6.

Residue calculus
Let z
0
be an isolated singularity of f(z), then there exists a certain
deleted neighborhood N

= {z : 0 < |z z
0
| < } such that f is
analytic everywhere inside N

. We dene
Res (f, z
0
) =
1
2i
_
C
f(z) dz,
where C is any simple closed contour around z
0
and inside N

.
1
Since f(z) admits a Laurent expansion inside N

, where
f(z) =

n=0
a
n
(z z
0
)
n
+

n=1
b
n
(z z
0
)
n
,
then
b
1
=
1
2i
_
C
f(z) dz = Res (f, z
0
).
Example
Res
_
1
(z z
0
)
k
, z
0
_
=
_
1 if k = 1
0 if k = 1
Res (e
1/z
, 0) = 1 since e
1/z
= 1 +
1
1!
1
z
+
2
2!
1
z
2
+ , |z| > 0
Res
_
1
(z 1)(z 2)
, 1
_
=
1
1 2
by the Cauchy integral formula.
2
Cauchy residue theorem
Let C be a simple closed contour inside which f(z) is analytic ev-
erywhere except at the isolated singularities z
1
, z
2
, , z
n
.
_
C
f(z) dz = 2i[Res (f, z
1
) + +Res (f, z
n
)].
This is a direct consequence of the Cauchy-Goursat Theorem.
3
Example
Evaluate the integral
_
|z|=1
z +1
z
2
dz
using
(i) direct contour integration,
(ii) the calculus of residues,
(iii) the primitive function logz
1
z
.
Solution
(i) On the unit circle, z = e
i
and dz = ie
i
d. We then have
_
|z|=1
z +1
z
2
dz =
_
2
0
(e
i
+e
2i
)ie
i
d = i
_
2
0
(1+e
i
) d = 2i.
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
z
+
1
z
2
, where the coecient of 1/z is seen to be 1. We have
Res
_
z +1
z
2
, 0
_
= 1,
and so
_
|z|=1
z +1
z
2
dz = 2iRes
_
z +1
z
2
, 0
_
= 2i.
(iii) When a closed contour moves around the origin (which is the
branch point of the function logz) in the anticlockwise direction,
the increase in the value of argz equals 2. Therefore,
_
|z|=1
z +1
z
2
dz = change in value of ln|z| +i argz
1
z
in
traversing one complete loop around the origin
= 2i.
5
Computational formula
Let z
0
be a pole of order k. In a deleted neighborhood of z
0
,
f(z) =

n=0
a
n
(z z
0
)
n
+
b
1
z z
0
+ +
b
k
(z z
0
)
k
, b
k
= 0.
Consider
g(z) = (z z
0
)
k
f(z).
the principal part of g(z) vanishes since
g(z) = b
k
+b
k1
(z z
0
) + b
1
(z z
0
)
k1
+

n=0
a
n
(z z
0
)
n+k
.
By dierenting (k 1) times, we obtain
b
1
= Res (f, z
0
) =
_

_
g
(k1)
(z
0
)
(k1)!
if g
(k1)
(z) is analytic at z
0
lim
zz
0
d
k1
dz
k1
_
(z z
0
)
k
f(z)
(k 1)!
_
if z
0
is a removable
singularity of g
(k1)
(z)
6
Simple pole
k = 1 : Res (f, z
0
) = lim
zz
0
(z z
0
)f(z).
Suppose f(z) =
p(z)
q(z)
where p(z
0
) = 0 but q(z
0
) = 0, q

(z
0
) = 0.
Res (f, z
0
) = lim
zz
0
(z z
0
)f(z)
= lim
zz
0
(z z
0
)
p(z
0
) +p

(z
0
)(z z
0
) +
q

(z
0
)(z z
0
) +
q

(z
0
)
2!
(z z
0
)
2
+
=
p(z
0
)
q

(z
0
)
.
7
Example
Find the residue of
f(z) =
e
1/z
1 z
at all isolated singularities.
Solution
(i) There is a simple pole at z = 1. Obviously
Res (f, 1) = lim
z1
(z 1)f(z) = e
1/z

