MATHEMATICS-III

CONTENTS

Linear ODE with variable coefficients and series solutions

Special functions

Complex function – Differentiation and integration

Power series expansions of complex functions and contour integration

Conformal mapping

TEXT BOOKS:

•Advanced Engineering Mathematics by Kreyszig, John Wiley & Sons.

•Higher Engineering Mathematics by Dr. B.S. Grewal, Khanna Publishers.

REFERENCES:

1.Complex Variables Principles And Problem Sessions By A.K.Kapoor, World

Scientific Publishers

2.Engineering Mathematics-3 By T.K.V.Iyengar andB.Krishna Gandhi Etc

3.A Text Book Of Engineering Mathematics By N P Bali, Manesh Goyal

4.Mathematics for Engineers and Scientists, Alan Jeffrey, 6th Edit. 2013, Chapman &

Hall/CRC

5.Advanced Engineering Mathematics, Michael Greenberg, Second Edition. Person

Education

6.Mathematics For Engineers By K.B.Datta And M.A S.Srinivas,Cengage Publications

Power Series Method

The power series method is the standard method for solving linear ODEs with variable coefficients. It gives solutions in the form of power series. These series can be used for computing values, graphing curves, proving formulas, and exploring properties of solutions. In this section we begin by explaining the idea of the power series method.

From calculus we remember that a power series (in powers of x − x0) is an infinite series of the form

(1)

Here, x is a variable. a0, a1, a2, … are constants, called the coefficients of the series. x0 is a constant, called the center of the series. In particular, if x0 = 0, we obtain a power series in powers of x

(2)

We shall assume that all variables and constants are real.

2

0 0 1 0 2 00

( ) ( ) ( ) .m

mm

a x x a a x x a x x

2 3

0 1 2 30

.m

mm

a x a a x a x a x

Then we collect like powers of x and equate the sum of the coefficients of each occurring power of x to zero, starting with the constant terms, then taking the terms containing x, then the terms in x2, and so on. This gives equations from which we can determine the unknown coefficients of (3) successively.

Idea and Technique of the Power Series Method (continued)

The nth partial sum of (1) is

where n = 0, 1, … .If we omit the terms of sn from (1), the remaining expression is

This expression is called the remainder of (1) after the term an(x − x0)n.

Theory of the Power Series Method

2

0 1 0 2 0 0(6) ( ) ( ) ( ) ( )n

n ns x a a x x a x x a x x

+1 +2

+1 0 +2 0(7) R ( )=a ( - ) + ( - ) + .n n

n n nx x x a x x

In this way we have now associated with (1) the sequence of the partial sums s0(x), s1(x), s2(x), …. If for some x = x1 this sequence converges, say,

then the series (1) is called convergent at x = x1, the number s(x1) is called the value or sum of (1) at x1, and we write

Then we have for every n,(8)If that sequence diverges at x = x1, the series (1) is called divergent at x = x1.

Theory of the Power Series Method (continued)

1 1 ( ) ( ),lim n

n

s x s x

1 1 00

( ) ( ) .m

mm

s x a x x

1 1 1( ) ( ) ( ).

n ns x s x R x

Where does a power series converge? Now if we choose x = x0 in (1), the series reduces to the single term a0 because the other terms are zero. Hence the series converges at x0.In some cases this may be the only value of x for which (1) converges. If there are other values of x for which the series converges, these values form an interval, the convergence interval. This interval may be finite, as in Fig. 105, with midpoint x0. Then the series (1) converges for all x in the interior of the interval, that is, for all x for which(10) |x − x0| < Rand diverges for |x − x0| > R. The interval may also be infinite, that is, the series may converge for all x.

Theory of the Power Series Method (continued)

Legendre’s Equation.

Legendre Polynomials Pn(x)

Legendre’s differential equation(1) (1 − x2)y” − 2xy’ + n(n + 1)y = 0 (n constant)is one of the most important ODEs in physics. It arises in numerous problems, particularly in boundary value problems for spheres.

The equation involves a parameter n, whose value depends on the physical or engineering problem. So (1) is actually a whole family of ODEs. For n = 1 we solved it in Example 3 of Sec. 5.1 (look back at it). Any solution of (1) is called a Legendre function.The study of these and other “higher” functions not occurring in calculus is called the theory of special functions.

