Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 10 ppt
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Solution 13.28
Convergence. If x
a
f(x) x
α
as x → 0 for some α > −1 then the integral
1
0
x
a
f(x) dx
will converge absolutely. If x
a
f(x) x
β
as x → ∞ for some β < −1 then the integral
∞
1
x
a
f(x)
will converge absolutely. These are sufficient conditions for the absolute convergence of
∞
0
x
a
f(x) dx.
Contour Integration. We put a branch cut on the positive real axis and choose 0 < arg(z) < 2π. We consider
the integral of z
a
f(z) on the contour in Figure ??. Let the singularities of f(z) occur at z
1
, . . . , z
n
. By the residue
theorem,
C
z
a
f(z) dz = ı2π
n
k=1
Res (z
a
f(z), z
k
) .
On the circle of radius , the integrand is o(
−1
). Since the length of C
is 2π, the integral on C
vanishes as
→ 0. On the circle of radius R, the integrand is o(R
−1
). Since the length of C
R
is 2πR, the integral on C
R
vanishes
as R → ∞.
The value of the integrand below the branch cut, z = x
e
ı2π
, is
f(x
e
ı2π
) = x
a
e
ı2πa
f(x)
In the limit as → 0 and R → ∞ we have
∞
0
x
a
f(x) dx +
0
−∞
x
a
e
ı2πa
f(x) dx = ı2π
n
k=1
Res (z
a
f(z), z
k
) .
734
∞
0
x
a
f(x) dx =
ı2π
1 −
e
ı2πa
n
k=1
Res (z
a
f(z), z
k
) .
Solution 13.29
In the interval of uniform convergence of th integral, we can differentiate the formula
∞
0
x
a
f(x) dx =
ı2π
1 −
e
ı2πa
n
k=1
Res (z
a
f(z), z
k
) ,
with respect to a to obtain,
∞
0
x
a
f(x) log x dx =
ı2π
1 −
e
ı2πa
n
k=1
Res (z
a
f(z) log z, z
k
) , −
4π
2
a
e
ı2πa
(1 −
e
ı2πa
)
2
n
k=1
Res (z
a
f(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
) ,
Differentiating the solution of Exercise 13.26 m times with resp ect to a yields
∞
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
)
,
Solution 13.30
Taking the limit as a → 0 ∈ Z in the solution of Exercise 13.26 yields
∞
0
f(x) dx = ı2π lim
a→0
n
k=1
Res (z
a
f(z), z
k
)
1 −
e
ı2πa
The numerator vanishes because the sum of all residues of z
n
f(z) is zero. Thus we can use L’Hospital’s rule.
∞
0
f(x) dx = ı2π lim
a→0
n
k=1
Res (z
a
f(z) log z, z
k
)
−ı2π
e
ı2πa
735
∞
0
f(x) dx = −
n
k=1
Res (f(z) log z, z
k
)
This suggests that we could have derived the result directly by considering the integral of f(z) log z on the contour
in Figure ??. We put a branch cut on the positive real axis and choose the branch arg z = 0. Recall that we have
assumed that f(z) has only isolated singularities and no singularities on the positive real axis, [0, ∞). By the residue
theorem,
C
f(z) log z dz = ı2π
n
k=1
Res (f(z) log z, z = z
k
) .
By assuming that f(z) z
α
as z → 0 where α > −1 the integral on C
will vanish as → 0. By assuming that
f(z) z
β
as z → ∞ where β < −1 the integral on C
R
will vanish as R → ∞. The value of the integrand below the
branch cut, z = x
e
ı2π
is f (x)(log x + ı2π). Taking the limit as → 0 and R → ∞, we have
∞
0
f(x) log x dx +
0
∞
f(x)(log x + ı2π) dx = ı2π
n
k=1
Res (f(z) log z, z
k
) .
Thus we corroborate the result.
∞
0
f(x) dx = −
n
k=1
Res (f(z) log z, z
k
)
Solution 13.31
Consider the integral of f(z) log
2
z on the contour in Figure ??. We put a branch cut on the positive real axis and
choose the branch 0 < arg z < 2π. Let z
1
, . . . z
n
be the singularities of f(z). By the residue theorem,
C
f(z) log
2
z dz = ı2π
n
k=1
Res
f(z) log
2
z, z
k
.