z=1
= e.
(ii) Since
e
1/z
= 1 +
1
z
+
1
2!z
2
+
has an essential singularity at z = 0, so does f(z). Consider
e
1/z
1 z
= (1+z +z
2
+ )
_
1 +
1
z
+
1
2!z
2
+
_
, for 0 < |z| < 1,
the coecient of 1/z is seen to be
1 +
1
2!
+
1
3!
+ = e 1 = Res (f, 0).
8
Example
Find the residue of
f(z) =
z
1/2
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
Res (f, 2) = lim
z2
d
dz
_
z
1/2
z
_
= lim
z2
_

z
1/2
2z
2
_
=
1
4

2
,
where the principal branch of 2
1/2
has been chosen (which is

2).
9
Example
Evaluate Res (g(z)f

(z)/f(z), ) if is a pole of order n of f(z), g(z)

is analytic at and g() = 0.
Solution
Since is a pole of order n of f(z), there exists a deleted neigh-
borhood {z : 0 < |z | < } such that f(z) admits the Laurent
expansion:
f(z) =
b
n
(z )
n
+
b
n1
(z )
n1
+ +
b
1
(z )
+

n=0
a
n
(z)
n
, b
n
= 0.
Within the annulus of convergence, we can perform termwise dif-
ferentiation of the above series
10
f

(z) =
nb
n
(z )
n+1

(n 1)b
n
(z )
n

b
1
(z )
2
+

n=0
na
n
(z )
n1
.
Provided that g() = 0, it is seen that
= lim
z
g(z)
(z )
_
nb
n
(z)
n+1

(n1)b
n
(z)
n

b
1
(z)
2
+

n=0
na
n
(z )
n1
_
b
n
(z)
n
+
b
n1
(z)
n1
+ +
b
1
z
+

n=0
a
n
(z )
n
= ng() = 0,
so that is a simple pole of g(z)f

(z)/f(z). Furthermore,
Res
_
g
f

f
,
_
= ng().
Remark
When g() = 0, becomes a removable singularity of gf

/f.
11
Example
Suppose an even function f(z) has a pole of order n at . Within the
deleted neighborhood {z : 0 < |z | < }, f(z) admits the Laurent
expansion
f(z) =
b
n
(z )
n
+ +
b
1
(z )
+

n=0
a
n
(z )
n
, b
n
= 0.
Since f(z) is even, f(z) = f(z) so that
f(z) = f(z) =
b
n
(z )
n
+ +
b
1
(z )
+

n=0
a
n
(z )
n
,
which is valid within the deleted neighborhood {z : 0 < |z +| < }.
Hence, is a pole of order n of f(z). Note that
Res (f(z), ) = b
1
and Res (f(z), ) = b
1
so that Res (f(z), ) = Res (f(z), ). For an even function, if
z = 0 happens to be a pole, then Res (f, 0) = 0.
12
Example
_
|z|=2
tanz
z
dz = 2i
_
Res
_
tanz
z
,

2
_
+Res
_
tanz
z
,

2
__
since the singularity at z = 0 is removable. Observe that

2
is a
simple pole and cos z = sin
_
z

2
_
, we have
Res
_
tanz
z
,

2
_
= lim
z

2
_
z

2
_
tanz
z
= lim
z

2
_
z

2
_
sinz
z
_

_
z

2
_
+
(
z

2
)
3
6
+
_
=
1

2
=
2

.
13
As tanz/z is even, we deduce that Res
_
tanz
z
,

2
_
=
2

using the
result from the previous example. We then have
_
|z|=2
tanz
z
dz = 0.
Remark
Let p(z) = sinz/z, q(z) = cos z, and observe that p
_

2
_
=
2

, q
_

2
_
=
0 and q

2
_
= 1 = 0, then
Res
_
tanz
z
,

2
_
= p
_

2
_
_
q

2
_
=
2

.
14
Example
Evaluate
_
C
z
2
(z
2
+
2
)
2
sinz
dz.
Solution
lim
z0
z
sinz
z
(z
2
+
2
)
2
=
_
lim
z0
z
sinz
_
_
lim
z0
z
(z
2
+
2
)
2
_
= 0
so that z = 0 is a removable singularity.
15
It is easily seen that z = i is a pole of order 2.
Res (f, i) = lim
zi
d
dz
[(z i)
2
f(z)]
= lim
zi
d
dz
_
z
2
(z +i)
2
sinz
_
= lim
zi
2z(z +i) sinz z
2
[(z +i) cos z +2sinz]
(z +i)
3
sin
2
z
=
2sinh +( cosh sinh)
4 sinh
2