Dividing (1) by 1 − x2, we obtain the standard form needed in Theorem 1 of Sec. 5.1 and we see that the coefficients −2x/(1 − x2) and n(n + 1)/(1 − x2) of the new equation are analytic at x = 0, so that we may apply the power series method. Substituting

(2)

and its derivatives into (1), and denoting the constant n(n + 1) simply by k, we obtain

By writing the first expression as two separate series we have the equation

0

( ) m

mm

y x a x

2 2 1

2 1 0

(1 ) ( 1) 2 0.m m m

m m mm m m

x m m a x x ma x k a x

2

2 2 1 0

( 1) ( 1) 2 0.m m m m

m m m mm m m m

m m a x m m a x ma x ka x

The reduction of power series to polynomials is a great advantage because then we have solutions for all x, without convergence restrictions. For special functions arising as solutions of ODEs this happens quite frequently, leading to various important families of polynomials. For Legendre’s equation this happens when the parameter n is a nonnegative integer because then the right side of (4) is zero for s = n, so that an+2 = 0, an+4 = 0, an+6 = 0, …. Hence if n is even, y1(x) reduces to a polynomial of degree n. If n is odd, the same is true for y2(x). These polynomials, multiplied by some constants, are called Legendre polynomials and are denoted by Pn(x).

Polynomial Solutions. Legendre Polynomials Pn(x)

Series Solutions

Near a Regular Singular Point, Part I

We now consider solving the general second order linear

equation in the neighborhood of a regular singular point x0.

For convenience, will will take x0 = 0.

Recall that the point x0 = 0 is a regular singular point of

iff

iff

0)()()(2

2

yxRdx

dyxQ

dx

ydxP

0at analytic are )()(

)(and)(

)(

)( 22 xxqxxP

xRxxxp

xP

xQx

on convergent ,)( and )(0

2

0

n

n

n

n

n

n xxqxqxxpxxp

Series Solutions

Near a Regular Singular Point, Part I

We now consider solving the general second order linear

equation in the neighborhood of a regular singular point x0.

For convenience, will will take x0 = 0.

Recall that the point x0 = 0 is a regular singular point of

iff

iff

0)()()(2

2

yxRdx

dyxQ

dx

ydxP

0at analytic are )()(

)(and)(

)(

)( 22 xxqxxP

xRxxxp

xP

xQx

on convergent ,)( and )(0

2

0

n

n

n

n

n

n xxqxqxxpxxp

Transforming Differential Equation

Our differential equation has the form

Dividing by P(x) and multiplying by x2, we obtain

Substituting in the power series representations of p and q,

we obtain

0)()()( yxRyxQyxP

0

2

0

,)(,)(n

n

n

n

n

n xqxqxxpxxp

0)()( 22 yxqxyxxpxyx

02

210

2

210

2 yxqxqqyxpxppxyx

Comparison with Euler Equations

Our differential equation now has the form

Note that if

then our differential equation reduces to the Euler Equation

In any case, our equation is similar to an Euler Equation but

with power series coefficients.

Thus our solution method: assume solutions have the form

0,0for ,)(0

0

2

210

n

nr

n

r xaxaxaxaaxxy

02

210

2

210

2 yxqxqqyxpxppxyx

02121 qqpp

000

2 yqyxpyx

Example 1: Regular Singular Point (1 of 13)

Consider the differential equation

This equation can be rewritten as

Since the coefficients are polynomials, it follows that x = 0 is

a regular singular point, since both limits below are finite:

012 2 yxyxyx

02

1

2

2

yx

yx

yx

2

1

2

1limand

2

1

2lim

2

2

020 x

xx

x

xx

xx

Example 1: Euler Equation (2 of 13)

Now xp(x) = -1/2 and x2q(x) = (1 + x )/2, and thus for

it follows that

Thus the corresponding Euler Equation is

As in Section 5.5, we obtain

We will refer to this result later.

