EE202 - EE MATH II

Jitkomut Songsiri

13. Residues and Its Applications

isolated singular points

residues
Cauchys residue theorem
applications of residues

13-1

Isolated singular points

z0 is called a singular point of f if
f fails to be analytic at z0
but f is analytic at some point in every neighborhood of z0
a singular point z0 is said to be isolated if f is analytic in some punctured disk
0 < |z z0| < 
centered at z0 (also called a deleted neighborhood of z0)
example: f (z) = 1/(z 2(z 2 + 1)) has the three isolated singular points at
z = 0,

Residues and Its Applications

z = j

13-2

Non-isolated singular points

1
example: the function
has the singular points
sin(/z)
z = 0,

1
z= ,
n

(n = 1, 2, . . .)

each singular point except z = 0 is isolated

0 is nonisolated since every punctured disk of 0 contains other singularities
for any > 0, we can find a positive integer n such that n > 1/
this means z = 1/n always lies in the punctured disk 0 < |z| <
Residues and Its Applications

13-3

Residues
assumption: z0 is an isolated singular point of f , e.g.,
there exists a punctured disk 0 < |z z0| < r0 throughout which f is analytic
consequently, f has a Laurent series representation
f (z) =

an(z z0)n +

n=0

b1
bn
+ +
+ ,
n
z z0
(z z0)

(0 < |z z0| < r0)

let C be any positively oriented simple closed contour lying in the disk
0 < |z z0| < r0
the coefficient bn of the Laurent series is given by
Z
f (z)
1
dz,
(n = 1, 2, . . .)
bn =
j2 C (z z0)n+1

Residues and Its Applications

13-4

the coefficient of 1/(z z0) in the Laurent expansion is obtained by

Z
f (z)dz = j2b1
C

b1 is called the residue of f at the isolated singular point z0, denoted by

b1 = Res f (z)
z=z0

this allows us to write

Z
f (z)dz = j2 Res f (z)
C

z=z0

which provides a powerful method for evaluating integrals around a contour

Residues and Its Applications

13-5

example: find

e1/z dz when C is the positive oriented circle |z| = 1

C

1/z 2 is analytic everywhere except z = 0; 0 is an isolated singular point

the Laurent series expansion of f is

f (z) = e

1/z 2

=1+

1
1
1
+
+
+
2
4
6
z
2!z
3!z

(0 < |z| < )

the residue of f at z = 0 is zero (b1 = 0), so the integral is zero

remark:
the analyticity of f within and on C is a sufficient condition for
R
f (z)dz to be zero; however, it is not a necessary condition
C
Residues and Its Applications

13-6

1
example: compute C
dz where C is circle |z + 2| = 1
3
z(z + 2)
R

f has the isolated singular points at 0 and 2

choose an annulus domain: 0 < |z + 2| < 2
on which f is analytic and contains C
f has a Laurent series on this domain and is given by
1
1
1
1
f (z) =
=

(z + 2 2)(z + 2)3
2 1 (z + 2)/2 (z + 2)3

X
X
1
(z + 2)n
(z + 2)n3
=
, (0 < |z + 2| < 2)
=
n+1
2(z + 2)3 n=0 2n
2
n=0
the residue of f at z = 2 is 1/23 which is obtained when n = 2
therefore, the integral is j2(1/23) = j/4 (check with the Cauchy formula)
Residues and Its Applications

13-7

Cauchys residue theorem

let C be a positively oriented simple closed contour
Theorem: if f is analytic inside and on C except for a finite number of singular
points z1, z2, . . . , zn inside C, then
Z
f (z)dz = j2
C

n
X
k=1

Res f (z)

z=zk

Proof.
since zk s are isolated points, we can find small
circles Ck s that are mutually disjoint
f is analytic on a multiply connected domain
from the Cauchy-Goursat theorem:
R
Pn R
f (z)dz = k=1 C f (z)dz
C
k