=
1
4 sinh
+
cosh
4 sinh
2

.
Recall that sini = i sinh and cos i = cosh. Hence,
_
C
z
2
(z
2
+
2
)
2
sinz
dz = 2iRes (f, i)
=
i
2
_

1
sinh
+
cosh
sinh
2

_
.
16
Theorem
If a function f is analytic everywhere in the nite plane except for a
nite number of singularities interior to a positively oriented simple
closed contour C, then
_
C
f(z) dz = 2iRes
_
1
z
2
f
_
1
z
_
, 0
_
.
17
We construct a circle |z| = R
1
which is large enough so that C is
interior to it. If C
0
denotes a positively oriented circle |z| = R
0
,
where R
0
> R
1
, then
f(z) =

n=
c
n
z
n
, R
1
< |z| < , (A)
where
c
n
=
1
2i
_
C
0
f(z)
z
n+1
dz n = 0, 1, 2, .
In particular,
2ic
1
=
_
C
0
f(z) dz.
How to nd 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
1
z
2
f
_
1
z
_
=

n=
c
n
z
n+2
=

n=
c
n2
z
n
, 0 < |z| <
1
R
1
,
so that
c
1
= Res
_
1
z
2
f
_
1
z
_
, 0
_
.
Remark
By convention, we may dene the residue at innity by
Res (f, ) =
1
2i
_
C
f(z) dz = Res
_
1
z
2
f
_
1
z
_
, 0
_
,
where all singularities in the nite plane are included inside C. With
the choice of the negative sign, we have

all
Res (f, z
i
) +Res (f, ) = 0.
19
Example
Evaluate
_
|z|=2
5z 2
z(z 1)
dz.
Solution
Write f(z) =
5z 2
z(z 1)
. For 0 < |z| < 1,
5z 2
z(z 1)
=
5z 2
z
1
1 z
=
_
5
2
z
_
(1 z z
2
)
so that
Res (f, 0) = 2.
20
For 0 < |z 1| < 1,
5z 2
z(z 1)
=
5(z 1) +3
z 1
1
1 +(z 1)
=
_
5 +
3
z 1
_
[1 (z 1) +(z 1)
2
(z 1)
3
+ ]
so that
Res (f, 1) = 3.
Hence,
_
|z|=2
5z 2
z(z 1)
dz = 2i [Res (f, 0) +Res (f, 1)] = 10i.
21
On the other hand, consider
1
z
2
f
_
1
z
_
=
5 2z
z(1 z)
=
5 2z
z
1
1 z
=
_
5
z
2
_
(1 +z +z
2
+ )
=
5
z
+3 +3z, 0 < |z| < 1,
so that
_
|z|=2
5z 2
z(z 1)
dz = 2iRes (f, )
= 2iRes
_
1
z
2
f
_
1
z
_
, 0
_
= 10i.
22
Evaluation of integrals using residue methods
A wide variety of real denite integrals can be evaluated eectively
by the calculus of residues.
Integrals of trigonometric functions over [0, 2]
We consider a real integral involving trigonometric functions of the
form
_
2
0
R(cos , sin) d,
where R(x, y) is a rational function dened 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 = e
i
, dz = ie
i
d = iz d,
cos =
e
i
+e
i
2
=
1
2
_
z +
1
z
_
,
sin =
e
i
e
i
2i
=
1
2i
_
z
1
z
_
.
The above integral can then be transformed into
_
2
0
R(cos , sin) d
=
_
|z|=1
1
iz
R
_
z +z
1
2
,
z z
1
2i
_
dz
= 2i
_
sum of residues of
1
iz
R
_
z +z
1
2
,
z z
1
2i
_
inside |z| = 1
_
.
24
Example
Compute I =
_
2
0
cos 2
2 +cos
d.
Solution
_
2
0
cos 2
2 +cos
d = i
_
|z|=1
1
2
_
z
2
+
1
z
2
_
2 +
1
2
_
z +
1
z
_
dz
z
= i
_
|z|=1
z
4
+1
z
2
(z
2
+4z +1)
dz.
The integrand has a pole of order two at z = 0. Also, the roots of
z
2
+4z +1 = 0, namely, z
1
= 2