If f(z) z
α
as z → 0 for some α > −1 then the integral on C
will vanish as → 0. f(z) z
β
as z → ∞ for some
β < −1 then the integral on C
R
will vanish as R → ∞. Below the branch cut the integrand is f(x)(log x + ı2π)
2
.
736
Thus we have
∞
0
f(x) log
2
x dx +
0
∞
f(x)(log
2
x + ı4π log x −4π
2
) dx = ı2π
n
k=1
Res
f(z) log
2
z, z
k
.
−ı4π
∞
0
f(x) log x dx + 4π
2
∞
0
f(x) dx = ı2π
n
k=1
Res
f(z) log
2
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
)
Solution 13.32
Convergence. We conside r
∞
0
x
a
1 + x
4
dx.
Since the integrand behaves like x
a
near x = 0 we must have (a) > −1. Since the integrand behaves like x
a−4
at
infinity we must have (a − 4) < −1. The integral converges for −1 < (a) < 3.
Contour Integration. The function
f(z) =
z
a
1 + z
4
has first order poles at z = (±1 ±ı)/
√
2 and a branch point at z = 0. We could evaluate the real integral by putting
a branch cut on the positive real axis with 0 < arg(z) < 2π and integrating f(z) on the contour in Figure 13.12.
Integrating on this contour would work because the value of the integrand below the branch cut is a constant times
the value of the integrand above the branch cut. After demonstrating that the integrals along C
and C
R
vanish in the
limits as → 0 and R → ∞ we would see that the value of the integral is a constant times the sum of the residues at
the four poles. However, this is not the only, (and not the best), contour that can be used to evaluate the real integral.
Consider the value of the integral on the line arg(z) = θ.
f(r
e
ıθ
) =
r
a
e
ıaθ
1 + r
4
e
ı4θ
737
C
R
C
ε
Figure 13.12: Possible path of integration for f(z) =
z
a
1+z
4
If θ is a integer multiple of π/2 then the integrand is a constant multiple of
f(x) =
r
a
1 + r
4
.
Thus any of the contours in Figure 13.13 can be used to evaluate the real integral. The only difference is how many
residues we have to calculate. Thus we choose the first contour in Figure 13.13. We put a branch cut on the negative
real axis and choose the branch −π < arg(z) < π to satisfy f(1) = 1.
We evaluate the integral along C with the Residue Theorem.
C
z
a
1 + z
4
dz = ı2π Res
z
a
1 + z
4
, z =
1 + ı
√
2
Let a = α + ıβ and z = r
e
ıθ
. Note that
|z
a
| = |(r
e
ıθ
)
α+ıβ
| = r
α
e
−βθ
.
738
C
C
C
C
C
C
R
ε
ε
R
ε
R
Figure 13.13: Possible Paths of Integration for f(z) =
z
a
1+z
4
The integral on C
vanishes as → 0. We demonstrate this with the maximum modulus integral bound.
C
z
a
1 + z
4
dz
≤
π
2
max
z∈C
z
a
1 + z
4
≤
π
2
α
e
π|β|/2
1 −
4
→ 0 as → 0
The integral on C
R
vanishes as R → ∞.
C
R
z
a
1 + z
4
dz
≤
πR
2
max
z∈C
R
z
a
1 + z
4
≤
πR
2
R
α
e
π|β|/2
R
4
− 1
→ 0 as R → ∞
739
The value of the integrand on the positive imaginary axis, z = x
e
ıπ/2
, is
(x
e
ıπ/2
)
a
1 + (x
e
ıπ/2
)
4
=
x
a
e
ıπa/2
1 + x
4
.
We take the limit as → 0 and R → ∞.
∞
0
x
a
1 + x
4
dx +
0
∞
x
a
e
ıπa/2
1 + x
4
e
ıπ/2
dx = ı2π Res
z
a
1 + z
4
,
e
ıπ/4
1 −
e
ıπ(a+1)/2
∞
0
x
a
1 + x
4
dx = ı2π lim
z→
e
ıπ/4
z
a
(z −
e
ıπ/2
)
1 + z
4
∞
0
x
a
1 + x
4
dx =
ı2π
1 −
e
ıπ(a+1)/2
lim
z→
e
ıπ/4
az
a
(z −
e
ıπ/2
) + z
a
4z
3
∞
0
x
a
1 + x
4
dx =
ı2π
1 −
e
ıπ(a+1)/2
e
ıπa/4
4
e
ı3π/4
∞
0
x
a
1 + x
4
dx =
−ıπ
2(
e
−ıπ(a+1)/4
−
e
ıπ(a+1)/4
)
∞
0
x
a
1 + x
4
dx =
π
4
csc
π(a + 1)
4
Solution 13.33
Consider the branch of f(z) = z
1/2
log z/(z + 1)
2
with a branch cut on the positive real axis and 0 < arg z < 2π. We
integrate this function on the contour i n Figure ??.
