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### Transcript of Slides 12 Handout

• Evaluation of integrals

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type I

Type I: Integrals of the form 2pi0

F (cos , sin ) d

If we take z = e i, then cos = 12(z + 1

z), sin = 1

2i(z 1

z) and d =

dz

iz.

Substituting for sin , cos and d the definite integral transforms intothe following contour integral 2pi

0

F (cos , sin ) d =

|z|=1

f (z) dz

where f (z) = 1iz

[F ( 12(z + 1

z), 1

2i(z 1

z))]

Apply Residue theorem to evaluate|z|=1

f (z) dz .

Lecture 18 Evaluation of integrals

• Example of Type I

Consider 2pi0

1

1 + 3(cos t)2dt.

2pi0

1

1 + 3(cos t)2dt =

|z|=1

1

1 + 3( 12(z + 1

z))2

dz

iz

= 4i|z|=1

z

3z4 + 10z2 + 3dz

= 4i|z|=1

z

3(z +

3i)(z 3i

)(z + i

3

)(z i

3

) dz= 4

3i

|z|=1

z

(z +

3i)(z 3i)(z + i

3

)(z i

3

) dz= 4

3i 2pii{Res(f , i

3) + Res(f , i

3)}.

Lecture 18 Evaluation of integrals

• Improper Integrals of Rational Functions

The improper integral of a continuous function f over [0,) is defined by 0

f (x)dx = limb

b0

f (x)dx

provided the limit exists.

If f is defined for all real x , then the integral of f over (,) isdefined by

f (x)dx = lim

a

0a

f (x)dx + limb

b0

f (x)dx

provided both limits exists.

There is another value associated with the improper integral f (x)dx

namely the Cauchy Principal value(P.V.) and it is given by

P. V.

f (x)dx := limR

RR

f (x)dx

provided the limit exists.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type II

If the improper integral f (x)dx converges, then P. V.

f (x)dx

exists and

f (x)dx = P. V.

f (x)dx .

The P. V. f (x)dx exists 6= the improper integral

f (x)dx

exists. Take f (x) = x .

However if f is an even function (i.e. f (x) = f (x) for all x R) thenP. V.

f (x)dx exists = the improper integral

f (x)dx exists and

their values are equal.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type II

Consider the rational function f (z) =P(z)

Q(z)where P(z) and Q(z) are

polynomials with real coefficients such that

Q(z) has no zeros in the real line

degree of Q(z) > 1+ degree of P(z)

then P. V.

f (x)dx can be evaluated using Cauchy residue theorem.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type II

Type II Consider the integral

1

(x2 + 1)2dx ,

To evaluate this integral, we look at the complex-valued function

f (z) =1

(z2 + 1)2

which has singularities at i and i . Consider the contour C like semicircle, theone shown below.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type II

Note that: C

f (z) dz =

aa

f (z) dz +

Arc

f (z) dz aa

f (z) dz =

C

f (z) dz

Arc

f (z) dz

Furthermore observe that

f (z) =1

(z2 + 1)2=

1

(z + i)2(z i)2 .

Then, by using Residue Theorem,C

f (z) dz =

C

1(z+i)2

(z i)2 dz = 2piid

dz

(1

(z + i)2

) z=i

=pi

2

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type II

Again, Arc

f (z) dz

api(a2 1)2 0 as a.So

1

(x2 + 1)2dx =

f (z) dz = lima+

aa

f (z) dz =pi

2.

Theorem

Suppose f is analytic in C except at a finite number ofsingular points in C \R. Let R denote the semi-circle |z | = Rin the upper half plane =(z) 0. If lim

zzf (z) = 0 then

limR

Rf (z)dz = 0.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type II

Exercise.

Show that P.V.

(x2 x + 2)dx(x4 + 10x2 + 9)

=5pi

12. [Poles are i ,3i .]

Show that

0

x2dx

(x2 + 9)(x2 + 4)2=

pi

200.

Show that

0

dx

x3 + 1=

2pi

3

3.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type III

Type III Integrals of the form

P. V.

P(x)

Q(x)cosmx dx or P. V.

P(x)

Q(x)sinmx dx ,

where

P(x),Q(x) are real polynomials and m > 0

Q(x) has no zeros in the real line

degree of Q(x) > degree of P(x)

then

P. V.

P(x)

Q(x)cosmx dx or P. V.

P(x)

Q(x)sinmx dx

can be evaluated using Cauchy residue theorem.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type III

Evaluate:

cosx

x2 + 1dx or

sinx

x2 + 1dx

Consider the integral

e ix

x2 + 1dx

We will evaluate it by expressing it as a limit of contour integrals along the

contour C that goes along the real line from a to a and thencounterclockwise along a semicircle centered at 0 from a to a. Take a > 1 sothat i is enclosed within the curve.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type III

Res

(e iz

z2 + 1, i

)= lim

zi(z i) e

iz

z2 + 1= lim

zie iz

z + i=

e

2i.