3 and z
2
= 2+

3, are simple
poles of the integrand.
25
Write f(z) =
z
4
+1
z
2
(z
2
+4z +1)
. Note that z
1
is inside but z
2
is outside
|z| = 1.
26
Res (f, 0) = lim
z0
d
dz
z
4
+1
z
2
+4z +1
= lim
z0
3z
3
(z
2
+4z +1) (z
4
+1)(2z +4)
(z
2
+4z +1)
2
= 4
Res (f, 2 +

3) =
z
4
+1
z
2

z=2+

3
_
d
dz
(z
2
+4z +1)

z=2+

3
=
(2 +

3)
4
+1
(2 +

3)
2

1
2(2 +

3) +4
=
7

3
.
I = (i)2i
_
Res (f, 0) +Res (f, 2 +

3)
_
= 2
_
4 +
7

3
_
.
27
Example
Evaluate the integral
I =
_

0
1
a b cos
d, a > b > 0.
Solution
Since the integrand is symmetric about = , we have
I =
1
2
_
2
0
1
a b cos
d =
_
2
0
e
i
2ae
i
b(e
2i
+1)
d.
The real integral can be transformed into the contour integral
I = i
_
|z|=1
1
bz
2
2az +b
dz.
The integrand has two simple poles, which are given by the zeros
of the denominator.
28
Let denote the pole that is inside the unit circle, then the other
pole will be
1

. The two poles are found to be

=
a
_
a
2
b
2
b
and
1

=
a +
_
a
2
b
2
b
.
Since a > b > 0, the two roots are distinct, and is inside but
1

is
outside the closed contour of integration. We then have
I =
1
ib
_
|z|=1
1
(z ) (z
1

)
dz
=
2i
ib
Res
_
_
1
(z ) (z
1

)
,
_
_
=
2i
ib
_

1

_
=

_
a
2
b
2
.
29
Integral of rational functions
_

f(x) dx,
where
1. f(z) is a rational function with no singularity on the real axis,
2. lim
z
zf(z) = 0.
It can be shown that
_

f(x) dx = 2i [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 C
R
and the diameter from R to R.
By the Residue Theorem
_
C
f(z) dz =
_
R
R
f(x) dx +
_
C
R
f(z) dz
= 2i [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 suces to show that as R ,
lim
R
_
C
R
f(z) dz = 0.
The modulus of the above integral is estimated by the modulus
inequality as follows:

_
C
R
f(z) dz

_

0
|f(Re
i
)| R d
max
0
|f(Re
i
)| R
_

0
d
= max
zC
R
|zf(z)|,
which goes to zero as R , since lim
z
zf(z) = 0.
32
Example
Evaluate the real integral
_

x
4
1 +x
6
dx
by the residue method.
Solution
The complex function f(z) =
z
4
1 +z
6
has simple poles at i,

3 +i
2
and

3 +i
2
in the upper half-plane, and it has no singularity on
the real axis. The integrand observes the property lim
z
zf(z) = 0.
We obtain
_

f(x) dx = 2i
_
Res(f, i) + Res
_
f,

3 +i
2
_
+Res
_
f,

3 +i
2
__
.
33
The residue value at the simple poles are found to be
Res(f, i) =
1
6z

z=i
=
i
6
,
Res
_
f,

3 +i
2
_
=
1
6z

z=

3+i
2
=

3 i
12
,
and
Res
_
f,

3 +i
2
_
=
1
6z

z=

3+i
2
=

3 +i
12
,
so that
_

x
4
1 +x
6
dx = 2i
_

i
6
+

3 i
12

3 +i
12
_
=
2
3
.
34
Integrals involving multi-valued functions
Consider a real integral involving a fractional power function
_

0
f(x)
x

dx, 0 < < 1,

1. f(z) is a rational function with no singularity on the positive real
axis, including the origin.
2. lim
z
f(z) = 0.
We integrate (z) =
f(z)
z

along the closed contour as shown.

