Calculus of Residue and Contour Integration

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Residue Theory

Definition.1. (Residue). Let f (z) has a non-removable isolated singularity at the point z0. Then f(z) has the Laurent series representation for all z in some punctured disk DR

*(z0) given

by f ( z )= ∑n=−∞

∞

an ( z−z0 )n

The coefficient a-1 of 1

z−z0 is called the residue of f(z) at z0 .

It is denoted by Res[f,z0] = a-1

Example .1.1.

Consider f (z) = e2/z

Then the Laurent series of f about the point z0 = 0is given by

¿1+ 21! z

+ 22

2 ! z2 +23

3! z3 +…,

The co-efficient of 1

z−z0 =

1z−0 =

1z is 2

Hence by definition of residue, residue of f (z) = e2/z at z0 = 0 is given by Res [f, z0] = 2

Example .1.2. Find residue of f (z) = 3

2 z+z2−z3 at z0 = 0

f(z) = 3

z (2+z−z2 ) = 3

z ( z+1 ) (2−z )

Now 3

z ( z+1 ) (2−z )= Az

+Bz+1

+ C

2−z

⇒A(z+1)(2-z) + Bz(2-z) + Cz (z+1) = 3

⇒A(-z2 +2 + z) +B(2z –z2) +C(z2 +z) = 3

⇒-A –B +C = 0---- (a)

A +2B +C = 0 ---- (b)

2A =3 ---- ( c) ⇒A = 3/2

2

(a) ⇒-B + C = A= 3/2

(b) ⇒2B + C = -A = -3/2

-----------------------------

Adding B + 2C =0⇒ B = - 2C

Put B = - 2C in -B + C = 3/2

⇒3C = 3/2⇒C= 1/2

Put C= 1/2, B = -2C = -1⇒ B = -1

Hence f(z) = = Az

+Bz+1

+ C

2−z =

32 z

- 1

z+1 +

12(2−z )

= 3

2 z - (1+z) -1 +

12

(2-z)-1 = 3

2 z - (1- z+ z2 -….) +

12

2-1(1− z2 )

−1

= 3

2 z - (1- z+ z2 -….) +

14

(1 + z2

+ ( z2 )2

+ ….)

= 3

2 z -

34

+9 z8

- …..

The residue of f at 0 is given by Res [f,0] = coefficient of 1z

= 32

Example 1.3. Find residue of f (z) = ez

z3 at z0 = 0

Laurent expansion of f(z) = 1z3 {1+z+ z2

2 !+ z3

3!…}

= 1

z3+ 1

z2+ 1z 2!

+ 13 !

…

The residue of f at 0 is given by Res [f,0] = coefficient of 1z

= 12

Contour integration

Contour integration is the process of calculating the values of a contour integral around a given

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contour in the complex plane.

The Cauchy integral formulae are useful in evaluating contour integrals over a simple closed

contour C where the integrand has the form f ( z )

( z−z0 )k and f is an analytic function

Example 1.

Evaluate ∫C

ydzalong the curve C : x = t-1, y = et-1, 2< t < 3

Solution.

Let z = x+iy ⇒ dz = dx + i dy

Given curve x = t-1 ⇒ dx = dt

And y = et-1 ⇒ dy = et-1dt

∫C

ydz = ∫2

3

et−1(dx+ idy) = ∫2

3

et−1(dt+i e t−1dt) = ∫2

3

et−1dt + ∫2

3

ie2 (t−1)dt

= e t−1 ] 32+ie2 t−2

2 ]32= e2 – e + ie4−e2

2 =

Example 2.

If C is the curve y = x3-3x2+4x-1 joining the points (1,1) and (2,3) then

find the value of ∫C

(12 z2−4 iz)dz

∫C

(12 z2−4 iz)dz = ∫1+i

2+3 i

(12 z2−4 iz)dz = 12 z3

3 - 4 i z2

2 ]2+3i1+i

= 4(2+3i)3 – 2i(2+3i)2 - 4(1+i)3 + 2i(1+i)2 = -156 +38i

Recall (i).(Cauchy’s integral Theorem)

Let D be any simply connected domain. Let C be any closed contour contained in D and f(z)

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analytic in D, then ∮C

f (z)dz = 0

Recall (ii).

For a function f(z) analytic in DR*(z0)and for any r with 0 < r < R, the Laurent series

coefficients of f(z) are given by

an=1

2πi∮C

f ( ξ )(ξ−z0 )n+1 dξfor n =0,±1 ,±2, ….. ------(I)

WhereC denotes the circle {z: |z-z0| = r} with positive orientation.

Put n = -1 in Equation (I) and replace C with any positively oriented simple closed contour C

containingz0, provided z0 is the still only singularity of f(z) that lies inside C,

then we obtain a-1 = 1

2πi∮C

f (z)dz.

We know that a-1 is the Res[f,z0]⇒ Res[f,z0] = 1

2πi∮C

f (z)dz

⇒∮C

f (z)dz = 2πi Res[f,z0]

If we know the Laurent series expansion for f(z), then using above equation we can evaluate contour integrals.

Example ii.1. Evaluate ∮C

e2z dzwhere C denotes the circle C = {z: |z| =1}with positive

orientation.

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Solution.

Let f(z) = e2z

From Example.1, we have Res [f, 0] = 2

Recall 2. gives us ∮C

f (z)dz = 2πi Res[f,z0]

Hence ∮C

e2z dz = 2πi Res[f,0] = 2πi (2) = 4πi

Theorem 1 (Cauchy's Residue Theorem).

Let D be a simply connected domain, and let C ⊂D be a closed positively oriented contour within and on the functionf(z) is analytic, except finite number of singular z1,z2,….,zn , then

∮C

f (z)dz = 2πi∑k=1

n

Res [ f , zk ]

Proof.

Let Ci be the neighborhood of zi, (i=1,2…n) lies inside C such that all Ciare disjoint.

Since each zi is a singular point of f and each Ci is a neighborhood of corresponding zi(i=1,2,..n), f is analytic in and on C except these neighborhoods Ci (i=1,2…n).

Then by Cauchy’s Theorem,(Recall 1)

∮C

f (z)dz - ∮C1

f (z)dz - …… - ∮Cn

f (z)dz = 0

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⇒∮C

f (z)dz = ∮C1

f (z)dz + …… + ∮Cn

f (z)dz

⇒∮C

f (z)dz = 2πiRes[f,z1] +….. + 2πi Res [f,zn] ( by Recall 2)

⇒ ∮C


n

Res [ f , zk ]

Note 1.1: The residue at z0depend only the coefficient a-1in the Laurent expansion, if f(z) has a

removable singularity at z0, then the Laurent expansion has no negative power term and

hencea-1=0 ⇒ Res[f,z0] = 0.

Theorem 2.(Residues at Poles).

(i) If f(z) has a simple pole at z0, then Res[f,z0] = lim ¿ z→ z0 ( z−z0 ) f (z)

(ii) If f(z) has a pole of order 2 at z0, then Res[f,z0] = lim ¿ z→ z0ddz

(( z−z0 )¿¿2¿¿ f (z ))¿¿

(iii) If f(z) has a pole of order 3 at z0, then Res[f,z0] = 12!

lim ¿ z→z0d2

d z2 (( z−z0 )¿¿3¿¿ f (z ))¿¿

(v) If f(z) has a pole of order k at z0,

then Res[f,z0] = 1

(k−1)!lim ¿ z→z0

dk−1

d zk−1 ( ( z−z0 )¿¿k¿¿ f (z))¿¿

Proof.

i) Suppose f(z) has a simple pole at z = z0, then the Laurent series expansion

f(z) = ∑n=0

∞

an ( z−z0 )n+a−1 ( z−z0 )−1

⇒ (z -z0)f(z) = (z-z0) ∑n=0

∞

an ( z−z0 )n+ a-1

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⇒ (z -z0)f(z) = (z-z0) ∑n=0

∞

an ( z−z0 )n + Res [f,z0]

Taking lim z→z0, both sides

¿ lim ¿ z→z 0 ( z−z0¿ f (z) = lim ¿ z→ z0 (z-z0) ∑n=0

∞

an ( z−z0 )n +lim ¿ z→ z0Res[f,z0]

= 0 + Res[f,z0]

Hence Res[f,z0] = ¿ lim ¿ z→z 0 ( z−z0¿ f (z)

v) Suppose f(z) has a pole of order k at z = z0, then the Laurent series expansion

f(z) = ∑n=0

∞

an ( z−z0 )n+a−1 ( z−z0 )−1+ a−2 ( z−z0)−2 + …… + a−k ( z−z0 )−k

Multiply both sides by (z-z0)k

(z-z0)kf(z) = a−k + ….. + a−1 ( z−z0 )k−1 + ∑n=0

∞

an ( z−z0 )n ( z−zo )k

Differentiate both sides k-1 times with respect to z,

dk−1

d zk−1 ( z−z0 )k f (z ) = 0+0+…..+ a−1 ( k−1 ) ! + dk−1

d zk−1 ¿

= a−1 ( k−1 ) ! + a0(z−z0)k ! + a1 ( z−z0 )2( (k+1 )!2! ) + ….

Taking lim z→z0, both sides

lim ¿ z→ z0dk−1

d zk−1 ( z− z0 )k f (z ) = lim ¿ z→ z0a−1 ( k−1 ) ! + 0 +…

= a−1 ( k−1 ) ! = Res [f,z0] (k-1)!

Hence Res [f,z0] = 1

(k−1 )!lim ¿ z→z0

dk−1

d zk−1 ( z−z0 )k f (z)

ii) and iii) are the particular case of v) (take k = 2 and k= 3)

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Example 2.1.

Find residue of f (z) = ez

z2−1 at z0 = 1

Solution.

Given f (z) = ez

z2−1 =

ez

(z−1)(z+1)

The poles of f(z) are z = 1 , z =-1 (simple poles)

Res [f,1] = lim ¿ z→1 ( z−1 ) f (z) = lim ¿ z→1( z−1 ) ez

( z−1 ) (z+1 ) = lim ¿ z→1

ez

( z+1 ) =

e2

Example 2.2 .

Find the residue of f(z) = (z8- ω8)-1, where ω is any complex constant.

Solution.

