Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 8 pot

Result 13.7.1 Integrals from Zero to Infinity. Let f(z) be a single-valued analytic func-
tion with only isolated singularities and no singularities on the positive, real axis, [0, ∞). Let
a ∈ Z. If the integrals exist then,

∞
0
f(x) dx = −
n

k=1
Res (f(z) log z, z
k
) ,

∞
0
x
a
f(x) dx =
ı2π
1 −
e
ı2πa
n

k=1
Res (z
a
f(z), z
k

) ,

∞
0
f(x) log x dx = −
1
2
n

k=1
Res

f(z) log
2
z, z
k

+ ıπ
n

k=1
Res (f(z) log z, z
k
) ,

∞
0
x
a
f(x) log x dx =

ı2π
1 −
e
ı2πa
n

k=1
Res (z
a
f(z) log z, z
k
)
+
π
2
a
sin
2
(πa)
n

k=1
Res (z
a
f(z), z
k
) ,

∞
0

x
a
f(x) log
m
x dx =
∂
m
∂a
m

ı2π
1 −
e
ı2πa
n

k=1
Res (z
a
f(z), z
k
)

,
where z
1
, . . . , z
n
are the singularities of f(z) and there is a branch cut on the positive real
axis with 0 < arg(z) < 2π.

654
13.8 Exploiting Symmetry
We have already used symmetry of the integrand to evaluate certain integrals. For f(x) an even function we were
able to evaluate

∞
0
f(x) dx by extending the range of integration from −∞ to ∞. For

∞
0
x
α
f(x) dx
we put a branch cut on the positive real axis and noted that the value of the integrand below the branch cut is a
constant multiple of the value of the function above the branch cut. This enabled us to evaluate the real integral with
contour integration. In this section we will use other kinds of symmetry to evaluate integrals. We will discover that
periodicity of the integrand will produce this symmetry.
13.8.1 Wedge Contours
We note that z
n
= r
n
e
ınθ
is periodic in θ with period 2π/n. The real and imaginary parts of z
n
are odd periodic
in θ with period π/n. This observation suggests that certain integrals on the positive real axis may be evaluated by
closing the path of integration with a wedge contour.

Example 13.8.1 Consider

∞
0
1
1 + x
n
dx
655
where n ∈ N, n ≥ 2. We can evaluate this integral using Result 13.7.1.

∞
0
1
1 + x
n
dx = −
n−1

k=0
Res

log z
1 + z
n
,
e
ıπ(1+2k)/n

= −

n−1

k=0
lim
z→
e
ıπ(1+2k)/n

(z −
e
ıπ(1+2k)/n
) log z
1 + z
n

= −
n−1

k=0
lim
z→
e
ıπ(1+2k)/n

log z + (z −
e
ıπ(1+2k)/n
)/z
nz
n−1


= −
n−1

k=0

ıπ(1 + 2k)/n
n
e
ıπ(1+2k)(n−1)/n

= −
ıπ
n
2
e
ıπ(n−1)/n
n−1

k=0
(1 + 2k)
e
ı2πk/n
=
ı2π
e
ıπ/n
n
2
n−1


k=1
k
e
ı2πk/n
=
ı2π
e
ıπ/n
n
2
n
e
ı2π/n
−1
=
π
n sin(π/n)
This is a bit grungy. To find a spiffier way to evaluate the integral we note that if we write the integrand as a function
of r and θ, it is periodic in θ with period 2π/n.
1
1 + z
n
=
1
1 + r
n
e
ınθ
The integrand along the rays θ = 2π/n, 4π/n, 6π/n, . . . has the same value as the integrand on the real axis. Consider

the contour C that is the boundary of the wedge 0 < r < R, 0 < θ < 2π/n. There is one singularity inside the
656
contour. We evaluate the residue there.
Res

1
1 + z
n
,
e
ıπ/n

= lim
z→
e
ıπ/n
z −
e
ıπ/n
1 + z
n
= lim
z→
e
ıπ/n
1
nz
n−1
= −
e

ıπ/n
n
We evaluate the integral along C with the residue theorem.

C
1
1 + z
n
dz =
−ı2π
e
ıπ/n
n
Let C
R
be the circular arc. The integral along C
R
vanishes as R → ∞.