35
The closed contour C consists of an innitely large circle and an
innitesimal circle joined by line segments along the positive x-axis.
36
(i) line segment from to R along the upper side of the positive
real axis: z = x, x R;
(ii) the outer large circle C
R
: z = Re
i
, 0 < < 2;
(iii) line segment from R to along the lower side of the positive
real axis
z = xe
2i
, x R;
(iv) the inner innitesimal circle C

in the clockwise direction

z = e
i
, 0 < < 2.
37
Establish: lim
R
_
C
R
(z) = 0 and lim
0
_
C

(z) = 0.

_
C
R
(z) dz

_
2
0
|(Re
i
)Re
i
| d 2 max
zC
R
|z(z)|

_
C

(z) dz

_
2
0
|(e
i
)| d 2 max
zC

|z(z)|.
It suces to show that z(z) 0 as either z or z 0.
1. Since lim
z
f(z) = 0 and f(z) is a rational function,
deg (denominator of f(z)) 1 + deg (numerator of f(z)).
Further, 1 < 1, z(z) = z
1
f(z) 0 as z .
2. Since f(z) is continuous at z = 0 and f(z) has no singularity at
the origin, z(z) = z
1
f(z) 0 f(0) = 0 as z 0.
38
The argument of the principal branch of z

is chosen to be 0 <
2, as dictated by the contour.
_
C
(z) dz =
_
C
R
(z) dz +
_
C

(z) dz
+
_
R

f(x)
x

dx +
_

R
f(xe
2i
)
x

e
2i
dx
= 2i [sum of residues at all the isolated singularities
of f enclosed inside the closed contour C].
By taking the limits 0 and R , the rst two integrals vanish.
The last integral can be expressed as

_

0
f(x)
x

e
2i
dx = e
2i
_

0
f(x)
x

dx.
Combining the results,
_

0
f(x)
x

dx =
2i
1 e
2i
[sum of residues at all the isolated
singularities of f in the nite complex plane].
39
Example
Evaluate
_

0
1
(1 +x)x

dx, 0 < < 1.

Solution
f(z) =
1
(1 +z)z

is multi-valued and has an isolated singularity at

z = 1. By the Residue Theorem,
_
C
1
(1 +z)z

dz
= (1 e
2i
)
_
R

dx
(1 +x)x

+
_
C
R
dz
(1 +z)z

+
_
C

dz
(1 +z)z

= 2i Res
_
1
(1 +z)z

, 1
_
=
2i
e
i
.
40
The moduli of the third and fourth integrals are bounded by

_
C
R
1
(1 +z)z

dz

2R
(R 1)R

0 as R ,

_
C

1
(1 +z)z

dz

2
(1 )

1
0 as 0.
On taking the limits R and 0, we obtain
(1 e
2i
)
_

0
1
(1 +x)x

dx =
2i
e
i
;
so
_

0
1
(1 +x)x

dx =
2i
e
i
(1 e
2i
)
=

sin
.
41
Example
Evaluate the real integral
_

e
x
1 +e
x
dx, 0 < < 1.
Solution
The integrand function in its complex extension has innitely many
poles in the complex plane, namely, at z = (2k + 1)i, k is any
integer. We choose the rectangular contour as shown
l
1
: y = 0, R x R,
l
2
: x = R, 0 y 2,
l
3
: y = 2, R x R,
l
4
: x = R, 0 y 2.
42
The chosen closed rectangular contour encloses only one simple pole
at z = i.
43
The only simple pole that is enclosed inside the closed contour C is
z = i. By the Residue Theorem, we have
_
C
e
z
1 +e
z
dz =
_
R
R
e
x
1 +e
x
dx +
_
2
0
e
(R+iy)
1 +e
R+iy
idy
+
_
R
R
e
(x+2i)
1 +e
x+2i
dx +
_
0
2
e
(R+iy)
1 +e
R+iy
idy
= 2i Res
_
e
z
1 +e
z
, i
_
= 2i
e
z
e
z

z=i
= 2ie
i
.
44
Consider the bounds on the moduli of the integrals as follows:

_
2
0
e
(R+iy)
1 +e
R+iy
idy

_
2
0
e
R
e
R
1
dy O(e
(1)R
),

_
0
2
e
(R+iy)
1 +e
R+iy
idy

_
2
0
e
R
1 e
R
dy O(e
R
).
As 0 < < 1, both e
(1)R
and e
R
tend to zero as R tends to
innity. Therefore, the second and the fourth integrals tend to zero
as R . On taking the limit R , the sum of the rst and
third integrals becomes
(1 e
2i
)
_

e
x
1 +e
x
dx = 2ie
i
;
so
_

e
x
1 +e
x
dx =
2i
e
i
e
i
=

sin
.
45
Example
Evaluate
_

0
1
1 +x
3
dx.
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
_
C
Log z
1 +z
3
dz,
where the branch cut is chosen to be along the positive real axis
whereby 0 Arg z < 2. Now
_
C
Log z
1 +z
3
dz =
_
R

lnx
1 +x
3
dx +
_

R
Log (xe
2i
)
1 +(xe
2i
)
3
dx
+
_
C
R
Log z
1 +z
3
dz +
_
C

Log z
1 +z
3
dz
46
= 2i
3

j=1
Res
_
Log z
1 +z
3
, z
j
_
,
where z
j
, j = 1, 2, 3 are the zeros of 1/(1 +z
3
). Note that

_
C

Log z
1 +z
3
dz

= O
_
ln
1
_
0 as 0;

_
C
R
Log z
1 +z
3
dz

= O
_
RlnR
R
3
_
0 as R .
Hence
lim
R
0
_
C
Log z
1 +z
3
dz =
_

0
lnx
1 +x
3
dx +
_
0

Log (xe
2i
)
1 +(xe
2i
)
3
dx
= 2i
_

0
1
1 +x
3
dx,
thus giving
_

0
1
1 +x
3
dx =
3

j=1
Res
_
Log z
1 +z
3
, z
j
_
.
47
The zeros of 1 + z
3
are = e
i/3
, = e
i
and = e
5i/3
. Sum of
residues is given by
Res
_
Log z
1 +z
3
,
_
+Res
_
Log z
1 +z
3
,
_
+Res
_
Log z
1 +z
3
,
_
=
Log
( )( )
+
Log
( )( )
+
Log
( )( )
= i
_

3
( ) +( ) +
5
3
( )
_
( )( )( )
=
2
3

3
.
Hence,
_

0
1
1 +x
3
dx =
2
3

3
.
48
Evaluation of Fourier integrals
A Fourier integral is of the form
_

e
imx
f(x) dx, m > 0,
1. lim
z
f(z) = 0,
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 > 0

_
C
R
f(z)e
iz
dz

_

0
|f(Re
i
)| |e
iRe
i
| R d
max
zC
R
|f(z)| R
_

0
e
Rsin
d
= 2R max
zC
R
|f(z)|
_
2
0
e
Rsin
d
2R max
zC
R
|f(z)|
_
2
0
e
R
2

d
= 2R max
zC
R
|f(z)|

2R
(1 e
R
),
which tends to 0 as R , given that f(z) 0 as R .
50
51
To evaluate the Fourier integral, we integrate e
imz
f(z) along the
closed contour C that consists of the upper half-circle C
R
and the
diameter from R to R along the real axis. We then have
_
C
e
imz
f(z) dz =
_
R
R
e
imx
f(x) dx +
_
C
R
e
imz
f(z) dz.
Taking the limit R , the integral over C
R
vanishes by virtue of
the Jordan Lemma.
Lastly, we apply the Residue Theorem to obtain
_

e
imx
f(x) dx = 2i [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
_

sin2x
x
2
+x +1
dx.
Solution
It is easy to check that f(z) =
1
z
2
+z+1
has no singularity along
the real axis and lim
z
1
z
2
+z+1
= 0. The integrand has two simple
poles, namely, z = e
2i
3
in the upper half-plane and e

