6.
Residue calculus
Let z0 be an isolated singularity of f (z), then there exists a certain
deleted neighborhood Nε = {z : 0 < |z − z0| < ε} such that f is
analytic everywhere inside Nε. We define
I
1
Res (f, z0) =
f (z) dz,
2πi C
where C is any simple closed contour around z0 and inside Nε.
1
Since f (z) admits a Laurent expansion inside Nε, where
f (z) =
∞
X
n=0
an(z − z0)n +
∞
X
n=1
bn(z − z0)−n,
then
I
1
f (z) dz = Res (f, z0).
b1 =
2πi C
Example
Res
1
, z0
k
(z − z0)
!
=
(
1 if k = 1
0 if k 6= 1
11
2 1
1/z
1/z
Res (e , 0) = 1 since e
=1+
+
+ ···
1! z
2! z 2
Res
, |z| > 0
!
1
1
,1 =
by the Cauchy integral formula.
(z − 1)(z − 2)
1−2
2
Cauchy residue theorem
Let C be a simple closed contour inside which f (z) is analytic everywhere except at the isolated singularities z1, z2, · · · , zn.
I
C
f (z) dz = 2πi[Res (f, z1) + · · · + Res (f, zn)].
This is a direct consequence of the Cauchy-Goursat Theorem.
3
Example
Evaluate the integral
I
using
z+1
dz
2
z
|z|=1
(i) direct contour integration,
(ii) the calculus of residues,
1
(iii) the primitive function log z − .
z
Solution
(i) On the unit circle, z = eiθ and dz = ieiθ dθ. We then have
I
2π
2π
z+1
−iθ
−2iθ
iθ
−iθ ) dθ = 2πi.
dz
=
(e
+e
)ie
dθ
=
i
(1+e
|z|=1 z 2
0
0
Z
Z
4
(ii) The integrand (z + 1)/z 2 has a double pole at z = 0. The
Laurent expansion in a deleted neighborhood of z = 0 is simply
1
1
+ 2 , where the coefficient of 1/z is seen to be 1. We have
z
z
Res
z+1
, 0 = 1,
2
z
and so
I
z+1
dz = 2πiRes
2
|z|=1 z
z+1
, 0 = 2πi.
2
z
(iii) When a closed contour moves around the origin (which is the
branch point of the function log z) in the anticlockwise direction,
the increase in the value of arg z equals 2π. Therefore,
I
z+1
1 in
dz
=
change
in
value
of
ln
|z|
+
i
arg
z
−
z
|z|=1 z 2
traversing one complete loop around the origin
= 2πi.
5
Computational formula
Let z0 be a pole of order k. In a deleted neighborhood of z0,
∞
X
b1
bk
an(z − z0) +
+ ··· +
,
f (z) =
k
z
−
z
(z
−
z
)
0
0
n=0
n
bk 6= 0.
Consider
g(z) = (z − z0)k f (z).
the principal part of g(z) vanishes since
g(z) = bk + bk−1(z − z0) + · · · b1(z − z0)k−1 +
∞
X
n=0
an(z − z0)n+k .
By differenting (k − 1) times, we obtain
 (k−1)
g
(z0 )




(k−1)!

"
#

k−1
k
d
(z − z0) f (z)
b1 = Res (f, z0) =
lim


z→z0 dz k−1

(k − 1)!



if g (k−1)(z) is analytic at z0
if z0 is a removable
singularity of g (k−1)(z)
6
Simple pole
k=1:
Suppose f (z) =
Res (f, z0) =
=
Res (f, z0) = lim (z − z0)f (z).
z→z0
p(z)
where p(z0) 6= 0 but q(z0) = 0, q ′ (z0) 6= 0.
q(z)
lim (z − z0)f (z)
z→z0
lim (z − z0)
z→z0
p(z0)
= ′
.
q (z0)
p(z0) + p′(z0)(z − z0) + · · ·
′′
0 ) (z − z )2 + · · ·
q ′(z0)(z − z0) + q (z
0
2!
7
Example
Find the residue of
at all isolated singularities.
e1/z
f (z) =
1−z
Solution
(i) There is a simple pole at z = 1. Obviously
(ii) Since
1/z Res (f, 1) = lim (z − 1)f (z) = −e = −e.
z→1
z=1
1
1
+
+ ···
2
z
2!z
has an essential singularity at z = 0, so does f (z). Consider
e1/z = 1 +
e1/z
1
1
2
= (1 + z + z + · · · ) 1 + +
+ ··· ,
2
1−z
z
2!z
the coefficient of 1/z is seen to be
1+
for 0 < |z| < 1,
1
1
+
+ · · · = e − 1 = Res (f, 0).
2!
3!
8
Example
Find the residue of
z 1/2
f (z) =
z(z − 2)2
at all poles. Use the principal branch of the square root function
z 1/2.
