Advanced Mathematical Methods for Scientists and Engineers Episode 2 Part 9 ppt

Since the nearest singularity is at z = −ı, the Laurent series will converge in the annulus 0 < |z − ı| < 2.
z
2
− 1 = ((z − ı) + ı)
2
− 1
= (z − ı)
2
+ ı2(z − ı) −2
ı/2
z + ı
=
ı/2
ı2 + (z − ı)
=
1/4
1 − ı(z − ı)/2
=
1
4
∞

n=0

ı(z − ı)
2

n
=
1

4
∞

n=0
ı
n
2
n
(z − ı)
n
This geometric series converges for |ı(z −ı)/2| < 1, or |z − ı| < 2. The series expansion of f(z) is
f(z) =
−ı/2
z − ı
− 2 + ı2(z − ı) + (z −ı)
2
+
1
4
∞

n=0
ı
n
2
n
(z − ı)
n
.
z

4
z
2
+ 1
=
−ı/2
z − ı
− 2 + ı2(z − ı) + (z −ı)
2
+
1
4
∞

n=0
ı
n
2
n
(z − ı)
n
for |z − ı| < 2
Laurent Series about z = ∞. Since the nearest singularities are at z = ±ı, the Laurent series will converge in
694
the annulus 1 < |z| < ∞.
z
4
z
2
+ 1

=
z
2
1 + 1/z
2
= z
2
∞

n=0

−
1
z
2

n
=
0

n=−∞
(−1)
n
z
2(n+1)
=
1

n=−∞
(−1)

n+1
z
2n
This geometric series converges for | − 1/z
2
| < 1, or |z| > 1. The series expansion of f(z) is
z
4
z
2
+ 1
=
1

n=−∞
(−1)
n+1
z
2n
for 1 < |z| < ∞
Solution 13.7
Method 1: Residue Theorem. We factor P (z). Let m be the number of roots, counting multipl ici ties, that lie
inside the contour Γ. We find a simple expression for P

(z)/P (z).
P (z) = c
n

k=1
(z − z

k
)
P

(z) = c
n

k=1
n

j=1
j=k
(z − z
j
)
695
P

(z)
P (z)
=
c

n
k=1

n
j=1
j=k
(z − z

j
)
c

n
k=1
(z − z
k
)
=
n

k=1

n
j=1
j=k
(z − z
j
)

n
j=1
(z − z
j
)
=
n

k=1

1
z − z
k
Now we do the integration using the residue theorem.
1
ı2π

Γ
P

(z)
P (z)
dz =
1
ı2π

Γ
n

k=1
1
z − z
k
dz
=
n

k=1
1
ı2π


Γ
1
z − z
k
dz
=

z
k
inside Γ
1
ı2π

Γ
1
z − z
k
dz
=

z
k
inside Γ
1
= m
Method 2: Fundamental Theorem of Calculus. We factor the polynomial, P (z) = c

n
k=1

(z −z
k
). Let m be
696
the number of roots, counting multiplicities, that lie inside the contour Γ.
1
ı2π

Γ
P

(z)
P (z)
dz =
1
ı2π
[log P (z)]
C
=
1
ı2π

log
n

k=1
(z − z
k
)


C
=
1
ı2π

n

k=1
log(z − z
k
)

C
The value of the logarithm changes by ı2π for the terms in which z
k
is inside the contour. Its value does not change
for the terms in which z
k
is outside the contour.
=
1
ı2π


z
k
inside Γ
log(z − z
k
)


C
=
1
ı2π

z
k
inside Γ
ı2π
= m
Solution 13.8
1.

C
e
z
(z − π) tan z
dz =

C
e
z
cos z
(z − π) sin z
dz
The integrand has first order poles at z = nπ, n ∈ Z, n = 1 and a double pole at z = π. The only pole inside
697
the contour occurs at z = 0. We evaluate the integral with the residue theorem.