So by residue theoremC

f (z) dz = (2pii)Res(f , i) = 2piie

2i= pie.

The contour C may be split into a straight part and a curved arc, so thatstraight

+

arc

= pie

and thus aa

e ix

x2 + 1dx = pie

arc

e iz

z2 + 1dz .

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type III

arc

e iz

z2 + 1dz

pi0

a e ia(cos+i sin )a2 1 d

aa2 1

pi0

ea sin d.

Hence, arc

e iz

z2 + 1dz

0 as aand

P.V.

cosx

x2 + 1dx = lim

a

aa

e ix

x2 + 1dx

= lima

[pie

arc

e iz

z2 + 1dz

]= pie.

Lecture 18 Evaluation of integrals

• Jordans Lemma

Lemma

Let f be analytic in C except for finite number of singular points inC \ R. Let R denote the semicircle |z | = R in the upper halfplane. If lim

R

(max|z|=R

|f (z)|)

= 0, then limR

R

f (z)e iazdz = 0.

The result follows from Jordans inequality: pi0

eR sin d 0).

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type IV

Type IV Integrals of the form 0

sin x

xdx

can be evaluated using Cauchy residue theorem.Before we discuss integrals of Type IV we need the following result.

Lemma: Suppose f has a simple pole at z = a on the real axis. If c is thecontour defined by c(t) = a + e

i(pit), t (0, pi) then

lim0

c

f (z)dz = ipiRes(f , a).

Proof: Since f has a simple pole at z = a, the Laurent series expansion of fabout z = a is of the form

f (z) =Res(f , a)

z a + g(z).

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type IV

Nowc

f (z)dz =

c

Res(f , a)

z a dz +c

g(z)dz

= Res(f , a) pi

0

ie i(pit)

e i(pit)dt

pi0

g(a + e i(pit))ie i(pit)dt

= ipiRes(f , a) pi

0

g(a + e i(pit))ie i(pit)dt.

Note that f has Laurent series expansion in 0 < |z a| < R for some R > 0.The function g is continuous on |z a| 0 for every < 0 < R. So|g(z)| < M on |z a| 0. So pi

0

g(a + e i(pit))ie i(pit)dt

Mpi 0 as 0.Hence

lim0

c

f (z)dz = ipiRes(f , a).

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type IV

Consider the integral 0

sin x

xdx .

Define f (z) = eiz

z, (z = 0 is a simple pole on the real axis).

Consider the contour C = [R,] [,R] .

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type IV

By Cauchys theoremC

e iz

zdz =

[R,]

e iz

zdz +

e iz

zdz +

[,R]

e iz

zdz +

e iz

zdz = 0.

But [R,]

e iz

zdz +

[,R]

e iz

zdz =

[,R]

e ix eixx

dx

So [,R]

e ix eixx

dx =

e iz

zdz

e iz

zdz = ipi

as 0 (by the previous Lemma ) and R (by Jordans inequality) andhence,

0

sin x

xdx =

pi

2.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type V

Integration along a branch cut: Consider the improper integral 0

xa

1 + xdx (0 < a < 1).

Define

f (z) =za

1 + z(|z | > 0, 0 < arg z < 2pi).

The functionza

1 + zis a multiple valued function with branch cut

arg z = 0 (positive real axis).

Consider the contour C = [+ i,R + i] R [R i, i] {}.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type V

By residue theorem([+i,R+i]

+

R

+

[Ri,i]

+

)f (z)dz = 2piiRes(f ,1) = 2piieiapi.

Since

f (z) =exp(a log z)

z + 1=

exp(a(ln r + i))re i + 1

,

where z = re i, it follows thatOn [+ i,R + i], 0 as 0,

f (z) =exp(a(ln r + i .0))

re i.0 + 1 r

a

1 + ras 0.

On [R i, i], 2pi as 0,

f (z) =exp(a(ln r + i .2pi))

re i.2pi + 1 r

a

1 + re2apii as 0.

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type V

But R

za

1 + zdz

RaR 1 2piR = 2piRR 1 1Ra 0 as R and

za

1 + zdz

a 1 2pi = 2pi1 1a 0 as 0.So

limR,0

( R

ra

1 + rdr +

R

ra

1 + re2apiidr

)= 2piieiapi

That is

(1 e2apii )

0

ra

1 + rdr = 2piieiapi

and hence 0

ra

1 + rdr =

2piieiapi

(1 e2apii ) =pi

sin api(0 < a < 1).

Lecture 18 Evaluation of integrals

• Evaluation of certain contour integrals: Type VI

Integration around a branch cut:Consider the improper integral

0

log x

1 + x2dx .

Define

f (z) =log z

1 + z2(|z | > 0, pi

2< arg z