Solution
The point z = 0 is not a simple pole since z 1/2 has a branch point
at this value of z and this in turn causes f (z) to have a branch point
there. A branch point is not an isolated singularity.
However, f (z) has a pole of order 2 at z = 2. Note that
d
Res (f, 2) = lim
z→2 dz
z 1/2
z
!
= lim
z→2
−
z 1/2
2z 2
!
1
=− √ ,
4 2
where the principal branch of 21/2 has been chosen (which is
√
9
2).
Example
Evaluate Res (g(z)f ′ (z)/f (z), α) if α is a pole of order n of f (z), g(z)
is analytic at α and g(α) 6= 0.
Solution
Since α is a pole of order n of f (z), there exists a deleted neighborhood {z : 0 < |z − α| < ε} such that f (z) admits the Laurent
expansion:
∞
X
bn
bn−1
b1
n,
a
(z−α)
f (z) =
+
+·
·
·+
+
n
(z − α)n (z − α)n−1
(z − α) n=0
bn 6= 0.
Within the annulus of convergence, we can perform termwise differentiation of the above series
10
∞
X
−nbn
(n − 1)bn
b1
n−1
f (z) =
−
−
·
·
·
−
+
na
(z
−
α)
.
n
n+1
n
2
(z − α)
(z − α)
(z − α)
n=0
Provided that g(α) 6= 0, it is seen that
′
P∞
(n−1)bn
b1
−nb
n
(z − α)
− (z−α)n − · · · −
+ n=0 nan(z − α)n−1
2
n+1
(z−α)
(z−α)
= lim g(z)
P∞
bn−1
z→α
b1
bn
n
+
+
·
·
·
+
+
a
(z
−
α)
n
n
n−1
n=0
z−α
(z−α)
(z−α)
= −ng(α) 6= 0,
so that α is a simple pole of g(z)f ′(z)/f (z). Furthermore,
Res
g
f′
f
!
, α = −ng(α).
Remark
When g(α) = 0, α becomes a removable singularity of gf ′/f .
11
Example
Suppose an even function f (z) has a pole of order n at α. Within the
deleted neighborhood {z : 0 < |z − α| < ε}, f (z) admits the Laurent
expansion
∞
X
bn
b1
n,
f (z) =
+
·
·
·
+
+
a
(z
−
α)
n
(z − α)n
(z − α)
n=0
bn 6= 0.
Since f (z) is even, f (z) = f (−z) so that
∞
X
bn
b1
n
+
·
·
·
+
+
f (z) = f (−z) =
a
(−z
−
α)
,
n
(−z − α)n
(−z − α) n=0
which is valid within the deleted neighborhood {z : 0 < |z + α| < ε}.
Hence, −α is a pole of order n of f (−z). Note that
Res (f (z), α) = b1
and
Res (f (z), −α) = −b1
so that Res (f (z), α) = −Res (f (z), −α). For an even function, if
z = 0 happens to be a pole, then Res (f, 0) = 0.
12
Example
I
tan z
dz = 2πi Res
|z|=2 z
tan z π
,
+ Res
z
2
π
z − 2 tan z
tan z π
,−
z
2
π
since the singularity at z = 0 is removable. Observe that
is a
2
π
, we have
simple pole and cos z = − sin z −
2
Res
tan z π
,
z
2
= limπ
z→ 2
= limπ
z→ 2
z
π
z − 2 sin z
"
#
3
π
(z− 2 )
π
z − z−2 +
+ ···
6
1
2
=
=− .
−π
π
2
13
tan z π
As tan z/z is even, we deduce that Res
,−
z
2
result from the previous example. We then have
I
=
2
using the
π
tan z
dz = 0.
z
|z|=2
Remark
π
Let p(z) = sin z/z, q(z) = cos z, and observe that p
2
π
0 and q ′
= −1 6= 0, then
2
Res
tan z π
,
z
2
,
π
=p
2
q′
π
2
=
2
π
= ,q
π
2
−2
.
π
14
=
Example
Evaluate
z2
dz.
2
2
2
C (z + π ) sin z
I
Solution
z
z
z
=
lim
z→0 sin z (z 2 + π 2)2
z→0 sin z
lim
z
z→0 (z 2 + π 2)2
lim
!
=0
so that z = 0 is a removable singularity.
15
It is easily seen that z = iπ is a pole of order 2.
d
[(z − iπ)2f (z)]
z→iπ dz
"
#
2
d
z
= lim
z→iπ dz (z + iπ)2 sin z
2z(z + iπ) sin z − z 2[(z + iπ) cos z + 2 sin z]
= lim
z→iπ
(z + iπ)3 sin2 z
cosh π
2 sinh π + (−π cosh π − sinh π)
1
=
−
+
.
=
2
2
4π sinh π
−4π sinh π
4π sinh π
Res (f, iπ) =
lim
Recall that sin iπ = i sinh π and cos iπ = cosh π. Hence,
z2
dz = 2πiRes (f, iπ)
2
2
2
C (z + π ) sin z
i
1
cosh π
=
−
+
.