C
e
z
cos z
(z − π) sin z
dz = ı2π Res

e
z
cos z
(z − π) sin z
, z = 0

= ı2π lim
z=0
z
e
z
cos z
(z − π) sin z
= −ı2 lim
z=0
z
sin z
= −ı2 lim
z=0
1
cos z
= −ı2


C
e
z
(z − π) tan z
dz = −ı2
2. The integrand has a first order poles at z = 0, −π and a second order pole at z = π inside the contour. The
value of the integral is ı2π times the sum of the residues at these points. From the previous part we know that
residue at z = 0.
Res

e
z
cos z
(z − π) sin z
, z = 0

= −
1
π
We find the residue at z = −π with the residue formula.
Res

e
z
cos z
(z − π) sin z
, z = −π

= lim
z→−π

(z + π)
e
z
cos z
(z − π) sin z
=
e
−π
(−1)
−2π
lim
z→−π
z + π
sin z
=
e
−π
2π
lim
z→−π
1
cos z
= −
e
−π
2π
698
We find the residue at z = π by finding the first few terms in the Laurent series of the integrand.
e
z

cos z
(z − π) sin z
=
(
e
π
+
e
π
(z − π) + O((z − π)
2
)) (1 + O((z − π)
2
))
(z − π) (−(z − π) + O((z − π)
3
))
=
−
e
π
−
e
π
(z − π) + O((z − π)
2
)
−(z − π)
2
+ O((z − π)

4
)
=
e
π
(z−π)
2
+
e
π
z−π
+ O(1)
1 + O((z − π)
2
)
=

e
π
(z − π)
2
+
e
π
z − π
+ O(1)


1 + O


(z − π)
2

=
e
π
(z − π)
2
+
e
π
z − π
+ O(1)
With this we see that
Res

e
z
cos z
(z − π) sin z
, z = π

=
e
π
.
The integral is

C
e

z
cos z
(z − π) sin z
dz = ı2π

Res

e
z
cos z
(z − π) sin z
, z = −π

+ Res

e
z
cos z
(z − π) sin z
, z = 0

+ Res

e
z
cos z
(z − π) sin z
, z = π



= ı2π

−
1
π
−
e
−π
2π
+
e
π


C
e
z
(z − π) tan z
dz = ı

2π
e
π
−2 −
e
−π

Cauchy Principal Value for Real Integrals
699
Solution 13.9

Consider the integral

1
−1
1
x
dx.
By the definition of improper integrals we have

1
−1
1
x
dx = lim
→0
+

−
−1
1
x
dx + lim
δ→0
+

1
δ
1
x
dx

= lim
→0
+
[log |x|]
−
−1
+ lim
δ→0
+
[log |x|]
1
δ
= lim
→0
+
log  − lim
δ→0
+
log δ
This limit diverges. Thus the integral diverges.
Now consider the integral

1
−1
1
x − ıα
dx
where α ∈ R, α = 0. Since the integrand is bounded, the integral exists.

1

−1
1
x − ıα
dx =

1
−1
x + ıα
x
2
+ α
2
dx
=

1
−1
ıα
x
2
+ α
2
dx
= ı2

1
0
α
x
2

+ α
2
dx
= ı2

1/α
0
1
ξ
2
+ 1
dξ
= ı2 [arctan ξ]
1/α
0
= ı2 arctan

1
α

700
Figure 13.8: The real and imaginary part of the integrand for several values of α.
Note that the integral exists for all nonzero real α and that
lim
α→0
+

1
−1
1

x − ıα
dx = ıπ
and
lim
α→0
−

1
−1
1
x − ıα
dx = −ıπ.
The integral exists for α arbitrarily close to zero, but diverges when α = 0. The real part of the integrand is an odd
function with two humps that get thinner and taller with decreasing α. The imaginary part of the integrand is an even
function with a hump that gets thinner and taller with decreasing α. (See Figure 13.8.)