C
R
1
1 + z
n
dz





≤
2πR
n
max
z∈C
R




1
1 + z
n




≤
2πR
n
1
R
n
− 1
→ 0 as R → ∞
We parametrize the contour to evaluate the desired integral.


∞
0
1
1 + x
n
dx +

0
∞
1
1 + x
n
e
ı2π/n
dx =
−ı2π
e
ıπ/n
n

∞
0
1
1 + x
n
dx =
−ı2π
e
ıπ/n
n(1 −

e
ı2π/n
)

∞
0
1
1 + x
n
dx =
π
n sin(π/n)
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13.8.2 Box Contours
Recall that
e
z
=
e
x+ıy
is periodic in y with period 2π. This implies that the hyperbolic trigonometric functions
cosh z, sinh z and tanh z are periodic in y with period 2π and odd periodic in y with period π. We can exploit this
property to evaluate certain integrals on the real axis by closing the path of integration with a box contour.
Example 13.8.2 Consider the integral

∞
−∞
1
cosh x
dx =


ı log

tanh

ıπ
4
+
x
2

∞
−∞
= ı log(1) − ı log(−1)
= π.
We will evaluate this integral using contour integration. Note that
cosh(x + ıπ) =
e
x+ıπ
+
e
−x−ıπ
2
= −cosh(x).
Consider the box contour C that is the boundary of the region −R < x < R, 0 < y < π. The only singularity of
the integrand in sid e the contour is a first order pole at z = ıπ/2. We evaluate the integral along C with the residue
theorem.

C
1

cosh z
dz = ı2π Res

1
cosh z
,
ıπ
2

= ı2π lim
z→ıπ/2
z − ıπ/2
cosh z
= ı2π lim
z→ıπ/2
1
sinh z
= 2π
658
The integrals along the sides of the box vanish as R → ∞.





±R+ıπ
±R
1
cosh z
dz





≤ π max
z∈[±R ±R+ıπ]




1
cosh z




≤ π max
y∈[0 π]




2
e
±R+ıy
+
e
∓R−ıy





=
2
e
R
−
e
−R
≤
π
sinh R
→ 0 as R → ∞
The value of the integrand on the top of the box is the negative of its value on the bottom. We take the limit as
R → ∞.

∞
−∞
1
cosh x
dx +

−∞
∞
1
−cosh x
dx = 2π

∞
−∞

1
cosh x
dx = π
13.9 Definite Integrals Involving Sine and Cosine
Example 13.9.1 For real-valued a, evaluate the integral:
f(a) =

2π
0
dθ
1 + a sin θ
.
What is the value of the integral for complex-valued a.
Real-Valued a. For −1 < a < 1, the integrand is bounded, hence the integral exists. For |a| = 1, the integrand
has a second order pole on the path of integration. For |a| > 1 the integrand has two first order poles on the path of
integration. The integral is divergent for these two cases. Thus we see that the integral exists for −1 < a < 1.
659
For a = 0, the value of the integral is 2π. Now consider a = 0. We make the change of variables z =
e
ıθ
. The real
integral from θ = 0 to θ = 2π becomes a contour integral along the unit circle, |z| = 1. We write the sine, cosine and
the differential in terms of z.
sin θ =
z − z
−1
ı2
, cos θ =
z + z
−1

2
, dz = ı
e
ıθ
dθ, dθ =
dz
ız
We write f(a) as an integral along C, the positively oriented unit circle |z| = 1.
f(a) =

C
1/(ız)
1 + a(z − z
−1
)/(2ı)
dz =

C
2/a
z
2
+ (ı2/a)z − 1
dz
We factor the denominator of the integrand.
f(a) =

C
2/a
(z − z
1

)(z − z
2
)
dz
z
1
= ı

−1 +
√
1 − a
2
a

, z
2
= ı

−1 −
√
1 − a
2
a

Because |a| < 1, the second root is outside the unit circle.
|z
2
| =
1 +
√

1 − a
2
|a|
> 1.
Since |z
1
z
2
| = 1, |z
1
| < 1. Thus th e pole at z
1
is inside the contou r and the pole at z
2
is outside. We evaluate the
contour integral with the residue theorem.
f(a) =