2
2
sinh π
sinh π
I
16
Theorem
If a function f is analytic everywhere in the finite plane except for a
finite number of singularities interior to a positively oriented simple
closed contour C, then
I
C
f (z) dz = 2πiRes
1
1
f
,0 .
2
z
z
17
We construct a circle |z| = R1 which is large enough so that C is
interior to it. If C0 denotes a positively oriented circle |z| = R0,
where R0 > R1, then
f (z) =
∞
X
cnz n,
R1 < |z| < ∞,
n=−∞
(A)
where
I
1
f (z)
cn =
dz
2πi C0 z n+1
n = 0, ±1, ±2, · · · .
In particular,
2πic−1 =
I
C0
f (z) dz.
How to find c−1? First, we replace z by 1/z in Eq. (A) such that
the domain of validity is a deleted neighborhood of z = 0.
18
Now
∞
∞
X
X
1
1
cn
cn−2
f
=
=
,
2
n+2
n
z
z
n=−∞ z
n=−∞ z
0 < |z| <
1
,
R1
so that
c−1 = Res
1
1
f
,0 .
2
z
z
Remark
By convention, we may define the residue at infinity by
I
1
1
1
Res (f, ∞) = −
f (z) dz = −Res
f
,0 ,
2
2πi C
z
z
where all singularities in the finite plane are included inside C. With
the choice of the negative sign, we have
X
all
Res (f, zi) + Res (f, ∞) = 0.
19
Example
Evaluate
I
5z − 2
dz.
|z|=2 z(z − 1)
Solution
Write f (z) =
5z − 2
. For 0 < |z| < 1,
z(z − 1)
so that
5z − 2
5z − 2 −1
2
=
= 5−
(−1 − z − z 2 − · · · )
z(z − 1)
z 1−z
z
Res (f, 0) = 2.
20
For 0 < |z − 1| < 1,
5z − 2
5(z − 1) + 3
1
=
z(z − 1)
z−1
1 + (z − 1)
3
= 5+
[1 − (z − 1) + (z − 1)2 − (z − 1)3 + · · · ]
z−1
so that
Res (f, 1) = 3.
Hence,
I
5z − 2
dz = 2πi [Res (f, 0) + Res (f, 1)] = 10πi.
|z|=2 z(z − 1)
21
On the other hand, consider
1
1
f
z2
z
5 − 2z 1
5 − 2z
=
z(1 − z)
z 1−z
5
=
− 2 (1 + z + z 2 + · · · )
z
5
=
+ 3 + 3z, 0 < |z| < 1,
z
=
so that
I
5z − 2
dz = −2πiRes (f, ∞)
|z|=2 z(z − 1)
1
1
, 0 = 10πi.
= 2πiRes
f
2
z
z
22
Evaluation of integrals using residue methods
A wide variety of real definite integrals can be evaluated effectively
by the calculus of residues.
Integrals of trigonometric functions over [0, 2π]
We consider a real integral involving trigonometric functions of the
form
Z 2π
0
R(cos θ, sin θ) dθ,
where R(x, y) is a rational function defined inside the unit circle
|z| = 1, z = x+iy. The real integral can be converted into a contour
integral around the unit circle by the following substitutions:
23
z = eiθ , dz = ieiθ dθ = iz dθ,
eiθ + e−iθ
1
1
cos θ =
=
z+
,
2
2
z
1
eiθ − e−iθ
1
z−
.
sin θ =
=
2i
2i
z
The above integral can then be transformed into
Z 2π
0
=
I
R(cos θ, sin θ) dθ
1
R
|z|=1 iz
z + z −1 z − z −1
dz
2i
"
!
#
−1
−1
1
z+z
z−z
= 2πi sum of residues of R
,
inside |z| = 1 .
iz
2
2i
2
,
!
24
Example
Compute I =
Z 2π
0
cos 2θ
dθ.
2 + cos θ
Solution
Z 2π
0
cos 2θ
dθ
2 + cos θ
1
1
2
I
dz
2 z + z2
= −i
|z|=1 2 + 1 z + 1 z
2
z
I
z4 + 1
= −i
|z|=1 z 2(z 2 + 4z + 1)
dz.
The integrand has a pole of order two at z = 0. Also, the roots of
√
√
2
z + 4z + 1 = 0, namely, z1 = −2 − 3 and z2 = −2 + 3, are simple
poles of the integrand.
25
z4 + 1
Write f (z) = 2 2
. Note that z1 is inside but z2 is outside
z (z + 4z + 1)
|z| = 1.