1
x − ıα

=
x
x
2
+ α
2
, 

1
x − ıα


=
α
x
2
+ α
2
Note that


1
0
1
x − ıα
dx → +∞ as α → 0
+
and


0
−1
1
x − ıα
dx → −∞ as α → 0
−
.
701
However,
lim
α→0



1
−1
1
x − ıα
dx = 0
because the two integrals above cancel each other.
Now note that when α = 0, the integrand is real. Of course the integral doesn’t converge for this case, but if we
could assign some value to

1
−1
1
x
dx
it would be a real number. Since
lim
α→0

1
−1


1
x − ıα

dx = 0,
This number should be zero.
Solution 13.10

1.
−

1
−1
1
x
2
dx = lim
→0
+


−
−1
1
x
2
dx +

1

1
x
2
dx

= lim
→0
+



−
1
x

−
−1
+

−
1
x

1


= lim
→0
+

1

− 1 − 1 +
1


The principal value of the integral does not exist.
702
2.

−

1
−1
1
x
3
dx = lim
→0
+


−
−1
1
x
3
dx +

1

1
x
3
dx

= lim
→0
+



−
1
2x
2

−
−1
+

−
1
2x
2

1


= lim
→0
+

−
1
2(−)
2
+
1
2(−1)
2

−
1
2(1)
2
+
1
2
2

= 0
3. Since f(x) is real analytic,
f(x) =
∞

n=1
f
n
x
n
for x ∈ (−1, 1).
We can rewrite the integrand as
f(x)
x
3
=
f
0
x
3
+

f
1
x
2
+
f
2
x
+
f(x) − f
0
− f
1
x − f
2
x
2
x
3
.
Note that the final term is real analytic on (−1, 1). Thus the principal value of the integral exists if and only if
f
2
= 0.
Cauchy Principal Value for Contour Integrals
Solution 13.11
We can write f(z) as
f(z) =
f
0

z − z
0
+
(z − z
0
)f(z) − f
0
z − z
0
.
Note that the second term is analytic in a neighborho od of z
0
. Thus it is bounded on the contour. Let M

be the
703
maximum modulus of
(z−z
0
)f(z)−f
0
z−z
0
on C

. By using the maximum modulus integral bound, we have






C

(z − z
0
)f(z) − f
0
z − z
0
dz




≤ (β − α)M

→ 0 as  → 0
+
.
Thus we see that
lim
→0
+

C

f(z) dz lim
→0
+


C

f
0
z − z
0
dz.
We parameterize the path of integration with
z = z
0
+ e
ıθ
, θ ∈ (α, β).
Now we evaluate the integral.
lim
→0
+

C

f
0
z − z
0
dz = lim
→0
+

β
α

f
0
e
ıθ
ıe
ıθ
dθ
= lim
→0
+

β
α
ıf
0
dθ
= ı(β − α)f
0
≡ ı(β − α) Res(f(z), z
0
)
This proves the result.
Solution 13.12
Let C
i
be the contour that is in den ted with circular arcs or radius  at each of the first order poles on C so as to enclose
these poles. Let A
1
, . . . , A
n

be these circular arcs of radius  centered at the points ζ
1
, . . . , ζ
n
. Let C
p
be the contour,
(not necessarily connected), obtained by subtracting each of the A
j
’s from C
i
.
Since the curve is C
1
, (or continuously differentiable), at each of the first order poles on C, the A
j
’s becomes
semi-circles as  → 0
+
. Thus

A
j
f(z) dz = ıπ Res(f(z), ζ
j
) for j = 1, . . . , n.
704
CONTINUE
Figure 13.9: The Indented Contour.
The principal value of the integral along C is

−

C
f(z) dz = lim
→0
+

C
p
f(z) dz
= lim
→0
+


C
i
f(z) dz −
n

j=1

A
j
f(z) dz

= ı2π

m


j=1
Res(f(z), z
j
) +
n

j=1
Res(f(z), ζ
j
)