C
2/a
z
2
+ (ı2/a)z − 1
dz
= ı2π
2/a
z
1
− z
2

= ı2π
1
ı
√
1 − a
2
660
f(a) =
2π
√
1 − a
2
Complex-Valued a. We note that the integral converges except for real-valued a satisfying |a| ≥ 1. On any closed
subset of C \ {a ∈ R | |a| ≥ 1} the integral is uniformly convergent. Thus except for the values {a ∈ R | |a| ≥ 1},
we can differentiate the integral with respect to a. f(a) is analytic in the complex plane except for the set of points
on the real axis: a ∈ (−∞. . . − 1] and a ∈ [1 . . . ∞). The value of the analytic function f(a) on the real axis for the
interval (−1 . . . 1) is
f(a) =
2π
√
1 − a
2
.
By analytic continuation we see that the value of f(a) in the complex plane is the branch of the function
f(a) =
2π
(1 − a
2
)
1/2

where f(a) is positive, real-valued for a ∈ (−1 . . . 1) and there are branch cuts on the real axis on the intervals:
(−∞. . . − 1] and [1 . . . ∞).
Result 13.9.1 For evaluating integrals of the form

a+2π
a
F (sin θ, cos θ) dθ
it may be useful to make the change of variables z =
e
ıθ
. This gives us a contour integral
along the unit circle about the origin. We can write the sine, cosine and differential in terms
of z.
sin θ =
z −z
−1
ı2
, cos θ =
z + z
−1
2
, dθ =
dz
ız
661
13.10 Infinite Sums
The function g(z) = π cot(πz) has simple poles at z = n ∈ Z. The residues at these points are all unity.
Res(π cot(πz), n) = lim
z→n
π(z −n) cos(πz)

sin(πz)
= lim
z→n
π cos(πz) − π(z − n) sin(πz)
π cos(πz)
= 1
Let C
n
be the square contour with corners at z = (n + 1/2)(±1 ± ı). Recall that
cos z = cos x cosh y −ı sin x sinh y and sin z = sin x cosh y + ı cos x sinh y.
First we bound the modulus of cot(z).
|cot(z)| =




cos x cosh y −ı sin x sinh y
sin x cosh y + ı cos x sinh y




=

cos
2
x cosh
2
y + sin
2

x sinh
2
y
sin
2
x cosh
2
y + cos
2
x sinh
2
y
≤

cosh
2
y
sinh
2
y
= |coth(y)|
The hype rbolic cotangent, coth(y), has a simple pole at y = 0 and tends to ±1 as y → ±∞.
Along the top and bottom of C
n
, (z = x ± ı(n + 1/2)), we bound the modulus of g(z) = π cot(πz).
|π cot(πz)| ≤ π


coth(π(n + 1/2))



662
Along the left and right sides of C
n
, (z = ±(n + 1/2) + ıy), the mo dul us of the function is bounded by a constant.
|g(±(n + 1/2) + ıy)| =




π
cos(π(n + 1/2)) cosh(πy) ∓ ı sin(π(n + 1/2)) sinh(πy)
sin(π(n + 1/2)) cosh(πy) + ı cos(π(n + 1/2)) sinh(πy)




= |∓ıπ tanh(πy)|
≤ π
Thus the modulus of π cot(πz) can be bounded by a constant M on C
n
.
Let f(z) be analytic except for isolated singularities. Consider the integral,

C
n
π cot(πz)f(z) dz.
We use the maximum modulus integral bound.






C
n
π cot(πz)f(z) dz




≤ (8n + 4)M max
z∈C
n
|f(z)|
Note that if
lim
|z|→∞
|zf(z)| = 0,
then
lim
n→∞

C
n
π cot(πz)f(z) dz = 0.
This implies that the sum of all residues of π cot(πz)f(z) is zero. Suppose further that f(z) is analytic at z = n ∈ Z.
The residues of π cot(πz)f(z) at z = n are f(n). This means
∞

n=−∞

f(n) = −( sum of the residues of π cot(πz)f(z) at the poles of f(z) ).
663
Result 13.10.1 If
lim
|z|→∞
|zf(z)| = 0,
then the sum of all the residues of π cot(πz)f(z) is zero. If in addition f(z) is analytic at
z = n ∈ Z then
∞

n=−∞
f(n) = −( sum of the resi dues of π cot(πz)f(z) at the poles of f(z) ).
Example 13.10.1 Consider the sum
∞