Given f(z) = (z8- ω8)-1 = 1

(z8−ω8)

The poles of f(z) are the zeros of z8-ω8 ⇒ zeros are given by z8-ω8 = 0

⇒ z8 = ω8 ⇒ z8 = ω8(cos2nπ +isin2nπ), n = 0,1,2,…7 ⇒ z8 = ω8 e2niπ

⇒ z = ω e2nπi/8 ⇒ z = ω enπi/4 , n = 0,1,2,…7

Hence z = ω enπi/4 , n = 0,1,2,…7 are the simple poles of f(z)

Let an = ω enπi/4, n = 0,1,2,…7

The residue of f (z) at z = an , is given by Res [f, an] = lim ¿ z→an ( z−an ) f (z )

=lim ¿ z→an ( z−an ) 1

( z8−an8)

Since it is not easy to factories (z8- an8) into eight factors , so we have to use L’Hospital’s rule

(that is differentiating Nr and Dr separately w.r.to z)

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= lim ¿ z→an1

8 z7 = 1

8an8 , n = 0,1,….7.

Example 2.3

Find the residue of 1

sinhπz

Solution.

Given f(z) = 1

sinhπz

The poles of f(z) are the zeros of sinhπz,

Also the zeros of sinhπz are z = ni, for all integer n (since sinhnπi = 0 for all n)

Hence Res [f,ni] = lim ¿ z→∋¿ ( z−¿ ) 1sinhπz

¿

By using L’Hospital’s rule

Res [f,ni] = lim ¿ z→∋¿1

πcoshπz¿

= 1

πcoshπni =

1πco sπn

(Since coshix= cosx)

= (-1)n/π (Since

Example 2.3. Find the residue of f(z) = π cot (πz )

z2 at z0 = 0

Solution.

Given f(z) = π cot (πz )

z2 = π cos (πz )z2 sin (πz )

Since z2 has a zero of order 2 at z0 = 0 and sin(πz) has a simple zero 1 at z0 = 0, we have

z2sin(πz) has a zero of order 3at z0 = 0 and πcos(πz) ≠ 0.

Hence f(z) has a pole of order 3 at z0 = 0 .

By part (iii) of Theorem .2, we have

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Res [f,0] = 12!

lim ¿ z→0d2

d z2 (( z−0 )¿¿3¿¿ f (z ))¿¿

= 12!

lim ¿ z→0d2

d z2 (z¿¿3¿¿π cot (πz )

z2 )¿¿

= 12!

lim ¿ z→0d2

d z2 (z¿¿ ¿¿ π cot (πz))¿¿ =

12!

lim ¿ z→0dd z

(¿¿¿ π cot (πz )−πzcsc2(πz) π)¿¿

= 12!

lim ¿ z→0dd z

(¿¿¿ π cot (πz )−π2 zcs c2(πz ))¿¿

=12!

lim ¿ z→0(¿−¿π csc2 (πz )π−π 2cs c2 (πz )+π2 z2csc (πz )csc (πz)cot (πz )π )¿¿

= 12!

lim ¿ z→0(¿−¿2π2 csc2 (πz )+2 π3 z csc2 (πz ) cot (πz ))¿¿

= 2π2

2 !lim ¿ z→0csc2(πz)(¿−¿1+π zcot (πz))¿¿

= π2 lim ¿ z→0

1

sin2 (πz )(¿−¿1+π z

cos (πz )sin (πz )

)¿¿

Res [f,0] = π2 lim ¿ z→0

(πzcos (πz )−sin (πz ))sin3 (πz )

As lim z → 0, LHS is indeterminate, so we have to use L’Hospital’s rule to evaluate the limit

(that is differentiating Nr and Dr separately w.r.to z)Res [f,0] = π

2 lim ¿ z→0(πcos (πz )−πzsin(πz)π−πcos (πz ))

3 sin2(πz )cos (πz )π

= π2 lim ¿ z→0

−π2 zsin(πz)3πsin2 (πz ) cos (πz )

= π2 lim ¿ z→0

−π z3sin (πz ) cos (πz )

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= π 2

3lim ¿ z→0

−πzsin (πz ) cos (πz )

= −π 2

3lim ¿ z→0 ( πz

sin (πz ) ) lim ¿ z→0( 1cos (πz ) )

= −π 2

31

lim ¿ z→0( sin (πz )πz )

lim ¿ z→0( 1cos (πz ) ) = −π 2

3(1)(1) = −π 2

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Example 2.4.

Find ∫C

dzz4+z3−2 z2 where C denotes the circle {z: |z| =3} with positive orientation.

Solution.

Let f(z) = 1

z4+z3−2 z2 = 1

z2(z¿¿2+z−2)¿ =

1

z2(z¿¿ +2)(z−1)¿

The singularities of f(z) that lie inside C are simple poles at the points z =1 and z= -2, and a pole of order 2 at z =0.

To find the Residue at z = 0 :

Res [f,0] =lim ¿ z→0ddz

(( z−0 )¿¿2¿¿ f (z ))¿¿ = lim ¿ z→0ddz

¿

= lim ¿ z→0ddz

¿= lim ¿ z→0ddz

¿

= lim ¿ z→0−2 z−1

(z2+z−2)2 = - 14

To find the Residue at z = 1 :

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Res [f,1] = lim ¿ z→1 ( z−1 ) f (z ) = lim ¿ z→1 ( z−1 ) 1

z2(z¿¿+2)(z−1)¿

= lim ¿ z→11

z2( z¿¿ +2)¿= ¿

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To find the Residue at z = -2 :

Res [f,-2] = lim ¿ z→−2 ( z+2 ) f (z ) = lim ¿ z→−2 ( z+2 ) 1

z2(z¿¿+2)(z−1)¿

= lim ¿ z→−21

z2(z−1) =

−112

By Cauchy’s residue theorem ∮C


n

Res [ f , zk ]

∫C

dzz4+z3−2 z2 = 2πi(Res[f,0] +Res[f,1]+Res[f,-2]) = 2πi(

−14

+ 13− 1

12¿ = 0

Example 2.5. Find ∫C

dzz4+4

where C denotes the circle {z: |z-1| =2 } with positive orientation

Solution.

Let f(z) = 1

z4+4

To find the poles of f(z) , we know that poles of f(z) is nothing but the zeros of z4 +4

Now we have to find the zeros of z4 +4

Put z4 +4 =0 ⇒z4 = -4 = 4i2 = (2i)2⇒ z2 = ±2i

Let z = a+ib ⇒ z2 = (a+ib)2 = a2 + 2iab – b2

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Supposez2= 2i ⇒a2–b2+ 2iab = 2i ⇒ a2 –b2=0 and ab = 1

⇒ a2 = b2and b = 1/a

⇒ a = ±b and b = 1/a

If a = b , then b = 1 ⇒ a,b=1

If a = -b , then b = -1⇒a = 1 , b = -1

The zeros are z = a+ib , 1+i, 1-i

Suppose z2= -2i ⇒a2 –b2+ 2iab = -2i ⇒ a2 –b2 =0 and ab = -1

⇒ a2 = b2 and b = -1/a

⇒ a = ±b and b = -1/a

If a = b , then b = -1 ⇒ a,b=-1

If a = -b , then b = 1⇒a = -1 , b =1

The zeros are z = a+ib ,-1-i, -1+i

Hence the poles of f(z) are 1±i, -1±i (simple poles)

The poles lie inside the circle {z: |z-1| =2} with positive orientation are 1±i

Res[f,1+i] = lim ¿ z→1+ i ( z−(1+i)) f (z) =lim ¿ z→1+ i (z−(1+i ) )( 1

z4+4)

As lim z →1+i, LHS is indeterminate, so we have to use L’Hospital’s rule to evaluate the limit

(that is differentiating Nr and Dr separately w.r.to z) = lim ¿ z→1+ i( 1

4 z3) =lim ¿ z→1+ i ( z

4 z4) =

1+ i4 (1+i )4

= 1+i

4 (−4 ) = 1+i−16

Similarly

Res[f,1-i] = 1−i−16

By Cauchy’s residue theorem ∮C


n

Res [ f , zk ]

14

∫C

dzz4+4

= 2πi(Res[f,1+i] +Res[f,1-i]) = 2πi(1+i−16

+ 1−i−16

¿ = - πi4

Result 3.

Let P(z) be a polynomial of degree at most 2. If a ,b and c are distinct complex numbers, then

f(z) = P ( z )

(z−a)(z−b)(z−c ) =

A(z−a) +

B(z−b) +

C(z−c)

Where A = Res [f,a] = P (a )

(a−b)(a−c)

B = Res [f,b] = P (b)

(b−a)(b−c )

C = Res [f,c] = P (c)

(c−a)(c−b)

Example 3.1.

Find the residue of f(z) = 3 z+2

z ( z−1 )(z−2) and express f(z) in partial fractions.

Solution.

In Result I, take a= 0, b = 1, c= 2 and P(z) = 3z+2.

The residues are

A = Res[f,0] = P(0)

(0−1)(0−2) = 1

B = Res [f,1] =P(1)

(1−0)(1−2) = -5

C = Res [f,2] = P(2)

(2−0)(2−1) = 4

The partial fraction expression of f(z)is given by

f(z) = A

(z−a) + B

(z−b) + C

(z−c)

= 1

(z−0) + −5

(z−1) + 4

(z−2)

15

= 1z

- 5

(z−1) + 4

(z−2)

Example 3.2. Find the residue of f (z) = 1

z4−1 and express in partial fractions.

f (z) = 1

z4−1 =

1

(z¿¿2−1)(z2+1)¿ =

1( z−1 )(z+1)(z+i)(z−i) =

A

z−1 +

B

z+1 +

C

z−i +

D

z+i

Where A = Res[f,1] = lim ¿ z→1(z−1) f (z) = lim ¿ z→1(z−1) 1( z−1 )(z+1)(z+i)(z−i)

= 1

2(1+i)(1−i) = 1

2(1−(−1)) = 14

B= Res[f,-1] = lim ¿ z→−1(z+1) f (z ) = lim ¿ z→−1(z+1) 1( z−1 )(z+1)(z+ i)(z−i)

= 1

−2(−1+i)(−1−i) = 1

2(−1−1)¿¿ =

−14

C= Res [f,i] = lim ¿ z→i(z−i) f (z ) = lim ¿ z→i(z−i) 1( z−1 )(z+1)(z+i)(z−i)

= 1

2i(i+1)(i−1) = 1

2i(−1−1) = −14 i

D=Res[f,-i] = = lim ¿ z→−i(z+i) f (z) = lim ¿ z→−i(z+i) 1( z−1 )(z+1)(z+i)(z−i)

= 1

−2i(−i+1)(−i−1) = 1

2i(1−(−1 )) = 14 i

f(z) = A

z−1 +

B

z+1 +

C

z−i +

D

z+i =

1

4 (z−1) -

14 (z¿¿+1)¿ -

14 i(z¿¿−i)¿ +

14 i(z¿¿ +i)¿

Result 4.