− ıπ
n

j=1
Res(f(z), ζ
j
)
−

C
f(z) dz = ı2π
m

j=1
Res(f(z), z
j
) + ıπ
n


j=1
Res(f(z), ζ
j
).
Solution 13.13
Consider
−

C
1
z − 1
dz
where C is the unit circle. Let C
p
be the circular arc of radius 1 that starts and ends a distance of  from z = 1. Let
C

be the negative, circular arc of radius  with center at z = 1 that joins the endpoints of C
p
. Let C
i
, be the union of
C
p
and C

. (C
p
stands for Principal value Contour; C
i

stands for Indented Contour.) C
i
is an indented contour that
avoids the first order pole at z = 1. Figure 13.9 shows the three contours.
Note that the principal value of the integral is
−

C
1
z − 1
dz = lim
→0
+

C
p
1
z − 1
dz.
705
We can calculate the integral along C
i
with Cauchy’s theorem. The integrand is analytic inside the contour.

C
i
1
z − 1
dz = 0
We can calculate the integral along C


using Result 13.3.1. Note that as  → 0
+
, the contour becomes a semi-circle,
a circular arc of π radians in the negative direction.
lim
→0
+

C

1
z − 1
dz = −ıπ Res

1
z − 1
, 1

= −ıπ
Now we can write the principal value of the integral along C in terms of the two known integrals.
−

C
1
z − 1
dz =

C
i

1
z − 1
dz −

C

1
z − 1
dz
= 0 − (−ıπ)
= ıπ
Integrals on the Real Axis
Solution 13.14
1. First we note that the integrand is an even function and extend the domain of integration.

∞
0
x
2
(x
2
+ 1)(x
2
+ 4)
dx =
1
2

∞
−∞

x
2
(x
2
+ 1)(x
2
+ 4)
dx
Next we close the path of integration in the upper half plane. Consider the integral along the boundary of the
706
domain 0 < r < R, 0 < θ < π.
1
2

C
z
2
(z
2
+ 1)(z
2
+ 4)
dz =
1
2

C
z
2
(z − ı)(z + ı)(z −ı2)(z + ı2)

dz
= ı2π
1
2

Res

z
2
(z
2
+ 1)(z
2
+ 4)
, z = ı

+ Res

z
2
(z
2
+ 1)(z
2
+ 4)
, z = ı2

= ıπ

z

2
(z + ı)(z
2
+ 4)




z=ı
+
z
2
(z
2
+ 1)(z + ı2)




z=ı2

= ıπ

ı
6
−
ı
3

=

π
6
Let C
R
be the circular arc portion of the contour.

C
=

R
−R
+

C
R
. We show that the integral along C
R
vanishes
as R → ∞ with the maximum modulu s bound.





C
R
z
2
(z
2

+ 1)(z
2
+ 4)
dz




≤ πR max
z∈C
R




z
2
(z
2
+ 1)(z
2
+ 4)




= πR
R
2
(R

2
− 1)(R
2
− 4)
→ 0 as R → ∞
We take the limit as R → ∞ to evaluate the integral along the real axis.
lim
R→∞
1
2

R
−R
x
2
(x
2
+ 1)(x
2
+ 4)
dx =
π
6

∞
0
x
2
(x
2

+ 1)(x
2
+ 4)
dx =
π
6
707
2. We close the path of integration in the upper half plane. Consider the integral along the boundary of the domain
0 < r < R, 0 < θ < π.

C
dz
(z + b)
2
+ a
2
=

C
dz
(z + b − ıa)(z + b + ıa)
= ı2π Res

1
(z + b − ıa)(z + b + ıa)
, z = −b + ıa

= ı2π
1
z + b + ıa





z=−b+ıa
=
π
a
Let C
R
be the circular arc portion of the contour.

