2 e3iz 2z 2 e3iz d z 3 e3iz lim = lim = + − res 2 ; i z→i (z + i)2 1! z→i dz (z + i)2 (z + i)2 (z + i)3 (z 2 + 1) 1 2(−1) 2i e−3 3i(−1)e−3 2(−1)e−3 e−3 = i· 3. −4i + 6i − = + − = 2e −i (2i)2 (2i)2 (2i)3 8 Then by the symmetry and the residuum formula, +∞ 2 +∞ 1 +∞ x2 cos 3x 1 x cos 3x x2 e3ix dx = Re 2 dx = 2 dx 2 2 −∞ (x2 + 1)2 2 (x2 + 1) 0 −∞ (x + 1) 1 1 z 2 e3iz π 1 Re 2πi · res Re 2πi · i · = = − 3. ; i = 2 3 2e 2 2e 2 (z 2 + 1). Example 5.11 Compute +∞ x sin x (a) dx, 2 + 1) (x2 + 4) (x 0. . +∞. (b) −∞. sin x dx. x2 + 4x + 5. In both cases the integrand satisﬁes the assumptions for the application of the residuum formula. (a) First we get by a decomposition, 1 1 1 1 1 = − . 3 x 2 + 1 3 x2 + 4 (x2 + 1) (x2 + 4) The integrand is even, so by the symmetry, followed by an application of the residuum formula, +∞ 1 +∞ x sin x x sin x dx = dx 2 + 1) (x2 + 4) 2 + 1) (x2 + 4) 2 (x (x 0 −∞ 1 +∞ x sin x 1 +∞ x sin x = dx − dx 6 −∞ x2 + 1 6 −∞ x2 + 4 z eiz z eiz 1 1 Im 2πi · res Im 2πi · res = ; i − ; 2i 6 6 z2 + 1 z2 + 4 z eiz z eiz π Re res ; i − res ; 2i = 2 3 z +1 z2 + 4 −1 i·i 2i ei·2i π e e−2 π ie Re −i· = Re − = 3 3 i+i 2i + 2i 2 2 1 π π 1 − = 2 (e − 1). = 6e 6 e e2 (b) The poles are z = −2 ± i, of which only z0 = −2 + i lies in the upper half plane. Then by the residuum formula, +∞ eiz sin x dx = Im 2πi res ; −2 + i 2 (z+2+i)(z+2−i) −∞ x +4x+5 π ei(−2+i) (Im π e−2i−1 = − sin 2. = Im 2πi · e −2 + i + 2 + i. 81 Download free eBooks at bookboon.com.

(88) Complex Funktions Examples c-7. Example 5.12 Prove that +∞ iax e dx = π e−a 2+1 x −∞. Improper integrals, where the integrand is a rational function times .... for a ≥ 0.. The claim is trivial for a = 0, because +∞ 1 dx = [Arctan x]+∞ −∞ = π. 2 −∞ x + 1 If a > 0, then the assumptions of using the residuum formula are satisﬁed, so iaz +∞ iax e ei a i e dx = 2πi · res ; i = 2πi · = π e−a . 2 z2 + 1 i+i −∞ x + 1 Remark 5.1 If instead a < 0, then we get by a complex conjugation and an application of the ﬁrst result that +∞ −iax +∞ iax e e dx = dx = π e−(−a) = π ea = π e−|a| , 2 2 −∞ x + 1 −∞ x + 1 so we have in general that +∞ iax e dx = π e−|a| , 2 −∞ x + 1. a ∈ R.. Example 5.13 Prove that +∞ π e−a cos x dx = 2 2 a −∞ x + a. ♦. for a > 0.. The conditions of convergence of the improper integrals and the legality of the application of the residuum formula are fulﬁlled. Then by the symmetry, +∞ sin x dx = 0, 2 + a2 x −∞ so . +∞ −∞. cos x dx = (Re) x2 + a2. . ∞ −∞. eix dx = 2πi · res 2 x + a2. . π e−a eiz ei·ia = . ; i a = 2πi · 2 2 z +a 2ia a. 82 Download free eBooks at bookboon.com.

(89) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.14 Compute for a, b ∈ R+ , . +∞. (a) −∞. . x sin ax dx, x2 + b2. +∞. (b) −∞. cos ax dx. x2 + b 2. In both cases the conditions of convergence of the improper integrals and the application of a residuum formula are fulﬁlled. Hence, because a, b > 0, (a) . +∞. −∞. x eiax z eiaz dx = Im 2πi lim 2 2 x→ib z + ib −∞ x + b −ab ib e = π e−ab . = Im 2πi · 2ib. x sin ax dx = Im x2 + b 2. . +∞. (b) . +∞. −∞. cos ax dx = (Re) x2 + b 2. . +∞ −∞. e−ab π eiax eiaz = 2π i · = e−ab . dx = 2πi lim 2 z→ib z + ib b +b 2ib. x2. We will turn your CV into an opportunity of a lifetime. Do you like cars? Would you like to be a part of a successful brand? We will appreciate and reward both your enthusiasm and talent. Send us your CV. You will be surprised where it can take you.. 83 Download free eBooks at bookboon.com. Send us your CV on www.employerforlife.com. Click on the ad to read more.

(90) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.15 Prove that the integral +∞ x sin x dx 4 +∞ 1 + x is convergent, and ﬁnd its value. We have an improper integral, where the integrand is a product of sin x and a real rational function without poles on the x-axis and with a zero of third order at ∞. From this we conclude that the integral is convergent, and its value is given by the residues at the poles in the upper half plane of the z eiz . We have more precisely, function 1 + z4 ' & +∞ z eiz −1 + i z eiz 1 + i x sin x + res ; √ ; √ dx = Im 2πi res 4 z4 + 1 z4 + 1 2 2 −∞ 1 + x & ' & ' & iz ' iz iz z eiz ze ze 1+i −1 + i ze + res = 2π Re = 2π Re res ; √ ; √ + z4 + 1 z4 + 1 4z 3 1+i 4z 3 −1+i 2 2 √ √ 2 2 & ' ' & 2 2 −1 + i 1+i π z z √ − i exp i = − Re i exp i √ + − eiz = 2π Re − eiz 2 4 4 1+i −1+i 2 2 √ √ 2 2 i 1 i 1 π · exp − √ − exp − √ · exp √ = − Re i exp − √ 2 2 2 2 2 1 1 1 1 π . · 2i sin √ = π exp − √ sin √ = − Re i exp − √ 2 2 2 2 2. Example 5.16 Prove that +∞ 1 1 π cos x +√ . dx = π exp − √ sin 4 4 2 2 −∞ 1 + x 1 does not have poles on the x-axis and that the 1 + x4 1 1 has a zero of order 4 at ∞. Since is a real rational function, we can obtain the factor 4 1 + x4 1+x value of the integral by a residuum formula. Now 1 + z 4 = 0 for " π π !# , p ∈ Z, z = exp i +p 2 4. We ﬁrst note that the integrand f (x) = cos x ·. so we get by the residuum formula, +∞ +∞ cos x eix dx = Re dx 4 4 −∞ 1 + x −∞ 1 + x iz iz " π # e e 3π + res . = Re 2πi res ; exp i ; exp i 4 4 1 + z4 1 + z4. 84 Download free eBooks at bookboon.com.

(91) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... All poles z0 with z04 = −1 are simple, so by Rule II, iz eiz0 e z0 eiz0 1 = res ; z = = − z0 eiz0 . 0 3 4 4 4 1+z 4z0 4z0 Finally, " π# 1 = √ (1 + i) exp i 4 2. og. 1 3π = √ (−1 + i), exp i 4 2. hence by insertion, ' & +∞ 1 1 1 1 1 cos x i √12 − √12 −i √12 − √12 √ √ √ √ + i e dx = Re 2πi − + − + i e 4 4 2 2 2 2 −∞ 1 + x ⎤ ⎡ 1 1 ! √ √ ⎢ πi − 2 · 2 1 i √1 −i √12 ⎥ · √ (1 + i)e 2 + (−1 + i)e = Re ⎣− · e ⎦ 2 2 & # " i #!' πi − √1 " √i √ − √i2 − √i2 2 2 2 = Re − √ e e −e +i e +e 2 2 ' ' & & 1 1 1 1 1 1 πi − √12 − √12 = Re π e · 2i · sin √ + i · 2 cos √ · √ sin √ + √ · cos √ = Re − √ · e 2 2 2 2 2 2 2 2 π 1 π 1 1 π − √12 − √12 = πe = πe . +√ sin √ · cos + cos √ · sin sin 4 4 4 2 2 2 Alternatively and slightly shorter, & +∞ !' 1 cos x i √12 − √12 −i √12 − √12 iπ i 3π 4 · e 4 e e dx = Re 2πi − + e 4 4 −∞ 1 + x “ & “ ” ” ' !, +π 1 1 1 π π 1 √ i π + √1 −i π −√ i√ −i √12 − √1 4+ 2 e 4 2 −e = Re π e 2 · = Re · e 2 ei 4 e 2 − e−i 4 e 2i 2i 1 1 π −√ . +√ = π e 2 sin 4 2. 85 Download free eBooks at bookboon.com.

(92) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.17 Prove that the improper integral # " +∞ sin x + π 4 dx 2 2 −∞ (x + 1) (x + 4) is convergent. Then ﬁnd the value of the integral. Since " 1 π# = √ (sin x + cos x), sin x + 4 2 1 is a real and even rational function with a zero of order 4 at ∞ and with (x2 + 1) (x2 + 4) no pole on the x-axis, the improper integral is convergent, and we can ﬁnd its value by a residuum formula, where we use that the integral of an odd function over a symmetric interval is 0, # " +∞ +∞ sin x + π +∞ 1 sin x cos x 4 dx = √ dx + dx 2 2 2 2 2 2 2 −∞ (x + 1) (x + 4) −∞ (x + 1) (x + 4) −∞ (x + 1) (x + 4) +∞ +∞ 1 1 1 1 1 1 cos x ix √ (Re) · · dx = − dx e = 0+ √ 3 x2 + 1 3 x2 + 4 2 2 −∞ (x2 + 1) (x2 + 4) −∞ iz −1 iz e e 2πi e 2πi e−2 (2e − 1)π √ √ res · = ; i − res ; 2i = − = √ . z2 + 1 z2 + 4 2i 4i 3 2 3 2 6 2 · e2. and since. Alternatively we may carry through the following computations, # ## " " " +∞ exp i x + π +∞ sin x + π 4 4 dx = Im dx 2 2 2 2 −∞ (x + 1) (x + 4) −∞ (x + 1) (x + 4) 6 5 π π ei(z+ 4 ) ei(z+ 4 ) ; i + res ; 2i . = Im 2πi res (z 2 +1) (z 2 +4) (z 2 +1) (z 2 +4) It follows from π π π e−1 i π ei(i+ 4 ) ei(z+ 4 ) ei(z+ 4 ) = ; i = lim = ·e 4, res z→i (z+i) (z 2 +4) (z 2 +1) (z 2 +4) 2i · 3 6i and. . res that . +∞ −∞. π. . ei(z+ 4 ) ; 2i (z 2 +1) (z 2 +4). π π −e−2 i π ei(2i+ 4 ) ei(z+ 4 ) = = ·e 4, 2 z→2i (z +1) (z + 2i) −3 · 4i 12i. = lim. " π# −1 sin x + e e−2 1+i 4 dx = Im 2πi − · √ (x2 + 1) (x2 + 4) 6i 12i 2 (2e − 1)π π · (2e − 1) 1 + i = √ = Im · √ . 2 6e 6 2 · e2 2. 86 Download free eBooks at bookboon.com.

(93) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.18 Given the function f (z) =. z2 . z4 + 4. 1) Find the singular points and their types in C ∪ {∞} for f (z). 2) Find the value of the following two complex line integrals, (a) f (z) dz, (b) f (z) dz. |z−4|=2. |z|=2. 3) Prove for every ω > 0 that . +∞. t4. −∞. π t2 eiωt dt = e−ω (cos ω − sin ω). 2 +4. 2. 1. –2. 0. 2. 4. 6. –1. –2. 1) Clearly, z = ∞ is a removable singularity (a zero of second order). The denominator z 4 + 4 has the zeros 1 + i,. −1 + i,. −1 − i,. 1 − i.. These are all simple pole of f (z). 2) a) Since there is no pole of f (z) inside the circle |z −4| = 2 (cf. the ﬁgure), it follows from Cauchy’s integral theorem that f (z) dz = 0. |z−4|=2. b) All singularities of f (z) lie inside the circle |z| = 2, and z = ∞ is a zero of second order. Hence, by reversing the direction of the curve, f (z) dz = − f (z) dz = −2πi · res(f ; ∞) = 0. |z|=2. |z|=2. 87 Download free eBooks at bookboon.com.

(94) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Alternatively, the residuum in a general pole z0 , for which z04 = −4, is given by 2 z02 1 z ; z = = , res 0 3 4 4z0 z +4 4z0 so f (z) dz |z|=2. 4 . . z2 ; zn 4 z +4. . . 1 1 1 1 + + + 1 − i −1 − i −1 + i 1 + i n=1 1 1 1 1 = 0. + − = 2πi − 1+i 1−i 1+i 1−i. = 2πi. res. . = 2πi. 3) Since the integrand has a zero of order 2 at ∞, and since there are no real singularities, the improper integral exists, and when ω > 0 its value can be found by the residues in the upper half plane, . 2 2 z z t2 iωt iωz iωz e dt = 2πi res e ; 1 + i + res e ; −1 + i 4 z4 + 4 z4 + 4 +∞ t + 4 eiω(−1+i) 2πi eiω(1+i) z 2 eiωz z 2 eiωz + = + lim = 2πi lim z→1+i 4z 3 z→−1+i 4z 3 4 1+i −1 + i 1 − i −ω iω 1 + i −ω −iω πi −ω 1 iω π ·e e − e · ·e e e − i eiω − e−iω − i e−iω · = = 2 2 2 2 2 iω eiω + e−iω π −ω π −ω e − e−iω e ·i ·i−i· = e · (cos ω − sin ω). = 2 2 2i 2 +∞. I joined MITAS because I wanted real responsibili� I joined MITAS because I wanted real responsibili�. Real work International Internationa al opportunities �ree wo work or placements. �e Graduate Programme for Engineers and Geoscientists. Maersk.com/Mitas www.discovermitas.com. � for Engin. M. Month 16 I was a construction M supervisor ina cons I was the North Sea supe advising and the N he helping foremen advi ssolve problems Real work he helping International Internationa al opportunities �ree wo work or placements ssolve p. 88 Download free eBooks at bookboon.com. Click on the ad to read more.

(95) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.19 (a) Given m > 0. Prove that the improper integral +∞ x2 eimx dx (3) 4 2 −∞ x + 6x + 25 is convergent, and ﬁnd its value. (b) What is the value of the improper integral (3), when m < 0 instead? x2 has a zero of order 2 at ∞, and the denominator is ≥ 25 for every x ∈ R. + 6x2 + 25 Hence, the improper integral is convergent, even for every m ∈ R, and when m ≥ 0 we can ﬁnd the value by a residuum formula. When the denominator is put equal to zero, 2 z 4 + 6z 2 + 25 = z 2 + 5 − (2z)2 = 0. (a) Clearly,. x2. we get z 2 = −3 ±. √. 9 − 25 = −3 ± 4i = (±1 + 2i)2 ,. so we have four simple poles, 1 + 2i,. −1 + 2i,. 1 − 2i,. −1 − 2i,. of which only the former two lie in the upper half plane. Hence, for m ≥ 0, +∞ z 2 eimz z 2 eimz x2 eimx dx = 2πi res , 1 + 2i + res , −1 + 2i 4 2 z 4 + 6z 2 + 25 z 4 + 6z 2 + 25 −∞ x + 6x + 25 2πi z 2 eimz z 2 eimz z eimz z eimz lim = 2πi lim + lim = + lim z→1+2i 4z 3 + 12z z→−1+2i 4z 3 + 12z z→1+2i z 2 + 3 z→−1+2i z 2 + 3 4 πi (1 + 2i)eim(1+2i) (−1 + 2i)eim(−1+2i) = + 2 1 − 4 + 4i + 3 1 − 4 − 4i + 3 π πi (1 + 2i)eim · e−2m − (−1 + 2i)e−im · e−2m = e−2m eim + e−im + 2i eim − e−im = 8 8 π π = e−2m {2 cos m + 2i · 2i sin m} = e−2m {cos m − 2 sin m}, 4 8 which is also true for m = 0, where +∞ π x2 dx = . 4 + 6x2 + 25 4 x −∞ (b) If m < 0, then we get by complex conjugation, +∞ +∞ π x2 eimx x2 ei|m|x dx = dx = · e−2|m| {cos |m| − 2 sin |m|}, 4 + 6x2 + 25 4 + 6x2 + 25 4 x x −∞ −∞ where we have used the result from (a) with |m| instead of m. Summing up we have for every m ∈ R, +∞ π x2 eimx dx = · e−2|m| {cos |m| − 2 sin |m|}. 4 2 + 25 4 x + 6x −∞. 89 Download free eBooks at bookboon.com.