n=−∞
1
(n + a)
2
, a ∈ Z.
By Result 13.10.1 with f(z) = 1/(z + a)
2
we have
∞

n=−∞
1
(n + a)
2
= −Res


π cot(πz)
1
(z + a)
2
, −a

= −π lim
z→−a
d
dz
cot(πz)
= −π
−π sin
2
(πz) − π cos
2
(πz)
sin
2
(πz)
.
∞

n=−∞
1
(n + a)
2
=
π

2
sin
2
(πa)
Example 13.10.2 Derive π/4 = 1 − 1/3 + 1/5 −1/7 + 1/9 − ···.
664
Consider the integral
I
n
=
1
ı2π

C
n
dw
w(w −z) sin w
where C
n
is the square with corners at w = (n + 1/2)(±1 ± ı)π, n ∈ Z
+
. With the substitution w = x + ıy,
|sin w|
2
= sin
2
x + sinh
2
y,
we see that |1/ sin w| ≤ 1 on C

n
. Thus I
n
→ 0 as n → ∞. We use the residue theorem and take the limit n → ∞.
0 =
∞

n=1

(−1)
n
nπ(nπ −z)
+
(−1)
n
nπ(nπ + z)

+
1
z sin z
−
1
z
2
1
sin z
=
1
z
− 2z

∞

n=1
(−1)
n
n
2
π
2
− z
2
=
1
z
−
∞

n=1

(−1)
n
nπ −z
−
(−1)
n
nπ + z

We substitute z = π/2 into the above expression to obtain
π/4 = 1 − 1/3 + 1/5 −1/7 + 1/9 −···
665

13.11 Exercises
The Residue Theorem
Exercise 13.1
Evaluate the following closed contour integrals using Cauchy’s residue theorem.
1.

C
dz
z
2
− 1
, where C is the contour parameterized by r = 2 cos(2θ), 0 ≤ θ ≤ 2π.
2.

C
e
ız
z
2
(z − 2)(z + ı5)
dz, where C is the p ositive circle |z| = 3.
3.

C
e
1/z
sin(1/z) dz, where C is the positive circle |z| = 1.
Hint, Solution
Exercise 13.2
Derive Cauchy’s integral formula from Cauchy’s residue theorem.

Hint, Solution
Exercise 13.3
Calculate the residues of the following functions at each of the poles in the fin ite part of the plane.
1.
1
z
4
− a
4
2.
sin z
z
2
3.
1 + z
2
z(z −1)
2
4.
e
z
z
2
+ a
2
666
5.
(1 − cos z)
2
z

7
Hint, Solution
Exercise 13.4
Let f(z) have a pole of order n at z = z
0
. Prove the Residue Formula:
Res(f(z), z
0
) = lim
z→z
0

1
(n − 1)!
d
n−1
dz
n−1
[(z − z
0
)
n
f(z)]

.
Hint, Solution
Exercise 13.5
Consider the function
f(z) =
z

4
z
2
+ 1
.
Classify the singularities of f(z) in the extended complex plane. Calculate the residue at each pole and at infinity. Find
the Laurent series expansions and their domains of convergence about the points z = 0, z = ı and z = ∞.
Hint, Solution
Exercise 13.6
Let P (z) be a polynomial none of whose roots lie on the closed contour Γ. Show that
1
ı2π

P

(z)
P (z)
dz = number of roots of P (z) which lie i nsi de Γ.
where the roots are counted according to their multiplicity.
Hint: From the fundamental theorem of algebra, it is always possible to factor P(z) in the form P (z) = (z −z
1
)(z −
z
2
) ···(z − z
n
). Using this form of P (z) the integrand P

(z)/P (z) reduces to a very simple expression.
Hint, Solution

667
Exercise 13.7
Find the value of

C
e
z
(z − π) tan z
dz
where C is the positively-oriented circle
1. |z| = 2
2. |z| = 4
Hint, Solution
Cauchy Principal Value for Real Integrals
Solution 13.1
Show that the integral

1
−1
1
x
dx.
is divergent. Evaluate the i ntegral

1
−1
1
x − ıα
dx, α ∈ R, α = 0.
Evaluate

lim
α→0
+

1
−1
1
x − ıα
dx
and
lim
α→0
−

1
−1
1
x − ıα
dx.
The integral exists for α arbitrarily close to zero, but diverges when α = 0. Plot the real and imaginary part of the
integrand. If one were to assign meaning to the integral for α = 0, what would the value of the integral be?
668
Exercise 13.8
Do the principal values of the following integrals exist?
1.