C
=

R
−R
+

C
R
. We show that the integral along C
R
vanishes
as R → ∞ with the maximum modulu s bound.






C
R
dz
(z + b)
2
+ a
2




≤ πR max
z∈C
R




1
(z + b)
2
+ a
2




= πR
1
(R − b)

2
+ a
2
→ 0 as R → ∞
We take the limit as R → ∞ to evaluate the integral along the real axis.
lim
R→∞

R
−R
dx
(x + b)
2
+ a
2
=
π
a

∞
−∞
dx
(x + b)
2
+ a
2
=
π
a
Solution 13.15

Let C
R
be the semicircular arc from R to −R in the upper half plane. Let C be the union of C
R
and the interval
708
[−R, R]. We can evaluate the principal value of the integral along C with Result 13.3.2.
−

C
f(x) dx = ı2π
m

k=1
Res (f(z), z
k
) + ıπ
n

k=1
Res(f(z), x
k
)
We examine the integral along C
R
as R → ∞.






C
R
f(z) dz




≤ πR max
z∈C
R
|f(z)|
→ 0 as R → ∞.
Now we are prepared to evaluate the real integral.
−

∞
−∞
f(x) dx = lim
R→∞
−

R
−R
f(x) dx
= lim
R→∞
−

C

f(z) dz
= ı2π
m

k=1
Res (f(z), z
k
) + ıπ
n

k=1
Res(f(z), x
k
)
If we close the path of integration in the lower half plane, the contour will be in the negative direction.
−

∞
−∞
f(x) dx = −ı2π
m

k=1
Res (f(z), z
k
) − ıπ
n

k=1
Res(f(z), x

k
)
Solution 13.16
We consider
−

∞
−∞
2x
x
2
+ x + 1
dx.
709
With the change of variables x = 1/ξ, this becomes
−

−∞
∞
2ξ
−1
ξ
−2
+ ξ
−1
+ 1

−1
ξ
2


dξ,
−

∞
−∞
2ξ
−1
ξ
2
+ ξ + 1
dξ
There are first order poles at ξ = 0 and ξ = −1/2 ± ı
√
3/2. We close the path of integration in the upper half plane
with a semi-circle. Since the integrand decays like ξ
−3
the integrand along the semi-circle vanishes as the radius tends
to infinity. The value of the integral is thus
ıπ Res

2z
−1
z
2
+ z + 1
, z = 0

+ ı2π Res


2z
−1
z
2
+ z + 1
, z = −
1
2
+ ı
√
3
2

ıπ lim
z→0

2
z
2
+ z + 1

+ ı2π lim
z→(−1+ı
√
3)/2

2z
−1
z + (1 + ı
√

3)/2

−

∞
−∞
2x
x
2
+ x + 1
dx = −
2π
√
3
Solution 13.17
1. Consider

∞
−∞
1
x
4
+ 1
dx.
The integrand
1
z
4
+1
is analytic on the real axis and has isolated singularities at the points

z = {
e
ıπ/4
,
e
ı3π/4
,
e
ı5π/4
,
e
ı7π/4
}.
Let C
R
be the semi-circle of radius R in the upper half plane. Since
lim
R→∞

R max
z∈C
R




1
z
4
+ 1






= lim
R→∞

R
1
R
4
− 1

= 0,
710
we can apply Result 13.4.1.

∞
−∞
1
x
4
+ 1
dx = ı2π

Res

1
z

4
+ 1
,
e
ıπ/4

+ Res

1
z
4
+ 1
,
e
ı3π/4

The appropriate residues are,
Res

1
z
4
+ 1
,
e
ıπ/4

= lim
z→
e

ıπ/4
z −
e
ıπ/4
z
4
+ 1
= lim
z→
e
ıπ/4
1
4z
3
=
1
4
e
−ı3π/4
=
−1 − ı
4
√
2
,
Res

1
z
4

+ 1
,
e
ı3π/4

=
1
4(
e
ı3π/4
)
3
=
1
4
e
−ıπ/4
=
1 − ı
4
√
2
,
We evaluate the integral with the residue theorem.