Residues and Its Applications

13-8

example: use the Cauchy residue theorem to evaluate the integral

Z

3(z + 1)
dz,
z(z

1)(z

3)
C

C is the circle |z| = 2, in counterclockwise

C encloses the two singular points of the integrand, so

Z

h
i
3(z + 1)
I=
dz = j2 Res f (z) + Res f (z)
f (z)dz =
z=0
z=1
C
C z(z 1)(z 3)
calculate Resz=0 f (z) via the Laurent series of f in 0 < |z| < 1
calculate Resz=1 f (z) via the Laurent series of f in 0 < |z 1| < 1
Residues and Its Applications

13-9

1
3
2
rewrite f (z) =
+
z z1 z3
the Laurent series of f in 0 < |z| < 1
1
3
2
1
2
f (z) = +

= +3(1+z+z 2+. . .) (1+(z/3)+(z/3)2+. . .)

z 1 z 3(1 z/3) z
3
the residue of f at 0 is the coefficient of 1/z, so Resz=0 f (z) = 1
the Laurent series of f in 0 < |z 1| < 1
1
3
1

f (z) =
1 + z 1 z 1 1 (z 1)/2
= 1 (z 1) + (z 1)2 + . . .

z1

1+

z1
+
2

z1
2

2
+ ...

the residue of f at 1 is the coefficient of 1/(z 1), so Resz=0 f (z) = 3

Residues and Its Applications

13-10

therefore, I = j2(1 3) = j4
alternatively, we can compute the integral from the Cauchy integral formula

Z
I=
C

3
2
1

+
dz
z z1 z3

= j2(1 3 + 0) = j4

Residues and Its Applications

13-11

Residue at infinity
f is said to have an isolated point at z0 = if
there exists R > 0 such that f is analytic for R < |z| <

singular points of

C is a positive oriented simple closed contour

Theorem: if f is analytic everywhere except for a finite number of singular

points interior to C, then
 

Z
1
1
f (z)dz = j2 Res 2 f
z=0 z
z
C
(see a proof on section 71, Churchill)
Residues and Its Applications

13-12

z3
example: find I = C
dz, C is the circle |z| = 2 (counterclockwise)
z(z 1)
R



2
I = j2 Res (1/z )f (1/z)
z=0


1 3z
= j2 Res
, j2 Res g(z)
z=0 z(1 z)
z=0
find the residue via the Laurent series of g in 0 < |z| < 1


1
3 (1 + z + z 2 + ) = Res g(z) = 1
write g(z) =
z=0
z
compare the integral with other methods .
Cauchy integral formula (write the partial fraction of f )
Cauchy residue theorem (have to find two residues; hence two Laurent series)
Residues and Its Applications

13-13

Principal part
f has an isolated singular point at z0, so f has a Laurent seires

b1
b2
bn
f (z) =
an(z z0) +
+
+ +
+
2
n
(z

z
)
(z

z
)
(z

z
)
0
0
0
n=0
n

in a punctured disk 0 < |z z0| < R

the portion of the series that involves negative powers of z z0

b1
b2
bn
+
+ +
+
(z z0) (z z0)2
(z z0)n
is called the principal part of f

Residues and Its Applications

13-14

Types of isolated singular points

three possible types of the principal part of f
no principal part
z2 z4
f (z) = cos z = 1 + + ,
2! 4!

(0 < |z| < )

finite number of terms in the principal part

1
1
1
f (z) = 2
= 2 + 1 z + z2 + ,
z (1 + z) z
z

(0 < |z| < 1)

infinite number of terms in the principal part

f (z) = e

Residues and Its Applications

1/z

1
1
1
=1+ +
+
+ ,
z 2!z 2 3!z 3

(0 < |z| < )

13-15

classify the number of terms in the principal part in a general form

none: z0 is called a removable singular point
f (z) =

an(z z0)n

n=0

finite (m terms): z0 is called a pole of order m

b1
b2
bm
f (z) =
an(z z0) +
+
+ +
2
m
z

z
(z

z
)
(z

z
)
0
0
0
n=0
n

infinite: z0 is said to be an essential singular point of f

f (z) =

X
n=0

Residues and Its Applications

an(z z0)n +

b2
bn
b1
+
+

+
+
2
n
z z0 (z z0)
(z z0)

13-16

examples:
f1(z) =
f2(z) =
f3(z) =
f4(z) =

z2 z4
cos z = 1 + +
2! 4!