If a repeated root occurs in partial fraction, and P(z) has degree of at most 2, then f(z) = P ( z)

( z−a )2(z−b) =

A

( z−a )2 +

B(z−a)

+ C

(z−b)

16

Where A = Res [(z-a)f(z),a]

B = Res [f, a]

C = Res [f, b]

Example 4.1. Find the residue of f(z) = z2+3 z+2z2(z−1)

and express in partial fraction.

Solution.

In Result II, take a = 0, b = 1 and P(z) = z2+3 z+2 , we have

f(z) = P (z)

(z−0)2(z−1) =

A

( z−0 )2 +

B(z−0) +

C(z−1)

Where A = Res [(z-0)f(z),0] = Res [ z z2+3 z+2z2(z−1)

,0] = Res [z2+3 z+2z (z−1)

,0]

= lim ¿ z→0(z−0)( z2+3 z+2z (z−1)

)= lim ¿ z→0( z2+3 z+2(z−1)

) = -2

B = Res [f, 0] = lim ¿ z→0ddz

(z−0 )2 z2+3 z+2z2(z−1) = lim ¿ z→0

ddz

z2 z2+3 z+2z2(z−1) =

lim ¿ z→0ddz

z2+3 z+2(z−1)

= lim ¿ z→0[ (2 z+3 ) ( z−1 )−(1 ) ( z2+3 z+2 ) ]

( z−1 )2

¿

¿ = lim ¿ z→0[ ( z2−2 z−5 ) ]

( z−1 )2

¿

¿= -5

C = Res [f, 1] = Res [z2+3 z+2z2(z−1)

,1] = lim ¿ z→1(z−1)( z2+3 z+2z2(z−1)

)

= lim ¿ z→1( z2+3 z+2

z2 ) = 6

f(z) = −2

( z−0 )2 +

−5(z−0) +

6(z−1) =

−2

z2 + −5z

+ 6

(z−1)

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Example 4.2. Find the residue of f(z) = 1

( z−1 )2(z−3)

Take P(z) = 1, a= 1, b=3

f(z) = P(z )

( z−1 )2(z−3) =

A

( z−1 )2 +

B

( z−1 ) +

C(z−3)

A = Res [(z-1)f(z),1] = Res [ (z-1) 1

( z−1 )2(z−3),1¿ = lim ¿ z→1(z−1)( 1

(z−1)( z−3))

= 1

(1−3) = 1

−2

B = Res [f,1] = lim ¿ z→1ddz

( z−1 )2 1

(z−1)2( z−3) = lim ¿ z→1 −1(1)( z−3 )2

= −14

C = Res [f,3] = lim ¿ z→3(z−3)( 1

( z−1 )2(z−3)) = lim ¿ z→3

1

( z−1 )2 =

14

EVALUATION OF REAL DEFINITE INTEGRALS

Cases of poles are not on the real axis.

Type I

Evaluation of the integral ∫0

2π

f (cosθsinθ )dθ where f(cosθ sinθ) is a real rational function of

sinθ,cosθ.

First we use the transformation z = eiθ = cosθ + i sinθ ------ ( a)

And 1z

= 1

eiθ = e-iθ = = cosθ - i sinθ -------- (b)

From (a) and (b) , we have cosθ = 12(z+ 1

z) , sinθ =

12i

(z−1z)

Now z = eiθ ⇒ dz = ieiθdθ ⇒ dθ = dziz

18

Hence ∫0

2π

f (cosθsinθ )dθ = ∫C

f [¿ 12 (z+ 1

z ) , 12 i

(z−1z)] dziz

¿

Where C, is the positively oriented unit circle |z| = 1

The LHS integral can be evaluated by the residue theorem and

∫C

f [¿ 12 (z+ 1

z ) , 12 i

(z−1z)] dziz

¿ = 2πi ∑ Res(zi) , where zi is any pole in the interior of the

circle |z| =1

Example I.1.

Evaluate ∫0

2π

e−cosθ cos (nθ+sinθ )dθ ,where nis a positive integer .

Solution.

Let I = ∫0

2π

e−cosθ¿¿

= ∫0

2π

e−cosθ e−i (nθ+sinθ ) dθ = ∫0

2π

e−cosθ−isinθe−i (nθ ) dθ

= ∫0

2π

e−(cosθ+isinθ)e−i (nθ) dθ = ∫0

2π

e−e iθ

e−i (nθ )dθ

Let z = eiθ, dz = ieiθ dθ ⇒ dθ = dz

ie iθ = dziz

and C denotes the unit circle |z| = 1

Therefore I = ∫C

e− z

zn ( dzzi ) = 1i∫C

e−zdzzn+1 = ∫

C

f ( z )dz where f(z) = e− z

i zn+1

By Cauchy’s residue theorem, ∫C

f ( z )dz = 2πi∑ Res[f,zk] where zk are the singularities(poles) of

f(z)

To find the poles of f(z) :

Since poles of f(z) = to the zeros of izn+1 ,and the only zero of izn+1 is z = 0 of order n+1

19

Hence the pole of f(z) is z =0 of order n+1

There are no poles on the real axis

To find the residue of f(z):

Res [f,0] = 1n !

lim ¿ z→0dn

d zn( z−0 )n+1

f ( z) =1n !

lim ¿ z→0dn

d zn( z−0 )n+1 e−z

i zn+1

= 1n !

lim ¿ z→0dn

d znzn+1 e−z

i zn+1 = 1n !

lℑz→0

dn

d zne− z

i =

1n !

lim ¿ z→0 (−1 )n e−z

i

= (−1 )n

n ! i

Hence ∑ Res[f,zk] = Res[f,0]= (−1 )n

n ! i

Therefore ∫C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi (−1 )n

n ! i = 2π

(−1 )n

n!

⇒ I = 2π (−1 )n

n!

⇒ ∫0

2π

e−cosθ¿¿ = 2π (−1 )n

n!

⇒ ∫0

2π

e−cosθ cos (nθ+sinθ )dθ – i∫0

2π

e−cosθ sin(nθ+si nθ)¿ dθ ¿ = 2π (−1 )n

n!

Equating real and imaginary parts,

∫0

2π

e−cosθ cos (nθ+sinθ )dθ = 2π (−1 )n

n!

And ∫0

2π

e−cosθ sin(nθ+sinθ)¿dθ ¿ = 0

Example I.2.

Prove that ∫0

2πdθ

a+bcosθ =

2π

√a2−b2 , a >b >0.

20

Solution.

Let I = ∫0

2πdθ

a+bcosθ

Put z = eiθ ⇒ dθ = dziz

and let C denotes the unit circle |z| = 1

Since z = eiθ = cosθ + isinθ and 1z

= cosθ – isinθ, we have cosθ = 12

(z + 1z¿

I = ∫0

2πdθ

a+bcosθ =

1i∫C

dz

z (a+ b2 (z+ 1

z )) = 1i∫C

dz

z a+ z2b2

+ b2

= 2bi∫C

dz

z2ab

+z2+1

= ∫C

f ( z )dz where f(z) = 2

bi(z2+2azb

+1)



f(z).

To find the poles of f(z):

Poles of f(z) = to the zeros of bi(z2+2azb

+1) and the zeros are given by bi(z2+2azb

+1) = 0

⇒ (z2+2azb

+1) =0 -------- (a)

⇒ z = −2ab

±√ 4 a2

b2−4

2

= −2ab

±√ 4 a2−4b2

b2

2

= −2ab

±2√a2−b2

2b =

−a±√a2−b2

b

⇒ z = −a+√a2−b2

b or

−a−√a2−b2

b are the simple poles of f(z)

Let α = −a+√a2−b2

b and β = −a−√a2−b2

b , these the roots of the equation (a)

Now the product of the roots αβ = 11

= 1

21

Now | αβ| = 1 ⇒ |α||β| = 1

Since a > b > 0, |β| = |−a−√a2−b2

b | = |a+√a2−b2

b | Here a > b ⇒ a2 > b2 ⇒ a2- b2>0⇒ √a2−b2 > 0 ⇒ a + √a2−b2 > a >b

⇒ a + √a2−b2 >b ⇒ a+√a2−b2

b > 1 ⇒ |a+√a2−b2

b | > 1

Hence z = α = −a+√a2−b2

b < 1 is the only simple pole lie inside the circle |z| = 1



Res [f,α] = lim ¿ z→α (z−α) f (z )

Here f(z) = 2

bi(z2+2azb

+1) and α,β are the factors of z2+2azb

+1

⇒ f(z) = 2

bi(z2+2azb

+1) = 2

bi(z−α)(z−β)

∴ Res [f,α] = lim ¿ z→α (z−α) f (z ) = lim ¿ z→α (z−α) 2bi(z−α )(z−β ) =

lim ¿ z→α2

bi(z−β) = 2

bi(α−β)

=

2

bi((−a+√a2−b2

b )−(−a−√a2−b2

b )) = 2

bib

(−a+√a2−b2+a+√a2−b2 ) = 2

2i √a2−b2 =

1

i√a2−b2

Hence ∑ Res[f,zk] = Res[f,α]= 1

i√a2−b2

22

∴∫C

f ( z )dz = 2πi∑ Res[f,zk] = 2 πi

i√a2−b2 = 2π

√a2−b2

∴ ∫0

2πdθ

a+bcosθ =

2π

√a2−b2

Example I.3.

Prove that ∫0

2πdθ

1+a2−2acosθ= 2π

1−a2 , 0 ≤ a < 1.

Solution

Let I = ∫0

2πdθ

1+a2−2acosθ

Let z = eiθ ⇒ dθ = dziz




(z + 1z¿

∴ I = ∫0

2πdθ

1+a2−2acosθ = ∫

C

dziz

1+a2−2a( 12 )(z+ 1

z ) =

1i∫C

dzz

(1+a2 )−a ( z2+1 )z

=

1i∫C

dzz

(1+a2 ) z−a ( z2+1 )z

= 1i∫C

dz

(1+a2 ) z−a ( z2+1 ) =

1i∫C

dzz+a2 z−az2−a

= −1ai

∫C

dz−za

−az+z2+1 =

−1ai

∫C

dz

(−za

+z2)+(−az+1)

23

= −1ai

∫C

dz

z (−1a

+z)−a(z−1a) =

−1ai

∫C

dz

(−1a

+z )(z−a) = ∫

C

f ( z )dz

Where f(z) =

−1

ai(−1a

+z )(z−a)



f(z).