26
d
z4 + 1
Res (f, 0) = lim
z→0 dz z 2 + 4z + 1
3z 3(z 2 + 4z + 1) − (z 4 + 1)(2z + 4)
= lim
= −4
2
2
z→0
(z + 4z + 1)
Res (f, −2 +
√
3) =
=
4
z + 1 z2 √
z=−2+ 3
√ 4
(−2 + 3) + 1
(−2 +
h
√
3)2
,
d 2
(z + 4z + 1)
√
dz
z=−2+ 3
1
7
√
·
=√ .
2(−2 + 3) + 4
3
I = (−i)2πi Res (f, 0) + Res (f, −2 +
!
7
= 2π −4 + √
.
3
√
i
3)
27
Example
Evaluate the integral
I=
Solution
Z π
1
dθ,
0 a − b cos θ
a > b > 0.
Since the integrand is symmetric about θ = π, we have
Z
Z 2π
1 2π
1
eiθ
I=
dθ =
dθ.
iθ
2iθ
2 0 a − b cos θ
+ 1)
0 2ae − b(e
The real integral can be transformed into the contour integral
I
1
I=i
dz.
2
|z|=1 bz − 2az + b
The integrand has two simple poles, which are given by the zeros
of the denominator.
28
Let α denote the pole that is inside the unit circle, then the other
1 . The two poles are found to be
pole will be α
α=
a−
q
a2 − b2
b
and
a+
1
=
α
q
a2 − b2
.
b
1 is
Since a > b > 0, the two roots are distinct, and α is inside but α
outside the closed contour of integration. We then have
I
1
1
I = −
dz
1)
ib |z|=1 (z − α) (z − α


2πi
1
Res 
, α
1
ib
(z − α) (z − α )
2πi
π
q
= −
=
.
1
2
2
ib α − α
a −b
= −
29
Integral of rational functions
Z ∞
where
−∞
f (x) dx,
1. f (z) is a rational function with no singularity on the real axis,
2. lim zf (z) = 0.
z→∞
It can be shown that
Z ∞
−∞
f (x) dx = 2πi [sum of residues at the poles of f in the upper
half-plane].
30
Integrate f (z) around a closed contour C that consists of the upper
semi-circle CR and the diameter from −R to R.
By the Residue Theorem
I
C
f (z) dz =
Z R
−R
f (x) dx +
Z
CR
f (z) dz
= 2πi [sum of residues at the poles of f inside C].
31
As R → ∞, all the poles of f in the upper half-plane will be enclosed
inside C. To establish the claim, it suffices to show that as R → ∞,
lim
I
R→∞ CR
f (z) dz = 0.
The modulus of the above integral is estimated by the modulus
inequality as follows:
Z
f
(z)
dz
CR
≤
≤
Z π
0
|f (Reiθ )| R dθ
max |f (Reiθ )| R
0≤θ≤π
= max |zf (z)|π,
Z π
0
dθ
z∈CR
which goes to zero as R → ∞, since lim zf (z) = 0.
z→∞
32
Example
Evaluate the real integral
x4
dx
6
−∞ 1 + x
Z ∞
by the residue method.
Solution
√
3+i
The complex function f (z) =
has
simple
poles
at
i,
1 + z6
2
√
− 3+i
and
in the upper half-plane, and it has no singularity on
2
the real axis. The integrand observes the property lim zf (z) = 0.
z→∞
We obtain
√
√
"
!
!#
Z ∞
3+i
− 3+i
f (x) dx = 2πi Res(f, i) + Res f,
+ Res f,
.
2
2
−∞
z4
33
The residue value at the simple poles are found to be
1 Res(f, i) =
6z Res
and
so that
i
=− ,
6
z=i
√
√
!
3+i
1 3−i
f,
=
=
,
√
2
6z 12
z= 3+i
2
√
!
− 3+i
1 Res f,
=
2
6z √
− 3+i
z= 2
√
3+i
=−
,
12
√
√
!
i
3−i
3+i
2π
dx
=
2πi
+
−
.
−
=
6
6
12
12
3
−∞ 1 + x
Z ∞
x4
34
Integrals involving multi-valued functions
Consider a real integral involving a fractional power function
Z ∞
f (x)
dx,
α
x
0
0 < α < 1,
1. f (z) is a rational function with no singularity on the positive real
axis, including the origin.
2. lim f (z) = 0.
z→∞
f (z)
We integrate φ(z) = α along the closed contour as shown.
z
35
The closed contour C consists of an infinitely large circle and an
infinitesimal circle joined by line segments along the positive x-axis.
36
(i) line segment from ε to R along the upper side of the positive
real axis: z = x, ε ≤ x ≤ R;
(ii) the outer large circle CR : z = Reiθ ,
0 < θ < 2π;
(iii) line segment from R to ε along the lower side of the positive
real axis
z = xe2πi ,
ε ≤ x ≤ R;
(iv) the inner infinitesimal circle Cε in the clockwise direction
z = εeiθ ,
0 < θ < 2π.
37
Establish:
lim
Z
φ(z) = 0
and
lim
Z
φ(z) = 0.