0,2/1,2/1,2/1 3221100 qqppqqp

0

2

0

,)(,)(n

n

n

n

n

n xqxqxxpxxp

020 2

00

2 yyxyxyqyxpyx

2/1,1011201)1(2 rrrrrrrxr

012 2 yxyxyx

Example 1: Differential Equation (3 of 13)

For our differential equation, we assume a solution of the form

By substitution, our differential equation becomes

or

012 2 yxyxyx

0

2

0 0

1

1)(

,)(,)(

n

nr

n

n n

nr

n

nr

n

xnrnraxy

xnraxyxaxy

0120

1

000

n

nr

n

n

nr

n

n

nr

n

n

nr

n xaxaxnraxnrnra

0121

1

000

n

nr

n

n

nr

n

n

nr

n

n

nr

n xaxaxnraxnrnra

Example 1: Combining Series (4 of 13)

Our equation

can next be written as

It follows that

and

0121

1

000

n

nr

n

n

nr

n

n

nr

n

n

nr

n xaxaxnraxnrnra

01)(121)1(21

10

n

nr

nn

r xanrnrnraxrrra

01)1(20 rrra

,2,1,01)(12 1 nanrnrnra nn

Example 1: Indicial Equation (5 of 13)

From the previous slide, we have

The equation

is called the indicial equation, and was obtained earlier when

we examined the corresponding Euler Equation.

The roots r1 = 1, r2 = ½, of the indicial equation are called the

exponents of the singularity, for regular singular point x = 0.

The exponents of the singularity determine the qualitative

behavior of solution in neighborhood of regular singular point.

0)1)(12(13201)1(2 20

0

0

rrrrrrraa

01)(121)1(21

10

n

nr

nn

r xanrnrnraxrrra

Example 1: Recursion Relation (6 of 13)

Recall that

We now work with the coefficient on xr+n :

It follows that

01)(121)1(21

10

n

nr

nn

r xanrnrnraxrrra

01)(12 1 nn anrnrnra

1,

112

1)(32

1)(12

1

2

1

1

nnrnr

a

nrnr

a

nrnrnr

aa

n

n

nn

Example 1: First Root (7 of 13)

We have

Starting with r1 = 1, this recursion becomes

Thus

2/1 and 1 ,1for ,

11211

1

rrnnrnr

aa n

n

1,

1211112

11

nnn

a

nn

aa nn

n

215325

13

012

01

aaa

aa

1,

!12753

)1(

etc,32175337

0

023

nnn

aa

aaa

n

n

Example 1: First Solution (8 of 13)

Thus we have an expression for the n-th term:

Hence for x > 0, one solution to our differential equation is

1,

!12753

)1( 0

n

nn

aa

n

n

1

0

1

1

00

0

1

!12753

)1(1

!12753

)1(

)(

n

nn

n

nn

rn

n

n

nn

xxa

nn

xaxa

xaxy

Example 1: Radius of Convergence for

First Solution (9 of 13)

Thus if we omit a0, one solution of our differential equation is

To determine the radius of convergence, use the ratio test:

Thus the radius of convergence is infinite, and hence the series

converges for all x.

0,

!12753

)1(1)(

1

1

xnn

xxxy

n

nn

10

132lim

)1(!13212753

)1(!12753limlim

111

1

nn

x

xnnn

xnn

xa

xa

n

nn

nn

nn

n

n

n

n

Example 1: Second Root (10 of 13)

Recall that

When r1 = 1/2, this recursion becomes

Thus

2/1 and 1 ,1for ,

11211

1

rrnnrnr

aa n

n

1,

122/1212/112/12

111

nnn

a

nn

a

nn

aa nnn

n

312132

11

012

01

aaa

aa

1,

!12531

)1(

etc,53132153

0

023

nnn

aa

aaa

n

n

Example 1: Second Solution (11 of 13)

Thus we have an expression for the n-th term:

Hence for x > 0, a second solution to our equation is

1,

!12531

)1( 0

n

nn

aa

n

n

1

2/1

0

1

2/1

02/1

0

0

2

!12531

)1(1

!12531

)1(

)(

n

nn

n

nn

rn

n

n

nn

xxa

nn

xaxa

xaxy

Example 1: Radius of Convergence for

Second Solution (12 of 13)

Thus if we omit a0, the second solution is

To determine the radius of convergence for this series, we can

use the ratio test:

Thus the radius of convergence is infinite, and hence the series

converges for all x.

10

12lim

)1(!11212531

)1(!12531limlim

111

1

nn

x

xnnn

xnn

xa

xa

n

nn

nn

nn

n

n

n

n

1

2/1

2!12531

)1(1)(

n

nn

nn

xxxy

Example 1: General Solution (13 of 13)

The two solutions to our differential equation are

Since the leading terms of y1 and y2 are x and x1/2, respectively,

it follows that y1 and y2 are linearly independent, and hence

form a fundamental set of solutions for differential equation.