Poles of f(z) = zeros of ai(−1a

+ z)(z−a) and

these zeros are given by ai(−1a

+ z)(z−a) = 0

⇒ (−1a

+z ) = 0 or (z−a) = 0

⇒ z = 1a

or z = a which are the simple poles of f(z)

Since 0≤ a < 1, 1a

> 1

Hence a<1 is the only pole lie inside the unit circle |z| =1



Res[f,a] = lim ¿ z→a(z−a) f (z) = lim ¿ z→a(z−a) −1

ai(−1a

+z )(z−a)

= lim ¿ z→a

−1

ai(−1a

+z) =

−1

ai(−1a

+a) = −1

i (−1+a2 ) = 1

i (a2−1 )

24

Hence ∑ Res[f,zk] = Res[f,a] = 1

i (a2−1 )

∴∫C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi

i (a2−1 ) = 2 π

(a2−1 )

⇒ I = 2 π

(a2−1 )

⇒ ∫0

2πdθ

1+a2−2acosθ =

2 π

(a2−1 )

Example I.4.

Evaluate ∫0

πadθ

a2+sin2θ

Solution.

First we have to change the limits to 0 to 2π from 0 to π for the given integral.Let I = ∫

0

πadθ

a2+sin2θ = ∫

0

πadθ

a2+ 1−cos 2θ2

= ∫0

π2adθ

2a2+1−cos 2θ

To change the limit , take 2θ = ϕ ⇒ 2dθ = dϕ If θ =0 , then ϕ = 0If θ= π, then ϕ = 2πHence I = ∫

0

2πadϕ

2a2+1−cosϕ

Put z = eiϕ ⇒ dϕ = dziz


Since z = eiϕ = cosϕ + isinϕ and 1z

= cosϕ – isinϕ, we have cosϕ = 12

(z + 1z¿

∴ I = ∫0

2πadϕ

2a2+1−cosϕ = ∫

C

adziz

2a2+1−12(z+

1z) = ∫

C

adziz

4 za2+2 z−( z2+1 )2 z

25

= 2ai∫C

dz4 z a2+2 z−z2−1

= −2ai

∫C

dz

z2−2 z (2a2+1 )+1 = ∫

C

f ( z )dz

Where f(z) = −2a

i( z¿¿2−2 z (2a2+1 )+1)¿



f(z).


Poles of f (z) = zeros of i(z2-2z(2a2+1)+1) , these zeros are given by i(z2-2z(2a2+1)+1) = 0

⇒ z2-2z(2a2+1)+1 = 0

⇒ z = 2(2a2+1)±√(2 (2a2+1 ))2−4 (1)(1)

2 (1) = 2(2a2+1)±2√( (2a2+1 ))2−1

2

= 2a2+1±√(2a2+1 )2−1 = 2a2+1±√4a4+1+4 a2−1

= 2a2+1±√4a4+4 a2 = 2a2+1±2a√a2+1

∴ z = 2a2+1+2a√a2+1 = α (say)

Or z = 2a2+1−2a √a2+1 = β (say)

Hence the poles of f(z) are α,β which are simple poles.

Now α,β are the roots of the equation z2-2z(2a2+1)+1 = 0

Product of the roots αβ = 1 ⇒ | αβ| = 1⇒|α||β| = 1

Clearly |α| = |2a2+1+2a√a2+1 | > 1 ⇒ |β| < 1

∴ the only pole lie inside the unit circle |z| =1 is β = 2a2+1−2a √a2+1


To find the residue of f(z) :

Res[f,β] = lim ¿ z→ β(z−β) f (z ) = lim ¿ z→ β(z−β) −2a

i(z¿¿2−2 z ( 2a2+1 )+1)¿

26

= lim ¿ z→ β(z−β) −2a

i(z−α)(z−β)= lim ¿ z→ β

−2a

i(z−α) =

−2a

i( β−α ) =

−2a

i(2a2+1−2a√a2+1−(2a2+1+2a√a2+1 ))

= −2a

i(2a2+1−2a√a2+1−2a2−1−2a√a2+1) =

−2a

i(−4 a√a2+1) =

1

2i √a2+1¿¿

Hence ∑ Res[f,zk] = Res[f,β] = 1

2i √a2+1¿¿

∴∫C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi

2i √a2+1¿¿ =

π

√a2+1¿¿

⇒ I = ∫0

πadθ

a2+sin2θ = π

√a2+1¿¿

Example I.5.

Evaluate ∫0

2πdθ

(a+bcosθ )2 ( a > 0, b > 0 ; a > b)

Solution

Let I = ∫0

2πdθ

(a+bcosθ )2

Take z = eiθ ⇒ dθ = dziz




(z + 1z¿

⇒ I = ∫C

dziz

(a+ b2(z+1

z))

2 = ∫C

dziz

( 2az+b ( z2+1 )2 z )

2 = 4i∫C

zdz

(2az+b(z2+1))2 =

4i∫C

zdz

( 2az+bz2+b¿)2

27

= ∫C

f ( z )dz where f(z) = 4 z

i (2az+b z2+b¿)2



f(z).


Poles of f(z) = zeros of i(2az+bz2+b)2 , these zeros are given by i(2az+bz2+b)2 = 0

⇒ bz2+2az +b = 0 ⇒ z2+2azb

+1 = 0

⇒ z = −2ab

±√(2ab

)2

−4

2 = −2a±√4 a2−4b2

2b = −a±√a2−b2

b

⇒ z = −a+√a2−b2

b = β (say)

Or z = −a−√a2−b2

b = α (say)

Hence the poles of f(z) are α,β both order 2

Since α,β are the roots of the equation bz2+2az +b = 0

Product of the root αβ = b/b = 1 ⇒ | αβ| = 1 ⇒ |α||β| = 1

Given a > b ⇒ a2 > b2 ⇒ a2- b2>0⇒ √a2−b2 > 0 ⇒ a + √a2−b2 > a >b

⇒ a + √a2−b2 >b ⇒ a+√a2−b2

b > 1 ⇒ |a+√a2−b2

b | > 1

Hence z = β = −a+√a2−b2

b < 1 is the only pole lie inside the circle |z| = 1


To fine the residue of f(z):

28

Res [f,β] = lim ¿ z→ βddz

( z−β )2 f (z ) = lim ¿ z→ β

ddz

( z−β )2 4 z

ib2( 2azb

+z2+1¿)2

= lim ¿ z→ βddz

( z−β )2 4 z

ib2¿¿¿¿ = lim ¿ z→ β

ddz

( z−β )2 4 z

ib2 ( z−α )2 ( z−β )2

= lim ¿ z→ βddz

4 z

i b2 ( z−α )2 = lim ¿ z→ β

4

ib2

ddz

z

( z−α )2 = lim ¿ z→ β

4

ib2 [ ( z−α )2−z 2 ( z−α )( z−α )4 ] =

= lim ¿ z→ β4

ib2(z−α )[ ( z−α )−z 2

( z−α )4 ] = lim ¿ z→ β4

ib2 [ ( z−α )−z2

( z−α )3 ] = 4

ib2 [ (β−α )−2β

(β−α )3 ]

= −4

i b2 [ α+β(β−α )3 ] =

−4

i b2 [ α+β(β−α )3 ] =

−4i b2 [ −a−√a2−b2

b+−a+√a2−b2

b

(−a+√a2−b2

b−−a−√a2−b2

b )3 ] =

−4i b2 [ −2a

b

(−a+√a2−b2

b+ a+√a2−b2

b )3 ]

=−4i b2 [ −2a

b

( 2√a2−b2

b )3 ] = [ 8ab3

8b3i (√a2−b2 )3 ] = a

i (√a2−b2 )3

Hence ∑ Res[f,zk] = Res[f,β] = a

i (√a2−b2 )3

∴∫C

f ( z )dz = 2πi∑ Res[f,zk] = 2πia

i (√a2−b2 )3 = 2πa

(√a2−b2 )3

⇒ I = ∫0

2πdθ

(a+bcosθ )2 =

2πa

(√a2−b2 )3

Type II.

29

Evaluation of the integral ∫−∞

∞

f ( x )dx where f(x) is a real rational function of the real

variable x.

If the rational function f(x) = g (x)h(x )

, then degree of h(x) exceeds that of g(x) and g(x) ≠ 0.To

find the value of the integral, by inventing a closed contour in the complex plane which includes the required integral. For this we have to close the contour by a very large semi-circle in the upper half-plane. Suppose we use the symbol “R” for the radius. The entire contour integral comprises the integral along the real axis from −Rto +Rtogether with the integral along the semi-circular arc. In the limit as R→∞the contribution from the straight line part approaches the required integral, while the curved section may in some cases vanish in the limit.

The poles z1,z2,….,zk of g (x)h(x )

, that lie in the upper half-plane

∫−∞

∞

f ( x )dx = ∫−∞

∞g(x )h (x)

dx = 2πi∑ Res[f,zk]

Example II.1

Using the residue of calculus compute ∫−∞

∞dx

(x2+1 ) ( x2+4 )

Solution

Consider the integral ∫C


( z2+1 ) ( z2+4 )

To find the poles of f(z) :

The poles of f(z) = zeros of (z2+1)(z2+4) , these zeros are given by (z2+1)(z2+4) = 0

⇒ z2+1 = 0 or z2+4 = 0

30

⇒ z2 = -1 = i2 ⇒ z = ± i

Or z2 = -4 = (2i)2⇒ z = ± 2i

Hence the poles of f(z) are ± i, ± 2i (all are simple poles)

And the poles z = i and z = 2i are the only poles lie inside the upper half of semi-circle.




f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R

f ( z )dz ------------- (a) ( on the real line –R to R (LR)+ the upper

half of the semi circle CR)


Res[f,i] = lim ¿ z→i(z−i) f (z ) = lim ¿ z→i(z−i) 1

( z2+1 ) ( z2+4 ) = = lim ¿ z→i(z−i) 1

( z+i )¿¿¿

= lim ¿ z→i1

( z+i ) (z+2 i)( z−2i) =

1

(i+ i) (i+2i)(i−2i) =

1(2 i)(3 i)(−i) =

16 i

= -i6

Res[f,2i] = lim ¿ z→2i(z−2 i) f (z) = lim ¿ z→2i(z−2 i) 1

( z2+1 ) ( z2+4 ) = =

lim ¿ z→2i(z−2 i) 1

( z+ i)¿¿¿

= lim ¿ z→2i1

(z−i) ( z+i )(z+2i) =

1(2i−i )(i+2 i)(2 i+2i) =

1(i)(3i)(4 i) =

−112i

= i

12

31

Consider |∫CR

f ( z )dz| = |∫CR

1

( z2+1 ) ( z2+4 )dz|≤ ∫

C R

¿ dz

( z2+1 ) ( z2+4 )∨¿≤

∫C R

¿dz∨ ¿¿ ( z2+1 ) ( z2+4 )∨¿

¿¿

≤ ∫C R

¿dz∨ ¿¿¿ ¿¿ ---------(b)