3
1
=
+ 1 + (z 2) + (z 2)3 +
(z 1)(z 2)
z2
1
1
1
2
=

+
1

z
+
z
+
z 2(1 + z) z 2 z
1
1
1
e1/z = 1 + +
+
+
2
3
z 2!z
3!z

0 is a removeable singular point of f1

2 is a pole of order 1 (or simple pole) of f2
0 is a pole of order 2 (or double pole) of f3
0 is an essential singular point of f4
note: for f2, f3 we can determine the pole/order from the denominator of f
Residues and Its Applications

13-17

Residue formula
if f has a pole of order m at z0 then
1
dm1
m
Res f (z) =
lim
(z

z
)
f (z)
0
z=z0
(m 1)! zz0 dz m1

Proof. if f has a pole of order m, its Laurent series can be expressed as

b2
bm
b1
+
+ +
f (z) =
an(z z0) +
2
m
(z

z
)
(z

z
)
(z

z
)
0
0
0
n=0
(z z0)mf (z) =

an(z z0)m+n + b1(z z0)m1 + b2(z z0)m2 + + bm

n=0

to obtain b1, we take the (m 1)th derivative and take the limit z z0
Residues and Its Applications

13-18

(z + 1)
example 1: find Resz=0 f (z) and Resz=2 f (z) where f (z) = 2
z (z 2)


d z+1
= 3/4
Res f (z) = lim
z=0
z0 dz
z2
z+1
Res f (z) = lim 2 = 3/4
z=2
z2 z

(0 is a double pole of f )

z+1
example 2: find Resz=0 g(z) where g(z) =
1 2z
g is analytic at 0 (0 is a removable singular point of g), so Resz=0 g(z) = 0
check . apply the results from the above two examples to compute
(z + 1)
dz,
2 (z 2)
z
C

C is the circle |z| = 3 (counterclockwise)

by using the Cauchy residue theorem and the formula on page 13-12
Residues and Its Applications

13-19

sometimes the pole order cannot be readily determined

sinh z
example 3: find Resz=0 f (z) where f (z) =
z4
use the Maclaurin series of sinh z


f (z) =

1
z
z

z
+
+
+
4
z
3! 5!


=

1
1
z
+
+
+
3
z
3!z 5!

0 is the third-order pole with residue 1/3!

here we determine the residue at z = 0 from its definition (the coeff. of 1/z )

no need to use the residue formula on page 13-18

Residues and Its Applications

13-20

when the pole order (m) is unknown, we can

assume m = 1, 2, 3, . . .
find the corresponding residues until we find the first finite value
1+z
example 4: find Resz=0 f (z) where f (z) =
1 cos z
assume m = 1
z(1 + z)
1 + 2z
Res f (z) = lim
= 0/0 = lim
= 1/0 = = (not 1st order)
z=0
z0 1 cos z
z0 sin z
assume m = 2
d
z0 dz

Res f (z) = lim

z=0

z (1 + z)
1 cos z


= 2 (finite)

0 is a double pole

note: use LH
opitals rule to compute the limit
Residues and Its Applications

13-21

Summary
R
many ways to compute a contour integral ( C f (z)dz)
parametrize the path (feasible when C is easily described)
use the principle of deformation of paths (if f is analytic in the region
between the two contours)
use the Cauchy integral formula (typically requires the partial fraction of f )
use the Cauchys residue theorem on page 13-8 (requires the residues at
singular points enclosed by C)
use the theorem of the residue at infinity on page 13-12 (find one residue at 0)
to find the residue of f at z0
read from the coeff of 1/(z z0) in the Laurent series of f
apply the residue formula on page 13-18
Residues and Its Applications

13-22

Application of the residue theorem

calculating real definite integrals

integrals involving sines and cosines
improper integrals
improper integrals from Fourier series
inversion of Laplace transforms