Therefore the general solution of the differential equation is

where y1 and y2 are as given above.

1

2/1

2

1

1

!12531

)1(1)(

!12753

)1(1)(

n

nn

n

nn

nn

xxxy

nn

xxxy

,0),()()( 2211 xxycxycxy

Shifted Expansions & Discussion

For the analysis given in this section, we focused on x = 0 as

the regular singular point. In the more general case of a

singular point at x = x0, our series solution will have the form

If the roots r1, r2 of the indicial equation are equal or differ by

an integer, then the second solution y2 normally has a more

complicated structure. These cases are discussed in Section 5.7.

If the roots of the indicial equation are complex, then there are

always two solutions with the above form. These solutions are

complex valued, but we can obtain real-valued solutions from

the real and imaginary parts of the complex solutions.

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zf

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zgzz

zgn

zz

zgzf

n

ysingularit essentiallimit the satisfies finite no if direction, any from (4)

as |)(|)( of pole a is if (3)

. thangreater order of pole a is then infinite, is if (2)

. than lessorder of pole a is then ,0 if (1)

number complex zero-non finite, a is and analytic is )(

)]()[(lim

is at order

of pole a has )(that for definition alternate An**

00

0

0

0

0

0

n

zzzfzfzz

nzza

nzza

azf

azfzz

zz n

zf

n

zz

1)(n pole simple a is ysingularit each

1}sinh

sinhcosh])2/1([{lim}

cosh

sinh])2/1([{lim

rule sHospital'l' Using

)2

1( )12(2

integer any is n )exp(])12(exp[exp

)exp(exp when ysingularit a has )(

)exp(exp

)exp(exp

cosh

sinh

tanh)( (2)

1 and 1at 1order of poles )1)(1(

2)(

1

1

1

1)( (1)

function the of iessingularit the Find :Ex

)2/1()2/1(

z

zzinz

z

zinz

inzniz

zniz

zzzf

zz

zz

z

z

zzf

zzzz

zzf

zzzf

inzinz

.approached is which from direction the oft independen and exists

)(limbut ed,undetermin )( of value the makes ySingularit

:essingularti Remove

0

0

z

zfzfzz

.0at ysingularit removable a has )(

so ,0 way the oft independen is 1)(lim

....!5!3

1........)!5!3

(1

)(

edundetermin 0/0)(lim :

0at ysingularit removable a is /sin)(that Show :Ex

0

5353

0

zzf

zzf

zzzzz

zzf

zfSol

zzzzf

z

z

zf

zf

/1 where ,0at )/1(

ofthat by given is infinityat )( ofbehavior The

zzf

nfzzf

z

fzzzf

zzf

bafzbzazf

zzfzzzf

bzazf

n

n

at ysingularit essential an has )(

)!()/1( exp)( (iii)

at 3order

of pole a has /1/1)/1( )1()( (ii)

at analytic is )( 0at analytic is

)/1( /1set )( (i)

exp)( (iii) and )1()( (ii)

)( (i) of infinityat behavior the Find :Ex

0

1

32

22

2

2

)(/1 of norder of pole a also is (iii)

zero. simple a called is 1,n if (ii)

n.order of zero a called is (i)

0)( and integer, positive a

is n if ,)()()( and 0)( If

0

0

0

0

00

zfzz

zz

zz

zg

zgzzzfzfn

Complex integral

dtdt

dxvidt

dt

dyuidt

dt

dyvdt

dt

dxu

vdxiudyivdyudx

idydxivudzzf

t

ttyytxx

tt

CCC C

C C

))(()(

is Bpoint

, is point A and )( ,)(

for ,parameter continuous real A2C

A 3C

y

x

1C

B

.at finishing and starting ,circle the

along ,/1)( of integralcomplex the Evaluate :Ex

RzR |z|

zzf

y

R

1C

x

t

. oft independen isresult calculated The

2sincos

cossin

by calculated also is integral The **

200

)sin)(sin

(coscos

cos)sin

()sin(cos1

sin ,

cos

,1

)( ,cos ,sin

20 ,sincos)(

2

0

2

0

2

0

2

0

2

0

2

0

2222

22

1

1

R

iidtdttiRtR

tiRtR

z

dz

iii

dttRR

titdtR

R

ti

tdtRR

tdttR

R

tdz

z

R

t

yx

yv

R

t

yx

xu

ivuyx

iyx

iyxzftR

dt

dytR

dt

dx

ttiRtRtz

C

C

3b3a3

2

C and C linesstraight two of up made Ccontour the (ii)