∞
−∞
1
x
4

+ 1
dx = ı2π

−1 − ı
4
√
2
+
1 − ı
4
√
2


∞
−∞
1
x
4
+ 1
dx =
π
√
2
711
2. Now consider

∞
−∞
x

2
(x
2
+ 1)
2
dx.
The integrand is analytic on the real axis and has second order poles at z = ±ı. Since the integrand decays
sufficiently fast at infinity,
lim
R→∞

R max
z∈C
R




z
2
(z
2
+ 1)
2





= lim

R→∞

R
R
2
(R
2
− 1)
2

= 0
we can apply Result 13.4.1.

∞
−∞
x
2
(x
2
+ 1)
2
dx = ı2π Res

z
2
(z
2
+ 1)
2
, z = ı


Res

z
2
(z
2
+ 1)
2
, z = ı

= lim
z→ı
d
dz

(z − ı)
2
z
2
(z
2
+ 1)
2

= lim
z→ı
d
dz


z
2
(z + ı)
2

= lim
z→ı

(z + ı)
2
2z − z
2
2(z + ı)
(z + ı)
4

= −
ı
4

∞
−∞
x
2
(x
2
+ 1)
2
dx =
π

2
3. Since
sin(x)
1 + x
2
712
is an odd function,

∞
−∞
cos(x)
1 + x
2
dx =

∞
−∞
e
ıx
1 + x
2
dx
Since
e
ız
/(1 + z
2
) is analytic except for simple poles at z = ±ı and the integrand decays sufficiently fast in the
upper half plane,
lim

R→∞

R max
z∈C
R




e
ız
1 + z
2





= lim
R→∞

R
1
R
2
− 1

= 0
we can apply Result 13.4.1.


∞
−∞
e
ıx
1 + x
2
dx = ı2π Res

e
ız
(z − ı)(z + ı)
, z = ı

= ı2π
e
−1
ı2

∞
−∞
cos(x)
1 + x
2
dx =
π
e
Solution 13.18
Consider the function
f(z) =
z

6
(z
4
+ 1)
2
.
The value of the function on the imaginary axis:
−y
6
(y
4
+ 1)
2
is a constant multiple of the value of the function on the real axis:
x
6
(x
4
+ 1)
2
.
713
Thus to evaluate the real integral we consider the path of integration, C, which starts at the origin, follows the real
axis to R, follows a circular path to ıR and then follows the imaginary axis back down to the origin. f(z) has se cond
order poles at the fourth roots of −1: (±1 ± ı)/
√
2. Of these only (1 + ı)/
√
2 lies inside the path of integration. We
evaluate the contour integral with the Residue Theorem. For R > 1:


C
z
6
(z
4
+ 1)
2
dz = ı2π Res

z
6
(z
4
+ 1)
2
, z =
e
ıπ/4

= ı2π lim
z→
e
ıπ/4
d
dz

(z −
e
ıπ/4

)
2
z
6
(z
4
+ 1)
2

= ı2π lim
z→
e
ıπ/4
d
dz

z
6
(z −
e
ı3π/4
)
2
(z −
e
ı5π/4
)
2
(z −
e

ı7π/4
)
2

= ı2π lim
z→
e
ıπ/4

z
6
(z −
e
ı3π/4
)
2
(z −
e
ı5π/4
)
2
(z −
e
ı7π/4
)
2

6
z
−

2
z −
e
ı3π/4
−
2
z −
e
ı5π/4
−
2
z −
e
ı7π/4


= ı2π
−ı
(2)(ı4)(−2)

6
√
2
1 + ı
−
2
√
2
−
2

√
2
2 + ı2
−
2
ı
√
2

= ı2π
3
32
(1 − ı)
√
2
=
3π
8
√
2
(1 + ı)
The integral along the circular part of the contour, C
R
, vanishes as R → ∞. We demonstrate this with the maximum
714
modulus integral bound.