(96) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.20 Find the Fourier transform of the function f (x) =. x2. i.e. compute fˆ(ξ) =. x+1 , + 2x + 2. +∞. −∞. x+1 e−ξx dx, x2 + 2x + 2. ﬁrst for ξ < 0, and then for ξ > 0. We see that f (z) =. z+1 z+1 P (z) = 2 = z + 1 + 2z + 2 (z + 1)2 + 1 Q(z). is a rational function, where 1) the polynomial Q(z) = (z + 1)2 + 1 of the denominator has the simple zeros z = −1 ± i, where none of these is lying on the real axis; 2) the polynomial of the denominator is of 1 degree bigger than the polynomial of the numerator; 3) if ξ < 0, then m = −ξ > 0. Hence, the conditions of convergence of the improper integral are satisﬁed for ξ < 0, and since −1 + i is the only (simple) pole in the upper half plane, the value of the improper integral is given by a residuum formula, +∞ z+1 x+1 −iξx −iξz ˆ e ·e dx = 2πi · res ; −1 + i f (ξ) = 2 (z + 1)2 + 1 −∞ x + 2x + 2 z+1 · e−iξz = πi · e−iξ(−1+i) = πi · eξ(1+i) , ξ < 0, = 2πi lim z→−1+i 2(z + 1) where we have applied Rule II. Now P (z) and Q(z) have real coeﬃcients, so if ξ > 0, then we get by complex conjugation, fˆ(ξ) =. +∞. −∞. x+1 e−iξx dx = x2 + 2x + 2. . +∞ −∞. Summing up, ⎧ ⎨ πi · eξ(1+i) = πi · e−|ξ|(1+i) ˆ f (ξ) = ⎩ −πi · eξ(−1+i) = −πi · e−|ξ|(1−i). x+1 eiξx dx = πi · e−ξ(1+i) = −πi · eξ(−1+i) . x2 + 2x + 2. for ξ < 0, for ξ > 0.. When ξ = 0, the integral does not converge.. 90 Download free eBooks at bookboon.com.

(97) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Remark 5.2 For ξ < 0 we have fˆ(ξ) = πi e−i|ξ|(1+i) = π e−|ξ| · i {cos |ξ| − i sin |ξ|} = π e−|ξ| {sin |ξ| + i cos |ξ|}, so by a complex conjugation when ξ > 0 we get all things considered, ⎧ for ξ < 0, ⎨ π e−|ξ| {sin |ξ| + i cos |ξ|}, ˆ ♦ f (ξ) = ⎩ π e−|ξ| {sin |ξ| − i cos |ξ|}, for ξ > 0. In a variant we may use the change of variable t = x + 1. Then we have the following calculation for ξ < 0: +∞ +∞ +∞ x+1 t t −iξx −iξ(t−1) iξ e e e−iξt dt dx = dt = e fˆ(ξ) = 2 2 2 −∞ x + 2x + 2 −∞ t + 1 −∞ t + 1 +z , z −iξz iξ −iξz e · e ; i = 2πi · e = πi · eiξ · eξ = πi e(1+i)ξ . = 2πi · eiξ · res z2 + 1 2z z=i. 91 Download free eBooks at bookboon.com. Click on the ad to read more.

(98) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Alternatively we may use for ξ > 0 another residuum formula, because the conditions of its use are still valid. We get +∞ z+1 x+1 −iξx −iξz e e dx = −2πi · res ; −1 − i fˆ(ξ) = 2 z 2 + 2z + 2 −∞ x + 2x + 1 z + 1 −iξz 1 −iξz e e = −2πi lim = −2πi lim z→−1−i 2z + 2 z→−1−i 2 = −πi e−iξ(−1−i) = −πi eξ(−1+i) .. Example 5.21 Given the function f by f (z) =. z eiz (z 2 + 1). 2.. 1) Find the singularities and their type of f in C ∪ {∞}. 2) Compute the complex line integral f (z) dz, CR. where CR denotes the simple closed curve, which consists of the half circle z = R eiθ ,. 0 ≤ θ ≤ π,. R > 1,. and the interval [−R, R] on the real axis. 3) Prove that the improper integral . +∞ 0. x sin x (x2 + 1). 2. dx. is convergent, and compute its value.. 1) Clearly, z = ±i are double poles. Furthermore, ∞ is an essential singularity. In fact, we have f (−iy) → +∞. for y → +∞,. and also f (x) → 0. for x → +∞,. so we can obtain at least two diﬀerent limit values for z → ∞.. 92 Download free eBooks at bookboon.com.

(99) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... 2) We have only the singularity z = i lying inside CR , so we get by a residuum formula, . f (z) dz CR. = 2πi · res . . z eiz (z 2 + 1). 2. ;i. d = 2πi lim z→i dz. . eiz i z eiz 2z eiz + − = 2πi lim z→i (z + i)2 (z + i)2 (z + i)3 −1 −1 −1 e πi e 2i e = 2πi = . − − 2e (2i)2 (2i)2 (2i)3. z eiz (z + i)2 . . 3) Since we have a zero of order 3 at inf inity, we get by taking the limit R → +∞ that +∞ π x sin x dx = Im lim f (z) dz = . 2 2 R→+∞ 2e −∞ (x + 1) CR Since the integrand is even, we ﬁnally get +∞ π x sin x 2 dx = 4e . 2 (x + 1) 0. no.1. Sw. ed. en. nine years in a row. STUDY AT A TOP RANKED INTERNATIONAL BUSINESS SCHOOL Reach your full potential at the Stockholm School of Economics, in one of the most innovative cities in the world. The School is ranked by the Financial Times as the number one business school in the Nordic and Baltic countries.. Stockholm. Visit us at www.hhs.se. 93 Download free eBooks at bookboon.com. Click on the ad to read more.

(100) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.22 Given the function f (z) =. z eiz (z 2 + 1). 2.. 1) Find the singular points and their types of f in C. 2) Let x1 , x2 , y1 denote any positive real numbers where y1 > 1, and let γ = γx1 ,x2 ,y1 denote the closed curve (run through in the positive sense), which surrounds the domain Ax1 ,x2 ,y1 = {z ∈ C | −x1 < Re(z) < x2 and 0 < Im(z) < y1 }. Prove that π f (z) dz = i . 2e γ 3) Prove that the improper integral +∞ x sin x 2 dx (x2 + 1) 0 is convergent and ﬁnd its value.. 1) The denominator has the two double zeros πi, and since the numerator is = 0 in these points, we conclude that ±i are double poles. 2) We see that +i is the only singularity inside γ, hence it follows by the residuum theorem that . f (z) dz. = 2πi res. z eiz. . d = 2πi lim z→i dz. . z eiz (z + i)2 . . 2 ; i (z 2 + 1) i z eiz 2z eiz eiz = 2πi lim + − z→i (z + i)2 (z + i)2 (z + i)3 2 π 1 2πi 1 i 2i · =i . = 2π i e−1 = + − 2e (2i)2 e 4 (2i)3 (2i)2. γ. 3) It follows from x (x2. + 1). 2. ∼. 1 |x|3. for |x| large,. that the improper integral is convergent. When we apply the parametric description z(t) = −x1 + it, 0 < t < y1 , for one part of γ we here get the estimate of the integrand, % % % z eiz % %% z %% e−t |x1 | % % % −t %· ≤" |f (z)| = % %≤ #2 e , % (z 2 + 1)2 % % z 2 + 1 % |z 2 +| 2 |x1 | − 1. 94 Download free eBooks at bookboon.com.

(101) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... and the line integral along this part of γ fulﬁls the estimate % y1 % % % % % ≤ " |x1 | # → 0 f (−x + it) i dt for x1 → +∞. 1 2 % % 2 0 |x1 | − 1 Analogously we get % % y1 % % %≤ " % f (x + it) i dt 2 % %. |x2 |. #2 → 0 |x2 | − 1. 0. for x2 → +∞.. 2. Finally, we get for the curvilinear part by choosing the parametric description z(t) = t + iy 1 , t ∈ [−x1 , x2 ] [ that % % % % z(t)| % % −y1 |f (z)| = % %e , 2 % (z(t)2 + 1) % so the corresponding line integral is estimated by % x2 % % % % f (z) dz %% ≤ constant · e−y1 → 0 for y1 → +∞. % −x1. Then by taking the limits x1 → +∞ and x2 → +∞ and y1 → +∞, +∞ π x eix 2 dx = i 2e . 2 −∞ (x + 1) We conclude from +∞ x eix −∞. that 0. +∞. (x2. + 1). 2. x sin x (x2. + 1). 2. . +∞. dx = −∞. dx =. . x cos x (x2. + 1). 2. +∞. dx + i −∞. . x sin x (x2. + 1). 2. +∞. dx = 2i. π . 4e. 95 Download free eBooks at bookboon.com. 0. x sin x (x2 + 1). 2. dx,.

(102) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... Example 5.23 Given the function f (z) =. z eiz (z 2 + 4). 2.. Denote by Γ = γ + C + the simple closed curve run through in the positive direction, consisting of γ , the line segment [− , + ] on the real axis and the half circle C + in the upper half plane of centrum 0 and radius . 1) Find the isolated singularities and their types of f in C. 2) Prove for > 2 that π f (z) dz = i 2 . 4e Γ 3) Prove that f (z) dz → 0 C+. as → +∞.. 4) Compute the improper integrals +∞ x eix og p.v. 2 dx 2 −∞ (x + 4). . +∞. x sin x (x2 + 4). 0. 2. dx.. 3. 2. 1. –3. –2. –1. 0. 1. 2. 3. –1. –2. Figure 11: The closed path of integration C and the two singularities ±2i. 1) The function f (z) has the two double poles ±2i. 2) When > 2, only the double pole 2i lies inside Γ . Hence by Cauchy’s residuum theorem, z eiz 1 d z eiz lim f (z) dz = 2πi res = 2πi · 2 ; 2i 1! z→2i dz (z + 2i)2 (z 2 + 4) Γ eiz i z eiz 2z eiz + − = 2πi lim z→2i (z + 2i)2 (z + 2i)2 (z + 2i)3 −2 π 1 1 e 2πi 1 i · 2i · e−2 4i · e−2 = i 2. + + = 2πi = − + − 4e e2 16 8 16 (4i)2 (4i)2 (4i)3. 96 Download free eBooks at bookboon.com.

(103) Complex Funktions Examples c-7. Improper integrals, where the integrand is a rational function times .... 3) A parametric description of C + may be chosen as z(t) eit , t ∈ [0, π], so we get the following estimate when > 2, % % π π % %. 2. | exp(i {cos t + i sin t})| % % ·. dt = f (z) dz % ≤ exp(− · sin t) dt % 2 2 % % C+ (varrho2 − 4) ( 2 − 4) 0 0 ≤. π 2 ( 2 − 4). 2. →0. for → +∞.. 4) Both the improper integrals are trivially absolutely convergent, so it is not necessary to write “p.v.” (= “principal value”) here. It follows by a residuum formula, where we use the limits above, +∞ π x eix z eiz dx = lim 2 2 dz = i · 4e2 , 2 2. →+∞ −∞ (x + 4) Γ (z + 4) and then by a reﬂection argument, 0. +∞. x sin x (x2 + 4). 2. dx = =. 1 2. . +∞. 1 2 dx = 2 Im 2 −∞ (x + 4) π ! π 1 Im i · 2 = 2 . 4e 8e 2 x sin x. . +∞. −∞. . x eix (x2 + 4). 97 Download free eBooks at bookboon.com. 2. dx. Click on the ad to read more.

(104) Complex Funktions Examples c-7. 6. Improper integrals,, where the integrand is a rational function times .... Improper integrals, where the integrand is a rational function times an exponential function. Example 6.1 Given a ∈ ]0, 1[, prove that . +∞. (a) −∞. . π eax , dx = sin πa ex + 1. +∞. (b) −∞. π cosh ax dx = πa . cosh x cos 2. eaz along a rectangle with the corners −R, R, R + 2πi and −R + 2πi, ez + 1 and then let R → +∞. The integral of (b) is found analogously, but it can also be derived from (a). Hint: Integrate the function. 6. 4. 2. –6. –4. –2. 0. 2. 6. 4. Figure 12: The curve C2π and the simple pole πi inside C2π .. (a) Since ez + 1 = 0 for z = πi + 2πi, p ∈ Z, it follows that z0 = πi is the only singularity inside CR for R > 0, and this singularity is clearly a simple pole. Then we get by the residuum theorem, az e 1 eaz eaz · eaπi = −2πi eaπi . dz = 2πi · res ; πi = 2πi lim = 2πi · z +1 z +1 z z→πi −1 e e e CR On the other hand, CR. eaz dz = ez + 1. . R. −R. eax dx+ ex + 1. 0. 2π. ea(R+iy) i dy +i eR+iy + 1. . −R R. ea(x+2πi) dx+ ex+2πi + 1. . 0. 2π. ea(−R+iy) i dy. e−R+iy + 1. Using that 0 < a < 1, it follows by some trivial estimates (though with a diﬀerent argument) that the second and the fourth integral tend to 0 for R → +∞. Furthermore, by some trivial estimates, each of the two remaining integrals converges for R → +∞, and we have −2πi eaπi. . +∞ ax +∞ ax eaz e e a·2πi dz = dx − e dx x+1 x+1 R→+∞ C ez + 1 e e −∞ −∞ R +∞ eax 2aπi = 1−e dx. x −∞ e + 1. =. lim. 98 Download free eBooks at bookboon.com.

(105) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Finally, by a rearrangement, +∞ ax 2πi eaπi π e π . dx = = = x+1 2aπi − 1 1 sin πa e e −∞ (eaπi − e−aπi ) 2i (b) It follows from a+1 1−a 2x 2x exp exp 2 2 = + , e2x + 1 e2x + 1 . eax + e−ax cosh ax e(a+1)x + e(1−a)x = x = −x cosh x e +e e2x + 1 and 0<. 1+a <1 2. and. 0<. 1−a < 1, 2. and (a) that . +∞ −∞. cosh ax dx = cosh x =. . +∞. 1. 1 e 2 (a+1)t dt + t 2 e +1. . +∞. 1. π 1 π e 2 (1−a)t 1 a+1 + · 1−a dt = · t 2 sin 2 π e +1 2 sin 2 π −∞ −∞ ⎫ ⎧ ⎬ π 1 π⎨ 1 π 1 1 # " π aπ + π aπ = = + πa . aπ aπ ⎭ ⎩ 2 cos 2 sin 2 + 2 sin 2 − 2 cos cos − 2 2 2. 1 2. 99 Download free eBooks at bookboon.com. Click on the ad to read more.