0 plane-half the

in || semicircle the of consisting Ccontour the (i)

along /1)( of integralcomplex the Evaluate :Ex

y

Rz

zzf

ca

xdx

xa

aii

titt

dttt

idttt

tdt

itt

i

dtiRRsiR

iRRdt

iRRtR

iRR

z

dz

sRsisRC

titRRtzC

izdzt

C

b

a

C

)(tan 2

)]2

(2

[2

0

|)]2/1

2/1(tan2[

2|)]221[ln(

2

1

221

1

221

12

1

1 term1st

)()(

10for )1( :

10for )1( : (ii)

/ 0

for but example, previous the in asjust is This (i)

1

22

10

110

2

1

0 2

1

0 2

1

0

1

0

1

0

3

3

3

2

y

R

bC3

1s

iR

aC3

R x

0t

path.different the oft independen is integral The

2)]

4(

4[0

|)2/1

2/1(tan|)]122[ln(

2

1

122

1

122

12

)1(

)]1()[1(

)1(

1 term 2nd

3

10

110

2

1

0 2

1

0 2

1

0 22

1

0

iz

dz

ii

siss

dsss

idsss

s

dsss

sisids

sis

i

C

examples. previous the in shown as C and C ,C path the

along )Re()( of integralcomplex the Evaluate :Ex

321

zzf

path.different the on depends integral The

)1(2

1)1(

2

1

)1()1)(1(

))(()()1(

:CCC (iii)

2)cossin(cos :C (ii)

)cossin(cos :C (i)

222

1

0

1

0

22

1

0

1

0

3b3a3

2

02

2

0

21

iRiRiR

dsisRdtitR

dsiRRsRdtiRRRt

Ri

dttiRtRtR

RidttiRtRtR

Cauchy theorem

C

dzzf

zfzf

0)(

Ccontour closed a on and withinpoint eachat

continuous is )( and function, analytic an is )( If'

apply. relations Riemann-Cauchy the and analytic is )(

0])(

[])()(

[

)()()(

and )(

)()(theorem divergence

ldemensiona-two by then C,contour closed a on

and within continuous are ),(

and ),(

If

zf

dxdyx

u

y

vidxdy

x

v

y

u

udyvdxivdyudxdzzfI

idydxdzivuzf

qdxpdydxdyy

q

x

p

y

yxq

x

yxp

RR

C C C

CR

.C and C along same the is )( of integral the then paths two

the by enclosed region the in and path each on function analytic

an is )( ifthat Show .C and C pathsdifferent two by

joined are planecomplex the in B and Apoints twos Suppose :Ex

21

21

zf

zf

A

1C

y

x

B

1 2

1 2 21

)()(

enclosingcontour closed a forms path

0)()()(

21

C C

C C CC

dzzfdzzf

RCC

dzzfdzzfdzzf

R

Cdzzfdzzfzf

C

C

)()( then contours two the between region the in analytic is )(

function the ifthat Show . with completely liesit that small lysufficient

being diagram, Argandthe in and contour closed twoConsider :Ex

y C

1C

2C

x

dzzfdzzfC

dzzfdzzfdzzfdzzf

dzzf

zf

C

CCC

)()( contour

ofthat as contour of direction the take If

)()()()(

0)(

analytic is )(

and , by bounded is area the

21

C

z fdzzfC

Rzzf

analytic. is )(0)(for , curve a by bounded

domain closed a in of function continuous a is )( if

:theorem sMorera'

Cauchy’s integral formula

Cdz

zz

zf

izfCz

Czf

000

)(

2

1)( then withinpoint a is and

contour closed a on and within analytic is )( If

C

)(2)(

)(

)exp( ),exp(for

)()(

0

02

0 0

2

0

0

0

00

zifdezfi

deie

ezfI

diidzizz

dzzz

zfdz

zz

zfI

i

i

i

i

C

0z

)(

)(

2

1)(

:functioncomplex a of derivative the of form integral The

20

0'

dzzz

zf

izf

C

dzzz

zf

i

nzf

dzzz

zf

i

dzzzhzz

zf

i

dzzzhzzh

zf

i

h

zfhzfzf

C n

n

C

Ch

Ch

h

10

0)(

20

000

000

00

00

'