ε→0 Cε
R→∞ CR
Z
Z 2π
φ(z) dz ≤
|φ(Reiθ )Reiθ | dθ ≤ 2π max |zφ(z)|
CR
z∈CR
0
Z
Z 2π
φ(z) dz ≤
|φ(εeiθ )|ε dθ ≤ 2π max |zφ(z)|.
Cε
z∈Cǫ
0
It suffices to show that zφ(z) → 0 as either z → ∞ or z → 0.
1. Since lim f (z) = 0 and f (z) is a rational function,
z→∞
deg (denominator of f (z)) ≥ 1 + deg (numerator of f (z)).
Further, 1 − α < 1, zφ(z) = z 1−αf (z) → 0 as z → ∞.
2. Since f (z) is continuous at z = 0 and f (z) has no singularity at
the origin, zφ(z) = z 1−αf (z) ∼ 0 · f (0) = 0 as z → 0.
38
The argument of the principal branch of z α is chosen to be 0 ≤ θ <
2π, as dictated by the contour.
I
C
φ(z) dz =
Z
φ(z) dz +
CR
Z R
Z
φ(z) dz
Cǫ
Z ǫ
f (x)
f (xe2πi )
+
dx +
dx
xα
ǫ
R xαe2απi
= 2πi [sum of residues at all the isolated singularities
of f enclosed inside the closed contour C].
By taking the limits ǫ → 0 and R → ∞, the first two integrals vanish.
The last integral can be expressed as
Z ∞
Z ∞
f (x)
f (x)
−2απi
−
dx
=
−e
dx.
α
2απi
α
x
0 x e
0
Combining the results,
Z ∞
f (x)
2πi
dx
=
[sum of residues at all the isolated
α
−2απi
x
1−e
0
singularities of f in the finite complex plane].
39
Example
Evaluate
Z ∞
0
1
dx,
α
(1 + x)x
0 < α < 1.
Solution
1
is multi-valued and has an isolated singularity at
α
(1 + z)z
z = −1. By the Residue Theorem,
f (z) =
I
1
dz
α
C (1 + z)z
Z R
Z
Z
dx
dz
dz
= (1 − e−2απi)
+
+
ǫ (1 + x)xα
CR (1 + z)z α
Cǫ (1 + z)z α
!
1
2πi
= 2πi Res
, −1 = απi .
(1 + z)z α
e
40
The moduli of the third and fourth integrals are bounded by
Z
2πR
1
−α
≤
dz
∼
R
→0
α
α
(R − 1)R
CR (1 + z)z
Z
1
2πǫ
1−α
dz
≤
∼
ǫ
→0
α
α
(1 − ǫ)ǫ
Cǫ (1 + z)z
as
R → ∞,
as
ǫ → 0.
On taking the limits R → ∞ and ǫ → 0, we obtain
Z ∞
(1 − e−2απi)
so
Z ∞
0
1
2πi
dx
=
;
α
απi
e
0 (1 + x)x
1
2πi
π
dx = απi
=
.
α
−2απi
(1 + x)x
e
(1 − e
)
sin απ
41
Example
Evaluate the real integral
eαx
dx,
x
−∞ 1 + e
Z ∞
Solution
0 < α < 1.
The integrand function in its complex extension has infinitely many
poles in the complex plane, namely, at z = (2k + 1)πi, k is any
integer. We choose the rectangular contour as shown
l1 : y = 0,
l2 : x = R,
l3 : y = 2π,
l4 : x = −R,
−R ≤ x ≤ R,
0 ≤ y ≤ 2π,
−R ≤ x ≤ R,
0 ≤ y ≤ 2π.
42
The chosen closed rectangular contour encloses only one simple pole
at z = πi.
43
The only simple pole that is enclosed inside the closed contour C is
z = πi. By the Residue Theorem, we have
eαz
dz =
z
C 1+e
I
2π eα(R+iy)
eαx
dx +
idy
x
R+iy
−R 1 + e
0 1+e
Z −R α(x+2πi)
Z 0 α(−R+iy)
e
e
+
dx +
idy
x+2πi
−R+iy
1+e
R
2π 1 + e
!
eαz
, πi
= 2πi Res
z
1+e
eαz = 2πi z = −2πieαπi.
e z=πi
Z R
Z
44
Consider the bounds on the moduli of the integrals as follows:
Z 2π α(R+iy)
Z 2π
e
eαR
−(1−α)R
≤
idy
dy
∼
O(e
),
R+iy
R
1
+
e
e
−
1
0
0
Z 0 α(−R+iy)
Z 2π
−αR
e
e
−αR
≤
idy
dy
∼
O(e
).