(106) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Example 6.2 Prove that +∞ π3 (ln x)2 dx = 1 + x2 8 0 by using the path of integration sketched on the ﬁgure ant then let R → +∞ and δ → 0+.. 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 –1.5. –1. –0.5. 0. 0.5. 1. 1.5. –0.2. Figure 13: The curve C√2, 1 and the simple pole i. 3. Let Log z denote the branch of the logarithm, which is given by ' ' π 3π , Im{Log z} ∈ − , 2 2 i.e. we choose the branch of the logarithm, for which the branch cut lies along the negative imaginary axis. Then 2. f (z) =. (Log z) 1 + z2. is analytic in the open upper half plane with the exception of the simple pole z = i. Therefore, if R > 1 and δ < 1, and we denote the curve by CR,δ , then 2 2 " π #2 (Log z) (Log i) π3 (Log z) = π · i , dz = 2πi · res ; i = 2πi · = − 2 2 i+i 1 + z2 4 CR,δ 1 + z which in particular shows that the value of the line integral is independent of R > 1 and δ < 1. The curve CR,δ is composed of the interval [δ, R], the circular arc CR , the interval [−R, −δ] and the circular arc Cδ (with obvious notations). If we put t = −x, then we get on the interval [−δ, −R], −δ −δ R 2 (Log x) ln |x| + iπ)2 (ln t + iπ)2 dx = dx = dt 2 2 1+x 1+x 1 + t2 −R −R δ R R R (ln t)2 ln t 1 2 dt + 2iπ dt − π dt. = 2 2 2 1+t δ δ 1+t δ 1+t. 100 Download free eBooks at bookboon.com.

(107) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... On the circular arc CR we put z = R eiθ , θ ∈ [0, π], and then 2. |Log z| = | ln R + iθ|2 = (ln R)2 + θ2 . We get the following estimate % % 2 % % (ln R)2 + π 2 (Log z) % % · πR → 0 dz % %≤ % CR 1 + z 2 % R2 − 1. for R → +∞.. Analogously we get the following estimate of the circular arc Cδ , % % 2 % (ln δ)2 + π 2 % (Log z) % % dz · πδ → 0 for δ → 0+, %≤ % % % Cδ 1 + z 2 1 − δ2 because. 2 1 ln 1 δ (ln δ)2 · δ = →0 for → +∞, i.e. for δ → 0 + . 1 δ δ Summing up we have for R > 1 and 0 < δ < 1, 2 π3 (Log z) − = dz 2 4 CR,δ 1 + z R R R (ln x)2 (ln t)2 ln t dx + dt + 2πi dt = 2 2 1 + t2 1 + x 1 + t δ δ δ R 2 2 1 (Log z) (Log z) 2 −π dt + dz + dz 2 2 2 δ 1+t CR 1 + z Cδ 1 + z R R 2 (ln x)2 dt (Log z) 2 dx − π + dz = 2 2 2 1 + x2 δ δ 1+t CR 1 + z R 2 (Log z) ln t + dz + 2iπ dt. 2 2 δ 1+t Cδ 1 + z . Then by a rearrangement, R R R 2 2 π2 (ln x)2 ln t dt (Log z) (Log z) 2 − dx = 2iπ dt = π − dz − dz. 2 2 2 2 2 1 + x2 4 δ δ 1+t δ 1+t CR 1 + z Cδ 1 + z Here the left hand side is separated in its real and imaginary part. This equation now holds for every R > 1 and δ ∈ ]0, 1[. The right hand side has a limit value for R → +∞ and δ → 0+, independent of each other, π2 ·. π3 π π3 − −0−0= , 2 4 4. π3 . Hence by separating hence the limit value of the left hand side must also exist, and it is equal to 4 into the real and the imaginary part we get +∞ +∞ π3 (ln x)2 ln x og dx = dx = 0. 2 1 + x2 1 + x 8 0 0. 101 Download free eBooks at bookboon.com.

(108) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Example 6.3 (a) Given the function √ √ 1 1 sinh z 1 tanh z √ . · = · √ F (z) = · √ z z z cosh z z Prove that F (z) is an analytic function in a domain Ω = C \ {zn | n ∈ N0 } , independent of the choice of the branch of the square root. (b) Find the poles {zn | n ∈ N0 } of F (z), as well as their orders. (c) Let Cp , p ∈ N, denote the simple, closed curve in the z-plane, which is composed of the line segment z = 1 + it, |t| ≤ p4 π 4 − 1, and the circular arc Γp : |z| = p2 π 2 ,. Re(z) ≤ 1.. Find for every ﬁxed t ≥ 0 the value of the line integral √ 1 1 ezt tanh z zt √ e F (z) dz = dz. 2πi Cp 2πi Cp z z (d) Given that | tanh w| ≤ 2. for w = pπ eiθ ,. θ ∈ R,. p ∈ N,. prove that for every ﬁxed t ≥ 0, √ ezt √ tanh z dz = 0. lim p→+∞ Γ z z p (e) Using that F (z) has an inverse Laplace transform given by 1+i ∞ +∞ 1 1 f (t) = ezt F (z) dz = e(1+i s)t F (1 + i s) ds, 2πi 1−i ∞ 2π −∞. t ≥ 0,. where the integral is convergent, ﬁnd f (t) expressed by a series and prove that this series is convergent for every t ≥ 0. √ 2 (a) We use that ( z) = z, no matter the choice of the branch of the square root. Then by some series expansions, cosh. +∞ √ z=. +∞ 1 √ 2n 1 zn = z (2n)! (2n)! n=0 n=0. and √ +∞ +∞ +∞ √ 2n+1 √ 1 sinh z 1 1 1 1 √ zn z = zn =√ =√ z (2n + 1)! z n=0 (2n + 1)! z z n=0 (2n + 1)! n=0. 102 Download free eBooks at bookboon.com.

(109) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... 10. 5. –10. –8. –6. –4. –2. 0. 2. –5. –10. Figure 14: The path of integration Cp for p = 1.. so we have indeed deﬁned a analytic function, which is independent of the choice of the branch of the square root. Notice in particular that √ √ sinh z and lim √ = 1. (4) lim cosh z = 1 z→0 z→0 z We therefore conclude that √ √ 1 tanh z 1 sinh z 1 √ √ √ F (z) = = · z z z z cosh z is analytic in a domain Ω, which does not contain z = 0 or the zeros of cosh √ (b) The zeros of cosh z are found in the following way, "π # √ z=i + pπ , p ∈ Z, 2. √. z.. thus z=−. "π 2. + pπ. #2. ,. p ∈ Z.. Then note that p and −p − 1, p ∈ N0 give the same z, so we can now replace Z by N0 When p is replaced by p − 1, then the singularities become z0 = 0. and. zp = −(2p − 1)2. π2 , 4. p ∈ N.. Then we determine the order of zp , p ∈ N0 . Since √ 1 sinh z 1 √ · √ , F (z) = · z cosh z z we conclude from (4) that z0 = 0 is a simple pole.. 103 Download free eBooks at bookboon.com.

(110) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... When zp = −(2p − 1)2. π2 , 4. p ∈ N,. we get √ sinh zp = 0 √ z p zp. og. cosh. √ zp = 0,. and since lim. z→zp. √ sinh zp √ √ 1 d cosh z = lim sinh z · √ = = 0, √ z→zp dz 2 zp 2 z. we conclude that every zp is a simple pole. 2 1 (c) Using that zp = − p − π 2 , it follows from Cauchy’s residuum theorem that 2 1 2πi. Cp. ezt F (z) dz =. 1 2πi. Cp. √ p ezt tanh z · √ dz = res ezt F (z) ; zp , z z n=0. because only z0 , z1 , . . . , zp lie inside Cp .. Excellent Economics and Business programmes at:. “The perfect start of a successful, international career.” CLICK HERE. www.rug.nl/feb/education. 104 Download free eBooks at bookboon.com. to discover why both socially and academically the University of Groningen is one of the best places for a student to be. Click on the ad to read more.

(111) Complex Funktions Examples c-7. Then by Rule Ia, . zt. res e F (z) ; z0. . Improper integrals,, where the integrand is a rational function times .... √ sinh z 1 1 √ = 1 · 1 · = 1, = lim e · √ · z→0 1 z cosh z zt. where we again have used (4). In the computation of res ezt F (z) ; zn ,. n ∈ N,. we shall use Rule II, because zn is a simple pole. We put √ √ ezt sinh z · √ and B(z) = cosh z, A(z) = z z and get by Rule II, A (zn ) = lim res ezt F (z) ; zn = B (zn ) z→zn. . √ ezt sinh z · √ · z z. 1 sinh. √. 1 z· √ 2 z. = lim. z→zn. 2 zn t 2 ezt = e , zn z. hence by insertion, 1 2πi. . ezt F (z) dz. Cp. (5). p . p 2 zn t res ezt F (z) ; zn = 1 + e z n=0 n=1 n 2 p 8 1 1 2 = 1− 2 exp − n − π t . π n=1 (2n − 1)2 2. =. √ (d) If z ∈ Γp , then |z| = p2 π 2 and | z| = pπ. According to the given formula, % √ % (6) %tanh z % ≤ 2 for |z| = p2 π 2 . We have on Γp that Re(z) ≤ 1 and |z| = p2 π 2 , so we get by (6) for every ﬁxed t ≥ 0 the following estimate, % % % zt % % % % e √ √ % et·1 ezt % % % √ tanh z dz % ≤ max % √ · tanh z %% · (Γp ) ≤ 3 3 · 2 · 2πp2 π 2 % z∈Γp z z % % Γp z z p π = thus (7). lim p∈N. p→+∞. Γp. 4 et →0 p. for p → +∞,. √ ezt √ tanh z dz = 0. z z. (e) We conclude from (5) that 1 2πi. . zt. e F (z) dz Cp. =. . √. 1 e F (z) dz + ezt F (z) dz √ 2πi Γp 1−i p4 π 4 −1 2 p 8 1 1 2 =1− 2 exp − n − π t , π n=1 (2n − 1)2 2 1 2πi. 1+i. p4 π 4 −1. zt. 105 Download free eBooks at bookboon.com.

(112) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... hence by a rearrangement, 2 1+i√p4 π4 −1 p 8 1 1 1 1 zt 2 e F (z) dz = 1− exp − p − π t − ezt F (z) dz. π 2 n=1 (2n−1)2 2 2πi Γp 2πi 1−i√p4 π4 −1 Then by (7) by taking the limit p → +∞, p ∈ N, 1 (8) f (t) := 2πi. . 1+i ∞ 1−i ∞. 2 +∞ 8 1 1 2 e F (z) dz = 1 − 2 exp − n − π t . π n=1 (2n − 1)2 2 zt. Clearly, 2 1 2 π t ≤1 exp − n − 2. for t ≥ 0 and n ∈ N,. so we have the estimate % +∞ % +∞ 2 % % 1 π2 1 1 % % 2 , exp − n − π t = ≤ % % % % 2 (2n − 1)2 (2n − 1)2 8 n=1 n=1 and the series is absolutely and uniformly convergent for t ≥ 0. Remark 6.1 This example is a simpliﬁed version of a problem connected with oil drilling in the North Sea. One wanted to ﬁnd the inverse Laplace transform of 1 tanh ϕ(z) F (z; λ, ω) = , z ϕ(z) where ϕ(z) = ϕ(z; λ, ω) = z −. ω , λ(z + ω). and where λ and ω are two positive parameters, which are ﬁxed by some practical measurements. The principles for solving this original problem are the same as the simpliﬁed example presented here, but one must admit that the computations are far more diﬃcult that in this special case, where ϕ(z) = z. ♦. Remark 6.2 All though it is not required we shall here also prove (5), i.e. (9) | tanh w| ≤ 2. for w = pπ · eiθ ,. θ ∈ R,. p ∈ N.. We ﬁrst introduce for p ∈ N a real auxiliary function ψp by (10) ψp (θ) = cosh(2pπ · cos θ) + cos(2pπ · sin θ),. θ ∈ R.. Then we prove that . (11) cos(2pπ · sin θ) ≥ 0. 1 for Arcsin 1 − 4p. ≤ |θ| ≤. π . 2. 106 Download free eBooks at bookboon.com.

(113) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Since cos(−u) = cos u, we may assume in (11) that ' & π 1 . , θ ∈ Arcsin 1 − 2 4p Since sin θ is increasing in this interval, we get ' + & , π 1 , 2pπ = 2pπ − , 2pπ , 2pπ sin θ ∈ 2pπ 1 − 2 4p + , π and since cos u ≥ 0, when u = 2pπ sin θ ∈ 2pπ − , 2pπ , we have proved (11). 2 Then we prove that ' & "π # 1 . (12) cosh(2pπ cos θ) ≥ cosh 8p − 1 for |θ| ∈ 0 , Arcsin 1 − 4p 2 ' & 1 , and using that cos θ is decreasing in this We may again assume that θ ∈ 0 , Arcsin 1 − 4p interval, it follows that . 1 1 = + 1 − sin2 Arcsin 1 − cos θ ≥ cos Arcsin 1 − 4p 4p . . 2 √ 1 1 8p − 1 1 = 1− 1− , + = = 1− 1− 2 4p 4p 2p 16p and since cosh is increasing in R+ , we get √ "π # 8p − 1 = cosh 8p − 1 . cos(2π cos θ) ≥ cosh 2pπ · 4p 2 Now, ψp (θ + π) = cosh(−2pπ cos θ) + cos(−2pπ sin θ) = ψp (θ), and cosh. "π 2. # 8p − 1 ≥ 2. for alle p ∈ N,. so we conclude in general by (11) and (12) that ⎧ ⎪ ⎨ cosh 0 + 0 = 1, ψp (θ) = cosh(2pπ cos θ) + cos(2pπ sin θ) ≥ " # ⎪ ⎩ cosh π √8p − 1 − 1 ≥ 1, 2 where at least one of the two estimates holds for any θ. Summing up we have proved that (13) ψp (θ) = cosh(2pπ cos θ) + cos(2pπ sin θ) ≥ 1.. 107 Download free eBooks at bookboon.com.

(114) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Then we use the deﬁnitions of the hyperbolic function of a complex variable, | tanh w|2. cos2 v − sin2 v | sinh w|2 cosh2 u − cos2 v =1− = 2 2 2 2 | cosh w| cosh u − sin v cosh u − 12 + 12 − sin2 v 2 cos 2v 2 cos 2v =1− = 1− . cosh 2u + cos 2v 2 cosh2 −1 + 1 − 2 sin2 v. =. Then put w = pπ eiθ = pπ cos θ + i pπ sin θ = u + iv, and apply (13) to get % % %tanh pπ eiθ %2 = 1 −. 2 2 cos(2pπ sin θ) ≤1+ ≤ 3, ψp (θ) cosh(2pπ cos θ) + cos(2pπ sin θ). thus % % √ %tanh pπ eiθ % ≤ 3 (< 2), and we have proved (9) with the even smaller constant. √ 3. ♦. In the past four years we have drilled. 89,000 km That’s more than twice around the world.. Who are we?. We are the world’s largest oilfield services company1. Working globally—often in remote and challenging locations— we invent, design, engineer, and apply technology to help our customers find and produce oil and gas safely.. Who are we looking for?. Every year, we need thousands of graduates to begin dynamic careers in the following domains: n Engineering, Research and Operations n Geoscience and Petrotechnical n Commercial and Business. What will you be?. careers.slb.com Based on Fortune 500 ranking 2011. Copyright © 2015 Schlumberger. All rights reserved.. 1. 108 Download free eBooks at bookboon.com. Click on the ad to read more.