We use the maximum modulus integral bound to show that the integral on C
ρ
vanishes as ρ → 0.
C
ρ
z
1/2
log z
(z + 1)
2
dz
≤ 2πρ max
C
ρ
z
1/2
log z
(z + 1)
2
= 2πρ
ρ
1/2
(2π −log ρ)
(1 − ρ)
2
→ 0 as ρ → 0
740
The integral on C
R
vanishes as R → ∞.
C
R
z
1/2
log z
(z + 1)
2
dz
≤ 2πR max
C
R
z
1/2
log z
(z + 1)
2
= 2πR
R
1/2
(log R + 2π)
(R − 1)
2
→ 0 as R → ∞
Above the branch cut, (z = x
e
ı0
), the integrand is,
f(x
e
ı0
) =
x
1/2
log x
(x + 1)
2
.
Below the branch cut, (z = x
e
ı2π
), we have,
f(x
e
ı2π
) =
−x
1/2
(log x + ıπ)
(x + 1)
2
.
Taking the limit as ρ → 0 and R → ∞, the residue theorem gives us
∞
0
x
1/2
log x
(x + 1)
2
dx +
0
∞
−x
1/2
(log x + ı2π)
(x + 1)
2
dx = ı2π Res
z
1/2
log z
(z + 1)
2
, −1
.
2
∞
0
x
1/2
log x
(x + 1)
2
dx + ı2π
∞
0
x
1/2
(x + 1)
2
dx = ı2π lim
z→−1
d
dz
(z
1/2
log z)
2
∞
0
x
1/2
log x
(x + 1)
2
dx + ı2π
∞
0
x
1/2
(x + 1)
2
dx = ı2π lim
z→−1
1
2
z
−1/2
log z + z
1/2
1
z
2
∞
0
x
1/2
log x
(x + 1)
2
dx + ı2π
∞
0
x
1/2
(x + 1)
2
dx = ı2π
1
2
(−ı)(ıπ) − ı
741
2
∞
0
x
1/2
log x
(x + 1)
2
dx + ı2π
∞
0
x
1/2
(x + 1)
2
dx = 2π + ıπ
2
Equating real and imaginary parts,
∞
0
x
1/2
log x
(x + 1)
2
dx = π,
∞
0
x
1/2
(x + 1)
2
dx =
π
2
.
Exploiting Symmetry
Solution 13.34
Convergence. The integrand,
e
az
e
z
−
e
−z
=
e
az
2 sinh(z)
,
has first order poles at z = ınπ, n ∈ Z. To study convergence, we split the domain of integration.
∞
−∞
=
−1
−∞
+
1
−1
+
∞
1
The principal value integral
−
1
−1
e
ax
e
x
−
e
−x
dx
exists for any a because the integrand has only a first order pole on the path of integration.
Now consider the integral on (1 . . . ∞).
∞
1
e
ax
e
x
−
e
−x
dx
=
∞
1
e
(a−1)x
1 −
e
−2x
dx
≤
1
1 −
e
−2
∞
1
e
(a−1)x
dx
This integral converges for a −1 < 0; a < 1.
742
Finally consid er the integral on (−∞. . . − 1).
−1
−∞
e
ax
e
x
−
e
−x
dx
=
−1
−∞
e
(a+1)x
1 −
e
2x
dx
≤
1
1 −
e
−2
−1
−∞
e
(a+1)x
dx
This integral converges for a + 1 > 0; a > −1.
Thus we see that the integral for I(a) converges for real a, |a| < 1.
Choice of Contour. Consider the contour C that is the boundary of the region: −R < x < R, 0 < y < π. The
integrand h as no singularities inside the contour. There are first order poles on the contour at z = 0 and z = ıπ. The
value of the integral along the contour is ıπ times the sum of these two residues.