)(

)(

2

!)( derivative nthFor

)(

)(

2

1

]))((

)(

2

1[lim

])11

()(

2

1[lim

)()(lim)(

.!

|)(|that show constant, some is

where circle, the on |)(| If .point the on centered

radius of circle a on and inside analytic is )(that Suppose :Ex

0)(

0

n

n

R

MnzfM

MzfzzR

Czf

nnC n

n

R

MnR

R

Mn

zz

dzzfnzf

!2

2

!|

)(

)(|

2

!|)(|

110

0)(

constant. a is )( then

allfor bounded and analytic is )( If :theorem sLiouville'

zfz

zf

zfzzfz

zzzf

zfzfRn

R

Mn!zf

n

n

constant )( allfor 0)( plane.- the inpoint

any as take may we , allfor analytic is )( S ince

0)(0|)(| let and 1set

|)(| :inequality sCauchy' Using

'

0

0'

0'

0)(

Taylor and Laurent series

n

n

nn

n

n zzn

zfzzazf

zzz

zf

)(!

)()()(

then C, insidepoint a is and ,point the on

centeredR radius of C circle a on and inside analytic is )( If

0

0

0)(

0

0

0

Taylor’s theorem:

!

)()(

!

)(2)(

2

1

)(

)()(

2

1)(

)(

2

1)(

)(11

in series geometric a as 1

expand

C on lies where )(

2

1)( so C, on and inside analytic is )(

0)(

0

00

)(

0

0

100

0

0 0

0

0

0 0

0

00

0

n

zfzz

n

zifzz

i

dz

fzz

id

z

zz

z

f

izf

z

zz

zzz

zz

z

dz

f

izfzf

nn

n

nn

n

C n

n

n

n

nC

n

n

C

202010

0

1

10

1

0

0

0

00

0

)()(.......)()(

)(

exp )( ,

.)()( series

Taylor a as expanded and at analytic is )()()( Then . on and

insidepoint other everyat analytic isbut at order of pole a has )( If

zzazzaazz

a

zz

a

zz

azf

eries Laurent sanded as a can bezfCinsidezallforThus

zzbzg

zzzfzzzgC

zzpzf

p

p

p

p

n

n

n

p

021

0

10

10

10

0)(

on centered and circles two

between region a in analytic is )()(

)(

)(

2

1

)(

)(

2

1

)(

)(

2

1

!

)( and

zzCC

Rzzazf

dzzz

zf

idz

zz

zg

ia

dzzz

zg

in

zgbba

n

n

n

npnn

n

n

npnn

0z

2C

1C

R

p

zfazzpzf

kaa

zzzf

zzm

zfmz-za

na

nazzzf

kpp

mm

n

n

ysingularit essential of valuelowest a find to impossible (ii)

)( of residue the called is ,at order of pole a has )(

0 allfor 0but 0 find to possible (i)

at analyticnot is )( If (2)

.at order of

zero a have to said is )( ,0 with )( is term

vanishing-nonfirst the ,0for 0 happen mayIt

.0for 0 all then ,at analytic is )( If )1(

10

0

0

0

0

n

n

n zzazf )()( 0

pole. eachat )( of residue the find and 3,order of pole

a is 2 and 1order of pole a is 0that verify Hence .2 and 0

iessingularit theabout )2(

1)( of seriesLaurent the Find :Ex

3

zf

zzzz

zzzf

.8/1 is 2at )( of residue the 3,order of pole a is 2

32

2

16

1

)2(8

1

)2(4

1

)2(2

1..

3216

1

8

1

4

1

2

1

...])2

()2

()2

()2

(1[2

1

)2/1(2

1)(

)2/1(2)2()2( 2set 2point (2)

1order of pole a is 0 ...32

5

16

3

16

3

8

1

...])2

(!3

)5)(4)(3()

2(

!2

)4)(3()

2)(3(1[

8

1

)2/1(8

1)(

0point (1)

2323

432

33

333

2

32

3

zzfz

z

zzz

zf

zzzz

zz

zz

zzz

zzzzf

z

001

1

10

0

0

1

101

1

101010

202010

0

1

0

at residue )]}()[()!1(

1{lim)(

limit the Take

)()!1()]()[(

...)(.......)()()(

...)()()(

......)(

)(

0

zzzfzzdz

d

mazR

zz

zzbamzfzzdz

d

zzazzaazfzz

zzazzaazz

a

zz

azf

m

m

m

zz

n

n

nm

m

m

mmm

m

m

m

How to obtain the residue ?