(115) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Example 6.4 Compute +∞ x + 3i 1 exp dx. 2 x2 + 9 −∞ x + 9 It follows from. 1 z + 3i , that this function can be extended analytically to −3i, so we get the = 2 z − 3i z +9. estimate % % % z + 3i % 4 1 % % % z 2 + 9 % = |z − 3i| ≤ |z|. for |z| ≥ 4,. hence % % % % 4 1 %exp z + 3i % ≤ exp ≤e ≤ exp % |z| z2 + 9 % |z − 3i| Then we estimate the integrand by % % % 1 z + 3i %% k % % z 2 + 9 exp z 2 + 9 % ≤ |z|2. for |z| ≥ 4.. for |z| ≥ 4.. The singularities are z = ±3i, where none of them lies on the real axis. We conclude that the improper integral is convergent and that its value can be found by a residuum formula, +∞ x + 3i 1 1 1 ; 3i . exp dx = 2πi res exp 2 x2 + 9 z2 + 9 z − 3i −∞ x + 9 The idea here is that the sum of the residues is 0. Since ∞ is a zero of second order, we have 1 1 ; ∞ = 0. res exp z2 + 9 z − 3i Now z = −3i is a simple pole, so i i 1 1 1 1 . = exp exp − ; −3i = res exp 2 6 6 6i −6i z +9 z − 3i The sum of the residues is zero, so it follows from the above that 1 i i 1 res . exp ; 3i = − exp z2 + 9 6 6 z − 3i Finally, by insertion +∞ i π i x + 3i i 1 . = exp exp dx = 2πi − exp 2 2 6 3 6 x +9 6 −∞ x + 9. Remark 6.3 We notice by separating the real and the imaginary part that it follows from this that +∞ 1 x 3 π 1 exp cos dx = cos , 2 2 2 6 x +9 x +9 3 −∞ x + 9 +∞ 1 x 3 π 1 ♦ exp sin dx = sin . 2+9 2+9 2+9 6 x x 3 x −∞. 109 Download free eBooks at bookboon.com.

(116) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Alternatively one may compute the residuum at z = 3i directly. We get by the change of variable w = z − 3i that 1 1 1 1 ;0 . exp ; 3i = res exp res w w(w + 6i) z − 3i z2 + 9 Here w0 = 0 is an essential singularity, so we must ﬁnd the Laurent series expansion and ﬁnd the 1 coeﬃcient a−1 of . When 0 < |w| < 6, then w +∞ +∞ +∞ " #p 1 1 1 1 1 1 1 1 1 1 1 p w · = · exp . = (−1) w 6i 1 + w n=0 n! wn w w(w + 6i) w 6i p=0 6i n=0 n! wn 6i It follows immediately that a−1 is the constant term inside the parenthesis, so a−1 is found by putting p = n, thus res. 1 exp z2 + 9. . 1 z − 3i. . ; 3i. =. +∞ i i 1 1 1 1 1 n , exp = − exp − = (−1) · · 6 6 6i 6i n=0 6i (6i)n n!. and we get as previously that +∞ 1 1 π i x + 3i π 1 . + i sin cos = exp exp dx = 2 6 6 3 6 x2 + 9 3 −∞ x + 9. American online LIGS University is currently enrolling in the Interactive Online BBA, MBA, MSc, DBA and PhD programs:. ▶▶ enroll by September 30th, 2014 and ▶▶ save up to 16% on the tuition! ▶▶ pay in 10 installments / 2 years ▶▶ Interactive Online education ▶▶ visit www.ligsuniversity.com to find out more!. Note: LIGS University is not accredited by any nationally recognized accrediting agency listed by the US Secretary of Education. More info here.. 110 Download free eBooks at bookboon.com. Click on the ad to read more.

(117) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... Example 6.5 Given the function f (z) =. eiz , cosh z. where cosh z =. ez + e−z . 2. Deﬁne for every R > 0 the den simple closed curve ΓR = Γ1R + Γ2R + Γ3R + Γ4R which is the sides of the rectangle shown on the ﬁgure.. Figure 15: The curve ΓR is composed of the four straight line segments: Γ1R = [−R, R] on the xaxis, Γ2R = R + i[0, π], parallel with the y-axis, Γ3R = [−R, R] + iπ parallel with the x-axis, and Γ4R = −R + i[0, π] parallel with the y-axis, and with the given sense of direction.. 1) Find all isolated singularities of f in C. Determine for each of them its type and its residuum. 2) Prove that " π# . f (z) dz = 2π exp − 2 ΓR 3) Prove that the line integrals along Γ2R and Γ4R tend to 0 for R → +∞. Hint: One may use that | cosh(x + iy)| = sinh2 x + cos2 y. 4) Prove that the improper integral 0. +∞. cos x dx cosh x. is convergent, and ﬁnd its value.. 111 Download free eBooks at bookboon.com.

(118) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... 1) The numerator and the denominator are both analytic in all of C, and the numerator is = 0 everywhere, so the singularities are given by the zeros of the denominator cosh z, i.e. ! π + pπ , p ∈ Z. zp = i 2 It follows from d cosh zp = sinh zp = 0, dz. for alle p ∈ Z,. that they are all simple pole of f (z). Finally, " res f , i. π + pπ 2. !#. # " π − pπ exp − e 2 = = i(−1)p sinh z z=zp " !# π = (−1)p−1 i exp − + pπ , 2 &. iz. '. p ∈ Z.. π 2) For every R > 0 the curve ΓR surrounds only the singularity z0 = i . 2 Then we use the residuum theorem, " π# " π# . = 2π · exp − f (z) dz = 2πi · res f , i 2 2 ΓR 3) The vertical line segment Γ2R (possibly Γ4R ) has e.g. the parametric description z(t) = R + it,. t ∈ [0, π],. so we obtain the estimate % % % % π % % π ei(R+it) % % e−t % % % · i dt%% ≤ dt. f (z) dz % = % % % % Γ2R 0 cosh(R + it) 0 | cosh(R + it)| From | cosh(R + it)| =. . sinh2 R + cos2 t ≥ | sinh R|,. we get the estimate % % % % π π 1 % % →0 dt = f (z) dz % ≤ % % Γ2R % | sinh R| 0 | sinh R|. for R → +∞.. We have only assumed in the argument above that R ∈ R, so we also have % % % % π % % f (z) dz % ≤ →0 for R → +∞. % % Γ4R % | sinh R|. 112 Download free eBooks at bookboon.com.

(119) Complex Funktions Examples c-7. Improper integrals,, where the integrand is a rational function times .... 1 2 x , so it follows from the estimate 2 % +∞ a cos x + b sin x %% C dx% ≤ 1 2 dx < +∞, cosh x −∞ 1 + 2 x. 4) Now cosh x ≥ 1 + % % % %. +∞. −∞. that the improper integral is convergent. When we return to the complex problem, then we get by the symmetry that . . R. f (z) dz = Γ1R. −R. eix dx = cosh x. . R. −R. cos x + i sin x dx = cosh x. . R −R. cos x dx, cosh x. and analogously, . . −R. f (z) dz = Γ3R. +R. ei(x+iπ) dx = +e−π cosh(x + iπ). . −R. because an integration of an odd function. R. her. cos x + i sin x dx = e−π cosh x. sin x cosh x. . R. −R. cos x dx, cosh x. over a symmetric interval [−R, R] is. always 0. Then we get by taking the limit R → +∞ in (2), R " π# +∞ cos x cos x = lim 2π exp − dx = 1 + e−π dx, R→+∞ −R cosh x 2 −∞ cosh x so. cos x being even, we get by a reﬂection argument that cosh x " π# " π# +∞ +∞ π exp − 2π exp − 1 1 cos x cos x 2 . 2 =π· "π# = dx = dx = −π −π 1 + e 2 (1 + e 2 2 cosh x cosh x ) 0 −∞ cosh 2. 113 Download free eBooks at bookboon.com.

(120) Complex Funktions Examples c-7. 7. Cauchy’s principal value. Cauchy’s principal value. Example 7.1 Compute the (important) improper integral . +∞ 0. sin x dx. x. sin z in the various solution formulæ, because It is not possible directly to apply the analytic function z it does not fulﬁl any of the inequalities required for the legality of some relevant residuum formula. Another problem is that we here only shall integrate along the positive real axis, i.e. not a “closed curve” in C∗ , and we cannot talk of a domain which is surrounded by the path of integration. Instead we shall rewrite the integrand by means of Euler’s formulæ. In order to avoid the singularity at the point 0 we shall integrate over an interval of the form [ε, R]. Then we get R −R ix R ix R −ε e−ix 1 1 sin x 1 R eix e e − dx = dx = dx. − + dx = 2i 2i 2i x x x x ε ε ε −ε −R ε If the right hand side has a limit value for ε → 0+ and R → +∞, then the limit of the left hand side does also exist, and we have +∞ ix +∞ 1 sin x e vp. dx = dx. 2i x x 0 −∞. .. 114 Download free eBooks at bookboon.com. Click on the ad to read more.

(121) Complex Funktions Examples c-7. Cauchy’s principal value. 1 The analytic function eiz has only the simple pole at z = 0, and this lies on the real axis, so it will z contribute to Cauchy’s principal value with the amount iz e ; 0 = πi. πi res z 1 1 1 Since we have the structure eiz = eimz of the integrand, where m = 1 > 0 and has a zero of ﬁrst z z z order at ∞, taking the limit R → +∞ will not cause any problem, and we have checked the conditions for the application of the residuum formula for Cauchy’s principal value. Finally, the integrand does not have any other singularity that z = 0, so we conclude that +∞ ix +∞ π 1 1 sin x e vp. · πi = . dx = dx = 2 2i 2i x x 0 −∞. Example 7.2 Compute dz vp. . 2 |z|=2 2z + 3z − 2. 2. 1. –2. –1. 0. 1. 2. –1. –2. Figure 16: The circle |z| = 2 with the evasive circular arc Γε around the point −2, and the singularity 1 inside the curve. 2 We ﬁrst note that the denominator 2z 2 + 3z − 2 is 0 for ⎧ √ ⎪ −2, −3 ± 5 ⎨ −3 ± 9 + 16 = = z= ⎪ 4 4 ⎩ 1. 2 1 We see that the pole z = −2 lies on the path of integration, while the pole z = lies inside the curve. 2 It follows by a decomposition that 1 1 = · 2z 2 + 3z − 2 2. 1 . 1 (z + 2) z − 2. =. 1 1 1 1 1 1 1 1 1 1 · , − · = · + · · · 1 1 5 5 z+2 5 z+2 2 2 5 z − z− − 2 2 2 2. 115 Download free eBooks at bookboon.com.

(122) Complex Funktions Examples c-7. and hence vp. |z|=2. dz 2z 2 + 3z − 2. Cauchy’s principal value. ⎫ ⎪ 1 ⎬ 1 vp. − z + 2⎪ |z|=2 5 ⎪ ⎩z − 1 ⎭ 2 1 dz 1 1 − lim dz , − 1 5 ε→0 Cε +Γε 5 |z|=2 Γε z + 2 z− 2 . =. =. ⎧ ⎪ 1⎨. where Cε = {z ∈ C | |z| = 2, |z + 2| ≥ ε} and Γε = {z ∈ C | |z + 2| = ε, |z| ≤ 2}. It follows from Cauchy’s integral theorem that dz = 0, z +2 Cε +Γε so we get the following reduced expression 2πi dz 1 1 + lim dz. = vp. 2 ε→0+ 5 5 Γε z + 2 |z|=2 2z + 3z − 2 We choose for Γε the following parametric description, z = −2 + ε · eiθ. for θ ∈ [Θ0 (ε), Θ1 (ε)] ,. where Θ0 (ε) → −. π 2. and Θ1 (ε) → +. π 2. for ε → 0,. and where the interval of the path of integration is run through in the opposite direction of the direction of the plane. Then we get by insertion and taking the limit, Θ2 (ε) i 2πi 1 2πi dz i ε eiθ dθ lim lim {Θ0 (ε) − Θ1 (ε)} + + = = vp. 2 iθ 5 ε→0+ Θ1 (ε) 5 ε→0+ 5 5 εe |z|=2 2z + 3z − 2 πi π π ! 2πi πi i 2πi = . − = − − + = 5 5 5 2 5 5 5 Example 7.3 Let C denote the square with the corners 1, i, −1, −i. Compute dz vp. . 4−1 z C It is obvious that the corners of C are the poles of the integrand, so for given ε > 0 we deﬁne the auxiliary curves Cε = {z ∈ C | z ∈ C, |z − a| ≥ ε, a = 1, i, −1, −i}. 116 Download free eBooks at bookboon.com.

(123) Complex Funktions Examples c-7. Cauchy’s principal value. 1. 0.5. –1. –0.5. 0.5. 1. –0.5. –1. Figure 17: The curves C and Cε with the arcs of evasion.. and Γa,ε = {z ∈ C | |z − a| = ε, z inside C},. a = 1, i, −1, −i,. of positive direction. Then vp. C. dz 4 z −1. ⎧ ⎫ ⎨ dz dz dz ⎬ = lim 4 = lim + ε z −1 ε ⎩ C −Γ −Γ −Γ z 4 − 1 a=1,i,−1,−i Γa,ε z 4 − 1 ⎭ ε 1,ε i,ε −1,ε −Γ−i,ε dz . = lim 4 ε→0 z −1 a=1,i,−1,−i Γa,ε. By a decomposition, 1 1 1 1 a = . = ; a · res 4 4 4 a=1,i,−1,−i z − a z−a z −1 z − 1 a=1,i,−1,−i Then we use the parametric descriptions + π, , Γa,ε : z = a + ε eiθ , θ ∈ Θ(a), Θ(a) + 2 in order to get dz vp. 4−1 z C. = =. a = 1, i, −1, −i,. Θ(a)+ π2 1 1 dz ε i eiθ = a a dθ 4 a=1,i,−1,−i Θ(a) 4 a=1,i,−1,−i Γa,ε z − a ε eiθ πi π 1 ai · = a = 0. 8 a=1,i,−1,−i 2 4 a=1,i,−1,−i. 117 Download free eBooks at bookboon.com.

(124) Complex Funktions Examples c-7. Cauchy’s principal value. Example 7.4 Compute +∞ dx . vp. 2 + 1) x (x −∞ The integrand f (z) =. z. 1 + 1). (z 2. is a rational function, and we have the three simple poles 0, i and −i. Of these, only 0 lies on the real axis, i.e. on the path of integration. Since z 3 f (z) =. z3 →1 z (z 2 + 1). for z → ∞,. there exists an R > 1, such that % % % % z3 % % for |z| ≥ R, % z (z 2 + 1) % ≤ 2 thus. % % % % 1 %≤ 2 % (14) % z (z 2 + 1) % |z|3. for |z| ≥ R.. Join the best at the Maastricht University School of Business and Economics!. Top master’s programmes • 3 3rd place Financial Times worldwide ranking: MSc International Business • 1st place: MSc International Business • 1st place: MSc Financial Economics • 2nd place: MSc Management of Learning • 2nd place: MSc Economics • 2nd place: MSc Econometrics and Operations Research • 2nd place: MSc Global Supply Chain Management and Change Sources: Keuzegids Master ranking 2013; Elsevier ‘Beste Studies’ ranking 2012; Financial Times Global Masters in Management ranking 2012. Visit us and find out why we are the best! Master’s Open Day: 22 February 2014. Maastricht University is the best specialist university in the Netherlands (Elsevier). www.mastersopenday.nl. 118 Download free eBooks at bookboon.com. Click on the ad to read more.

(125) Complex Funktions Examples c-7. . +∞. −∞. +∞. dx exists and that it can be computed by a residuum formula, x (x2 + 1) 1 1 dx = 2πi res ; i + πi res ; 0 , z (z 2 + 1) z (z 2 + 1) x (x2 + 1). It follows that vp.. vp.. Cauchy’s principal value. −∞. because z = i lies to the left of the path of integration seen in its direction, and because its weight is 2πi, while the residuum at the pole z = 0 on the x-axis roughly speaking is halved with only part going to the upper half plane and the other half to the lower half plane. In the present case it suﬃces to convince oneself that the integral is convergent, because the integrand is an odd function, so the only possible value is 0, i.e. +∞ dx = 0. vp. 2 + 1) x (x −∞ For completeness we compute res. . 1 ;i z (z 2 + 1). 1 1 1 = lim z = 2 = − , z→i 2z 2i 2. 1 and Q(z) = z 2 + 1, and z 1 1 ; 0 = lim 2 = 1, res 2 z→0 z + 1 z (z + 1). (regel II),. where P (z) =. (regel Ia),. and we have (control), +∞ 1 1 dx vp. = 2πi res ; i + πi res ;0 2 z (z 2 + 1) z 2 (z 2 + 1) −∞ x (x + 1) 1 + πi = 0. = 2πi · − 2. Example 7.5 Compute +∞ dx vp. . 3 −∞ x (x + 1) The integrand f (z) =. 1 z (z 3 + 1). is a rational function with a zero of fourth order at ∞. It is analytic in all of the complex plane except for the simple poles √ √ 3 1 3 1 , +i , +i 0, −1, 2 2 2 2. 119 Download free eBooks at bookboon.com.