The integrals along the vertical sides of the contour vani sh as R → ∞.
R+ıπ
R
e
az
e
z
−
e
−z
dz
≤ π max
z∈(R R+ıπ)
e
az
e
z
−
e
−z
≤ π
e
aR
e
R
−
e
−R
→ 0 as R → ∞
−R+ıπ
−R
e
az
e
z
−
e
−z
dz
≤ π max
z∈(−R −R+ıπ)
e
az
e
z
−
e
−z
≤ π
e
−aR
e
−R
−
e
R
→ 0 as R → ∞
743
Evaluating the Integral. We take the limit as R → ∞ and apply the residue theorem.
∞
−∞
e
ax
e
x
−
e
−x
dx +
−∞+ıπ
∞+ıπ
e
az
e
z
−
e
−z
dz
= ıπ Res
e
az
e
z
−
e
−z
, z = 0
+ ıπ Res
e
az
e
z
−
e
−z
, z = ıπ
∞
−∞
e
ax
e
x
−
e
−x
dx +
−∞
∞
e
a(x+ıπ
e
x+ıπ
−
e
−x−ıπ
dz = ıπ lim
z→0
z
e
az
2 sinh(z)
+ ıπ lim
z→ıπ
(z −ıπ)
e
az
2 sinh(z)
(1 +
e
ıaπ
)
∞
−∞
e
ax
e
x
−
e
−x
dx = ıπ lim
z→0
e
az
+az
e
az
2 cosh(z)
+ ıπ lim
z→ıπ
e
az
+a(z −ıπ)
e
az
2 cosh(z)
(1 +
e
ıaπ
)
∞
−∞
e
ax
e
x
−
e
−x
dx = ıπ
1
2
+ ıπ
e
ıaπ
−2
∞
−∞
e
ax
e
x
−
e
−x
dx =
ıπ(1 −
e
ıaπ
)
2(1 +
e
ıaπ
)
∞
−∞
e
ax
e
x
−
e
−x
dx =
π
2
ı(
e
−ıaπ/2
−
e
ıaπ/2
)
e
ıaπ/2
+
e
ıaπ/2
∞
−∞
e
ax
e
x
−
e
−x
dx =
π
2
tan
aπ
2
Solution 13.35
1.
∞
0
dx
(1 + x
2
)
2
=
1
2
∞
−∞
dx
(1 + x
2
)
2
We apply Result 13.4.1 to the integral on the real axis. First we verify that the integrand vanishes fast enough
in the upper half plane.
lim
R→∞
R max
z∈C
R
1
(1 + z
2
)
2
= lim
R→∞
R
1
(R
2
− 1)
2
= 0
744
Then we evaluate the integral with the residue theorem.
∞
−∞
dx
(1 + x
2
)
2
= ı2π Res
1
(1 + z
2
)
2
, z = ı
= ı2π Res
1
(z −ı)
2
(z + ı)
2
, z = ı
= ı2π lim
z→ı
d
dz
1
(z + ı)
2
= ı2π lim
z→ı
−2
(z + ı)
3
=
π
2
∞
0
dx
(1 + x
2
)
2
=
π
4
2. We wish to evaluate
∞
0
dx
x
3
+ 1
.
Let the contour C be the boundary of the region 0 < r < R, 0 < θ < 2π/3. We factor the denominator of the
integrand to see that the contour encloses the simple pole at
e
ıπ/3
for R > 1.
z
3
+ 1 = (z −
e
ıπ/3
)(z + 1)(z −
e
−ıπ/3
)
745
We calculate the residue at that point.
Res
1
z
3
+ 1
, z =
e
ıπ/3
= lim
z→
e
ıπ/3
(z −
e
ıπ/3
)
1
z
3
+ 1
= lim
z→
e
ıπ/3
1
(z + 1)(z −
e
−ıπ/3
)
=
1
(
e
ıπ/3
+1)(
e
ıπ/3
−
e
−ıπ/3
)
= −
e
ıπ/3
3
We use the residue theorem to evaluate the integral.
C
dz
z
3
+ 1
= −
ı2π
e
ıπ/3
3
Let C
R
be the circular arc portion of the contour.