Let z = Reiθ, dz = iReiθdθ⇒ |dz| = |iReiθdθ| = R dθ (∵ |i| =1 =|eiθ|)If z = -R, then Reiθ = -R ⇒ eiθ = -1⇒ θ = πIf z = R, then Reiθ = R ⇒ eiθ = 1⇒ θ = 0Hence (b) ⇒ |∫CR

f ( z )dz|≤ ∫0

πRdθ

(R2−1 ) (R2−4 ) =

R

(R2−1 ) (R2−4 ) ∫0π

dθ = Rπ

(R2−1 ) (R2−4 )

As R → ∞, Rπ

(R2−1 ) (R2−4 ) → 0 ⇒ ∫C R

f ( z )dz → 0

Now as R→ ∞, (a) ⇒ ∫C

f ( z )dz = ∫−∞

∞

f ( x )dx + 0 =∫−∞

∞

f ( x )dx

Where f(x) = 1

(x2+1 ) (x2+4 )

Hence ∫−∞

∞1

(x2+1 ) ( x2+4 )dx = ∫

−∞

∞

f ( x )dx = ∫C

f ( z )dz = 2πi∑ Res[f,zk]

= 2πi{Res[f,i] + Res[f,2i]} = 2πi{−i6

+ i

12 }=2πi(

−i12

¿= π6

Example II.2

Using the residue of calculus compute ∫−∞

∞dx

(x2+4 )3

Solution

32



( z2+4 )3


Poles of f(z) = zeros of (z2+4)3, these zeros are given by (z2+4)3 = 0

⇒ z2 +4 = 0 ⇒ z2 = -4 ⇒ z2 = (2i)2⇒ z = ±2i ⇒ z = 2i or z = -2i Hence the poles of f(z) are z = 2i , z = -2i , both of order 3

The only pole lie inside the upper half of the semi-circle is z = 2i of order 3




f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R

f ( z )dz ( on the real line –R to R (LR)+ the upper half of the semi

circle CR)


Res[f,2i] = 1

(2 )!lim ¿ z→2 i

d2

d z2 ( z−2 i )3 f (z ) = 12

lim ¿ z→2 id2

d z2 ( z−2 i )3 1

( z2+4 )3

= 12

lim ¿ z→2 id2

d z2 ( z−2 i )3 1¿¿¿ ¿ =

12

lim ¿ z→2 id2

d z2 ( z−2 i )3 1

( z−2 i )3 ( z+2 i )3

= 12

lim ¿ z→2 id2

d z2

1

( z+2 i )3 =

12

lim ¿ z→2 idd z

dd z ( 1

( z+2 i )3 ) = 12

lim ¿ z→2 idd z (−3 ( z+2i )2

( z+2i )6 )

33

= 12

lim ¿ z→2 idd z ( −3

( z+2 i )4 ) = 12

lim ¿ z→2 i( 3 (4 ) ( z+2 i )3

( z+2i )8 ) = 122

lim ¿ z→2 i( 1

( z+2i )5 )= 6( 1

(2 i+2 i )5 ) = 6

(4 i )5 = 6

1024 i =

3

512i

Consider |∫CR

f ( z )dz| = |∫CR

1

( z2+4 )3dz|≤ ∫

C R

¿ dz

( z2+4 )3∨¿≤ ∫

C R

¿dz∨ ¿¿ ( z2+4 )3∨¿

¿¿

≤ ∫C R

¿dz∨ ¿¿¿¿ ¿¿ ---------(b)


f ( z )dz|≤ ∫0

πRdθ

(R2−4 )3 =

R

(R2−4 )3 ∫0

π

dθ = Rπ

(R2−4 )3

As R → ∞, Rπ

(R2−4 )3 → 0 ⇒ ∫C R

f ( z )dz → 0

Now as R→ ∞, (a) ⇒ ∫C

f ( z )dz = ∫−∞

∞

f ( x )dx + 0 =∫−∞

∞

f ( x )dx

Where f(x) = 1

(x2+4 )3

Hence ∫−∞

∞1

(x2+4 )3dx = ∫

−∞

∞

f ( x )dx = ∫C


= 2πi{Res[f,2i]} = 2πi{ 3

i512 }=

3π256

Example II.3

34

Prove that ∫−∞

∞ (x¿¿2−x+2)dx(x4+10 x2+9 )

¿ = 5π12

Solution


f ( z )dz where f(z) = (z¿¿2−z+2)( z4+10 z2+9 )

¿


Poles of f(z) = zeros of z4+10z2+9 , these zeros are given by z4+10z2+9 =0

⇒ z4+z2+9z2+9=0 ⇒ (z2+1)(z2+9) = 0

⇒ z2 = -1 = i2 or z2 = -9 = (3i)2

⇒ z = ±i or z = ±3iHence the poles of f(z) are i,-i,3i,-3i (all are simple poles)

The poles that are lying the upper half of the semi-circle are i,3i




f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R

f ( z )dz ( on the real line –R to R (LR)+ the upper half of the

semi-circle CR)


35

Res[f,i] = lim ¿ z→i(z−i) f (z ) = lim ¿ z→i(z−i)(z¿¿2−z+2)( z4+10 z2+9 )

¿ =

lim ¿ z→i(z−i)(z¿¿2−z+2)

( z−i¿(z+ i)(z+3 i)(z−3 i)¿

= lim ¿ z→i(z¿¿2−z+2)

(z+ i)(z+3 i)(z−3 i)¿ =

(i¿¿2−i+2)(i+i)(i+3 i)(i−3 i)

¿ = 1−i

(2 i)(4 i)(−2 i) = 1−i1 6 i

Res[f,3i] = lim ¿ z→3i(z−3i) f (z ) = lim ¿ z→3i(z−3i)(z¿¿2−z+2)( z4+10 z2+9 )

¿

= lim ¿ z→3i(z−3i)(z¿¿2− z+2)

( z−i¿(z+i)(z+3i)(z−3 i)¿ = lim ¿ z→3i

(z¿¿2−z+2)( z−i¿(z+i)(z+3 i)

¿ =

((3 i)¿¿2−(3 i)+2)(3 i−i¿(3 i+ i)(3 i+3i)

¿

= −7−3 i

(2i¿(4 i)(6 i) = −(7+3i)−48 i

= 7+3 i48i

Consider |∫CR

f ( z )dz| = ¿≤ ∫C R

¿( z¿¿2−z+2)dz

( z4+10 z2+9 )∨¿¿≤

∫C R

¿(z¿¿2−z+2)∨¿dz∨ ¿¿ ( z4+10 z2+9 )∨¿

¿¿¿

≤ ∫C R

¿(z¿¿2−z+2)∨¿dz∨ ¿¿¿¿ ¿¿¿ ≤ ∫

C R

(¿ z∨¿2−¿ z∨+2)∨dz∨ ¿¿¿¿ ¿¿¿ ---------(b)


f ( z )dz|≤ ∫0

πR2dθ

(R2−1 )(R2−9) - ∫

0

πRdθ

(R2−1 )(R2−9) + ∫

0

π2dθ

(R2−1 )(R2−9) =

R2

(R2−1 )(R2−9)∫0

π

dθ - R

(R2−1 )(R2−9)∫0

π

dθ + 2

(R2−1 )(R2−9)∫0

π

dθ

36

= R2π

(R2−1 )(R2−9) -

Rπ

(R2−1 )(R2−9) +

2π

(R2−1 )(R2−9)

As R → ∞, R2π

(R2−1 )(R2−9) → 0 ,

Rπ

(R2−1 )(R2−9) → 0 and

2π

(R2−1 )(R2−9) → 0 ⇒ ∫

C R

f ( z )dz

→ 0

Now as R→ ∞, (a) ⇒ ∫C

f ( z )dz = ∫−∞

∞

f ( x )dx + 0 =∫−∞

∞

f ( x )dx

Where f(x) = (x¿¿2−x+2)(x4+10 x2+9 )

¿

Hence ∫−∞

∞ (x¿¿2−x+2)(x4+10x2+9 )

dx ¿ = ∫−∞

∞

f ( x )dx = ∫C


= 2πi {1−i1 6 i

+ 7+3 i48 i } = 2πi {3−3i+7+3i

48i } =10π24

= 5π12

Example II.4

Evaluate ∫0

∞dx

x4+a4

Solution

Let us take ∫−∞

∞dx

x4+a4



z4+a4


Poles of f(z) = zeros of z4+a4 , these zeros are given by z4+a4 =0

⇒ z4 = - a4 ⇒ z4 = a4eiπ (∵ eiπ = -1)

⇒ z4 = a4eiπ ei2nπ (∵ ei2nπ = 1)

⇒ z4 = a4eiπ+2nπi = a4ei(2n+1)π⇒ z = a ei(2n+1)π/4 , n = 0,1,2,3

37

If n=0, z = a eiπ/4 = a( cosπ4+i sin

π4

¿ = a( 1

√2+ i

√2¿ = α (say)

If n = 1, z = a ei3π/4 = a( cos3π4

+i sin3 π4

¿ = a( −1

√2+ i

√2¿ = β (say)

If n = 2, z = a ei5π/4 = a( cos5π4

+i sin5 π4

¿ = -a( 1

√2+ i

√2¿ = γ(say)

If n=3, z = a ei7π/4 = a( cos7 π4

+i sin7 π4

¿ = a( 1

√2− i

√2¿ = δ(say)

The poles lying inside the upper hemi circle are aeiπ/4 =α , aei3π/4 = β (both are simple poles)There are no poles on the real axis



f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R


semi-circle CR)


Res[f,α] =lim ¿ z→α (z−α) f (z ) = lim ¿ z→α (z−α) 1

z4+a4 It is difficult to solve while factoring 1

z4+a4 and taking limit, so we will use L’Hospital rule (that is differentiating Nr and Dr separately w.r.to z)= lim ¿ z→α

1

4 z3 = 1

4 α3 = α

4 α4 Now α = a eiπ/4 ⇒ α4 = a4eiπ ⇒ α4 =- a4 (∵eiπ = -1)