(126) Complex Funktions Examples c-7. Cauchy’s principal value. 1. 0.5. –0.5. –1. 0. 0.5. 1. –0.5. –1. Figure 18: The four poles, of which two are lying on the path of integration, i.e. the real axis.. cf. the ﬁgure. We conclude from z 4 f (z) =. z4 →1 z (z 3 + 1). for z → ∞,. that there exists an R > 0, such that % 4 % %z f (z)% ≤ 2,. dvs. |f (z)| ≤. 2 |z|4. for |z| ≥ R.. Now, a = 4 > 1, so Cauchy’s principal value exists. In fact, we have simple poles on the x-axis, and a > 1 in (15). The value is given by the residuum formula √ +∞ 3 1 1 dx = 2πi res ; +i vp. 3 + 1) 3 + 1) 2 2 z (z x (x −∞ 1 1 +πi res ; 0 + πi res ; −1 . z (z 3 + 1) z (z 3 + 1) Then by Rule Ia, 1 1 ; 0 = lim 3 = 1. res z→0 z + 1 z (z 3 + 1) The other two poles satisfy the equation z03 = −1. Putting P (z) =. 1 z. and. Q(z) = z 3 + 1,. then P (z) and Q(z) are analytic in a neighbourhood of z0 , and since P (z) 1 , = Q(z) z (z 3 + 1) it follows from Rule II that 1 1 P (z) P (z0 ) 1 1 1 1 res ; z0 = ; z0 = res = · = · 3 =− . z (z 3 + 1) z0 3z02 Q(z) 3 z0 3 Q (z0 ). 120 Download free eBooks at bookboon.com.

(127) Complex Funktions Examples c-7. Cauchy’s principal value. √ 1 3 , so it follows that at This is true for both z0 = −1 and for z0 = + i 2 2 √ +∞ 3 1 dx 1 = 2πi res ; +i vp. 3 3 2 z (z + 1) 2 −∞ x (x + 1) 1 1 +πi res ; 0 + πi res ; −1 z (z 3 + 1) z (z 3 + 1) 1 2 1 1 = 0. = πi − + 1 − + πi + πi − = 2πi − 3 3 3 3 Note also that since the integrand is real, the result shall also be real. Example 7.6 Compute +∞ dx vp. . 6 −∞ x − 1. 1. 0.5. –1. –0.5. 0.5. 1. –0.5. –1. Figure 19: The poles of the integrand. Two of these, ±1 lie on the path of integration.. The integrand f (z) = z 6 f (z) =. 1 is a rational function with a zero of order 6 at ∞, i.e. z6 − 1. z6 →1 −1. z6. for z → ∞.. Hence there exists an R > 0, such that % % % 1 % %≤ 2 f (z)| = %% 6 for |z| ≥ R. z − 1 % |z|6 Since f (z) has only simple poler and da a = 6 > 1, it follows that Cauchy’s principal value exists and is given by the following residuum formula, √ √ +∞ 1 3 1 1 3 dx 1 + res = 2πi res ; +i ; − +i vp. 6 2 z6 − 1 2 2 z6 − 1 2 −∞ x − 1 1 1 +πi res ; 1 + res ; −1 . z6 − 1 z6 − 1. 121 Download free eBooks at bookboon.com.

(128) Complex Funktions Examples c-7. Cauchy’s principal value. If z06 = 1, then we have by Rule II, 1 1 z0 1 1 ; z0 = 5 = = z0 . res 6 6 6z0 6 z0 6 z −1 We can compute all four residues by this rule, so √ √ +∞ 1 3 1 3 1 1 dx + πi · {1 + (−1)} + − +i +i = 2πi vp. 6−1 6 2 2 2 6 2 x −∞ √ π 2π 3 1 √ = −√ . = 2πi · · i 3 + 0 = − 6 6 3 As a very weak control we see that since the integrand is real, the result is also real.. > Apply now redefine your future. - © Photononstop. AxA globAl grAduAte progrAm 2015. axa_ad_grad_prog_170x115.indd 1. 19/12/13 16:36. 122 Download free eBooks at bookboon.com. Click on the ad to read more.

(129) Complex Funktions Examples c-7. Cauchy’s principal value. Example 7.7 1) Prove that the Cauchy principal value +∞ i a x e −1 dx K(a) = vp. 2 x −∞ exists for every real number a, and show that K(a) = −π|a|. 2) Compute the integral +∞ cos ax − cos bx dx, x2 −∞ expressed by K(a) and K(b).. 1) First note that the integrand ei a z − 1 z2 has only the pole z = 0 (and ∞ as an essential singularity). % numerator has a zero of ﬁrst % The order at z = 0, hence z = 0 is a pole of ﬁrst order. From %ei a z % ≤ 1 for a ≥ 0 and Im(z) ≥ 0, follows the estimate % iaz % %e 2 − 1 %% % for |z| ≥ R and Im(z) ≥ 0, % z2 % ≤ R2 where we also have assumed that a ≥ 0. Thus the conditions of the existence of Cauchy’s principal value are fulﬁlled for a ≥ 0, and it is given by a residuum formula, iiaz +∞ i a x e 1 e −1 −1 d iaz lim e K(a) = vp. dx = πi · res ; 0 = πi· −1 2 2 z→0 1! x z dz −∞ = π i lim i a ei a z = −π a = −π|a|. z→0. If a < 0, i.e. a = −|a|, we get by a complex conjugation and the result above that . +∞. K(a) = vp. −∞. ei a z − 1 dx = vp. x2. . +∞. −∞. ei|a| − 1 dx = K(|a|) = −π|a|. x2. Summing up, . +∞. K(a) = K(|a|) = vp. −∞. ei a x − 1 dx = −π|a|, x2. a ∈ R.. The result is real, so K(a). =. +∞. vp. Re . −∞ +∞. = −∞. ei a x − 1 dx x2. . . +∞. = vp. −∞. cos(ax) − 1 dx x2. cos(ax) − 1 dx = −π|a|. x2. 123 Download free eBooks at bookboon.com.

(130) Complex Funktions Examples c-7. Cauchy’s principal value. cos(ax) − 1 The numerator cos(ax) − 1 has a zero of at least second order at 0, so the integrand x2 has a removable singularity at x = 0, and we can remove “vp.” in front of the integral, and we have the estimate 2 cos(ax) − 1 ≤ 2 2 x x. for |x| ≥ 1.. 2) Clearly, the zero of the numerator at x = 0 has order 2, so the singularity at 0 is removable. Since % % % cos(ax) − cos(bx) % %≤ 2 % for x = 0, % x2 % x2 and the integrand is continuous with a continuous extension to 0, we conclude that the improper integral exists and that it is given by . +∞ −∞. = = = =. −ε +∞ cos(ax) − cos(bx) cos(ax) − cos(bx) dx = lim dx + 2 ε→0+ x x2 −∞ ε −ε +∞ −ε +∞ cos(ax) − 1 cos(bx) − 1 lim + dx − lim + dx 2 ε→0+ −∞ ε→0+ −∞ x x2 ε ε +∞ +∞ cos(ax) − 1 cos(bx) − 1 vp. dx − vp. dx 2 x x2 −∞ −∞ +∞ +∞ cos(ax) − 1 cos(bx) − 1 dx − dx 2 x x2 −∞ −∞ K(a) − K(b) = −π(|a| − |b|) = π(|b| − |a|).. Example 7.8 1) Find the poles, their order and their residuum for the function f (z) =. (z −. Log z . − 2)(z − 3). 1)2 (z. 2) Use the calculus of residues to ﬁnd Cauchy’s principal value of the integral . +∞. vp.. f (x) dx, −∞. and then compute the integral 0 dx . 2 (x − 2)(x − 3) (x − 1) −∞. 1) We have a branch cut along R− ∪ {0}, so it only makes sense to ﬁnd the poles of the function outside this half line. It follows immediately that z = 2 and z = 3 are simple poles. Furthermore, (z − 1)2 has a zero of order 2, while Log z has a zero of order 1. Hence, z = 1 is also a simple pole.. 124 Download free eBooks at bookboon.com.

(131) Complex Funktions Examples c-7. Cauchy’s principal value. The residuum at z = 1 is computed by considering z = 1 as a pole of at most order 2, 1 1 d 1 d Log z + = lim Log z · lim res(f ; 1) = z→1 z(z − 2)(z − 3) dz (z − 2)(z − 3) 1! z→1 dz (z − 2)(z − 3) 1 1 = . = 0+ 2 1 · (1 − 2) · (1 − 3) Then we get res(f ; 2) =. Log 2 = − ln 2, (2 − 1)2 · (2 − 3). and res(f ; 3) =. Log 3 1 = ln 3. (3 − 1)2 · (3 − 2) 4. 4. 3. 2. 1. –4. –2. 2. 4. Figure 20: The path of integration CR,ε = C4, 13 . 2) Choose the path of integration CR,ε as the one given on the ﬁgure, where 0 < ε < 1 < 3 < R. We shall allow the path of integration to pass through the simple poles at z = 1, 2 and 3, and they contribute to the integral with πi times their residues. We shall further assume that the part of the path of integration which runs along the negative and real axis, actually lies in the upper half plane above the branch cut. It follows from these assumptions that 1 1 − ln 2 + ln 3 . f (z) dz = π {res(f ; 1) + res(f ; 2) + res(f ; 3)} = πi 4 2 CR,ε We get the following estimate along the circular arc |z| = R, z = R eiθ , for R → ∞, % % π % % ln R + π Log z % dz %% ≤ · π R → 0. % 2 (z − 2)(z − 3) 2 (R − 2)(R − 3) (R − 1) (z − 1) θ=0 Along the circular arc |z| = ε, i.e. z = ε ei θ , we get the following estimate for ε → 0+, % % π % % | ln ε| + π Log z % dz %% ≤ · πε % 2 1 − ε)2 (2 − ε)(3 − ε) θ=0 (z − 1) (z − 2)(z − 3) ε | ln ε| + πε = π· (2 − ε)(3 − ε) → 0, (1 − ε)2. 125 Download free eBooks at bookboon.com.

(132) Complex Funktions Examples c-7. Cauchy’s principal value. where we have used that ε| ln ε| → 0 for ε → 0+ due to the rules of magnitudes. Hence we conclude by taking the limits ε → 0+ and R → +∞ that +∞ 1 1 − ln 2 + ln 3 . f (x) dx = π i vp. 4 2 −∞ This implies that πi. 1 1 − ln 2 + ln 3 4 2 . R Log x Log x = lim lim dx + dx 2 2 ε→0+ R→+∞ −R (x − 1) (x − 2)(x − 3) ε (x − 1) (x − 2)(x − 3) −ε R −ε ln |x| dx = lim lim dx + iπ . + 2 2 ε→0+ R→+∞ −R ε (x − 1) (x − 2)(x − 3) −R (x − 1) (x − 2)(x − 3) −ε. Then by taking the imaginary part and then the limits, 0 1 1 dx = − ln 2 + ln 3. 2 (x + 2)(x − 3) 4 2 (x − 1) −∞. 126 Download free eBooks at bookboon.com. Click on the ad to read more.

(133) Complex Funktions Examples c-7. Cauchy’s principal value. As a check we see that the latter result can also be derived by a decomposition and a simple integration. It follows from 1 (x − 1)2 (x − 2)(x − 3) 1 A 1 = · + x−1 (1 − 2)(1 − 3) (x − 1)2 1 1 1 1 + + · · 2 2 (2 − 1) (2 − 3) x − 2 (3 − 1) (3 − 2) x − 3 1 1 1 1 1 A , + = − + 2 (x − 1)2 x−1 x−2 4 x−3 that 1 1 1 A + − x−1 x−2 4 x−3 1 1 1 2 − (x − 2)(x − 3) = − · = 2 2 (x − 1) (x − 2)(x − 3) 2 (x − 1) 2(x − 1)2 (x − 2)(x − 3) −x + 3 + 1 −(x − 1)(x − 3) + x − 3 + 2 = = 2 2(x − 1)(x − 2)(x − 3) 2(x − 1) (x − 2)(x − 3) −x + 4 = 2(x − 1)(x − 2)(x − 3) 1 −3 + 4 1 −2 + 4 1 −1 + 4 · + · + · = 2(1 − 2)(1 − 3) x − 1 2(2 − 1)(2 − 3) x − 2 2(3 − 1)(3 − 2) x − 3 1 1 1 1 3 , + · − · = 4 x−1 x−2 4 x−3 3 , and then by insertion and a usual integration, 4 0 0 1 1 1 1 1 3 1 dx dx + − = + 2 2 (x−1)2 4 x−1 x−2 4 x−3 −∞ (x − 1) (x − 2)(x − 3) −∞ '0 & 1 3 1 1 = lim − + ln |x−1|−ln |x−2| + ln |x−3| R→−∞ 4 2 x−1 4 R % % 1 1 %% (x−1)3 (x−3) %% 1 1 1 1 + 0 − ln 2 + ln 3 − lim + ln − = − · % x→−∞ 4 2 x−1 4 % (x−2)4 2 (−1) 1 1 − ln 2 + ln 3, = 4 2. hence A =. in accordance with the previous result.. 127 Download free eBooks at bookboon.com.

(134) Complex Funktions Examples c-7. Cauchy’s principal value. Example 7.9 Compute √ +∞ ei x π dx. vp. 2 −∞ x (x + π) Then prove that the improper integral √ +∞ sin ( π x) dx 2 −∞ x (x + π) exists and ﬁnd its value. The integrand √. ei z π f (z) = z (z 2 + π) has the three singularities (they are all simple poles), √ √ 0, i π, −i π,. % √ % % % and it is analytic in any other point of C. If Im(z) ≥ 0, then %ei z π % ≤ 1, and since % % 2 % % % √ % % % 3 % %z f (z)% = %%ei z π %% · % z % z2 + π % ≤ C where R >. √. (15) |f (z)| ≤. for |z| ≥ R and Im(z) ≥ 0,. π is ﬁxed, we conclude that C |z|3. for |z| ≥ R and Im(z) ≥ 0,. (note that we cannot here allow Im(z) < 0). Remark 7.1 By a more careful analysis, which shall not be given here, one can show that one can choose C=. R2 , −π. R2. because % % % % % % % % % % % % z2 % % z2 + π − π % % %=% % = %1 − π % = % π − 1% % % % % z2 + π % % z2 + π % % % 2 2 z +π z +π is maximum for z = ±i R. ♦ The pole 0 on the real axis is simple, and we have a = 3 > 1 in (15). This implies that Cauchy’s principal value exists and that it can be computed by a residuum formula, √ √ √ +∞ √ ei x π ei z π ei z π dx = 2π i res ; i π + π i res ;0 vp. 2 z (z 2 + π) z (z 2 + π) −∞ x (x + π) √. √. 2π e · e−π 1 ei z π ei z π √ + π i lim 2 = 2π i lim = + π i · = i · 1 − e−π . √ z→0 z + π π −2π z→i π z (z + i π). 128 Download free eBooks at bookboon.com.