C
dz
z
3
+ 1
=
R
0
dx
x
3
+ 1
+
C
R
dz
z
3
+ 1
−
R
0
e
ı2π/3
dx
x
3
+ 1
= (1 +
e
−ıπ/3
)
R
0
dx
x
3
+ 1
+
C
R
dz
z
3
+ 1
We show that the integral along C
R
vanishes as R → ∞ with the maximum modulus integral bound .
C
R
dz
z
3
+ 1
≤
2πR
3
1
R
3
− 1
→ 0 as R → ∞
We take R → ∞ and solve for the desired integral.
1 +
e
−ıπ/3
∞
0
dx
x
3
+ 1
= −
ı2π
e
ıπ/3
3
∞
0
dx
x
3
+ 1
=
2π
3
√
3
746
Figure 13.14: The semi-circle contour.
Solution 13.36
Method 1: Semi-Circle Contour. We wish to evaluate the integral
I =
∞
0
dx
1 + x
6
.
We note that the integrand is an even function and express I as an integral over the whole real axis.
I =
1
2
∞
−∞
dx
1 + x
6
Now we will evaluate the integral using contour integration. We close the path of integration in the upper half plane.
Let Γ
R
be the semicircular arc from R to −R in the upper half plane. Let Γ be the union of Γ
R
and the interval
[−R, R]. (See Figure 13.14.)
We can evaluate the integral along Γ with the residue theorem. The integrand has first order poles at z =
e
ıπ(1+2k)/6
,
747
k = 0, 1, 2, 3, 4, 5. Three of these poles are in the upper half plane. For R > 1, we have
Γ
1
z
6
+ 1
dz = ı2π
2
k=0
Res
1
z
6
+ 1
,
e
ıπ(1+2k)/6
= ı2π
2
k=0
lim
z→
e
ıπ(1+2k)/6
z −
e
ıπ(1+2k)/6
z
6
+ 1
Since the numerator and denominator vanish, we apply L’Hospital’s rule.
= ı2π
2
k=0
lim
z→
e
ıπ(1+2k)/6
1
6z
5
=
ıπ
3
2
k=0
e
−ıπ5(1+2k)/6
=
ıπ
3
e
−ıπ5/6
+
e
−ıπ15/6
+
e
−ıπ25/6
=
ıπ
3
e
−ıπ5/6
+
e
−ıπ/2
+
e
−ıπ/6
=
ıπ
3
−
√
3 − ı
2
− ı +
√
3 − ı
2
=
2π
3
Now we examine the integral along Γ
R
. We use the maximum modulus integral bound to show that the value of the
748
integral vanishes as R → ∞.
Γ
R
1
z
6
+ 1
dz
≤ πR max
z∈Γ
R
1
z
6
+ 1
= πR
1
R
6
− 1
→ 0 as R → ∞.
Now we are prepared to evaluate the original real integral.
Γ
1
z
6
+ 1
dz =
2π
3
R
−R
1
x
6
+ 1
dx +
Γ
R
1
z
6
+ 1
dz =
2π
3
We take the limit as R → ∞.
∞
−∞
1
x
6
+ 1
dx =
2π
3
∞
0
1
x
6
+ 1
dx =
π
3
We would get the same result by closing the path of integration in the lower half plane. Note that in this case the
closed contour would be i n the negative direction.
Method 2: Wedge Contour. Consider the contour Γ, which starts at the origin, goes to the point R along the
real axis, then to the point R
e
ıπ/3
along a circle of radius R and then back to the origin along the ray θ = π/3. (See
Figure 13.15.)
We can evaluate the integral along Γ with the residue theorem. The integrand has one first order pole inside the
749
Figure 13.15: The wedge contour.
contour at z =
e
ıπ/6
. For R > 1, we have
Γ
1
z
6
+ 1
dz = ı2π Res
1
z
6
+ 1
,
e
ıπ/6
= ı2π lim
z→
e
ıπ/6
z −
e
ıπ/6
z
6
+ 1
Since the numerator and denominator vanish, we apply L’Hospital’s rule.
= ı2π lim
z→
e
ıπ/6
1
6z
5
=
ıπ
3
e
−ıπ5/6
=
π
3
e
−ıπ/3
Now we examine the integral along the circular arc, Γ
R
. We use the maximum modulus integral bound to show that
750
the value of the integral vanishes as R → ∞.
