Course 214
Applications of Cauchy’s Residue Theorem
Second Semester 2008
David R. Wilkins
c David R. Wilkins 1989–2008
Copyright Contents
8 Applications of Cauchy’s Residue Theorem
i
81
8
Applications of Cauchy’s Residue Theorem
Lemma 8.1 Let R be a positive real number, and let f be a continuous
complex-valued function defined everywhere on the semicircle SR , where
SR = {z ∈ C : |z| = R and Im[z] ≥ 0}.
Suppose that there exists a non-negative real number M (R) such that |f (z)| ≤
M (R) for all z ∈ SR . Then
Z
πM (R)
isz
≤
f
(z)e
dz
s
σR
for all s > 0, where σR : [0, π] → C is the path with [σR ] = SR defined such
that σR (θ) = Reiθ for all θ ∈ [0, π].
Proof It follows from the definition of the path integral that
Z
Z π
isz
f (z)e dz =
f (σR (θ))eisσR (θ) σR0 (θ) dθ
σR
Z0 π
f (Reiθ )eiRs cos θ−Rs sin θ iReiθ dθ.
=
0
Therefore
Z
isz
f (z)e
σR
Z
dz ≤ R
π
|f (Reiθ )||eiRs cos θ−Rs sin θ | dθ
0
Z π
≤ RM (R)
e−Rs sin θ dθ.
0
Now sin θ ≥ 2θ/π when 0 ≤ θ ≤ π/2, and therefore
Z π
Z π
2
2
−2Rsθ
π
−Rs sin θ
e
dθ ≤
e π dθ ≤
.
2Rs
0
0
Also
Z
π
−Rs sin θ
e
Z
dθ =
π
2
π
2
e−Rs sin θ dθ.
0
(This follows on making the substitution that replaces θ by π − θ.) Therefore
Z π
π
e−Rs sin θ dθ ≤
.
Rs
0
It follows that
Z
σR
isz
f (z)e
πM (R)
dz ≤
,
s
as required.
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Example We shall apply Cauchy’s Residue Theorem (Theorem 6.16) and
Lemma 8.1 in order to evaluate
Z ∞
eisx
dx
2
2
−∞ x + a
when s > 0.
Let a be a positive real number, let R be a real number satisfying R > a,
and let σR : [0, π] → C be the path that sends θ ∈ [0, π] to Reiθ . (Thus σR (θ)
traverses a semicircle of radius R in the upper half of the complex plane
from R to −R as θ increases from 0 to π. Now it follows from the Triangle
Inequality that |z 2 | ≤ |z 2 + a2 | + |a2 |, and thus |z 2 + a2 | ≥ |z|2 − a2 for all
complex numbers z, and therefore
1 1
z 2 + a2 ≤ R2 − a2
for all complex numbers z satisfying |z| ≥ R. It now follows from Lemma 8.1
that
Z
isz
e
π
≤
dz
2
2
2
s(R − a2 )
σR z + a
for all real numbers s and R satisfying s > 0 and R > a. Therefore
Z
eisz
lim
dz = 0.
R→+∞ σ z 2 + a2
R
Now the function f has poles at ia and −ia. Moreover
eisz
eisz
e−sa
=
lim
=
,
lim (z − ia) 2
z→ia
z→ia z + ia
z + a2
2ia
eisz
as a simple
z 2 + a2
pole at ia with residue e−sa /2ia. Thus if we apply Cauchy’s Residue Theorem
(Theorem 6.16) in order to evaluate the path integral of this function around
the boundary of the set
and therefore the meromorphic function that sends z to
{z ∈ C : |z| ≤ R and Im[z] ≥ 0},
we find that
Z
lim
R→+∞
R
−R
eisx dx
dx + lim
R→+∞
x2 + a2
Z
σR
82
e−sa
πe−sa
eisz
=
2πi
×
=
z 2 + a2
2ia
a
when s > 0. If we then take the limit of the left hand side of this identity as
R → +∞, we find that
Z ∞
πe−sa
eisx
dx
=
.
2
2
a
−∞ x + a
when s > 0. This formula does not hold when s ≤ 0. And indeed, if we take
the complex conjugate of the above identity we find that
Z ∞ −isx
e
πe−sa
dx
=
.
2
2
a
−∞ x + a
when s > 0. It follows that
Z ∞
−∞
eisx
πe−|s|a
dx
=
.
x2 + a2
a
when |s| =
6 0. This identity also holds when s = 0, though this does not
follow from the above calculations.
Example We can also evaluate the above integral by applying Cauchy’s
Residue Theorem to the path integral taken around the boundary of a rectangle in the complex plane with vertices at −R, R, R + iM and −R + iM ,
where R and M are large positive real numbers.
Let a be a real number, and let R and M be real numbers satisfying
R > a and M > a. The inequality
1 1
z 2 + a2 ≤ R2 − a2
is satisfied for all complex numbers z for which |z| > a. It follows from this
that
Z
Z M
Z M
isz
1
1
e
is(R−iy)
dz ≤
|e
| dy = 2
e−sy dy
2 + a2
2 − a2
2
−
a
z
R
R
0
0
[R,R+iM ]
M
≤
.