(135) Complex Funktions Examples c-7. Cauchy’s principal value. Now √ 1 sin ( π x) =√ , 2 x→0 x (x + π) π lim. √ sin ( π · x) so we conclude that the integrand is continuous, so x (x2 + π) % % √ √ 1% +∞ % % sin ( π x) % % sin ( π x) % 1 % dx ≤ % dx + % % dx, % x (x2 + π) % % x (x2 + π) % 2+π x −∞ −1 |x|≥1 because we have for |x| ≥ 1, % % √ % sin ( π x) % % % ≤ 1, x ∈ R. % % x It follows from the continuity that the former integral exists, and the latter integral is of Arctan type, i.e. in particular convergent. Hence we conclude that the improper integral √ +∞ sin ( π x) dx 2 −∞ x (x + π) is convergent and that its value is given by −ε +∞ √ √ +∞ sin ( π x) sin ( π x) dx = lim dx . vp. + 2 ε→0+ x (x2 + π) −∞ x (x + π) −∞ ε We have . (16)i 1 − e. −π. . =. +∞. vp. −∞ +∞. =. vp. −∞. √ √ √ +∞ ei x π cos (x π) + i sin (x π) dx = vp. dx x (x2 + π) x (x2 + π) −∞ √ √ √ +∞ +∞ cos (x π) sin (x π) sin (x π) dx + i dx = 0 + i dx, 2 2 x (x2 + π) −∞ x (x + π) −∞ x (x + π). because “vp” is superﬂuous on the sine integral according to the above. If one wants to be particular careful, then notice that we have by the deﬁnition, −ε +∞ √ √ +∞ cos (x π) cos (x π) dx = lim dx = 0, vp. + 2 ε→0+ x (x2 + π) −∞ x (x + π) −∞ ε because the integrand is an odd function in x, and because the improper integrals +∞ · · · dx clearly exist. ε Then we conclude from (16) that . +∞ −∞. √ sin (x π) dx = 1 − e−π . x (x2 + π). 129 Download free eBooks at bookboon.com. −ε −∞. · · · dx and.

(136) Complex Funktions Examples c-7. 8. Sum of special types of series. Sum of special types of series. Example 8.1 Find the sum of the series +∞ . 1. . n=−∞. 1 n− 2. 2 ,. and then derive the value of the important sum +∞ . 1 . (2n + 1)2 n=0 −2 1 c z− , it is obvious that f (z) satisﬁes an estimate of the type |f (z)| ≤ 2 |z|2 1 / Z is the only pole, the conditions for the application of some residuum for |z| ≥ 1. Since z0 = ∈ 2 formula are satisﬁed. The auxiliary function Putting f (z) =. cot(πz) g(z) := 2 , 1 z− 2. z =. 1 , 2. Need help with your dissertation? Get in-depth feedback & advice from experts in your topic area. Find out what you can do to improve the quality of your dissertation!. Get Help Now. Go to www.helpmyassignment.co.uk for more info. 130 Download free eBooks at bookboon.com. Click on the ad to read more.

(137) Complex Funktions Examples c-7. Sum of special types of series. 1 , so we may apply Rule I with q = 2. This gives 2. has at most a double pole at z0 = ⎞. ⎛. ⎜ cot(πz) 1⎟ 1 d ⎟ ⎜ lim ; res ⎜ ⎟= 2 2 ⎠ (2 − 1)! z→ 12 dz ⎝ 1 z− 2. . 1 z− 2. . 2 g(z). = lim1 − 1 + cot2 (πz) π = −π. z→ 2. Finally, by insertion into the residuum formula for the sum of a series of this type, ⎛ ⎞ +∞ . ⎜ cot(πz) 1⎟ ⎜ ⎟ 2 2 = −π · res ⎜ 2 ; ⎟ = π . 2 ⎝ ⎠ 1 1 n− z− 2 2. . −∞. 1. It follows by a small rearrangement that +∞ . π2 =. 1. . n=−∞. n−. 1 2. 2 =. +∞ n=1. 1. . n−. 1 2. 2 +. +∞ n=0. 1. . n+. 1 2. 2 = 2. +∞ . 4 , (2n + 1)2 n=0. and then ﬁnally +∞ . π2 1 . = 2 (2n + 1) 8 n=0. Remark 8.1 Since any number m ∈ Z can be written as m = 2r (2n + 1), and since the series +∞ 1 n2 n=1. =. for uniquely determined r ∈ N0 and n ∈ N0 ,

(138) +∞ 1 n=1 2 is absolutely convergent, it is easy to derive that the sum is given by n. +∞ +∞ 1 1 1 1 1 + + + ··· 2 2 2 2 2 2 n=0 (2n + 1) (2n + 1) (2 ) n=0 (2n + 1)2 n=0 +∞ . =. 1+. 1 1 1 + + 3 + ··· 4 42 4. +∞. 1 = (2n + 1)2 n=0. 1 1−. 1 4. ·. π2 π2 = , 8 6. which is a very important result in the applications. ♦. 131 Download free eBooks at bookboon.com.

(139) Complex Funktions Examples c-7. Sum of special types of series. Example 8.2 Find for every given constant a > 0 the sum of the inﬁnite series +∞ n=0. n2. 1 . + a2. We can obviously for e.g. |z| ≥ 1 ﬁnd a constant c > 0, such that we have the estimate % % % 1 % %≤ c , |f (z)| = %% 2 z + a2 % |z|2 which proves one of the assumptions. Since f (z) has only the two simple poles z = ±ia ∈ / Z, the other assumption for the application of the residuum formula is also fulﬁlled. Since cot(±iπa) = 0, it suﬃces to compute the residues 1 1 1 1 and res . ; ia = ; −ia = − res 2 2 2 2 z +a 2ia 2ia z +a Then we get by the residuum formula, +∞ 1 1 1 cot(−iπa) = −π cot(iπa) − 2ia n2 + a2 2ia n=−∞ = −. π cosh(πa) π π cos(iπa) = · coth(πa). =− · · a ia i · sinh(πa) ia sin(iπa). Thus +∞ . +∞ 1 1 1 π 1 1 coth(πa). = + = 2+ 2 + a2 2 2 + a2 2a 2 2a 2a n n n=−∞ n=0. If a = 1, then we get in particular +∞ 1 π 1 = + coth π. 2 2 2 n n=0. Example 8.3 Let a > 0 denote a constant. Find the sum of the alternating series +∞ (−1)n . n2 + a2 n=0. 1 is the same as in Example 8.2, so we have already checked + a2 the assumptions of the relevant residuum formula in Example 8.2. The only diﬀerence is that the auxiliary factor cot(πz) has been replaced by 1/ sin(πz), so it follows immediately by insertion into the residuum formula that +∞ 1 π 1 π 1 1 1 (−1)n 1 . = · =− · · − = −π · 2 + a2 sinh(πa) a sin(iπa) ia sin(−iπa) 2ai sin(iπa) n 2ai n=−∞ The underlying function f (z) =. z2. 132 Download free eBooks at bookboon.com.

(140) Complex Funktions Examples c-7. Sum of special types of series. Since (−1)n (−1)−n = , (−n)2 + a2 n2 + a2 it easily follows that +∞ +∞ 1 1 1 (−1)n 1 π (−1)n . · = 2+ = 2+ 2 + a2 2 + a2 sinh(πa) 2a 2 2a 2a n n n=−∞ n=0. If in particular a = 1, then +∞ π 1 (−1)n . = + 2+1 2 sinh π 2 n n=0. Remark 8.2 Even if one may use the theory to ﬁnd the sum of many convergent series, where the term has the structure of a rational function in n, one should not be misled to believe that this is true for every series of this type. The simplest example is +∞ 1 n3 n=1. (≈ 1, 202),. the exact value of which is still unknown. ♦ Example 8.4 Let a ∈ R+ \ N. Find the sum of the series +∞ . 1 . 2 − a2 n n=0 The degree of the denominator is precisely 2 larger than the degree of the numerator, so the series +∞ . 1 2 − a2 n n=−∞ is convergent when 2a ∈ / N, and the value is given by +∞ . +∞ k 1 1 1 = 2 + = −π cot (aj π) res (f ; aj ) a2 n2 − a2 n2 − a2 n=−∞ n=0 j=1 1 1 ; −a (z+a)(z−a) ; a + cot(−aπ) res = −π cot(aπ) res (z+a)(z −a) 2 π 1 1 cos(−aπ) = − cot(aπ), cos(aπ) + = −π a −2a 2a. hence by a rearrangement, +∞ n=0. n2. 1 π 1 cot(aπ). =− 2 − 2 2a 2a −a. 133 Download free eBooks at bookboon.com.

(141) Complex Funktions Examples c-7. Sum of special types of series. The expression (the series) +∞ . 1 2 − a2 n n=0 1 is continuous in a ∈ R+ \N, so this formula also holds for a = p+ , p ∈ N0 . Since cot 2 0, p ∈ N0 , we obtain in this case +∞ . 1 1 2 , 2 = − 2 = − 2 = − (2p + 1)2 1 1 1 n=0 2 2 p+ 2 p+ n − p+ 2 2 2 . 1. Alternatively, cot(az) res ; ±a =0 z 2 − a2. 1 for a = p + , 2. . 1 p+ 2. π =. p ∈ N0 .. p ∈ N0 ,. because the singularity is then removable.. Brain power. By 2020, wind could provide one-tenth of our planet’s electricity needs. Already today, SKF’s innovative knowhow is crucial to running a large proportion of the world’s wind turbines. Up to 25 % of the generating costs relate to maintenance. These can be reduced dramatically thanks to our systems for on-line condition monitoring and automatic lubrication. We help make it more economical to create cleaner, cheaper energy out of thin air. By sharing our experience, expertise, and creativity, industries can boost performance beyond expectations. Therefore we need the best employees who can meet this challenge!. The Power of Knowledge Engineering. Plug into The Power of Knowledge Engineering. Visit us at www.skf.com/knowledge. 134 Download free eBooks at bookboon.com. Click on the ad to read more.

(142) Complex Funktions Examples c-7. Sum of special types of series. Example 8.5 Find the sum of the series +∞ n=0. n2. 1 . + 4n + 5. First we note that n2 + 4n + 5 = (n + 2)2 + 1. We use that +∞ . 1 π 1 = + coth π 2+1 2 2 n n=0 is the sum of a known series, so by a small rearrangement, +∞ . +∞ +∞ +∞ 1 1 π 1 1 1 + = = = −1 − = −1 + coth π. 2 2 2 2 2 n=0 n + 1 2 n + 4n + 5 n=0 (n + 2) + 1 n=2 n + 1 n=0. Example 8.6 Find the sum of the series +∞ . 1 . 2 + 1) (2n + 1) (n n=−∞ The polynomial of the denominator is of degree 3, and it does not have any zero in i Z. We therefore conclude that the series is convergent, and its sum can be found by a residuum formula. The poles are i,. −i,. 1 − , 2. thus +∞ . 1 2 + 1) (2n + 1) (n n=−∞ = −π cot(i π) · res. 1 ;i (z 2 + 1) (2z + 1) 1 ; −i + cot(−i π) · res (z 2 + 1) (2z + 1) 1 cot(π z) ;− +res 2 (z 2 + 1) (2z + 1) cos(−iπ) 1 cos(iπ) 1 +0 + = −π sin(−iπ) (−i−i)(−2i+1) sin(iπ) (i+i)(2i+1) 1 1 cosh π 1 cosh π 1 · · + · · = −π sinh π (−i) (−2i)(1−2i) sinh π i 2i(1+2i) π 1 + 2i + 1 − 2i 1 π 1 π = coth π. + = coth π · = coth π · 5 5 2 1 + 2i 1 − 2i 2. 135 Download free eBooks at bookboon.com.

(143) Complex Funktions Examples c-7. Sum of special types of series. 1 1 has a simple pole at z = − , so the auxiliary function (z 2 + 1) (2z + 1) 2 1 cot(πz) has a removable singularity for z = − . This is in agreement with that the residuum 2 (z 2 + 1) (2z + 1) 1 of the auxiliary function is 0 at z = − . ♦ 2 Remark 8.3 The function. Example 8.7 Find the sum of the series +∞ . 2n + 1 2 + 1) (3n + 1) (n n=−∞ The corresponding analytic function is f (z) =. 1 · 3. 2z + 1 , 1 (z 2 + 1) z + 3. which has a zero of second order at ∞, and the simple poles 1 a1 = − , 3. a2 = i,. a3 = −i,. / Z. It follows that the series is convergent with the sum where 2aj ∈ " π# 1 2n + 1 + cot(iπ)res(f ; i) + cot(−iπ)res(f ; −i) . res f ; − = −π cos − 3 3 (n2 + 1) (3n + 1) n=−∞ +∞ . Here, 2 & ' 1 2z + 1 1 1 −3 + 1 = , = = · 3 z 2 + 1 z=− 1 10 3 1 +1 3 9 ' & 2z + 1 −1 − 7i 2i + 1 res(f ; i) = , = = (z + i)(3z + 1) z=i 20 2i(3i + 1) . 1 res f ; − 3. . & res(f ; −i) =. 2z + 1 (z − i)(3z + 1). ' = z=−i. −1 + 7i −2i + 1 . = 20 −2i(−3i + 1). Finally, cot(iπ) =. cosh π cos(iπ) = −i coth π, = −i sinh π sin(iπ). and cot(−iπ) = i coth π,. 136 Download free eBooks at bookboon.com.

(144) Complex Funktions Examples c-7. Sum of special types of series. so the sum is −1 + 7i −1 − 7i 1 1 2n + 1 √ + i coth π · − i coth π · = −π − · 20 20 (n2 + 1) (2n + 1) 3 10 n=−∞ √ π 3 7π 7i 1 √ + i coth π · − · coth π. + = =π 10 30 10 10 3 +∞ . Example 8.8 Prove that +∞ . +∞. 1 (−1)n π3 (−1)n . = = 2 −∞ (2n + 1)3 (2n + 1)3 32 n=0 When we split the sum and change the variable n = −m − 1, i.e. m = −n − 1, then +∞ . (−1)n (2n + 1)3 n=−∞. =. =. +∞ . −1 +∞ +∞ (−1)n (−1)n (−1)n (−1)−m−1 + = + (2n + 1)3 n=−∞ (2n + 1)3 (2n + 1)3 m=0 (−2m − 2 + 1)3 n=0 n=0 +∞ . +∞ +∞ (−1)n (−1)m−4 (−1)n + = 2 , 3 3 (2n + 1) (2m + 1) (2n + 1)3 n=0 m=0 n=0. and the ﬁrst equality follows. 1 We have a triple pole at z = − , and the series is clearly convergent, so we obtain by a residuum 2 formula, +∞ . (−1)n = −π res (2n + 1)3 n=−∞. . 1 1 ;− 3 2 (2z + 1) sin(πz). .. A small rearrangement gives 1. 1 1 = · 3 (2z + 1) sin πz 8. z+. 1 2. 3 ·. 1 , sin πz. so +∞ . (−1)n (2n + 1)3 n=−∞. π cos πz 1 π 1 d d2 lim π lim = = − · 16 z→− 12 dz sin2 πz 8 2! z→− 12 dz 2 sin πz π3 π2 π2 π sin πz cos2 πz − 2π · 0 = = = lim 1 −π · − . − 2π · (−1) 16 z→− 2 16 16 sin2 πz sin3 πz. Summing up, +∞ . +∞ 1 π3 (−1)n (−1)n . = = 2 n=−∞ (2n + 1)3 (2n + 1)3 32 n=0. 137 Download free eBooks at bookboon.com.

(145) Complex Funktions Examples c-7. Sum of special types of series. Example 8.9 Find +∞ . (−1)n . (2n + 1)4 n=0 We see that z = − f (z) =. 1 is a four-tuple pole of 2. 1 1 1 · = 4 , (2z + 1)4 16 1 z+ 2. 138 Download free eBooks at bookboon.com. Click on the ad to read more.