1
−1
1
x

2
dx,
2.

1
−1
1
x
3
dx,
3.

1
−1
f(x)
x
3
dx.
Assume that f(x) is real analytic on the interval (−1, 1).
Hint, Solution
Cauchy Principal Value for Contour Integrals
Exercise 13.9
Let f(z) have a first order pole at z = z
0
and let (z −z
0
)f(z) be analytic in some neighborhood of z
0
. Let the contour
C


be a circular arc from z
0
+ e
ıα
to z
0
+ e
ıβ
. (Assume that β > α and β − α < 2π.) Show that
lim
→0
+

C

f(z) dz = ı(β − α) Res(f(z), z
0
)
Hint, Solution
Exercise 13.10
Let f (z) be analytic inside and on a simple, closed, positive contour C, except for isolated singularities at z
1
, . . . , z
m
inside the contour and first order poles at ζ
1
, . . . , ζ
n
on the contour. Further, let the contour be C

1
at the locations of
these first order poles. (i.e., the contour does not have a corner at any of the first order poles.) Show that the principal
value of the integral of f(z) along C is
−

C
f(z) dz = ı2π
m

j=1
Res(f(z), z
j
) + ıπ
n

j=1
Res(f(z), ζ
j
).
Hint, Solution
669
Exercise 13.11
Let C be the unit circle. Evaluate
−

C
1
z − 1
dz

by indenting the contour to exclude the first order p ole at z = 1.
Hint, Solution
Integrals on the Real Axis
Exercise 13.12
Evaluate the following improper integrals.
1.

∞
0
x
2
(x
2
+ 1)(x
2
+ 4)
dx =
π
6
2.

∞
−∞
dx
(x + b)
2
+ a
2
, a > 0
Hint, Solution

Exercise 13.13
Prove Result 13.4.1.
Hint, Solution
Exercise 13.14
Evaluate
−

∞
−∞
2x
x
2
+ x + 1
.
Hint, Solution
Exercise 13.15
Use contour integration to evaluate the integrals
670
1.

∞
−∞
dx
1 + x
4
,
2.

∞
−∞

x
2
dx
(1 + x
2
)
2
,
3.

∞
−∞
cos(x)
1 + x
2
dx.
Hint, Solution
Exercise 13.16
Evaluate by contour integration

∞
0
x
6
(x
4
+ 1)
2
dx.
Hint, Solution

Fourier Integrals
Exercise 13.17
Suppose that f(z) vanishes as |z| → ∞. If ω is a (positive / negative) real number and C
R
is a sem i-c ircle of radius
R in the (upper / lower) half plane then show that the integral

C
R
f(z)
e
ıωz
dz
vanishes as R → ∞.
Hint, Solution
Exercise 13.18
Evaluate by contour integration

∞
−∞
cos 2x
x − ıπ
dx.
671
Hint, Solution
Fourier Cosine and Sine Integrals
Exercise 13.19
Evaluate

∞

−∞
sin x
x
dx.
Hint, Solution
Exercise 13.20
Evaluate

∞
−∞
1 − cos x
x
2
dx.
Hint, Solution
Exercise 13.21
Evaluate

∞
0
sin(πx)
x(1 − x
2
)
dx.
Hint, Solution
Contour Integration and Branch Cuts
Exercise 13.22
Evaluate the following integrals.
1.


∞
0
ln
2
x
1 + x
2
dx =
π
3
8
2.

∞
0
ln x
1 + x
2
dx = 0
672
Hint, Solution
Exercise 13.23
By methods of contour integration find

∞
0
dx
x
2

+ 5x + 6
[ Recall the trick of considering

Γ
f(z) log z dz with a suitably chosen contour Γ and branch for log z. ]
Hint, Solution
Exercise 13.24
Show that

∞
0
x
a
(x + 1)
2
dx =
πa
sin(πa)
for −1 < (a) < 1.
From this derive that

∞
0
log x
(x + 1)
2
dx = 0,

∞
0

log
2
x
(x + 1)
2
dx =
π
2
3
.
Hint, Solution
Exercise 13.25
Consider the integral
I(a) =