)(

)(

)(

1lim)(

)(

)(lim)(

)(

)()(lim)(

0)( and at zero-non

and analytic is )( ,)(

)()( and at simple a has )( If (2)

)]()[(lim)(1 pole simple aFor (1)

0'

0

'00

00

0

00

0

00

000

0

zh

zg

zhzg

zh

zzzg

zh

zgzzzR

zhz

zgzh

zgzfzzzf

zfzzzRm

zzzzzz

zz

.point theat )1(

exp)( function the of residue the

evaluate Hence .at )( of )( residue thefor expression

general a deriving ,about )( of seriesLaurent the gconsiderin

By .point theat order of pole a has )(that Suppose :Ex

22

00

0

0

izz

izzf

zzzfzR

zzf

zzmzf

e

ie

ie

i

iiR

iziz

iziz

i

iz

iz

dz

dzfiz

dz

d

iz

iziziziz

iz

z

izzf

2]

)2(

2

)2([

!1

1)(

exp)(

2exp

)(]

)(

exp[)]()[(

:at polefor

and at 2order of poles )()(

exp

)1(

exp)(

1

3

1

2

322

2

2222

20.14 Residue theorem

0z

C

C

ninnin

nin

mn

nn

Cmn

n

ii

C

n

mn

n

iadzzfI

iidn

ni

eidein

deiadzzzaI

deidzezz

dzzfdzzfI

zzazf

zzmzf

1

2

0

20

)1(1)1(2

0

1

)1(2

0

10

0

0

0

2)(

2 1for

0|)1(

1for

)(

set

)()(

)()(

at order of pole a has )(

Residue theorem:

C within poles itsat )( of residues the of sum the is

2)(

C within poles ofnumber finite afor except analytic, and

Ccontour closed a on and within continuous is )(

zfR

Ridzzf

zf

j

j

j

jC

C 'C

Ccontour open an along )( of integral The zfIC

0z

112

121100

0

101

0

2010

0

101

0

2 2contour closed afor

)()1

(lim)(lim

0)(lim

)()()(

except )( of ysingularit nothat enough small chosen is

)arg( and ||

gsurroundinneighbour some within analytic is )(

)()()(

at pole simple a has )( if

2

1

iaI

iadeie

adzzfI

dzz

dzzzadzzdzzfI

zzzf

zzzz

zz

zzazzf

zzzf

i

iC

C

CC C

20.16 Integrals of sinusoidal functions

dzizdz

ziz

z

izdF

1

2

0

),1

(2

1sin ,)

1(

2

1cos

circleunit in expset )sin,(cos

0for cos2

2cos Evaluate :Ex

2

0 22

abd

abbaI

dz

a

bz

b

azz

z

ab

i

a

b

b

azzz

dzz

ab

i

ababzzbzaz

idzz

zzabba

dzizzz

dabba

zzzznnn

))((

)1(

2)1)((

)1(

2

)(

)1(2

1

)(2

12

))((2

1

cos2

2cos

)(2

12cos )(

2

1cos

2

4

22

4

2222

4

122

122

22

22

)(

2]

)([

22

)(

)(

)//()/(

1)/(]

)/)(/(

1)/[(lim)/(

1 ,/at pole (2)

//})/()/(

)]//(2)[1)(1(

)/)(/(

4{lim

]})/)(/(

1[

!1

1{lim)0(

2 ,0at pole (1)

)]()[()!1(

1{lim)( : Residue

circleunit the within / and 0at poles double

))((

1

2

222

2

22

4422

22

44

2

4

2

4

/

22

43

0

2

42

0

01

1

0

2

4

0

abb

a

abab

ba

ab

ba

ab

iiI

abab

ba

abbaba

ba

abzbazz

zbazbaR

mbaz

abbaabzbaz

abbazz

abzbaz

z

abzbazz

zz

dz

dR

mz

zfzzdz

d

mzR

bazzdz

a

bz

b

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z

ab

iI

baz

z

z

m

m

m

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C

Some infinite integrals

dxxf )(

. as zero to tends

along integral the ,) on || of maximum(2|)(|for

2)(

exist both )( and )( (3)