38

∴ Res [f,α] = −α

4 a4 = - ae iπ4

4a4 = - e iπ

4

4 a3

Now β = a ei3π/4 ⇒ β4 = a4ei3π ⇒ β4 =- a4 (∵ei3π = -1)Similarly , Res [f,β] = 1

4 β3 = β

4 β4 = - ae 3iπ4

4a4 = - e 3iπ

4

4 a3

Consider |∫CR

f ( z )dz| = |∫CR

1

z4+a4 dz|≤ ∫

C R

¿ dzz4+a4 ∨¿≤ ∫

C R

¿dz∨ ¿¿ z 4+a4∨¿

¿¿

≤ ∫C R

¿dz∨ ¿¿ z∨¿4−a4 ¿¿ ---------(b)


f ( z )dz|≤ ∫0

πRdθ

(R4−a4 ) ≤

R

(R4−a4 )∫0

π

dθ ≤ Rπ

(R4−a4 )

As R → ∞, Rπ

(R4−a4 )→ 0 ⇒ ∫C R

f ( z )dz → 0

Now as R→ ∞, (a) ⇒ ∫C

f ( z )dz = ∫−∞

∞

f ( x )dx + 0 =∫−∞

∞

f ( x )dx

Where f(x) = 1

(x4+a4 )

Hence ∫−∞

∞1

(x4+a4 )dx = ∫

−∞

∞

f ( x )dx = ∫C


= 2πi { - e iπ4

4 a3 - e i3π

4

4a3 } = - 2πi

4 a3 { e iπ4 +¿ e i3π

4 } = - πi

2a3 {( 1

√2+ i

√2¿ +( −1

√2+ i

√2¿ }

39

= 2π

2√2a3 = π

√2a3

We know that ∫−∞

∞1

(x4+a4 )dx = 2∫

0

∞1

(x4+a4 )d x

⇒ ∫0

∞1

(x4+a4 )dx =

12∫−∞

∞1

(x4+a4 )dx =

π

2√2a3

Type III.

Evaluation of the integral ∫−∞

∞

f ( x ) sinmxdx , ∫−∞

∞

f ( x ) cosmxdx where m > 0 and f(x) is a real

rational function of the real variable x.

If the rational function f(x) = g (x)h(x )

, then degree of h(x) exceeds that of g(x) and g(x) ≠ 0.

Let g(x) and h(x) be polynomials with real coefficients, of degree p and q, respectively, where q ≥ p+1.

If h(x) ≠0 for all real x, and m is a real number satisfying m > 0, then

∫−∞

∞g(x )h (x)

cos mxdx= limR→∞

∫−R

Rg (x)h(x )

cosmxdx and ∫−∞

∞g(x )h (x)

sinmx dx= limR→∞

∫−R

Rg (x)h(x )

sinmxdx

We know that Euler’s formula e imx = cos mx + i sin mx , where cos mx = Re[e imx]

and sin mx = Im[e imx] , m is a positive real.

We have ∫−∞

∞g(x )h (x)

eimx dx = ∫−∞

∞g(x )h (x)

cos mxdx + i ∫−∞

∞g(x )h (x)

sinmx dx

Here we are going to use the complex function f(z) = g (z)h(z )

eimz to evaluate the given integral.

∫−∞

∞g(x )h (x)

cos mxdx = Re {2πi ∑ Res[f,zk]} and

40

∫−∞

∞g(x )h (x)

sinmx dx = Im {2πi ∑ Res[f,zk]}, where z1,z2,…..zk are the poles lies on the upper half

of the semi-circle.

Lemma III.1.(Jordan’s Lemma)

If f (z) → 0 uniformly as z →∞, then limR→∞∫C1

e imz f (z )dz = 0, (m > 0) where C1 denotes the semi-

circle |z| = R, I(z) > 0.

Proof.

Given f (z) → 0 uniformly as z →∞⇒ given 𝜀 > 0 , ∃ a R0 > 0 such that | f(z) – 0| < 𝜀 , ∀ R ≥ R0 That is | f(z)| < 𝜀 , ∀ R ≥ R0 --------(a)Let |z| = R which is the semi-circle Put z = Reiθ ⇒ dz = R eiθ i dθ ⇒ dz = izdθ, 0≤ θ ≤ πNow e imz = eℑ eiθ = e imR(cosθ+isinθ) = e imRcosθ−mRsinθ = e imRcosθ e−mRsinθ

⇒ ¿e imR(cosθ+isinθ) | = ¿e imRcosθ∨¿ ¿e−mRsinθ∨¿ = e−mRsinθ ( ∵¿e imRcosθ∨¿ = 1) ----- (b)

We know that sinθθ

is monotonically decreases as θ increases from 0 to π2

.

If 0 ≤ θ ≤ π2

, then sin ( π

2)

π2

≤ sinθθ

41

⇒ 1π2

≤ sinθθ

⇒ 2π

≤ sinθθ

⇒ sinθ ≥ 2θπ

⇒ - 2θπ

≥ - sinθ⇒ -

mR2θπ

≥ - mRsinθ⇒ e−mR2θ

π ≥ e−mRsinθ ------ ( c)

From (a), (b) and (c),

|∫C1

e imz f ( z )dz| ≤ ∫C1

¿ eimz f ( z )dz∨¿

≤∫C1

¿ eimz∨|f ( z )|∨dz∨¿

≤ ε∫0

π

e−mRsinθ Rdθ ( ∵ |dz| = |izdθ| = |z|dθ = Rdθ) ≤ 2𝜀R ∫

0

π2

e−m Rsinθ

dθ

≤ 2𝜀R ∫0

π2

e−mR2θ

π dθ

≤ 2𝜀R [ e−mR2θ

π

−2mRπ

] π20

≤ 2𝜀R{(e¿¿−mR ) π−2mR

¿ - (e0) π−2mR

} ≤ 2 εRπ

2mR{¿¿

≤ επm

{−e−mR+1 } ≤ επm

(∵{−e−mR+1 }<1¿

42

|∫C1

e imz f ( z )dz| ≤ επm

= 𝜀’ (say)As lim R → ∞ , |∫C1

e imz f ( z )dz−0| ≤ 𝜀’

⇒ ∫C1

eimz f ( z )dz → 0 as R → ∞Example III.1.Use the method of contour integration to prove that ∫

0

∞cosmx

(x2+a2 )dx =

π e−ma

2a and

∫0

∞sinmx

(x2+a2 )dx = 0

Solution.


f ( z )dz where f(z) = eimz

( z2+a2 )


Poles of f(z) = zeros of z2+a2 , these zeros are given by z2+a2 =0

⇒ z2 = - a2 ⇒ z = (ai)2 ⇒ z = ± ai ⇒ z = ai or z = -ai

Poles of f (z) are ai, -ai (both are simple poles)There are no poles on the real axis

The only pole lie inside the upper half of semi-circle is z = ai



f(z).

43

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R


semi-circle CR).


Res[f,ai] = lim ¿ z→ai(z−ai) f (z ) = lim ¿ z→ai(z−ai) eimz

( z2+a2 ) =

lim ¿ z→ai(z−ai) eimz

(z−ai¿(z+ai)

= lim ¿ z→aie imz

(z+ai) =

e imai

(ai+ai) =e−ma

2ai

Now lim ¿ z→∞1

( z2+a2 ) = 0

∴ by Jordan’s lemma , lim ¿R→∞∫CR

eimz

( z2+a2 )dz = 0

As R → ∞ ,

∫C

f ( z )dz = ∫−∞

∞

f ( x )dx + lim ¿R→∞∫CR

eimz

( z2+a2 )dz = ∫

−∞

∞eimx

(x2+a2 )dx +0

= ∫−∞

∞e imx

(x2+a2 )dx

Hence ∫−∞

∞e imx

(x2+a2 )dx = ∫

C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi e−ma

2ai = π e

−ma

a

⇒ ∫−∞

∞cosmx+isinmx

(x2+a2 )dx =

π e−ma

a

⇒ ∫−∞

∞cosmx

(x2+a2 )dx + i∫

−∞

∞sinmx

(x2+a2 )dx =

π e−ma

a


∫−∞

∞cosmx

(x2+a2 )dx = π e

−ma

a and ∫

−∞

∞sinmx

(x2+a2 )dx = 0

44

⇒ ∫0

∞cosmx

(x2+a2 )dx =

12∫−∞

∞cosmx

(x2+a2 )dx =

π e−ma

2a

Example III.2.Apply the calculus of residue to evaluate ∫

−∞

∞cosx

(x2+a2 )(x2+b2)dx , (a > b > 0)

Solution.Consider the integral ∫

C

f ( z )dz where f(z) = e iz

( z2+a2 )(z2+b2)


Poles of f(z) = zeros of (z2+a2) (z2+b2), these zeros are given by (z2+a2) (z2+b2) =0

⇒ z2 = - a2 or z2

= - b2 ⇒ z = (ai)2 or z = (bi)2 ⇒ z = ± ai or z = ± bi ⇒ z = ai or z = -ai or z =bi or z = -biPoles of f (z) are ai,-ai, bi,-bi (all are simple poles)There are no poles on the real axis.

The poles lie inside the upper half of semi-circle are z = ai , z = bi



f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R


semi-circle CR).


45

Res[f,ai] = lim ¿ z→ai(z−ai) f (z ) = lim ¿ z→ai (z−ai ) e iz

( z2+a2 )(z2+b2)

=lim ¿ z→ai(z−ai) e iz

( z+ai )(z−ai)(z+bi)(z−bi) = lim ¿ z→ai

e iz

( z+ai )(z+bi)(z−bi)

= e iai

(ai+ai )(ai+bi)(ai−bi) =

−e−a

2ai(a2−b2) =

e−a

2ai(b2−a2)

Similarly,

Res[f,bi] = lim ¿ z→bi(z−bi) f (z) = e−b

2bi(a2−b2)

Now lim ¿ z→∞ 1

( z2+a2 )(z2+b2) = 0

By Jordan’s Lemma,

lim ¿R→∞∫CR

eiz

( z2+a2 )(z2+b2)dz = 0

As R → ∞ ,

∫C

f ( z )dz = ∫−∞

∞


eiz

( z2+a2 ) ( z2+b2 )dz = ∫

−∞

∞eix

(x2+a2 ) (x2+b2 )dx +0

= ∫−∞

∞eix

(x2+a2 ) (x2+b2 )dx

Hence ∫−∞

∞eix

(x2+a2 ) (x2+b2 )dx = ∫

C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi ¿ +e−b

2bi(a2−b2)¿

= 2πi(be−a−ae−b)

2abi(b2−a2) =

π (be−a−ae−b)ab(b2−a2)

⇒ ∫−∞

∞cosx+isinx

(x2+a2 ) (x2+b2 )dx =



∫−∞

∞cosx

(x2+a2 ) (x2+b2 )dx =


and ∫−∞

∞sinx

(x2+a2 ) (x2+b2 )dx = 0

46

Example III.3

Evaluate ∫−∞

∞xcosxx2+4

dx and ∫−∞

∞xsinxx2+4

dx

Solution


f ( z )dz where f(z) = z eiz

z2+4


Poles of f(z) = zeros of (z2+4), these zeros are given by (z2+ 4) =0

⇒ z2 = - 22 ⇒ z = (2i)2 ⇒ z = ± 2i

⇒ z = 2i or z = -2i Poles of f (z) are 2i,-2i (both are simple poles)There are no poles on the real axis.