2i
3
in the lower
half-plane. By virtue of the Jordan Lemma, we have
53
_

sin2x
x
2
+x +1
dx = Im
_

e
2ix
x
2
+x +1
dx = Im
_
C
e
2iz
z
2
+z +1
dz,
where C is the union of the innitely large upper semi-circle and its
diameter along the real axis.
Note that
_
C
e
2iz
z
2
+z +1
dz = 2i Res
_
e
2iz
z
2
+z +1
, e
2i
3
_
= 2i
e
2iz
2z +1

z=e
2i
3
= 2i
e
2ie
2i
3
2e
2i
3
+1
.
Hence,
_

sin2x
x
2
+x +1
dx = Im
_
_
_
_
2i
e
2ie
2i
3
2e
2i
3
+1
_
_
_
_
=
2

3
e

3
sin1.
54
Example
Show that
_

0
sinx
2
dx =
_

0
cos x
2
dx =
1
2
_

2
.
Solution
55
0 =
_
C
e
iz
2
dz =
_
R
0
e
ix
2
dx +
_
4
0
e
iR
2
e
2i
iRe
i
d
+
_
0
R
e
ir
2
e
i/2
e
i/4
dr.
Rearranging
_
R
0
(cos x
2
+i sinx
2
) dx = e
i/4
_
R
0
e
r
2
dr
_
/4
0
e
iR
2
cos 2R
2
sin2
iRe
i
d.
Next, we take the limit R . We recall the well-known result
e
i

4
_

0
e
r
2
dr =

2
e
i/4
=
1
2
_

2
+
i
2
_

2
.
Also, we use the transformation 2 = and observe sin
2

, 0

2
,
to obtain
56

_
/4
0
e
iR
2
cos 2R
2
sin
2

iRe
i
d

_
/4
0
e
R
2
sin2
Rd
=
R
2
_
/2
0
e
R
2
sin
d

R
2
_
/2
0
e
2R
2
/
d
=

4R
(1 e
R
2
) 0 as R .
We then obtain
_

0
(cos x
2
+i sinx
2
) dx =
1
2
_

2
+i
1
2
_

2
so that
_

0
cos x
2
dx =
_

0
sinx
2
dx =
1
2
_

2
.
57
Example
Evaluate
_

0
ln(x
2
+1)
x
2
+1
dx.
Hint: Use Log(i x) +Log(i +x) = Log(i
2
x
2
) = ln(x
2
+1) +i.
Solution
Consider
_
C
Log(z +i)
z
2
+1
dz around C as shown.
58
The only pole of
Log(z +i)
z
2
+1
in the upper half plane is the simple
pole z = i. Consider
2iRes
_
Log(z +i)
z
2
+1
, i
_
= 2i lim
zi
(z i)Log(z +i)
(z i)(z +i)
= Log 2i = ln2 +

2
2
i.
_
C
R
Log(z +i)
z
2
+1
dz = O
_
(lnR)R
R
2
_
0 as R
_
R
0
Log(i x)
x
2
+1
dx +
_
R
0
Log(x +i)
x
2
+1
dx +
_
C
R
Log(z +i)
z
2
+1
dz = ln2 +

2
2
i
and Log(i x) +Log(i +x) = ln(x
2
+1) +i.
From
_

0
ln(x
2
+1)
x
2
+1
dx +
_

0
i
x
2
+1
dx = ln2 +

2
2
i and
_

0
dx
1 +x
2
=

2
so that
_

0
ln(x
2
+1)
x
2
+1
dx = ln2.
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 x
0
inside the interval. The integral of f(x)
over the interval [a, b] is an improper integral, which may be dened
as
_
b
a
f(x) dx = lim

1
,
2
0
_
_
x
0

1
a
f(x) dx +
_
b
x
0
+
2
f(x) dx
_
,
1
,
2
> 0.
In many cases, the above limit exists only when
1
=
2
, and does
not exist otherwise.
60
Example
Consider the following improper integral
_
2
1
1
x 1
dx,
show that the Cauchy principal value of the integral exists, then nd
the principal value.
Solution
Principal value of
_
2
1
1
x 1
dx exists if the following limit exists.
lim
0
+
_
_
1
1
1
x 1
dx +
_
2
1+
1
x 1
dx
_
= lim
0
+
_
_
ln|x 1|

1
1
+ln|x 1|

2
1+
_
_
= lim
0
+
[(ln ln2) +(ln1 ln)] = ln2.
Hence, the principal value of
_
2
1
1
x 1
dx exists and its value is
ln2.
61
Lemma
If f has a simple pole at z = c and T
r
is the circular arc dened by
T
r
: z = c +re
i
(
1