−R+iy
−R
1
+
e
1
−
e
2π
0
As 0 < α < 1, both e−(1−α)R and e−αR tend to zero as R tends to
infinity. Therefore, the second and the fourth integrals tend to zero
as R → ∞. On taking the limit R → ∞, the sum of the first and
third integrals becomes
2απi
(1 − e
so
eαx
απi
dx
=
−2πie
;
)
x
−∞ 1 + e
Z ∞
eαx
2πi
π
dx
=
=
.
x
απi
−απi
e
−e
sin απ
−∞ 1 + e
Z ∞
45
Example
Evaluate
Z ∞
0
1
dx.
3
1+x
Solution
Since the integrand is not an even function, it serves no purpose to
extend the interval of integration to (−∞, ∞). Instead, we consider
the branch cut integral
I
Log z
dz,
3
C 1+z
where the branch cut is chosen to be along the positive real axis
whereby 0 ≤ Arg z < 2π. Now
I
Log z
dz =
3
C 1+z
ǫ Log (xe2πi )
ln x
dx +
dx
3
2πi
3
)
ǫ 1+x
R 1 + (xe
I
I
Log z
Log z
+
dz
+
dz
3
3
CR 1 + z
Cǫ 1 + z
Z R
Z
46
= 2πi
3
X
Res
j=1
Log z
, zj ,
1 + z3
where zj , j = 1, 2, 3 are the zeros of 1/(1 + z 3). Note that
I
Cǫ
I
CR
Hence
lim
R→∞
ǫ→0
Log z
1 + z3
Log z
1 + z3
I
ǫ
ln
ǫ
−→ 0 as ǫ → 0;
dz = O
1
R
ln
R
dz = O
−→ 0 as R → ∞.
R3
Log z
dz =
3
C 1+z
Z ∞
0
0 Log (xe2iπ )
ln x
dx +
dx
3
2iπ
3
1+x
)
∞ 1 + (xe
= −2πi
Z
Z ∞
0
1
dx,
3
1+x
thus giving
Z ∞
0
3
X
1
dx = −
Res
3
1+x
j=1
Log z
, zj .
3
1+z
47
The zeros of 1 + z 3 are α = eiπ/3, β = eiπ and γ = e5πi/3. Sum of
residues is given by
Log z
Log z
Log z
Res
, α + Res
, β + Res
,γ
1 + z3
1 + z3
1 + z3
Log α
Log β
Log γ
=
+
+
(αh− β)(α − γ)
(β − α)(β − γ)
(γ
− α)(γ − β)
i
π (β − γ) + π(γ − α) + 5π (α − β)
2π
3
3
= −i
=− √ .
(α − β)(β − γ)(γ − α)
3 3
Hence,
Z ∞
2π
1
√ .
dx
=
3
0 1+x
3 3
48
Evaluation of Fourier integrals
A Fourier integral is of the form
Z ∞
−∞
eimx f (x) dx,
m > 0,
1. lim f (z) = 0,
z→∞
2. f (z) has no singularity along the real axis.
Remarks
1. The assumption m > 0 is not strictly essential. The evaluation
method works even when m is negative or pure imaginary.
2. When f (z) has singularities on the real axis, the Cauchy principal
value of the integral is considered.
49
Jordan Lemma
We consider the modulus of the integral for λ > 0
Z
iλz
f
(z)e
dz
CR
≤
Z π
0
iθ
|f (Reiθ )| |eiλRe | R dθ
≤ max |f (z)| R
z∈CR
Z π
0
Z π
2
= 2R max |f (z)|
z∈CR
≤ 2R max |f (z)|
z∈CR
= 2R max |f (z)|
z∈CR
e−λR sin θ dθ
0
Z
π
2
0
e−λR sin θ dθ
−λR 2θ
π
e
dθ
π
(1 − e−λR ),
2Rλ
which tends to 0 as R → ∞, given that f (z) → 0 as R → ∞.
50
51
To evaluate the Fourier integral, we integrate eimz f (z) along the
closed contour C that consists of the upper half-circle CR and the
diameter from −R to R along the real axis. We then have
I
C
eimz f (z) dz =
Z R
−R
eimxf (x) dx +
Z
CR
eimz f (z) dz.
Taking the limit R → ∞, the integral over CR vanishes by virtue of
the Jordan Lemma.
Lastly, we apply the Residue Theorem to obtain
Z ∞
−∞
eimxf (x) dx = 2πi [sum of residues at all the isolated
singularities of f in the upper half-plane]
since C encloses all the singularities of f in the upper half-plane as
R → ∞.
52
Example
Evaluate the Fourier integral
Z ∞
sin 2x
dx.
2
−∞ x + x + 1
Solution
1
It is easy to check that f (z) = z 2+z+1
has no singularity along
1
the real axis and lim z 2+z+1
= 0. The integrand has two simple
z→∞
2πi
e 3
− 2πi
3
poles, namely, z =
in the upper half-plane and e
half-plane. By virtue of the Jordan Lemma, we have
in the lower
53
∞
e2iz
sin 2x
e2ix
dx = Im
dx = Im
dz,
−∞ x2 + x + 1
−∞ x2 + x + 1
C z2 + z + 1
where C is the union of the infinitely large upper semi-circle and its
diameter along the real axis.