∞
0
x
a
1 + x
2
dx.
1. For what values of a does the integral exist?
2. Evaluate the integral. Show that
I(a) =
π
2 cos(πa/2)
3. Deduce from your answer in part (b) the results

∞

0
log x
1 + x
2
dx = 0,

∞
0
log
2
x
1 + x
2
dx =
π
3
8
.
673
You may assume that it is valid to differentiate under the integral sign.
Hint, Solution
Exercise 13.26
Let f(z) be a single-valued analytic function with only isolated singularities and no singularities on the positive real
axis, [0, ∞). Give sufficient conditions on f(x) for absolute convergence of the integral

∞
0
x
a
f(x) dx.

Assume that a is not an integer. Evaluate the integral by considering the integral of z
a
f(z) on a suitable contour.
(Consider the branch of z
a
on which 1
a
= 1.)
Hint, Solution
Exercise 13.27
Using the solution to Exercise 13.26, evaluate

∞
0
x
a
f(x) log x dx,
and

∞
0
x
a
f(x) log
m
x dx,
where m is a positive integer.
Hint, Solution
Exercise 13.28
Using the solution to Exercise 13.26, evaluate


∞
0
f(x) dx,
i.e. examine a = 0. The solution will suggest a way to evaluate the integral with contour integration. Do the contour
integration to corroborate the value of

∞
0
f(x) dx.
Hint, Solution
674
Exercise 13.29
Let f(z) be an analytic function with only isolated singularities and no singularities on the positive real axis, [0, ∞).
Give sufficient conditions on f (x) for absolute convergence of the integral

∞
0
f(x) log x dx
Evaluate the integral with contour integration.
Hint, Solution
Exercise 13.30
For what values of a does the following integral exist?

∞
0
x
a
1 + x
4

dx.
Evaluate the integral. (Consider the branch of x
a
on which 1
a
= 1.)
Hint, Solution
Exercise 13.31
By considering the integral of f(z) = z
1/2
log z/(z + 1)
2
on a suitable contour, show that

∞
0
x
1/2
log x
(x + 1)
2
dx = π,

∞
0
x
1/2
(x + 1)
2
dx =

π
2
.
Hint, Solution
Exploiting Symmetry
Exercise 13.32
Evaluate by contour integration, the principal value integral
I(a) = −

∞
−∞
e
ax
e
x
−
e
−x
dx
675
for a real and |a| < 1. [Hint: Consider the contour that is the boundary of the box, −R < x < R, 0 < y < π, but
indented around z = 0 and z = ıπ.
Hint, Solution
Exercise 13.33
Evaluate the following integrals.
1.

∞
0
dx

(1 + x
2
)
2
,
2.

∞
0
dx
1 + x
3
.
Hint, Solution
Exercise 13.34
Find the value of the integral I
I =

∞
0
dx
1 + x
6
by considering the contour integral

Γ
dz
1 + z
6
with an appropriately chosen contour Γ.

Hint, Solution
Exercise 13.35
Let C be the boundary of the sector 0 < r < R, 0 < θ < π/4. By integrating
e
−z
2
on C and letting R → ∞ show
that

∞
0
cos(x
2
) dx =

∞
0
sin(x
2
) dx =
1
√
2

∞
0
e
−x
2
dx.

Hint, Solution
676
Exercise 13.36
Evaluate

∞
−∞
x
sinh x
dx
using contour integration.
Hint, Solution
Exercise 13.37
Show that

∞
−∞
e
ax
e
x
+1
dx =
π
sin(πa)
for 0 < a < 1.
Use this to derive that

∞
−∞

cosh(bx)
cosh x
dx =
π
cos(πb/2)
for −1 < b < 1.
Hint, Solution
Exercise 13.38
Using techniques of contour integration find for real a and b:
F (a, b) =

π
0
dθ
(a + b cos θ)
2
What are the restrictions on a and b if any? Can the result be applied for complex a, b? How?
Hint, Solution
Exercise 13.39
Show that

∞
−∞
cos x
e
x
+
e
−x
dx =

π
e
π/2
+
e
−π/2
[ Hint: Begin by considering the integral of
e
ız
/(
e
z
+
e
−z
) around a rectangle with vertices: ±R, ±R + ıπ.]
Hint, Solution
677