The only pole lie inside the upper half of semi-circle is z = 2i



f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R


semi-circle CR).


Res[f,2i] = lim ¿ z→2i(z−2 i) f (z) = lim ¿ z→2i(z−2 i) zeiz

z2+4 =

lim ¿ z→2i(z−2 i) zeiz

(z+2 i)(z−2 i)

47

=lim ¿ z→2izeiz

(z+2i) =

2 ie i2i

(2 i+2i) = e

−2

2

Now lim ¿ z→∞ z

( z2+4 ) = 0


lim ¿R→∞∫CR

zeiz

( z2+4 )dz = 0

As R → ∞ ,

∫C

f ( z )dz = ∫−∞

∞


z eiz

( z2+4 )dz = ∫

−∞

∞x eix

(x2+4 )dx +0

Hence ∫−∞

∞xe ix

(x2+4 )dx = ∫

C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi ¿ ¿ = e-2πi ⇒ ∫

−∞

∞xcosx+ixsinx

(x2+4 )dx = e-2πi


∫−∞

∞xcosx

(x2+4 )dx = 0and ∫

−∞

∞sinx

(x2+4 )dx = e-2π

Example III.4

Evaluate ∫0

∞cosmx

x4+x2+1dx (m > 0)

Solution


f ( z )dz where f(z) = e imz

z4+z2+1


Poles of f(z) = zeros of (z4+z2+1), these zeros are given by z4+z2+1=0

z4+- z2 + z2+ z2 +1=0 ⇒z2 (z2 +1) + z2 +1 = z2

⇒ (z2 +1)2 - z2 = 0 ⇒ (z2 +1-z)( z2 +1+z) = 0

48

⇒ z2 +1-z = 0 or z2 +1+z = 0

⇒ z = 1+√3 i2

,or z = 1−√ 3i

2, or z =

−1+√3 i2

,or z = −1−√ 3i

2

Poles of f (z) are1+√3 i2

,1−√ 3i2

, −1+√3 i

2 , −1−√ 3i

2 (all are simple poles)

There are no poles on the real axis.

The poles lie inside the upper half of semi-circle is z = 1+√3 i

2 = α (say) and z = −1+√3 i

2

= β(say)



f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R

f ( z )dz (on the real line –R to R (LR) + the upper half of the

semi-circle CR).


Res[f,α] = lim ¿ z→α (z−α) f (z ) = lim ¿ z→α (z−α) e imz

(z−α )(z−β)(z+α )(z+β ) =

lim ¿ z→αe imz

(z−β )(z+α)(z+β)

= eimα

(α−β )(α+α )(α+β) =

e imα

(α−β )(2α)(α+β) =

e imα

(α−β )(2α)(α+β) =

eℑ (1+√3 i)

2

[ 1+√3 i2

−−1+√3 i2

]¿¿

49

=e

ℑ (1+√3i )2

(2 ( 1+√3 i )

2)( 2√3 i

2 ) = e

ℑ (1+√3 i )2

(√3 i−3) = e

ℑ2 e

−√ 3m2

(√3 i−3)

Similary,

Res[f,β] = e

ℑ (−1+√3 i)2

[(−1+√3 i )

2−1+√3 i

2] (

2 (−1+√3 i )2

)[(1+√3 i )

2+

(−1+√3 i)2

] =

eℑ (−1+√3 i)

2

(−1 ) (2 (−1+√3 i) )2

(2√3i )

eℑ (−1+√3i )

2

(−1)(2 (−1+√3 i )

2)( 2√3 i

2 ) = e

ℑ(−1+√3 i )2

(√3 i+3) = e

−ℑ2 e

−√ 3m2

(√3 i+3)

Now lim ¿ z→∞ 1

( z4+z2+1 ) = 0


lim ¿R→∞∫CR

e imz

( z4+z2+1 )dz = 0

As R → ∞ ,

∫C

f ( z )dz = ∫−∞

∞


e imz

( z4+z2+1 )dz = ∫

−∞

∞e imx

(x4+x2+1 )dx +0

Hence ∫−∞

∞e imx

(x4+x2+1 )dx = ∫

C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi [ eℑ2 e

−√3m2

(√3 i−3 )+ e

−ℑ2 e

−√3m2

(√3 i+3 ) ] = 2πi [ e−√3m

2 {(3+√3 i)eℑ2 +(√3i−3 ) e

−ℑ2 }

(√3i−3 ) (√3 i+3 ) ] =

2πi [ e−√3m2 {(3+√3 i )(cos (m2 )+ isin(m2 ))+(√3 i−3 ) (cos (m2 )−isin(m2 ))}

(−3−9 ) ]

50

= 2πi [ e−√3m

2

−12

{(3+√3 i )(cos (m2 )+ isin(m2 ))+(√3 i−3 ) (cos (m2 )−isin(m2 ))}]=

2πi [ e−√3m2

−12

{2√3 i cos(m2 )+6 isin (m2 )} ] = −4πe

−√ 3m2 ¿¿ = πe

−√3m2 ¿¿

⇒ ∫−∞

∞cosmx+isinmx

(x4+x2+1 )dx = πe

−√3m2 ¿¿


∫−∞

∞cosmx

(x4+x2+1 )dx = πe

−√3m2 ¿¿

and ∫−∞

∞sinmx

(x4+x2+1 )dx = 0

Hence ∫0

∞cosmx

(x4+x2+1 )dx =

12∫−∞

∞cosmx

(x4+x2+1 )dx = πe

−√3m2 ¿¿

Example III.5

Prove that ∫0

∞cosmx

(a2+x2 )2dx= π

2a3 (1+ma) e−ma (m > 0, a > 0)

Solution


f ( z )dz where f(z) = e imz

(a2+z2 )2


Poles of f(z) = zeros of (z2+a2 )2, these zeros are given by (z2+a2 )2=0

⇒ z2 = - a2 ⇒ z = (ai)2 ⇒ z = ± ai ⇒ z = ai (twice)or z = -ai(twice)

Poles of f (z) are ai, -ai (both are order 2)

51


The only pole lie inside the upper half of semi-circle is z = ai (order 2)



f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R


semi-circle CR).


Res[f,ai] = lim ¿ z→aiddz

( z−ai )2 f (z) = lim ¿ z→aiddz

( z−ai )2 e imz

(a2+z2 )2

= lim ¿ z→aiddz

( z−ai )2 eimz

( z+ai )2 ( z−ai )2= lim ¿ z→ai

ddz

eimz

( z+ai )2 =

lim ¿ z→ai[ ( z+ai )2 ℑe imz−eimz 2 ( z+ai ) ]

( z+ai )4

= lim ¿ z→ai[ ( z+ai ) ℑe imz−eimz 2 ]

( z+ai )3 =

[ (ai+ai ) ℑe imai−eimai2 ](ai+ai )3

= [−2ame−ma−2e−ma ]

(2ai )3 =

−2e−ma (ma+1 )−8a3i

= e−ma (ma+1 )

4a3 i

Now lim ¿ z→∞1

( z2+a2 )2 = 0

52

∴ by Jordan’s lemma , lim ¿R→∞∫CR

e imz

( z2+a2 )2dz = 0

As R → ∞ ,

∫C

f ( z )dz = ∫−∞

∞


e imz

( z2+a2 )2dz = ∫

−∞

∞eimx

(x2+a2 )2dx +0

= ∫−∞

∞eimx

(x2+a2 )2dx

Hence ∫−∞

∞eimx

(x2+a2 )2dx = ∫

C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi e−ma (ma+1 )

4a3 i= π e−ma(ma+1)

2a3

⇒ ∫−∞

∞cosmx+isinmx

( x2+a2 )2dx =

π e−ma(ma+1)2a3

⇒ ∫−∞

∞cosmx

(x2+a2 )2dx + i∫

−∞

∞sinmx

(x2+a2 )2dx =

π e−ma(ma+1)2a3


∫−∞

∞cosmx

(x2+a2 )2dx =

π e−ma(ma+1)2a3 and ∫

−∞

∞sinmx

(x2+a2 )2dx = 0

⇒ ∫0

∞cosmx

(x2+a2 )2dx =

12∫−∞

∞cosmx

(x2+a2 )2 dx = π e−ma(ma+1)

4 a3

Note:III.1

z = reiθ ⇒ r = |z| and θ= arg (z)

logz = Log r+i arg(z)

If z = x+iy , r = (x2+y2)1/2 , θ= arg (z) = arg (x+iy) = tan-1(y/x)

log(x+i) = log (x2+1)1/2 + iarg(x) = log (x2+1)1/2 +0 = log (x2+1)1/2

Example III.6

Prove that ∫0

∞log (1+x2)

1+ x2 dx=πlog 2

53

Solution


f ( z )dz where f(z) = log (z+i)z2+1


Poles of f(z) = zeros of (z2+1 ), these zeros are given by (z2+1 )=0

⇒ z2 = - 1 ⇒ z = (i)2 ⇒ z = ± i ⇒ z = i or z = -i

Poles of f (z) are i, -i (both are simple poles)There are no poles on the real axis.

The only pole lie inside the upper half of semi-circle is z = i



f(z).

Now ∫C

f ( z )dz = ∫−R

R

f ( x )dx + ∫C R


semi-circle CR).