(146) Complex Funktions Examples c-7. Sum of special types of series. so the sum is computed by a residuum formula, +∞ . (−1)n (2n + 1)4 n=0. =. ⎛. ⎞. +∞ ⎜ 1 π (−1)n ⎜ res = − ⎜ 2 n=−∞ (2n + 1)4 2 ⎝. cot(πz) 1⎟ ⎟ 4 ; − ⎟ 2 ⎠ 1 16 · z + 2. ⎛. ⎞. ⎜ cot(πz) π 1⎟ π 1 d3 ⎜ ⎟ lim 1 3 cot(πz) res ⎜ ; − ⎟=− · 4 32 2⎠ 32 3! z→− 2 dz ⎝ 1 z+ 2 cos(πz) π2 1 d π d2 lim 1 2 −π · = lim −2π = − dz sin3 (πz) 6 · 32 z→− 2 dz 6 · 32 z→− 12 sin2 (πz) π4 π sin(πz) π3 cos2 (πz) = lim 1 − . = − − 3π · 3 4 96 z→− 2 96 sin (πz) sin (πz) = −. Remark 8.4 It follows that j +∞ +∞ +∞ +∞ 1 1 1 1 1 (−1)n 1 (−1)n = 1 + + + + + · · · = 24 n=0 (2n + 1)4 24 44 84 164 n4 (2n + 1)4 n=1 n=0 j=0 =. 1 1−. 1 24. ·. 16 π 4 π4 π4 · = = . 15 96 96 90. ♦. 139 Download free eBooks at bookboon.com.

(147) Complex Funktions Examples c-7. Sum of special types of series. Example 8.10 Find the sum S of the series +∞ . 1. . 1 n+ 2. n=0. 2 .. It is well-known that +∞ . π2 1 , = 2 (2n + 1) 8 n=0 hence +∞ . 1. . n=0. n+. 1 2. 2 = 4. +∞ . π2 π2 1 = . =4· 2 (2n + 1) 8 2 n=0. Example 8.11 Prove that the series +∞ n=−∞. . 1 n− 4. 1 n−. 3 4. . is convergent. Then ﬁnd the sum of the series. Finally, ﬁnd +∞ . 1 , (4n − 1)(4n − 3) n=1 where we only sum over the positive integers. 3 1 is a polynomial of second degree, which is not 0 for any n ∈ Z. Hence, the n− Here, n − 4 4 series is convergent and the sum is given by a residuum formula, ⎧ ⎛ ⎞⎫ ⎞ ⎛ ⎪ ⎪ ⎪ ⎪ +∞ ⎨ ⎜ ⎟⎬ ⎟ ⎜ 3 1 cot(πz) cot(πz) 1 ⎟ ⎟ ⎜ = −π res ⎜ ; ⎠ ; ⎠ + res ⎝ ⎝ ⎪ 3 3 1 3 1 1 4 ⎪ 4 ⎪ ⎪ n=−∞ n − ⎭ ⎩ z− z− z− n− z− 4 4 4 4 4 4 ⎧ ⎫ ⎫ ⎧ 3π ⎪ π ⎪ ⎪ ⎪ ⎨ ⎬ ⎨ cot cot cot(πz) cot(πz) ⎬ 4 4 = −π lim1 + lim3 = −π + = −π(−2 − 2) = 4π. 1 ⎪ 1 ⎪ ⎪ ⎪ z→ 4 ⎩z→ 4 z − 3 ⎭ ⎩ −1 z− ⎭ 4 4 2 2 Then we note that 0 +∞ +∞ 1 1 1 = = (4n − 1)(4n − 3) (−4n − 1)(−4n − 3) n=0 (4n + 3)(4n + 1) n=−∞ n=0 =. +∞ . +∞ 1 1 = , (4{n + 1} − 1)(4{n + 1} − 3) n=1 (4n − 1)(4n − 3) n=0. 140 Download free eBooks at bookboon.com.

(148) Complex Funktions Examples c-7. Sum of special types of series. hence +∞ . 1 (4n − 1)(4n − 3) n=1. +∞. =. 1 1 2 −∞ (4n − 1)(4n − 3). =. +∞ 1 2 · 4 · 4 n=−∞. n−. 1 4. . 1 n−. 1 4. n−. 3 4. =. π 4π = . 8 32. Alternatively, use Leibniz’s series, +∞ π (−1)n = Arctan 1 = , 4 2n + 1 n=0. where we have added parentheses in a convergent series, which is always possible without destroying the convergence or the limit value. Then +∞ 1 1 1 1 1 1 π 1 1 (−1)n + ··· − + ··· + − + − + = − = 4n − 3 4n − 1 9 11 5 7 4 1 3 2n + 1 n=0 +∞ +∞ +∞ 1 1 1 (4n − 1) − (4n − 3) = = − , =2 4n − 1 4n − 3 (4n − 1)(4n − 3) (4n − 1) · (4n − 3) n=1 n=1 n=1 hence +∞ . π 1 = . 8 (4n − 1)(4n − 3) n=1 Finally +∞ −∞. n−. 1 4. 1 n−. 3 4. = 16. +∞ π 1 1 = 16 · 2 = 16 · 2 · = 4π. 8 (4n − 1)(4n − 3) (4n − 1)(4n − 3) n=−∞ n=1 +∞ . 141 Download free eBooks at bookboon.com.

(149) Complex Funktions Examples c-7. Sum of special types of series. Example 8.12 Find the sum of the series +∞ n=−∞. 1 n2 −. 1 9. .. Also, ﬁnd the sum of the series +∞ . 1 . (3n)2 − 1 n=1 Let f (z) =. P (z) = Q(z). 1 z2 −. 1 9. where P (z) = 1. 1 Q(z) = z 2 − . 9. and. The denominator has a degree which is 2 bigger than the degree of the numerator, and the zeros of 1 / Z. We conclude that the series is convergent, and that is sum can be the denominator are z = ± ∈ 3 found by a residuum formula, ⎧ ⎛ ⎞ ⎛ ⎞⎫ ⎪ ⎪ +∞ ⎨ 1 ⎟⎬ 1 ⎜ cot(πz) ⎜ cot(πz) 1 ⎟ = −π res ⎝ ; ⎠ + res ⎝ ;− ⎠ 1 1 3 1 ⎪ 3 ⎪ ⎭ ⎩ n=−∞ n2 − z2 − z2 − 9 9 9 ⎫ ⎧ " π# π ⎪ ⎪ ⎬ ⎨ cot cot − √ π 3π 3 3 · 2 cot = −π 3. = −π + =− 2 2 ⎪ ⎪ 3 2 ⎭ ⎩ − 3 3 Then by a small rearrangement, +∞ √ −π 3 = n=−∞. 1 n2 −. 1 9. =−. +∞ +∞ 1 1 1 +2 = −9 + 18 , 2−1 1 1 9n n=1 n2 − n=1 9 9. so √ +∞ 1 π 3 1 1 ≈ 0, 1977. = = − 18 2 (3n)2 − 1 n=1 9n2 − 1 n=1 +∞ . 142 Download free eBooks at bookboon.com.

(150) Complex Funktions Examples c-7. Sum of special types of series. Example 8.13 Given the function f (z) =. 2z 2 + (2i − 1)z − i (16z 4 − 1) (z 2 + 1). 2.. (a) Determine the singular points in C ∪ {∞} of f (z) and their types. Then ﬁnd the residuum of f (z) at z = ∞. (b) Compute the complex line integral f (z) dz. |z|=3. (c) Find the sum of the series +∞ n=1. 1 16n4. −1. .. 1. 0.5. –1. –0.5. 0.5. 1. –0.5. –1. Figure 21: The four simple zeros and the two double zeros of the denominator of f (z).. (a) The denominator 2 2 16z 4 − 1 z 2 + 1 = 4z 2 + 1 4z 2 − 1 z 2 + 1 1 1 is zero for z = ±i (double zeros), and for z = ± and z = ± i (simple zeros). The numerator 2 2 can be written 2z 2 + (2i − 1)z − i = 2 z 2 + iz − (z + i) = (2z − 1)(z + i), 1 and z = −i. We therefore see that f (z) can be i.e. the numerator has the simple zeros z = 2 reduced in the following way, f (z). = =. 2z 2 + (2i − 1)z − i (16z 4. −. 1) (z 2. 2. =. (2z − 1)(z + i) (4z 2 + 1) (2z − 1)(2z + 1)(z + i)2 (z − i)2. + 1) 1 . (2z + i)(2z − i)(2z + 1)(z + i)(z − i)2. 143 Download free eBooks at bookboon.com.

(151) Complex Funktions Examples c-7. Sum of special types of series. It follows that z = i is a double pole, that 1 i i z = −i, − , , − 2 2 2 1 are simple poles, and that z = is a removable singularity. Finally, we have a zero of order 6 in 2 ∞. In particular, res(f ; ∞) = 0. (b) All ﬁnite singularities lie inside |z| = 3, so f (z) dz = − f (z) dz = −2πi res(f ; ∞) = 0, |z|=3. |z|=3. 7. denotes a closed line integral with the opposite direction of the orientation of the plane, where 7 7 =− . i.e.. Challenge the way we run. EXPERIENCE THE POWER OF FULL ENGAGEMENT… RUN FASTER. RUN LONGER.. RUN EASIER…. READ MORE & PRE-ORDER TODAY WWW.GAITEYE.COM. 1349906_A6_4+0.indd 1. 22-08-2014 12:56:57. 144 Download free eBooks at bookboon.com. Click on the ad to read more.

(152) Complex Funktions Examples c-7. Sum of special types of series. (c) We have 16z 4 − 1 = 0 for z=. i , 2. 1 − , 2. 1 , 2. i − , 2. where none of the roots belongs to Z. Furthermore, we have a zero of order 4 at ∞, so the series is convergent, and we can use a residuum formula, +∞ n=1. 1 16n4. =. −1. +∞ +∞ 1 1 1 1 1 1 · + − = 2 n=−∞ 16n4 − 1 2 16 · 04 − 1 2 32 n=−∞. ⎛ =. ⎞. 1 4 1 4 n − 2. 3 ⎜ cot(πz) π 1 1 n⎟ ⎜ ⎟ − res ⎜ 4 ; i ⎟ . 2 32 n=0 2 ⎝ ⎠ 1 z4 − 2. Then consider the function cot(πz) 4 1 4 z − 2 It follows that ⎛. for zn =. 1 n i . 2. 1 1 and − are removable singularities, and since also 2 2 ⎞. ⎜ cot(πz) ⎟ cot (πz ) zn cot (πzn ) zn cot (πzn ) ⎜ ⎟ n res ⎜ = 4 zn cot (πzn ) , = = 4 ; z n ⎟ = 3 4 1 4zn 4zn ⎝ ⎠ 1 4 · 4 z − 16 2 we ﬁnd for zn = ± ⎛. i that 2 ⎞. π ⎜ cot(πz) cosh i i⎟ i ⎜ ⎟ 2 = 2 coth π , = 2i · res ⎜ 4 ; ± ⎟ = 4 · · cot π π 2 2 2⎠ 2 ⎝ 1 sinh z4 − 2 2 hence by insertion, +∞ n=1. 1 16n4. −1. =. π π! 1 π π π 1 = − coth ≈ 0, 0718. 2 coth + 2 coth − 2 8 2 2 2 2 32. 145 Download free eBooks at bookboon.com.

(153) Complex Funktions Examples c-7. Sum of special types of series. Example 8.14 Given the function g(z) =. f (z) (z 2 + 1). 2,. where f (z) is analytic in a neighbourhood of the points z = ±i. Furthermore, it is also given that f (−i) = −f (i) = 0,. f (−i) = f (i).. and. 1) Show that g(z) has poles at the points z = ±i, and indicate in both cases the order of the pole. 2) Prove that res(g(z); i) = res(g(z); −i). 3) Show that the series +∞ n=0. 1 (n2. + 1). 2. is convergent and ﬁnd its sum. 2 1) The denominator z 2 + 1 = (z −i)2 (z +i)2 has the double roots ±i, and since f (−i) = −f (i) = 0, and f (z) is analytic in a neighbourhood of the points z = ±i, we conclude that g(z) =. f (z) (z 2. + 1). 2. has double poles at z = ±i. 2) Then we ﬁnd that d 1 lim res(g(z); i) = 1! z→i dz. . f (z) (z + i)2. . = lim. z→i. f (z) f (z) −2 (z + i)3 (z + i)2. . 1 i = − f (i) − f (i), 4 4. and res(g(z); −i). 1 f (z) f (z) f (z) d lim = lim − 2 z→−i (z − i)2 1! z→−i dz (z − i)2 (z − i)3 1 i i 1 = − f (−i) − f (−i) = − f (i) − f (i) = res(g(z); i). 4 4 4 4. =. 1. 2 are z = ±i (double poles), and none of them lies in Z. Since we have a (z 2 + 1) zero of order 4 > 1 at ∞, it follows that the series is convergent, and we can ﬁnd the sum by a residuum formula, +∞ 1 cot(πz) cot(πz) = −π res . 2 ; i + res 2 ; −i 2 2 (z 2 + 1) (z 2 + 1) n=−∞ (1 + n ). 3) The poles of. 146 Download free eBooks at bookboon.com.

(154) Complex Funktions Examples c-7. Sum of special types of series. If we put f (z) = cot(πz), then f (z) = −. 1 , sin2 (πz). and f (i) = cot(πi) = −i coth π = 0, f (−i) = cot(−πi) = i coth π = −f (i), π π π =− = f (−i), = f (i) = − 2 (i sinh π)2 sin (πi) sinh2 π which is precisely the case of (1) and (2). Hence we get +∞ n=−∞. . 1 (1 + nn ). 2. . 1 · {f (i) + i f (i)} = −π · 2 res = −2π · − 2 ; i 4 (z 2 + 1) π π2 π π 1 + i · (−i coth π) = + coth π. = 2 2 2 2 sinh π 2 sinh π cot(πz). This e-book is made with. SetaPDF. . SETA SIGN. PDF components for PHP developers. www.setasign.com 147 Download free eBooks at bookboon.com. Click on the ad to read more.

(155) Complex Funktions Examples c-7. Sum of special types of series. Then ﬁnally, +∞ n=0. 1 (1 + n2 ). =. 2. =. 1 2. +∞ n=0. 1 2. (1 + n2 ). +. 0 n=−∞. . 1 2. (1 + n2 ). 1 = 2. . +∞ n=−∞. . 1 (1 + n2 ). 2. +1. 1 1 π π2 · + coth π + . 2 2 4 4 sinh π. Example 8.15 Prove that f (t) =. +∞ . 1 (πn + t)2 n=−∞. is convergent for every ﬁxed t ∈ C \ {pπ | p ∈ Z}. Then ﬁnd f (t) for every t ∈ C \ {pπ | p ∈ Z}, expressed by elementary functions. We deﬁne for every ﬁxed t ∈ C \ {pπ | p ∈ Z}, i.e. for − f (z; t) =. t ∈ / Z, a function F (z; t) by π. 1 . (πz + t)2. t . Then F (z; t) is analytic for z ∈ C \ − π Since for z = 0, ⎛ z 2 F (z; t) =. ⎞2. 2. z 1 ⎜ 1 ⎟ 1 = 2⎝ ⎠ → 2 2 t π π (πz + t) 1+ πz. for z → ∞,. % % 1 we conclude that there exists a constant Rt for every c > 2 , such that %z 2 F (z; t)% ≤ c for |z| ≥ Rt , π i.e. |F (z; t)| ≤ If −. c |z|2. for |z| ≥ Rt .. t ∈ / Z, then it follows directly that the series π. f (t) =. +∞ . 1 (πn + t)2 n=−∞. 148 Download free eBooks at bookboon.com.