2
),
then
lim
r0
+
_
T
r
f(z) dz = i(
2

1
)Res (f, c).
In particular, for the semi-circular arc S
r
lim
r0
+
_
S
r
f(z) dz = iRes (f, c).
62
Proof
Since f has a simple pole at c,
f(z) =
a
1
z c
+

k=0
a
k
(z c)
k
. .
g(z)
, 0 < |z c| < R for some R.
For 0 < r < R,
_
T
r
f(z) dz = a
1
_
T
r
1
z c
dz +
_
T
r
g(z) dz.
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,

_
T
r
g(z) dz

M arc length of T
r
= Mr(
2

1
)
and so
lim
r0
+
_
T
r
g(z) dz = 0.
Finally,
_
T
r
1
z c
dz =
_

2

1
1
re
i
ire
i
d = i
_

2

1
d = i(
2

1
)
so that
lim
r0
+
_
T
r
f(z) dz = a
1
i(
2

1
) = Res (f, c)i(
2

1
).
64
Example
Compute the principal value of
_

xe
2ix
x
2
1
dx.
Solution
The improper integral has singularities at x = 1. The principal
value of the integral is dened to be
lim
R
r
1
,r
2
0
+
_
_
1r
1
R
+
_
1r
2
1+r
1
+
_
R
1+r
2
_
xe
2ix
x
2
1
dx.
65
Let
I
1
=
_
S
r
1
ze
2iz
z
2
1
dz
I
2
=
_
S
r
2
ze
2iz
z
2
1
dz
I
R
=
_
C
R
ze
2iz
z
2
1
dz.
Now, f(z) =
ze
2iz
z
2
1
is analytic inside the above closed contour.
By the Cauchy Integral Theorem
_
_
1r
1
R
+
_
1r
2
1+r
1
+
_
R
1+r
2
_
xe
2ix
x
2
1
dx +I
1
+I
2
+I
R
= 0.
66
By the Jordan Lemma, and since
z
z
2
1
0 as z , so
lim
R
I
R
= 0.
Since z = 1 are simple poles of f,
lim
r
1
0
+
I
1
= iRes (f, 1) = i lim
z1
(z +1)f(z)
= (i)e
2i
/2.
Similarly, lim
r
2
0
+
I
2
= iRes (f, 1) =
ie
2i
2
.
PV
_

xe
2ix
x
2
1
dx =
ie
2i
2
+
ie
2i
2
= i cos 2.
67
Poisson integral formula
f(z) =
1
2i
_
C
f(s)
s z
ds.
Here, C is the circle with radius r
0
centered at the origin. Write
s = r
0
e
i
and z = re
i
, r > r
0
. We choose z
1
such that |z
1
| |z| = r
2
0
and both z
1
and z lie on the same ray so that
z
1
=
r
2
0
r
e
i
=
r
2
0
z
=
ss
z
.
68
Since z
1
lies outside C, we have
f(z) =
1
2i
_
C
f(s)
_
1
s z

1
s z
1
_
ds
=
1
2
_
2
0
_
s
s z

s
s z
1
_
f(s) d.
The integrand can be expressed as
s
s z

1
1 s/z
=
s
s z
+
z
s z
=
r
2
0
r
2
|s z|
2
and so f(re
i
) =
r
2
0
r
2
2
_
2
0
f(r
0
e
i
)
|s z|
2
d.
69
Now |s z|
2
= r
2
0
2r
0
r cos( ) + r
2
> 0 (from the cosine rule).
Taking the real part of f, where f = u +iv, we obtain
u(r, ) =
1
2
_
2
0
r
2
0
r
2
r
2
0
2r
0
r cos( ) +r
2
. .
P(r
0
,r,)
u(r
0
, ) d, r < r
0
.
Knowing u(r
0
, ) on the boundary, u(r, ) is uniquely determined.
The kernel function P(r
0
, r, ) is called the Poisson kernel .
P(r
0
, r, ) =
r
2
0
r
2
|s z|
2
= Re
_
s
s z
+
z
s z
_
= Re
_
s
s z
+
z
s z
_
= Re
_
s +z
s z
_
which is harmonic for |z| < r
0
.
70