Residues and Its Applications

13-23

Definite integrals involving sines and cosines

we consider a problem of evaluating definite integrals of the form
Z

F (sin , cos )d
0

since varies from 0 to 2, we can let be an argument of a point z

z = ej

(0 2)

this describe a positively oriented circle C centered at the origin

make the substitutions
z z 1
sin =
,
j2
Residues and Its Applications

z + z 1
cos =
,
2

dz
d =
jz
13-24

this will transform the integral into the contour integral



Z
F
C

z z 1 z + z 1
,
j2
2

dz
jz

the integrand becomes a function of z

if the integrand reduces to a rational function of z, we can apply the Cauchys
residue theorem
example:
Z
0

d
5 + 4 sin

Z
Z
dz
dz
=
=
,
g(z)dz
2 + j5z 2
(zz 1 ) jz
2z
C 5+4
C
C
2j


Z
dz
= j2
Res g(z) = 2/3
=
z=j/2
C 2(z + 2j)(z + j/2)
1

where C is the positively oriented circle |z| = 1

Residues and Its Applications

13-25

the above idea can be summarized in the following theorem

Theorem: if F (cos , sin ) is a rational function of cos and sin which is

finite on the closed interval 0 2, and if f is the function obtained from
F (, ) by the substitutions
z + z 1
cos =
,
2

z z 1
sin =
j2

then
Z

F (cos , sin ) d = j2
C

X
k

f (z)
z=zk jz

Res

where the summation takes over all zk s that lie within the circle |z| = 1

Residues and Its Applications

13-26

Z
example: compute I =
0

cos 2
d, 1 < a < 1
2
1 2a cos + a

make change of variables

ej2 + ej2 z 2 + z 2 z 4 + 1
=
=
cos 2 =
2
2
2z 2
2
2
az

(a
+ 1)z + a
2
1
2
1 2a cos + a = 1 2a(z + z )/2 + a =
z

we have

R 2
0

F ()d =

f (z)
dz
C jz

R
C

g(z)dz where

(z 4 + 1)z
(z 4 + 1)
g(z) =
=
2
2
2
jz 2z (az (a + 1)z + a) j2z 2(1 az)(z a)
we see that only the poles z = 0 and z = a lie inside the unit circle C
Residues and Its Applications

13-27

therefore, the integral becomes

Z
I=

g(z)dz = j2 Res g(z) + Res g(z)

C

z=0

z=a

note that z = 0 is a double pole of g(z), so

d 2
1 a2 + 1
Res g(z) = lim (z g(z)) =
z=0
z=0 dz
j2
a2
1
a4 + 1
2
Res g(z) = lim (z a)g(z) =
z=a
z=a
j2 a (1 a2)
2a2
hence, I =
1 a2

Residues and Its Applications

13-28

Improper integrals
lets first consider a well-known improper integral
Z
I=

dx
=
2
1 + x

of course, this can be evaluated using the inverse tangent function

we will derive this kind of integral by means of contour integration
some poles of the integrand lie in the upper half plane
let CR be a semicircular contour with radius R
Z

Z
f (x)dx +

and show that

Residues and Its Applications

f (z)dz = j2
CR

R
CR

X
k

Res f (z)

z=zk

f (z)dz 0 as R
13-29

Theorem: if all of the following assumptions hold

1. f (z) is analytic in the upper half plane except at a finite number of poles
2. none of the poles of f (z) lies on the real axis
M
3. |f (z)| k when z = Rej ; M is a constant and k > 1
R
then the real improper integral can be evaluated by a contour integration, and


Z
sum of the residues of f (z) at the poles
f (x)dx = j2
which lie in the upper half plane

assumption 2: f is analytic on C1
R
assumption 3: C f (z)dz 0 as R
R

Residues and Its Applications

13-30

Proof. consider a semicircular contour with radius R large enough to include all
the poles of f (z) that lie in the upper half plane
from the Cauchys residue theorem
Z
i
hX
f (z)dz = j2
Res f (z) at all poles within C1 CR
C1 CR