(156) Complex Funktions Examples c-7. Sum of special types of series. is convergent, and we can ﬁnd its sum by a residuum formula, +∞ t F (n; t) = −π res cot(πz)F (z; t) ; − f (t) = π n=−∞ 2 1 t t cot(πz) cot(πz) d lim = −π · z+ ;− · = −π res 1! z→− πt dz π π (πz + t)2 (πz + t)2 1 π d π lim t − 2 cot(πz) = − = − 2 lim t π z→− π π z→− π dz sin (πz) 1 1 = , = sin2 t sin2 (−t) 2 t , and not where we have applied Rule I with q = 2. Note that we shall use the factor z + π (πz + t)2 , in the denominator. It is of course also possible directly to prove the convergence. Example 8.16 (a) Prove that f (t) =. +∞ . t 2 − π 2 n2 t n=−∞. is convergent for every ﬁxed t ∈ C \ {pπ | p ∈ Z}. (b) Express f (t) for every t ∈ C \ {pπ | p ∈ Z}, by elementary functions. (c) Finally, ﬁnd g(t) =. +∞ n=1. t2. t , − π 2 n2. t ∈ C \ {pπ | p ∈ Z},. expressed by elementary functions.. (a) Since t2 − π 2 n2 = 0 for every n ∈ Z, when t ∈ C \ {pπ | p ∈ Z}, it follows that deﬁned for n ∈ Z. Furthermore, % % % % |t| t % % for |n| ≥ Nt , % t2 − π 2 n2 % ≤ n2. t is t2 − π 2 n2. so we conclude that % +∞ % % % % +∞ % N +∞ % % % % % % t |t| t t % % % % % %+2 < +∞, ≤ ≤ % % % % % % 2 2 2 2 2 2 2 2 2 %n=−∞ t − π n % n=−∞ t − π n n2 t −π n n=−N. n=N +1. and we see that we could give a direct proof of the convergence. Alternatively we check the assumptions of the residuum formula, because they at the same time assures the convergence, and we also obtain the sum.. 149 Download free eBooks at bookboon.com.

(157) Complex Funktions Examples c-7. Sum of special types of series. Consider for every ﬁxed t ∈ C \ {pπ | p ∈ Z} the function t t =− 2 · 2 2 π −π z. 1 2 . t z2 − π t t t / Z, and that F (z) is a . Since we have assumed that ± ∈ ;− This is analytic for z ∈ C \ π π π % % % % % t % %t% % % rational function with a zero of second order at ∞, there exist constants c > % 2 % and T > %% %%, π π such that c for |z| ≥ R, |F (z; t)| ≤ 2 |z| F (z; t) =. t2. and the conditions for the convergence and the residuum formula are fulﬁlled. Hence +∞ . F (n; t) =. n=−∞. +∞ . t = f (t) 2 − π 2 n2 t −∞. is convergent. t are at most simple poles, so we bet by the residuum formula and Rule II, π ⎧ ⎛ ⎞ ⎛ ⎞⎫ ⎪ ⎪ ⎪ ⎪ ⎪ +∞ ⎟ ⎜ t ⎟⎪ ⎬ ⎨ ⎜ t t t t cot(πz) cot(πz) ⎟ ⎜ ⎟ ⎜ f (t) = = −π ; + res − ; − res − ⎟ ⎜ ⎟ ⎜ 2 2 ⎪ π⎠ π ⎠⎪ t2 − π 2 n2 ⎝ π2 ⎝ π2 t t ⎪ ⎪ n=−∞ ⎪ ⎪ z2 − z2 − ⎭ ⎩ π π cot(πz) cot(πz) t + lim t = −π · − 2 · limt 2z 2z π z→ π z→− π ⎫ ⎧ t t ⎪ ⎪ ⎪ ⎪ cot π · cot π · − ⎬ ! π t π t ⎨ π π · cot t = cot t, · cot t + = · + · = t ⎪ t 2t π 2t π ⎪ ⎪ ⎪ 2· ⎭ ⎩ 2· − π π. (b) Now, ±. and we have proved that +∞ . cot t =. n=−∞. t2. t , − π 2 n2. t ∈ C \ {pπ | p ∈ Z}.. (c) Since F (−z) = F (z), it follows from (b) for t ∈ C \ {pπ | p ∈ Z} that +∞ −1 1 1 t t t = + 2 − π 2 n2 2 − π 2 n2 2 − π 2 n2 2 2 t t t n=−∞ n=1 n=1 +∞ t 1 1 1 1 t . · cot t − = − = t 2 n=−∞ t2 − π 2 n2 2 t2 − π 2 02 2. g(t) =. +∞ . 150 Download free eBooks at bookboon.com.

(158) Complex Funktions Examples c-7. Sum of special types of series. Example 8.17 Find the sum of the series +∞ . 1 . 4+4 n n=−∞ The corresponding analytic function f (z) =. 1 z4 + 4. has the simple poles {1 + i, −1 + i, −1 − i, 1 − i}, none of which lies in Z. Furthermore, f (z) is a rational function with a zero of fourth order at ∞, hence the series is convergent, and its sum is given by a residuum formula, cot(πz) 1 = −π ; z res 0 . z4 + 4 z4 + 4 4 n=−∞ +∞ . z0 =−4. Since z04 = −4 for every pole z0 , it follows by Rule II that cot(πz) 1 1 z0 ; z0 ; z0 = cot (πz0 ) · 3 = 4 · cot (πz0 ) = cot (πz0 ) res res z4 + 4 z4 + 4 4z0 4z0 1 = − z0 cot (πz0 ) . 16. www.sylvania.com. We do not reinvent the wheel we reinvent light. Fascinating lighting offers an infinite spectrum of possibilities: Innovative technologies and new markets provide both opportunities and challenges. An environment in which your expertise is in high demand. Enjoy the supportive working atmosphere within our global group and benefit from international career paths. Implement sustainable ideas in close cooperation with other specialists and contribute to influencing our future. Come and join us in reinventing light every day.. Light is OSRAM. 151 Download free eBooks at bookboon.com. Click on the ad to read more.

(159) Complex Funktions Examples c-7. Sum of special types of series. Then by insertion, π cot(πz) 1 = π ; z = res z0 cot (πz0 ) 0 16 4 z4 + 4 z4 + 4 4 n=−∞ +∞ . z0 =−4. = = =. z0 =−4. π {(1 + i) cot(π + iπ) + (1 − i) cot(π − iπ) + (−1 + i) cot(−π + iπ) + (−1 − i) cot(−π − iπ)} 16 π π {2(1 + i) cot(π + iπ) + 2(1 − i) cot(π − iπ)} = {(1 + i) cot(iπ) + (1 − i) cot(−iπ)} 8 16 π cosh π π = · coth π, · 2i · 4 i sinh π 8. thus +∞ n=−∞. n4. π 1 = · coth π. 4 +4. Remark 8.5 In a variant we have the following estimates for e.g. |z| ≥ 2, 1 1 1 ≤ 4 = 4· |z 4 + 4| |z| − 4 |z|. |f (z)| =. so in particular, C =. 1 4 1− 4 |z|. ≤. 1 · |z|4. 1 4 1− 16. =. 4 1 · , 3 |z|4. 4 and a = 4 ≥ 2 for |z| ≥ 2. ♦ 3. Example 8.18 Compute the sum of the series +∞ n=−∞. n+. 1 3. 1 n+. 2 3. .. If we put f (z) = . 1 z+ 3. 1 z+. 2 3. ,. 2 1 2 1 / Z. Furthermore, then f (z) is analytic in C \ − , − , where − , − ∈ 3 3 3 3 % 2 % %z f (z)% → 1 for z → ∞, . 152 Download free eBooks at bookboon.com.

(160) Complex Funktions Examples c-7. Sum of special types of series. so we have checked the conditions for the convergence of the series and the sum can be found by a residuum formula. Hence, ⎧ ⎛ ⎞ ⎞⎫ ⎛ ⎪ ⎪ ⎪ ⎪ +∞ ⎨ ⎜ ⎬ ⎜ 1⎟ 2⎟ cot(πz) cot(πz) 1 ⎟ ⎟ ⎜ ⎜ ; − ⎠ +res ⎝ ;− ⎠ = −π res ⎝ ⎪ 2 2 2 1 1 1 3 3 ⎪ ⎪ ⎪ n=−∞ n+ ⎩ ⎭ z+ z+ n+ z+ z+ 3 3 3 3 3 3 ⎫ ⎧ ⎫ ⎧ " π# 2π ⎪ π⎪ π ⎪ ⎪ ⎪ ⎪ cot − ⎬ ⎬ ⎨ ⎨ cot − cot − cot 3 3 = π · 3 · 2 · cot π 3 + 3 + = −π = −π 1 ⎪ 1 2 1 ⎪ 1 2 ⎪ ⎪ 3 ⎪ ⎩ ⎪ − ⎭ − + ⎭ ⎩ − + 3 3 3 3 3 3 √ 1 = π · 3 · 2 · √ = 2 3 π, 3 where we have used that − +∞ −∞. . 2 1 and − are simple poles of f (z). Thus we have proved that 3 3. √ 1 = 2 3 π. 2 1 n+ n+ 3 3. 360° thinking. .. 360° thinking. .. 360° thinking. .. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. Discover the truth at www.deloitte.ca/careers. © Deloitte & Touche LLP and affiliated entities.. © Deloitte & Touche LLP and affiliated entities.. Discover the truth153 at www.deloitte.ca/careers Click on the ad to read more Download free eBooks at bookboon.com. © Deloitte & Touche LLP and affiliated entities.. D.

(161) Complex Funktions Examples c-7. Sum of special types of series. Example 8.19 1) Determine the singular points in C of f , deﬁned by (17) f (z) =. 1 , z 2p (ez − 1). p ∈ N,. and ﬁnd the residuum of f at every singular point. Hint: When res(f (z); 0) shall be found one may without proof apply the following Taylor series expansion +∞. Bn z = zn ez − 1 n!. for |z| < 2π,. b=0. where the left hand side of the equality is replaced by 1 for z = 0. 2) Let Kn denote in the (x, y)-plane the boundary of the square [−rn , rn ] × [−rn , rn ] ,. where rn = π + 2nπ, and n ∈ N.. Prove for every (ﬁxed) p ∈ N that f (z) dz → 0 for n → +∞ Kn. on N,. where f is the function given by (17). 3) Apply Cauchy’s residuum theorem on the square with the boundary Kn and then apply the results of (1) and (2) above, and the limit n → +∞ to prove that +∞ 2p 1 p+1 (2π) B2p = (−1) n2p 2(2p)! n=1. for every p ∈ N.. Prove also that B2 , B4 , . . . , B2p , . . . have alternating signs.. 1) The singular points are z = 2i n π, n ∈ Z. Of these, z = 0 is a (2n + 1)-tuple pole, while all the others are simple poles. 2) Putting rn = π + 2nπ we get the following estimates, % % % %. Kn. ≤. ≤. % % dz % 2p z z (e − 1) % rn. . dy 2p. rn. dy 2p. −r +iy − 1| |rn + iy| − 1| −rn |−rn + iy| |e n rn rn dx dx + + 2p x+irn 2p x−irn − 1| − 1| −rn |x + irn | |e −rn |x − irn | |e 2rn 2rn 2rn 2rn 1 2p−1 + + →0 2p + 2p = 2 rn 2p 1 2p 1 |rn | |rn | |rn | · |rn | · 2 2 −rn. |ern +iy. +. for n → +∞.. 154 Download free eBooks at bookboon.com.

(162) Complex Funktions Examples c-7. Sum of special types of series. 3) Now, z = 2inπ is a simple pole for n ∈ Z \ {0}, so we have the following computation of the residuum, res(f (z); 2inπ). 1 1 1 1 = lim 2p · z z→2iπ e z z 2p d (ez − 1) dz 1 1 1 = (−1)p · . (2inπ)2p (2π)2p n2p. =. lim. z→2inπ. = Since +∞ Bn n z = z , ez − 1 n=0 n!. it follows by a division by z 2p+1 , +∞ 1 Bn −n−2p−1 = z . 2p z z (e − 1) n=0 n!. We ﬁnd the term. a−1 by choosing n = 2p, so z. res(f (z); 0) = a−1 =. B2p . (2p)!. Then by Cauchy’s residuum theorem, 1 2πi. f (z) dz = Kn. n . res(f (z); 2ikπ) =. k=−n. n B2p 1 1 + 2 · (−1)p · , (2π)2p k 2p (2p)! k=1. hence by a rearrangement, n 2p (−1)p 1 1 2p p+1 (2π) B2p · (2π) , = · f (z) dz + (−1) 2πi Kn k 2p 2 2(2p)!. n ∈ N.. k=1. Since p ∈ N, the left hand side converges, when n → +∞. Then by (2) it follows from taking this limit, +∞ (2π)2p B2p 1 = (−1)p+1 · 2p n 2(2p)! n=1. for every p ∈ N.. The left hand side is always positive, so the factor (−1)p+1 on the right hand side causes that ⎧ for p odd, ⎨ >0 B2p ⎩ <0 for p even, and the sequence B2 , B4 , . . . , B2p , . . . has alternating sign.. 155 Download free eBooks at bookboon.com.

(163) Complex Funktions Examples c-7. Sum of special types of series. Example 8.20 Given the function 1 . z 2 sin z. f (z) =. 1) Find all the isolated singularities in C of the function f , and determin the type for each of them. 2) Find in a neighbourhood of z0 = 0 the principal part of the Laurent series of f , i.e. that part of the series which contains terms of the type bn , zn. n > 0.. (Hint. Use the Taylor series of sin z with z0 = 0 as expansion point). 3) Find the residues in the isolated singularities of f . 4) Denote by N a positive integer, and let CN denote the curve run through in a positive sense, which is bounding the square % ' & % 1 1 % π . π, N + z = x + iy % x, y ∈ − N + 2 2 7 Compute CN f (z) dz. 5) When z = x + iy, then | sin z|2 = sin2 x + sinh2 y. It follows that | sin z| ≥ | sin x| Prove that f (z) dz → 0 CN. and. | sin z| ≥ | sinh y|.. for N → +∞.. 6) Prove that +∞ π2 (−1)n+1 . = n2 12 n=1. 1) We have poles at z = pπ, p ∈ Z. When p = 0, we see that the pole z = 0 is a triple pole; any other pole is simple. 2) Now, f (z) is an odd function with the triple pole at z = 0, so the principal part must have the structure a−3 1 a−1 = 3 + + ··· , 2 z sin z z z hence sin z + a−1 · z sin z + terms of order > 3 z z2 = a−3 · 1 − + · · · + a−1 z 2 − · · · + · · · 6 1 = a−3 + a−1 − a−3 z 2 + · · · . 6. 1 = a−3 ·. 156 Download free eBooks at bookboon.com.

(164) Complex Funktions Examples c-7. Sum of special types of series. Then it follows by the identity theorem that a−3 = 1. og. a−1 =. 1 , 6. so the principal part is 1 1 1 . + 3 6 z z 3) We have in the triple pole z0 = 0, 1 1 ; 0 = a−1 = . res 2 6 z sin z When zp = pπ, p ∈ Z \ {0}, the pole is simple, thus (−1)p 1 1 1 1 = ; pπ = lim 2 · · 2. res 2 z→pπ z cos z p z sin z π2. We will turn your CV into an opportunity of a lifetime. Do you like cars? Would you like to be a part of a successful brand? We will appreciate and reward both your enthusiasm and talent. Send us your CV. You will be surprised where it can take you.. 157 Download free eBooks at bookboon.com. Send us your CV on www.employerforlife.com. Click on the ad to read more.

(165) Complex Funktions Examples c-7. Sum of special types of series. Figure 22: The curve CN for N = 2 and the singularities inside.. 4) Using Cauchy’s residuum theorem, f (z) dz = CN. N p=−N. N 2 (−1)n 1 . res(f (z); pπ) = + 2 6 π n=1 n2. 5) Then we have the estimates % % % dz %% % % % ≤ (2N + 1) · % ΓN z 2 sin z %. 1 1 N+ 2. →0. 2. for N → +∞,. π2. where ΓN is anyone of the vertical line segments of CN . In instead ΓN is one of the horizontal segments, then % % % 1 dz %% 1 % →0 · % ≤ (2N + 1) · % 2 2 % ΓN z sin z % 1 1 2 π sinh N + N+ π 2 2 for N → +∞. Summing up we get lim. N →+∞. f (z) dz = CN. +∞ 2 (−1)n 1 + 2 = 0. 6 π n=1 n2. Finally, by a rearrangement, +∞ π2 (−1)n+1 . = n2 12 n=1. 158 Download free eBooks at bookboon.com.

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