Γ
R
1
z
6
+ 1
dz
≤
πR
3
max
z∈Γ
R
1
z
6
+ 1
=
πR
3
1
R
6
− 1
→ 0 as R → ∞.
Now we are prepared to evaluate the original real integral.
Γ
1
z
6
+ 1
dz =
π
3
e
−ıπ/3
R
0
1
x
6
+ 1
dx +
Γ
R
1
z
6
+ 1
dz +
0
R
e
ıπ/3
1
z
6
+ 1
dz =
π
3
e
−ıπ/3
R
0
1
x
6
+ 1
dx +
Γ
R
1
z
6
+ 1
dz +
0
R
1
x
6
+ 1
e
ıπ/3
dx =
π
3
e
−ıπ/3
We take the limit as R → ∞.
1 −
e
ıπ/3
∞
0
1
x
6
+ 1
dx =
π
3
e
−ıπ/3
∞
0
1
x
6
+ 1
dx =
π
3
e
−ıπ/3
1 −
e
ıπ/3
∞
0
1
x
6
+ 1
dx =
π
3
(1 − ı
√
3)/2
1 − (1 + ı
√
3)/2
∞
0
1
x
6
+ 1
dx =
π
3
Solution 13.37
First note that
cos(2θ) ≥ 1 −
4
π
θ, 0 ≤ θ ≤
π
4
.
751
Figure 13.16: cos(2θ) and 1 −
4
π
θ
These two functions are plotted in Figure 13.16. To prove this inequality analytically, note that the two functions are
equal at the endpoints of the interval and that cos(2θ) is concave downward on the interval,
d
2
dθ
2
cos(2θ) = −4 cos(2θ) ≤ 0 for 0 ≤ θ ≤
π
4
,
while 1 − 4θ/π is linear.
Let C
R
be the quarter circle of radius R from θ = 0 to θ = π/4. The i ntegral along this contour vanishes as
R → ∞.
C
R
e
−z
2
dz
≤
π/4
0
e
−(R
e
ıθ
)
2
Rı
e
ıθ
dθ
≤
π/4
0
R
e
−R
2
cos(2θ)
dθ
≤
π/4
0
R
e
−R
2
(1−4θ/π)
dθ
=
R
π
4R
2
e
−R
2
(1−4θ/π)
π/4
0
=
π
4R
1 −
e
−R
2
→ 0 as R → ∞
752
Let C be the boundary of the domain 0 < r < R, 0 < θ < π/4. Since the integrand is analytic inside C the integral
along C is zero. Taking the limit as R → ∞, the integral from r = 0 to ∞ along θ = 0 is equal to the integral from
r = 0 to ∞ along θ = π/4.
∞
0
e
−x
2
dx =
∞
0
e
−
“
1+ı
√
2
x
”
2
1 + ı
√
2
dx
∞
0
e
−x
2
dx =
1 + ı
√
2
∞
0
e
−ıx
2
dx
∞
0
e
−x
2
dx =
1 + ı
√
2
∞
0
cos(x
2
) − ı sin(x
2
)
dx
∞
0
e
−x
2
dx =
1
√
2
∞
0
cos(x
2
) dx +
∞
0
sin(x
2
) dx
+
ı
√
2
∞
0
cos(x
2
) dx −
∞
0
sin(x
2
) dx
We equate the imaginary part of this equation to see that the integrals of cos(x
2
) and sin(x
2
) are equal.
∞
0
cos(x
2
) dx =
∞
0
sin(x
2
) dx
The real part of the equation then gives us the desired identity.
∞
0
cos(x
2
) dx =
∞
0
sin(x
2
) dx =
1
√
2
∞
0
e
−x
2
dx
Solution 13.38
Consider the box contour C that is the boundary of the rectangle −R ≤ x ≤ R, 0 ≤ y ≤ π. There is a removable
753
singularity at z = 0 and a first order pole at z = ıπ. By the residue theorem,
−
C
z
sinh z
dz = ıπ Res
z
sinh z
, ıπ
= ıπ lim
z→ıπ
z(z − ıπ)
sinh z
= ıπ lim
z→ıπ
2z −ıπ
cosh z
= π
2
The integrals along the side of the box vanish as R → ∞.
±R+ıπ
±R
z
sinh z
dz
≤ π max
z∈[±R,±R+ıπ]
z
sinh z
≤ π
R + π
sinh R
→ 0 as R → ∞
The value of the integrand on the top of the box is
x + ıπ
sinh(x + ıπ)
= −
x + ıπ
sinh x
.