2
s(R − a2 )
Similarly
Z
eisz
M
≤
dz
,
2
2
2
z +a
s(R − a2 )
Z
2M e−sM
eisz
≤
dz
.
z 2 + a2 M 2 − a2
[−R,−R+iM ]
and
[−R+iM,R+iM ]
83
If we take, for example, M = R in these inequalities, and let R tend to +∞,
we find that
Z
eisz
lim
dz = 0,
R→+∞ [R,R+iR] z 2 + a2
Z
eisz
lim
dz = 0,
R→+∞ [−R,−R+iR] z 2 + a2
Z
eisz
dz = 0.
lim
R→+∞ [−R+iR,R+iR] z 2 + a2
Also, Cauchy’s Residue Theorem ensures that
Z
Z
πe−sa
eisz
eisz
=
dz
+
dz
2
2
2
2
a
[−R,R] z + a
[R,R+iR] z + a
Z
Z
eisz
eisz
dz
−
dz
−
2
2
2
2
[−R+iR,R+iR] z + a
[−R,−R+iR] z + a
when s > 0. It follows that
Z ∞ −isx
Z
e
eisz
πe−sa
dx
=
lim
dz
=
2
2
R→+∞ [−R,R] z 2 + a2
a
−∞ x + a
when s > 0.
Example Let α be a real number satisfying 0 < α < 1. We evaluate the
integral
Z ∞
xα
dx
x(x + 1)
0
through an application of Cauchy’s Residue Theorem. Let
D = C \ {x ∈ R : x ≤ 0},
so that D is the open set obtained on removing the negative real axis from the
complex plane, let log: D → C denote the principal branch of the logarithm
that sends reiθ to log r + iθ for all real numbers r and θ satisfying r > 0 and
−π < θ < π, and let z α = exp(α log z) for all z ∈ D. Then the function f
zα
that sends z ∈ D \{1} to
is a meromorphic function on D. The only
z(z − 1)
pole of this function that lies within the open set D is a simple pole at z = 1
with residue 1. Let R and η be real numbers satisfying R > 1 and 0 < η < 1,
84
p
and let θη ∈ [ 34 π, π] be determined such that −R + iη = R2 + η 2 exp(iθη ).
It follows from Cauchy’s Residue Theorem (Theorem 6.16) that
Z
Z
f (z) dz −
f (z) dz
[−R+iη,−η+iη]
αη
Z
Z
−
f (z) dz +
f (z) dz
[−R−iη,−η−iη]
βR,η
= 2πi,
where αη : [− 34 π, 34 π] → C is the path from −η − iη to −η + iη that sends
√
t ∈ [− 43 π, 34 π] to 2 ηeit , and βR,η : [−θη , θη ] → C is the path from −R − iη
p
2
2 it
to −R + iη that sends t ∈ [−θη , θη ] to
√ R + η e . [Thus αη (t) traverses
a three-quarters of a circle of radius 2 η about zero in the anti-clockwise
direction as t increases from − 34 π to 43 π, and βR,η (t) traverses most of a
p
circle of radius R2 + η 2 about zero as t increases from −θη to θη .] Now the
inequality α > 0 ensures that
Z
lim
f (z) dz = 0.
η→0
Also
αη
Z
Z
lim
f (z) dz =
η→0
βR,η
f (z) dz,
σR
where σR : [−π, π] → C is the path that sends t ∈ [−π, π] to Reit . It follows
that
Z
Z
Z
2πi −
f (z) dz = lim+
f (z) dz −
f (z) dz
η→0
σR
[−R+iη,−η+iη]
0
Z
=
=
[−R−iη,−η−iη]
Z
0
f (x + iη) dx −
lim
Z 0
η→0+
−R
f (x − iη) dx
−R
lim (f (x + iη) − f (x − iη)) dx
+
−R η→0
Z R
lim (f (−x + iη) − f (−x − iη)) dx
=
0
η→0+
Now
lim+ f (−x + iη) =
η→0
=
(−x + iη)α
η→0 (−x + iη)(−x − 1 − iη)
lim+ exp(α log(−x + iη))
lim+
η→0
x(x + 1)
85
exp(α(log x + iπ))
exp(α log x) exp(iπα))
=
x(x + 1)
x(x + 1)
α
x
= eiπα
.
x(x + 1)
=
Similarly
lim exp(α log(−x − iη))
lim f (−x − iη) =
η→0+
x(x + 1)
exp(α(log x − iπ))
exp(α log x) exp(−iπα))
=
=
x(x + 1)
x(x + 1)
α
x
= e−iπα
.
x(x + 1)
η→0+
It follows that
lim+ (f (−x + iη) − f (−x − iη)) =
eiπα − e−iπα
η→0
=
Therefore
R
Z
2i sin(πα)
0
xα
x(x + 1)
2ixα sin πα
x(x + 1)
xα
dz = 2πi −
x(x + 1)
Z
f (z) dz.
σR
Now the inequality α < 1 ensures that
Z
lim
f (z) dz = 0.
R→+∞
It follows that
Z +∞
0
σR
xα
dz = lim
R→+∞
x(x + 1)
Z
when 0 < α < 1.
86
0
R
xα
π
dz =
x(x + 1)
sin πα