Res[f,i] = lim ¿ z→i(z−i) f (z ) =lim ¿ z→i(z−i)log (z+i)z2+1

= lim ¿ z→i(z−i)log (z+i)

(z−i)(z+i)

= lim ¿ z→ilog (z+i)

(z+i) = log (i+i)

(i+i) = log 2 i

2i =

log (22 )12+ i tan−1( 2

0)

2 i =

log 2+itan−1(∞)2 i

(using Note III.1)

= log 2+ iπ

22i

54

Now lim ¿ z→∞log(z+i)z2+1

= lim ¿ z→∞log(z+i)

(z+i)(z−i) = lim ¿ z→∞1

(z−i)

lim ¿ z→∞log(z+i)(z+i)

Consider lim ¿ z→∞1

(z−i) = 0

Consider lim ¿ z→∞log (z+i)

( z+i) it is undetermined, so we have to use L’Hospital’s rule

lim ¿ z→∞

1z+i1

= lim ¿ z→∞1z+ i

= 0

Hence lim ¿ z→∞log(z+i)z2+1

= 0

⇒ lim ¿ z→∞∫C R

log(z+i)z2+1

= 0 ⇒ lim ¿R→∞∫CR

log (z+i)z2+1

= 0 (∵ |z| = R)

As R → ∞ ,

∫C

f ( z )dz = ∫−∞

∞


log (z+i)z2+1

dz = ∫−∞

∞log (x+i)x2+1

dx +0

= ∫−∞

∞log (x2+1 )

12

x2+1dx = ∫

−∞

∞12

log (x2+1)

x2+1dx (By using the Note III.1)

= 12∫−∞

∞log (x2+1)

x2+1dx = ∫

0

∞log (x2+1)

x2+1dx

Hence ∫0

∞log (x2+1)

x2+1dx = ∫

C

f ( z )dz = 2πi∑ Res[f,zk] = 2πi log 2+ iπ

22i

=πlog2 +iπ 2

2

Equating real part ∫0

∞log (x2+1)

x2+1dx = πlog2

Case of poles are on the real axis.

55

Type IV

If the rational function f(z) = g (z)h(z ) , then degree of h(z) exceeds that of g(z) and g(z) ≠ 0.

Suppose h(z) has simple zeros on the real axis ( that is simple poles of f(z) on the real axis) , let it be a1,a2,…ak

and h(z) has zeros inside the upper half of semi-circle ( that is poles of f(z) inside the upper half of semi-circle), let it be b1,b2,…bs,

then ∫−∞

∞

f ( x )dx = πi∑ Res[f,ak] + 2πi∑ Res[f,bs] , where k = 1,2,….k and s = 1,2,…s

Where C1,C2,….Ck are the semi circles and b1,b2,…bs are lie upper half of these semi circles.

Example IV.1.

Evaluate ∫−∞

∞x

x3−8dx

Solution.


f ( z )dz where f(z) = z

z3−8


Poles of f(z) = zeros of (z3- 8 ), these zeros are given by (z3- 8)=0

⇒ z3 = 8 ⇒ z3 = (2)3 ⇒ z = 2

Since z-2 is a factor of z3- 8 , z3- 8 = (z-2)(z2+2z+4) = 0

⇒ z2+2z+4 = 0 ⇒ z = −2±√4−16

2 =

−2±√−4∗32

= −2±2 i√3

2 = -1±i√3

Poles of f (z) are 2 , -1+i√3 and -1-i√3 (all are simple poles)

56

Pole lie on the real axis z =2

Pole lie inside the upper half of semi-circle z = -1+i√3


Res[f,2] = lim ¿ z→2(z−2) f (z) = lim ¿ z→2(z−2) z

z3−8 =

lim ¿ z→2(z−2) z(z−2)(z+1−i √3)(z+1+i √3)

= lim ¿ z→2z

(z+1−i √3)( z+1+i √3) =

2(2+1−i√3)(2+1+ i√3)

= 2

(3−i√3)(3+i √3)

=2

(9+3) = 16

Res[f, -1+i√3] = lim ¿ z→−1+i √3 (z−(−1+ i√3)) f (z )

= lim ¿ z→−1+i √3 (z+1−i √3) z

z3−8 =

lim ¿ z→−1+i √3 (z+1−i √3) z(z−2)(z+1−i√3)(z+1+i √3)

=lim ¿ z→−1+i √3z

(z−2)(z+1+i √3) =

−1+i √3(−1+i √3−2)(−1+ i√ 3+1+i√3)

=

−1+ i√ 3(−3+i √3)(2 i √3) =

−1+i √3(−6 i √3−6)

=−1+i √3

−6(i √3+1) =

(−1+i √3)(1−i√3)−6(i √3+1)(1−i√3)

= −(−1+i √3)(−1+i √3)

−6(1+3) =

(1−3−2 i√3)24

=(−2−2i √3)

24

= −2(1+i √3)

24 =

−(1+i √3)12

57


∞

f ( x )dx = πi∑ Res[f,ak] + 2πi∑ Res[f,bs] where ak ‘s are the poles lie on real

axis and bs ‘s are the poles lie inside the upper half of semi-circle.

∫−∞

∞x

x3−8dx = πi∑ Res[f,ak] + 2πi∑ Res[f,bs] = πi(

16¿ + 2πi(

−(1+i √3)12

¿

= πi6

- πi6

-πi √3 i

6 =

√3 π6

Type V

If the rational function f(z) = g (z)h(z )

, then degree of h(z) exceeds that of g(z) and g(z) ≠ 0.

Suppose h(z) has simple zeros on the real axis ( that is simple poles of f(z) on the real axis) , let it be a1,a2,…ak

and h(z) has zeros inside the upper half of semi-circle ( that is poles of f(z) inside the upper half of semi-circle), let it be b1,b2,…bs,

Let m be a positive real number and if f(z) = eimz g (z)h(z)

, then

∫−∞

∞

cosmxg ( x )h ( x )

dx = Re ∫−∞

∞

cosmx f ( x )dx

= ℜ[2 πi∑i=1

s

Res [ f , b i ]] + ℜ[πi∑j=1

k

Res [ f , aj ]]And

∫−∞

∞

sinmxg ( x )h (x )

dx = Img ∫−∞

∞

sinmx f ( x )dx

= Img[2πi∑i=1

s

Res [ f , bi ]] + Img[ πi∑j=1

k

Res [ f , aj ] ] Where b1,b2,…bs, are the poles of f(z) that lie in the upper half of the semi-circles

C1,C2,….Ck .

Example V.1.

58

Prove that ∫−∞

∞cos x

(x−1)(x2+4 )dx= π

10(−1e2 −2 sin 1) and

∫−∞

∞sinx

(x−1)(x2+4 )dx=π

5(−1e2 +cos1)

Solution.


f ( z )dz where f(z) = e iz

(z−1)(z2+4)


Poles of f(z) = zeros of (z-1)(z2+4), these zeros are given by (z-1)(z2+4) =0

⇒ z-1 = 0 or z2+4 =0 ⇒ z = 1 or z2 = -4 = (2i)2

⇒ z = 1 or z = 2i or z = -2iPoles of f(z) are z = 1 , z = 2i , z = -2i (all are simple poles)

The only pole lie on the real axis is z = 1

The only pole lie inside the semi-circle is z = 2i


Res[f,1] = lim ¿ z→1(z−1) f (z) = lim ¿ z→1(z−1) e iz

(z−1)(z2+4) = lim ¿ z→1

eiz

(z2+4 )

= e i

(1+4) = e

i

5

Res[f,2i] = lim ¿ z→2i(z−2 i) f (z) = lim ¿ z→2i(z−2 i) eiz

(z−1)(z+2i)(z−2 i)

59

= lim ¿ z→2ie iz

(z−1)(z+2 i) =

e i2 i

(2 i−1)(2i+2 i) =

e−2

(2 i−1)(4 i)

We know that

∫−∞

∞

cosmxg ( x )h ( x )

dx =

ℜ∫−∞

∞

cosmx f ( x )dx = ℜ[2 πi∑i=1

s

Res [ f , b i ]] + ℜ[πi∑j=1

k

Res [ f , aj ]]And ∫

−∞

∞

sinmxg ( x )h ( x )

dx =

Img∫−∞

∞

sinmx f ( x )dx = Img[2πi∑i=1

s


k

Res [ f , aj ] ]Where ak’s are the poles lie on the real axis and bs’s are the poles lie inside the upper half of the semi-circle.

∫−∞

∞cosx

(x−1)(x2+4 )dx = Re ∫

−∞

∞

cos x f ( x )dx = Re ∫−∞

∞

cos xe i x

(x−1)(x2+4)dx = ℜ [2πi Res [ f ,2i ] ]

+ ℜ [ πi Res [ f ,1 ] ]

= ℜ[2 πie−2

(2 i−1)(4 i) ] + ℜ[πi e i

5 ]= ℜ[π e−2

(2 i−1)(2) ] + ℜ[πi( cos1+isin15

)]= ℜ[π e−2(2 i+1)

2(2i−1)(2 i+1) ] + ℜ[ πicos1−π sin 15 ] = ℜ[π e−2(2i+1)

2(−4−1) ] - πsin15

= ℜ[π e−2(2i+1)−10 ] -

πsin15

= [ π e−2

−10 ] - πsin15

= −π10

[e−2+2 sin1 ]

∫−∞

∞sin x

(x−1)(x2+4 )dx = Img ∫

−∞

∞

sinx f ( x )dx = Img ∫−∞

∞

cosxe ix

(x−1)(x2+4 )dx

60

= Img[2πi∑i=1

s


k

Res [ f , aj ] ] = Img[π e−2(2 i+1)

2(2 i−1)(2i+1) ] + Img[ πicos1−πsin15 ]

= Img[ π e−2(2 i+1)−10 ] +

π cos15

= πe−2 2−10

+ πcos 1

5 = π

e−2

−5 +

πcos 15

= π5

[−e−2+cos1 ]

Example V.2.

Prove that ∫0

∞sinm x

xdx=π

2

Solution.


f ( z )dz where f(z) = eim z

z


Poles of f(z) = zeros of (z), this zero is given by z =0

The only pole of f(z) is z = 0 simple and lie on real axis


Res[f,0] = lim ¿ z→0(z−0) f (z) = lim ¿ z→0 zeiz m

z = e0 = 1


∞

cosmxg ( x )h ( x )

dx =

ℜ∫−∞

∞

cosmx f ( x )dx = ℜ[2 πi∑i=1

s

Res [ f , b i ]] + ℜ[πi∑j=1

k

Res [ f , aj ]]

61

And ∫−∞

∞

sinmxg ( x )h ( x )

dx =

Img∫−∞

∞

sinmx f ( x )dx = Img[2πi∑i=1

s


k

Res [ f , aj ] ]Where ak’s are the poles lie on the real axis and bs’s are the poles lie inside the upper half of the semi-circle.

∴ ∫−∞

∞sinmx

xdx = Img∫

−∞

∞

sinmx f ( x )dx = Img∫−∞

∞

sinmxe imx

xdx

= Img[2πi∑i=1

s


k

Res [ f , aj ] ]= Img [πi Res [ f ,0 ] ] = Img [πi(1)] = πHence ∫

0

∞sinmx

xdx =

12∫−∞

∞sinmxx

dx = π2

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Calculus of Residue and Contour Integration

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