(to apply this, f (z) cannot have singular points on C1, i.e., the real axis)
the integral along the real axis is our desired integral
Z

lim

Z
f (x)dx + lim

Z
f (z)dz = lim

CR

f (z)dz
C1 CR

hence, it suffices to show that

Z
lim
f (z)dz = 0 by using |f (z)| M/Rk , where k > 1
R

CR

Residues and Its Applications

13-31

 apply the modulus of the integral and use |f (z)| M/Rk

Z





 M
M R

f (z)dz  k length of CR =
R
Rk
CR

hence, limR

R
CR

f (z)dz = 0 if k > 1

remark: an example of f (z) that satisfies all the conditions in page 13-30
p(x)
f (x) =
,
q(x)

p and q are polynomials

q(x) has no real roots and deg q(x) deg p(x) + 2

(relative degree of f is greater than or equal to 2)

Residues and Its Applications

13-32

example: show that

Z
f (z)dz = 0
CR

as R where CR is the arc z = Rej , 0

f (z) = (z + 2)/(z 3 + 1)
|z + 2| |z| + 2 = R + 2,
hence, |f (z)|

R+2
R3 1

(relative degree of f is 2)
|z 3 + 1| ||z 3| 1| = |R3 1|

and apply the modulus of the integral

Z
 Z


1 + R22
R
+
2
 f (z)dz 
|f (z)|dz 3
R =


R

1
R R12
C
C
the upper bound tends to zero as R

Residues and Its Applications

13-33

 f (z) = 1/(z 2 + 2z + 2)
z 2 + 2z + 2 = (z (1 + j))(z (1 j)) , (z z0)(z z0)
hence, |z z0| ||z| |z0|| = R |1 + j| = R
|z z0| ||z| |z0|| = R

2 and similarly,

then it follows that

2

|z + 2z + 2| (R

2)

|f (z)|
(R 2)2

Z
 Z



1
 f (z)dz 

|f
(z)|dz

R
=


2
(R 2)
C
C
(1 R2 )2
the upper bound tends to zero as R
Residues and Its Applications

13-34

dx
example: compute I =
2
1 + x
Z

1
define f (z) =
and create a contour C = C1 CR as on page 13-29
2
1+z
R
relative degree of f is 2, so C f (z)dz = 0 as R
R

f (z) has poles at z = j and z = j (no poles on the real axis)

only the pole z = j lies in the upper half plane
by the residues theorem
j2

I
Res f (z) =

z=zk

Z
f (z)dz =

f (z)dz
f (x)dx +
} | CR {z }
| R {z
=I as R

=0 as R

I = j2 Res f (z) = j2 lim (z j)f (z) =

z=j

Residues and Its Applications

zj

13-35

example: compute

x2
I=
dx
2 + a2 )(x2 + b2 )
(x

z2
define f (z) = 2
and create C = C1 CR as on page 13-29
2
2
2
(z + a )(z + b )
R
relative degree of f is 2, so C f (z)dz = 0 as R
R

f (z) has poles at z = ja and z = jb

(no poles on the real axis)

only the poles z = ja and z = jb lie in the upper half plane

by the residues theorem
j2

Res f (z) =

z=zk

f (z)dz =
C

f (z)dz
f (x)dx +
} | CR {z }
| R {z
=I as R

I = j2 Res f (z) + Res f (z) = j2

z=ja

Residues and Its Applications

z=jb

Z
=0 as R

a
b

+
=
j2(a2 b2) j2(b2 a2)
a+b
13-36

Improper integrals from Fourier analysis

we can use residue theory to evaluate improper integrals of the form

Z
f (x) cos mx dx,

f (x) sin mx dx,

or

ejmxf (x) dx

we note that ejmz is analytic everywhere, moreover

|ejmz | = ejm(x+jy) = emy < 1

for all y in the upper half plane

therefore, if |f (z)| M/Rk with k > 1, then so is |ejmz f (z)|

hence, if f (z) satisfies the conditions in page 13-30 then
Z

ejmxf (x)dx = j2

Residues and Its Applications

jmz

sum of the residues of e

f (z) at the poles
which lie in the upper half plane

13-37

denote

jmz

sum of the residues of e

f (z) at the poles
which lie in the upper half plane

S=

and note that S can be complex

by comparing the real and imaginary part of the integral
Z

jmx

(cos mx + j sin mx)f (x)dx = j2S

f (x)dx =

we have

cos mxf (x) dx = Re(j2S) = 2 Im S

sin mxf (x) dx = Im(j2S) = 2 Re S

Residues and Its Applications

13-38

example: compute I =

ejmz
1+z 2

define f (z) =

cos mx dx
1 + x2

and create C = C1 CR as on page 13-29

relative degree of f is 2, so

R
CR

f (z)dz = 0 as R

f has poles at z = j and z = j (no poles on the real axis)