Z ∞
Z
I
Note that
I
e2iz
C z2 + z + 1
Hence,
dz = 2πi Res
e2iz
z2 + z + 1
2πi
,e 3
!
2πi
3
2iz
2ie
e
e
= 2πi
= 2πi
.
2πi
2πi
2z + 1 2e 3 + 1
z=e 3

2πi
2ie 3

√


sin 2x
e
2
−
3 sin 1.
2πi
 = − √ πe
dx
=
Im


2πi
−∞ x2 + x + 1
3
3
2e
+1
Z ∞
54
Example
Show that
Z ∞
Z ∞
r
1 π
2
2
sin x dx =
cos x dx =
.
2
2
0
0
Solution
55
0=
I
iz 2
C
e
dz =
Z R
0
+
ix2
e
Z 0
dx +
Z
π
4
0
iR2 e2iθ
e
iReiθ dθ
2 eiπ/2 iπ/4
ir
e
e
dr.
R
Rearranging
Z R
0
(cos x2+i sin x2) dx = eiπ/4
Z R
0
−r2
e
dr−
Z π/4
0
2
2
eiR cos 2θ−R sin 2θ iReiθ dθ.
Next, we take the limit R → ∞. We recall the well-known result
√
r
r
Z ∞
π
2
1
i
π
π
π
i4
iπ/4
−r
e
e
=
+
.
e
dr =
2
2
2
2
2
0
2φ
π
Also, we use the transformation 2θ = φ and observe sin φ ≥
,0 ≤ φ ≤ ,
π
2
to obtain
56
Z
π/4
iR2 cos 2θ−R2 sin2 θ
iθ
e
iRe dθ 0
≤
Z π/4
0
2
e−R sin 2θ R dθ
R π/2 −R2 sin φ
=
e
dφ
2 0
Z
R π/2 −2R2φ/π
dφ
≤
e
2 0
π
−R2
=
) −→ 0 as R → ∞.
(1 − e
4R
Z
We then obtain
r
Z ∞
r
1 π
1 π
2
2
(cos x + i sin x ) dx =
+i
2 2
2 2
0
so that
Z ∞
Z ∞
r
1 π
.
cos x2 dx =
sin x2 dx =
2 2
0
0
57
Example
Z ∞
ln(x2 + 1)
Evaluate
dx.
2
x +1
0
Hint: Use Log(i − x) + Log(i + x) = Log(i2 − x2) = ln(x2 + 1) + πi.
Solution
I
Log(z + i)
Consider
dz around C as shown.
2
z +1
C
58
Log(z + i)
in the upper half plane is the simple
2
z +1
pole z = i. Consider
The only pole of
Log(z + i)
,i
2
z +1
(z − i)Log(z + i)
π2
2πi lim
= πLog 2i = π ln 2 +
i.
z→i (z − i)(z + i)
2
2πiRes
=
Z
!
!
Log(z + i)
(ln R)R
dz = O
→ 0 as R → ∞
2
2
z +1
R
CR
Z R
Z R
Z
Log(i − x)
Log(x + i)
Log(z + i)
π2
dx +
dx +
dz = π ln 2 +
i
2
2
2
x +1
x +1
z +1
2
0
0
CR
and Log(i − x) + Log(i + x) = ln(x2 + 1) + πi.
Z ∞
Z ∞
Z ∞
ln(x2 + 1)
πi
π2
dx
π
From
dx +
dx = π ln 2 +
i and
=
x2 + 1
2
2
0
0 x2 + 1
0 1 + x2
so that
Z ∞
ln(x2 + 1)
dx = π ln 2.
2
x
+
1
0
59
Cauchy principal value of an improper integral
Suppose a real function f (x) is continuous everywhere in the interval
[a, b] except at a point x0 inside the interval. The integral of f (x)
over the interval [a, b] is an improper integral, which may be defined
as
Z b
a
f (x) dx =
lim
ǫ1 ,ǫ2 →0
"Z
x0 −ǫ1
a
f (x) dx +
Z b
x0 +ǫ2
#
f (x) dx ,
ǫ1, ǫ2 > 0.
In many cases, the above limit exists only when ǫ1 = ǫ2, and does
not exist otherwise.
60
Example
Consider the following improper integral
Z 2
1
dx,
−1 x − 1
show that the Cauchy principal value of the integral exists, then find
the principal value.
Solution
Principal value of
Z 2
1
dx exists if the following limit exists.
1 x−1
lim
ǫ→0+
"Z
−1
Z 2
1
1
dx +
dx
x−1
1+ǫ x − 1
1−ǫ
2 

= lim ln |x − 1|
+ ln |x − 1|
ǫ→0+
−1
1+ǫ
=

1−ǫ
lim [(ln ǫ − ln 2) + (ln 1 − ln ǫ)] = − ln 2.