Taking the limit as R → ∞,
∞
−∞
x
sinh x
dx + −
∞
−∞
−
x + ıπ
sinh x
dx = π
2
.
Note that
−
∞
−∞
1
sinh x
dx = 0
as there is a first order pole at x = 0 and the integrand is odd.
∞
−∞
x
sinh x
dx =
π
2
2
754
Solution 13.39
First we evaluate
∞
−∞
e
ax
e
x
+1
dx.
Consider the rectangular contour in the positive direction with corners at ±R and ±R + ı2π. With the maximum
modulus integral bound we see that the integrals on the vertical sides of the contour vanish as R → ∞.
R+ı2π
R
e
az
e
z
+1
dz
≤ 2π
e
aR
e
R
−1
→ 0 as R → ∞
−R
−R+ı2π
e
az
e
z
+1
dz
≤ 2π
e
−aR
1 −
e
−R
→ 0 as R → ∞
In the limit as R tends to infinity, the integral on the rectangular contour is the sum of the integrals along the top and
bottom sid es.
C
e
az
e
z
+1
dz =
∞
−∞
e
ax
e
x
+1
dx +
−∞
∞
e
a(x+ı2π)
e
x+ı2π
+1
dx
C
e
az
e
z
+1
dz = (1 −
e
−ı2aπ
)
∞
−∞
e
ax
e
x
+1
dx
The only singularity of the integrand inside the contour is a first orde r pole at z = ıπ. We use the residue theorem to
evaluate the integral.
C
e
az
e
z
+1
dz = ı2π Res
e
az
e
z
+1
, ıπ
= ı2π lim
z→ıπ
(z −ıπ)
e
az
e
z
+1
= ı2π lim
z→ıπ
a(z −ıπ)
e
az
+
e
az
e
z
= −ı2π
e
ıaπ
755
We equate the two results for the value of the contour integral.
(1 −
e
−ı2aπ
)
∞
−∞
e
ax
e
x
+1
dx = −ı2π
e
ıaπ
∞
−∞
e
ax
e
x
+1
dx =
ı2π
e
ıaπ
−
e
−ıaπ
∞
−∞
e
ax
e
x
+1
dx =
π
sin(πa)
Now we derive the value of,
∞
−∞
cosh(bx)
cosh x
dx.
First make the change of variables x → 2x in the previous result.
∞
−∞
e
2ax
e
2x
+1
2 dx =
π
sin(πa)
∞
−∞
e
(2a−1)x
e
x
+
e
−x
dx =
π
sin(πa)
Now we set b = 2a − 1.
∞
−∞
e
bx
cosh x
dx =
π
sin(π(b + 1)/2)
=
π
cos(πb/2)
for −1 < b < 1
Since the cosine is an even function, we also have,
∞
−∞
e
−bx
cosh x
dx =
π
cos(πb/2)
for −1 < b < 1
Adding these two equations and dividing by 2 yields the desired result.
∞
−∞
cosh(bx)
cosh x
dx =
π
cos(πb/2)
for −1 < b < 1
756
Solution 13.40
Real-Valued Parameters. For b = 0, the integral has the value: π/a
2
. If b is nonzero, then we can write the integral
as
F (a, b) =
1
b
2
π
0
dθ
(a/b + cos θ)
2
.
We define the new parameter c = a/b and the function,
G(c) = b
2
F (a, b) =
π
0
dθ
(c + cos θ)
2
.
If −1 ≤ c ≤ 1 then the integrand has a double pole on the path of integration. The integral diverges. Otherwise
the integral exists. To evaluate the integral, we extend the range of integration to (0 2π) and make the change of
variables, z =
e
ıθ
to integrate along the unit circle in the complex plane.
G(c) =
1
2
2π
0
dθ
(c + cos θ)
2
For this change of variables, we have,
cos θ =
z + z
−1
2
, dθ =
dz
ız
.
G(c) =
1
2
C
dz/(ız)
(c + (z + z
−1
)/2)
2
= −ı2
C
z
(2cz + z
2
+ 1)
2
dz
= −ı2
C
z
(z + c +
√
c
2
− 1)
2
(z + c −
√
c
2
− 1)
2
dz
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