).|| as 0)(that is

condition sufficient (a as zero to tends on ||

of maximum the times , radius of semicircle a on (2)

axis. real the on is which of none poles, ofnumber finite a

forexcept ,0Im plane,-halfupper the in analytic is )( (1)

:properties following the has )(

0

0

R

ΓfRdzzf

Ridxxf

dxxfdxxf

zzzf

RΓf

RRΓ

zzf

zf

j

j

y

R 0 R

x

real is )(

Evaluate :Ex0 422

aax

dxI

77770 422

7

3

4

1

4

442422

422

422422422

422422422

32

5

32

10

2

1

32

10)

32

5(2

)(

32

5)

2(

!3

)6)(5)(4(

)2(

1 is oft coefficien the

)2

1()2(

1

)2(

1

)(

1 0 ,set

plane-halfupper theat only

,at 4order of poles 0)(

)()(0

)(

as )()()(

aaI

aa

ii

ax

dx

a

i

a

i

a

a

i

aiaiazaiz

aiz

aizaz

ax

dx

az

dz

az

dz

Raz

dz

ax

dx

az

dz

CΓ

Γ

R

RC

R 0 R

ai

R 0 R - 0z

y

x

For poles on the real axis:

zero is integral the , as /1 thanfaster vanishes )( If

Re)(Re)(

Re Reset )(for (2)

)( at pole a has )(for (1)

)()()(

)()()()()(

contour closed afor

)()()(

0 as defined integral, the of value Principal

2

10

0

0

0

0

RRzf

difdzzf

didzzdzzf

aidzzfzzdzzf

dzzfdzzfdxxfP

dzzfdxxfdzzfdxxfdzzf

C

dxxfdxxfdxxfP

ii

iiθ

Γ

R

R

R

z

z

RC

R

z

z

R

R

R

Jordan’s lemma

contourar semicircul the is , as 0)(

then ,0 (3)

plane-halfupper the in || as 0|)(| of maximum the )2(

0Im in poles ofnumber

finite afor except plane-halfupper the in analytic is )( )1(

RdzzfeI

m

zzf

z

zf

imz

0 0 as

)1(2

0 ,on )(of maximum e

2

)(

)sinexp()exp(

2//sin1 ,2/0for

2/

0

)/2(

2

0

sin

0

sin

IMR

m

Me

m

MdeMRI

MRR |z||z |fM is th

dθeMR

deMR||dz|zf|eI

|θmR||imz|

mRmR

π/ θmR

Γ

π θmRimz

2/

)(f /2f

sinf

a R 0 R -

0 real, cos

of value principal the Find :Ex

madx

ax

mx

madxax

mxPmadx

ax

mxP

eaaidxax

eP

dzaz

eR

dzaz

edx

ax

edz

az

edx

ax

e

dzaz

eI

|z||az|

dzaz

eI

imaimx

imz

imzR

a

imximza

R

imx

C

imz

C

imz

cossin

and sincos

and 0

0 0 and As

0

as 0)( and plane,-halfupper

the in pole no 0 integral theConsider

11

1

Integral of multivalued functions

y

x

R

B A

D Copposite. and equalnot are

to joining CD and ABlines two along

values its d,multivalue is integrand The .0 and

let Weotigin. theat ispoint branch Single

, as such functions dMultivalue2/1

Rzz

R

Lnzz

0for )(

:Ex0 2/13

axax

dxI

2/5

2/1

2

2

2/13

3

13

13

2/12/2/12/1

2/13

8

3

!2

1lim

])(

1)[(

)!13(

1lim)(

)( and ,at pole (2)

integralcontour the to oncontributi no make circles two the

and 0 as 0)( ,)()( integrand the (1)

a

iz

dz

d

zazaz

dz

daR

iaeaaaz

R|z|zfzazzf

az

az

i

2/50 2/13

2/50 2/13

2/5

0

, )22/1(2/132,0 2/13

20

2/5

8

3

)(

4

3

)()

11(

4

3

)()(

CD line along , ABline along

0 and 0 and

)8

3(2

axax

dx

axax

dx

e

aexaxe

dx

xax

dx

xezxez

dzdz

a

iidzdzdzdz

i

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