the pole z = j lies in the upper half plane
by residues theorem

j2

I
Res f (z) =

z=zk

Z
f (z)dz =

f (x)dx +
f (z)dz
| R {z
} | CR {z }
=I as R

Residues and Its Applications

Z
=0 as R

13-39

 therefore,

ejmz
ejmx
dx = j2 Res
2
z=j 1 + z 2
1
+
x

(z j)ejmz
m
= j2 lim
=
e
zj
1 + z2
our desired integral can be obtained by
Z

cos mx dx
1 + x2

Z
sin mx dx
2
1 + x

Residues and Its Applications

= Re(em) = em,
= Im(em) = 0

13-40

Summary of improper integrals

the examples of f we have seen so far are in the form of

f (x) =

p(x)
q(x)

where p, q are polynomials and deg p(x) deg q(x) + 2

Residues and Its Applications

13-41

the assumption on the degrees of p, q is sufficient to guarantee that

Z

f (z)ejaz dz = 0

(a > 0)

CR

as R where CR is the arc z = Rej , 0

we can relax this assumption to consider function f such as

z
,
z 2 + 2z + 2

1
z+1

(relative degree is 1)

and obtain the same result by making use of Jordans inequality

Residues and Its Applications

13-42

Jordan inequality
for R > 0,

R sin

d <
R

Proof.
sin 2/,
eR sin
Z
0

e2R/ , R > 0, 0

/2

eR sin d

2R

the last line is another form of the Jordan inequality

because the graph of y = sin is symmetric about the line = /2
Residues and Its Applications

13-43

R
z
example: let f (z) = 2
show that C f (z)ejaz dz = 0 for a > 0 as
R
z + 2z + 2
R
first note that |ejaz | = |eja(x+jy)| = |ejax eay | = eay < 1
similar to page 13-34, we see that |f (z)| R/(R
Z




CR

(since a > 0)

2)2 , MR and

 Z

jaz
f (z)e dz 

R =
2
2)
CR (R
(1 R2 )2

which does not tend to zero as R

however, for z that lies on CR, i.e., z = Rej
f (z)e

jaz

= f (z)e

Residues and Its Applications

jaRej

= f (z)ejaR(cos +j sin ) = f (z)eaR sin ejaR cos

13-44

 if we find an upper bound of the integral, and use Jordans inequality:

Z

Z





jaz
aR sin
jaR cos
j



f (z)e dz  = 
f (z)e
e
jRe d

CR
0
Z


aR sin
jaR cos
j 


f (z)e
e
jRe d
0
Z
eaR sin d
= RMR
0

<

MR
a

the final term approach 0 as R because MR 0

conclusion: then we can apply the residues theorem to integrals like
Z
x cos(ax)
dx
2
x + 2x + 2

Residues and Its Applications

13-45

Inversion of Laplace transforms

recall the definitions:
Z

F (s) , L[f (t)] ,

f (t)estdt

f (t) = L

[F (s)] =

1
j2

a+j

F (s)estds

aj

Theorem: suppose F (s) is analytic everywhere except at the poles

p1, p2, . . . , pn,
all of which lie to the left of the vertical line Re(s) = a

(a convergence factor)

if |F (s)| MR and MR 0 as s through the half plane Re(s) a then

L1[F (s)] =

n
X
i=1

Residues and Its Applications

Res F (s)est

s=pi

13-46

Proof sketch.
parametrize C1 and C2 by
C1 = {z | z = a + jy, R y R }



3
C2 = z | z = a + Rej ,

2
2

1. create a huge semicircle that is large enough to contain all the poles of F (s)
2. apply the Cauchys residue theorem to conclude that
Z
C1

estF (s)ds = j2

n
X
k=1

Res [estF (s)]

s=pk

estF (s)ds

C2

3. prove that the integral along C2 is zero when the circle radius goes to
Residues and Its Applications