ǫ→0+
Hence, the principal value of
− ln 2.
#
Z 2
1
dx exists and its value is
−1 x − 1
61
Lemma
If f has a simple pole at z = c and Tr is the circular arc defined by
Tr : z = c + reiθ
(θ1 ≤ θ ≤ θ2),
then
lim
r→0+
Z
Tr
f (z) dz = i(θ2 − θ1)Res (f, c).
In particular, for the semi-circular arc Sr
lim
Z
r→0+ Sr
f (z) dz = iπRes (f, c).
62
Proof
Since f has a simple pole at c,
∞
X
a−1
f (z) =
+
ak (z − c)k ,
z−c
k=0
|
Z
{z
g(z)
0 < |z − c| < R
for some R.
}
Z
Z
1
For 0 < r < R,
f (z) dz = a−1
dz +
g(z) dz.
Tr
Tr z − c
Tr
63
Since g(z) is analytic at c, it is bounded in some neighborhood of
z = c. That is,
|g(z)| ≤ M
for
|z − c| < r.
For 0 < r < R,
and so
Z
g(z) dz ≤ M · arc length of Tr = M r(θ2 − θ1)
Tr
lim
Z
r→0+ Tr
g(z) dz = 0.
Finally,
Z
so that
θ2 1
θ2
1
iθ
dz =
ire dθ = i
dθ = i(θ2 − θ1)
iθ
z
−
c
re
Tr
θ1
θ1
lim
r→0+
Z
Z
Tr
Z
f (z) dz = a−1i(θ2 − θ1) = Res (f, c)i(θ2 − θ1).
64
Example
Compute the principal value of
xe2ix
dx.
2
−∞ x − 1
Z ∞
Solution
The improper integral has singularities at x = ±1. The principal
value of the integral is defined to be
lim
R→∞
r1,r2 →0+
Z −1−r
1
−R
+
Z 1−r
2
−1+r1
+
Z R
1+r2
!
xe2ix
dx.
2
x −1
65
Let
ze2iz
dz
I1 =
Sr 1 z 2 − 1
Z
ze2iz
I2 =
dz
2
Sr 2 z − 1
Z
ze2iz
IR =
dz.
2
z
−
1
CR
Z
ze2iz
Now, f (z) = 2
is analytic inside the above closed contour.
z −1
By the Cauchy Integral Theorem
Z −1−r
1
−R
+
Z 1−r
2
−1+r1
+
Z R
1+r2
!
xe2ix
dx + I1 + I2 + IR = 0.
2
x −1
66
By the Jordan Lemma, and since
z
→ 0 as z → ∞, so
2
z −1
lim IR = 0.
R→∞
Since z = ±1 are simple poles of f ,
lim I1 = −iπRes (f, −1) = −iπ
r1
→0+
lim (z + 1)f (z)
z→−1
= (−iπ)e−2i/2.
−iπe2i
Similarly, lim I2 = −iπRes (f, 1) =
.
+
2
r2 →0
xe2ix
iπe−2i
iπe2i
PV
dx =
+
= iπ cos 2.
2
2
2
−∞ x − 1
Z ∞
67
Poisson integral formula
I
f (s)
1
ds.
f (z) =
2πi C s − z
Here, C is the circle with radius r0 centered at the origin. Write
s = r0eiφ and z = reiθ , r > r0. We choose z1 such that |z1| |z| = r02
and both z1 and z lie on the same ray so that
r02 iθ
r02
ss
z1 = e =
= .
r
z
z
68
Since z1 lies outside C, we have
I
!
1
1
1
f (s)
ds
−
2πi C
s − z s − z1
!
Z 2π
1
s
s
=
f (s) dφ.
−
2π 0
s − z s − z1
f (z) =
The integrand can be expressed as
r02 − r2
s
1
s
z
−
=
+
=
s − z 1 − s/z
s−z
s−z
|s − z|2
r02 − r2 2π f (r0eiθ )
iθ
and so f (re ) =
dφ.
2
2π
|s − z|
0
Z
69
Now |s − z|2 = r02 − 2r0r cos(φ − θ) + r2 > 0 (from the cosine rule).
Taking the real part of f , where f = u + iv, we obtain
r02 − r2
1 2π
u(r0, φ) dφ,
u(r, θ) =
2π 0 r02 − 2r0r cos(φ − θ) + r2
Z
|
{z
P (r0 ,r,φ−θ)
r < r0.
}
Knowing u(r0, φ) on the boundary, u(r, θ) is uniquely determined.
The kernel function P (r0, r, φ − θ) is called the Poisson kernel.
r02 − r2
s
z
P (r0 , r, φ − θ) =
= Re
+
2
|s − z|
s−z
s−z
s
z
= Re
+
s−z s−z
s+z
= Re
which is harmonic for |z| < r0.
s−z
70