C
R
z
6
(z
4
+ 1)
2
dz




≤
πR
4
max
z∈C
R

z
6
(z
4
+ 1)
2

=
πR

4
R
6
(R
4
− 1)
2
→ 0 as R → ∞
Taking the limit R → ∞, we have:

∞
0
x
6
(x
4
+ 1)
2
dx +

0
∞
(ıy)
6
((ıy)
4
+ 1)
2
ı dy =
3π

8
√
2
(1 + ı)

∞
0
x
6
(x
4
+ 1)
2
dx + ı

∞
0
y
6
(y
4
+ 1)
2
dy =
3π
8
√
2
(1 + ı)
(1 + ı)


∞
0
x
6
(x
4
+ 1)
2
dx =
3π
8
√
2
(1 + ı)

∞
0
x
6
(x
4
+ 1)
2
dx =
3π
8
√
2
Fourier Integrals

Solution 13.19
We know that

π
0
e
−R sin θ
dθ <
π
R
.
715
First take the case that ω is positive and the semi-ci rcle is in the upper half plane.





C
R
f(z)
e
ıωz
dz




≤






C
R
e
ıωz
dz




max
z∈C
R
|f(z)|
≤

π
0



e
ıωR
e
ıθ
R
e

ıθ



dθ max
z∈C
R
|f(z)|
= R

π
0


e
−ωR sin θ


dθ max
z∈C
R
|f(z)|
< R
π
ωR
max
z∈C
R
|f(z)|
=

π
ω
max
z∈C
R
|f(z)|
→ 0 as R → ∞
The procedure is almost the same for negative ω.
Solution 13.20
First we write the integral in terms of Fourier integrals.

∞
−∞
cos 2x
x − ıπ
dx =

∞
−∞
e
ı2x
2(x − ıπ)
dx +

∞
−∞
e
−ı2x
2(x − ıπ)
dx

Note that
1
2(z−ıπ)
vanishes as |z| → ∞. We close the former Fourier integral in the upper half plane and the latter in
the lower half plane. There is a first order pole at z = ıπ in the upper half plane.

∞
−∞
e
ı2x
2(x − ıπ)
dx = ı2π Res

e
ı2z
2(z − ıπ)
, z = ıπ

= ı2π
e
−2π
2
There are no singularities in the lower half plane.

∞
−∞
e
−ı2x
2(x − ıπ)
dx = 0

716
Thus the value of the original real integral is

∞
−∞
cos 2x
x − ıπ
dx = ıπ
e
−2π
Fourier Cosine and Sine Integrals
Solution 13.21
We are considering the integral

∞
−∞
sin x
x
dx.
The integrand is an entire function. So it doesn’t appear that the residue theorem would directly apply. Also the
integrand is unbounded as x → +ı∞ and x → −ı∞, so closing the integral in the upper or lower half plane is not
directly applicable. In order to proceed, we must write the integrand in a different form. Note that
−

∞
−∞
cos x
x
dx = 0
since the integrand is odd and has only a first order pole at x = 0. Thus


∞
−∞
sin x
x
dx = −

∞
−∞
e
ıx
ıx
dx.
Let C
R
be the semicircular arc in the upper half plane from R to −R. Let C be the closed contour that is the union
of C
R
and the real interval [−R, R]. If we close the path of integration with a semicircular arc in the upper half plane,
we have

∞
−∞
sin x
x
dx = lim
R→∞

−


C
e
ız
ız
dz −

C
R
e
ız
ız
dz

,
provided that all the integrals exist.
The integral along C
R
vanishes as R → ∞ by Jordan’s lemma. By the residue theorem for principal values we have
−

e
ız
ız
dz = ıπ Res

e
ız
ız
, 0


= π.
717