13-47

choose a and R: choose the center and radius of the circle

a > 0 is so large that all the poles of F (s) lie to the left of C1
a>

max

k=1,2,...,n

Re(pk )

R > 0 is large enough so that all poles of F (s) are enclosed by the semicircle
if the maximum modulus of p1, p2, . . . , pn is R0 then
k, |pk a| |pk | + a R0 + a

pick R > R0 + a

C1 = {z | z = a + jy, R y R }



3
C2 = z | z = a + Rej ,

2
2

Residues and Its Applications

13-48

integral along C2 is zero

C1 = {z | z = a + jy, R y R }



C2 = z | z = a + Rej ,
2
2

for s = a + Rej and ds = jRej d, the integral becomes


 Z 3/2




estF (s)ds = 
eat eRt cos +jRt sin F (a + Rej )Rjej d
 /2

C2

Residues and Its Applications

13-49

 apply the modolus of the integral

Z




C2

 Z

st
e F (s)ds

3/2 


eateRt cos ejRt sin F (a + Rej )Rjej  d

/2

since |F (s)| MR for s that lies on C2

Z




C2


Z

estF (s)ds MRReat

3/2

eRt cos d

/2

make change of variable = /2 and apply the Jordan inequality


Z
at

M
e
R
Rt
sin

at
st
e
d <
e F (s)ds MRRe
t
C2
{z
}
|0

</Rt

the last term approaches zero as R because MR 0 (by assumption)

Residues and Its Applications

13-50

example: find L

s
[F (s)] where F (s) = 2
and c > 0
(s + c2)2


C2 = z | z = a + Re ,

3
2

poles of F (s) are s = jc so we choose a > 0

the semicircle must enclose all the pole
so we have R > a + c

first we verifty that |F (s)| MR and MR 0 as s for s on C2

we note that |s| = |a + Rej | a + R and |s| |a R| = R a
since |s2 + c2| ||s|2 c2| (R a)2 c2 > 0, then
|F (s)| =
Residues and Its Applications

|s|
(R + a)

, MR 0
2
2
2
2
2
2
|s + c |
[(R a) c ]

as R
13-51

therefore, we can apply the theorem on page 13-46

L

[F (s)] =

sest
sest
+ Res 2
Res [e F (s)] = Res 2
s=sk
s=jc (s + c2 )2
s=jc (s + c2 )2
st

poles of F (s) are s = jc (double poles)


  st

st
st
d
se
e (1 + ts)
2se
st
Res e F (s) = lim
=

s=jc
sjc ds (s + jc)2
(s + jc)2
(s + jc)3 s=jc
tejct
=
j4c

  st

st
st
se
e
(1
+
ts)
2se
d
=

Res estF (s) = lim

s=jc
sjc ds (s jc)2
(s jc)2
(s jc)3 s=jc
tejct
=
j4c
hence L

[F (s)] =

Residues and Its Applications

t
jct
4jc (e

jct

t sin ct
)=
2c
13-52

example: find L

1
[F (s)] where F (s) =
(s + a)2 + b2

F (s) has poles at s = a jb (simple poles)

L1[F (s)] =

Res

s=a+jb

estF (s) +

Res

s=ajb

estF (s)

(provided that |F (s)| MR and MR 0 as s on C2 ... please check .)

e(a+jb)t
est
=
Res = lim
s=a+jb
s=a+jb s + a + jb
j2b
est
e(ajb)t
Res = lim
=
s=ajb
s=ajb s + a jb
j2b
at jbt
jbt
at
e
(e

e
)
e
sin(bt)
hence, L1[F (s)] =
=
2jb
b

Residues and Its Applications

13-53

References
Chapter 6-7 in
J. W. Brown and R. V. Churchill, Complex Variables and Applications, 8th
edition, McGraw-Hill, 2009
Chapter 7 in
T. W. Gamelin, Complex Analysis, Springer, 2001
Chapter 22 in
M. Dejnakarin, Mathematics for Electrical Engineering, CU Press, 2006

Residues and Its Applications

13-54