DEPARTMENT OF MATHEMATI CS [ YEAR OF ESTABLISHMENT – 1997 ] DEPARTMENT OF MATHEMATICS, CVRCE

DEPARTMENT OF MATHEMATICS, CVRCE

DEPARTMENT OF MATHEMATICS

[YEAR OF ESTABLISHMENT –

1997]

MATHEMATICS - II

● LAPLACE TRANSFORMS● FOURIER SERIES● FOURIER TRANSFORMS● VECTOR DIFFERENTIAL CALCULUS● VECTOR INTEGRAL CALCULUS● LINE, DOUBLE, SURFACE, VOLUME INTEGRALS

● BETA AND GAMMA FUNCTIONS

FOR BTECH SECOND SEMESTER COURSE [COMMON TO ALL BRANCHES OF ENGINEERING]

DEPARTMENT OF MATHEMATICS, CVRCE

TEXT BOOK: ADVANCED

ENGINEERING MATHEMATICS BY ERWIN KREYSZIG

[8th EDITION]

MATHEMATICS - II

DEPARTMENT OF MATHEMATICS, CVRCE

FOURIER INTEGRAL [chapter – 10.8]

LECTURE :16

DEPARTMENT OF MATHEMATICS, CVRCE

LAYOUT OF LECTURE

EXISTENCE OF FOURIER INTEGRAL

INTRODUCTION &

MOTIVATION

FROM FOURIER

SERIES TO FOURIER INTEGRAL

APPLICATIONS

FOURIER SINE AND COSINE

INTEGRALS

SOME PROBLEMS

INTRODUCTION & MOTIVATION

FOURIER SERIES ARE POWERFUL TOOLS IN TREATING VARIOUS PROBLLEMS INVOLVING PERIODIC FUNCTIONS. HOWEVER, FOURIER SERIES ARE NOT APPLICABLE TO MANY PRACTICAL PROBLEMS SUCH AS A SINGLE PULSE OF AN ELECTRICAL SIGNAL OR MECHANICAL FORCE VIBRATION WHICH INVOLVE NONPERIODIC FUNCTIONS. THIS SHOWS THAT METHOD OF FOURIER SERIES NEEDS TO BE EXTENDED. HERE WE START WITH THE FOURIER SERIES OF AN ARBITRARY PERIODIC FUNCTION fL OF PERIOD 2L AND THEN LET L SO AS TO DEVELOP FOURIER INTEGRAL OF A NON-PERIODIC FUNCTION.

Jean Baptiste Joseph Fourier (Mar21st 1768 –May16th 1830) French Mathematician & Physicist

FROM FOURIER SERIES TO FOURIER INTEGRAL

Let fL (x) be an arbitrary periodic function whose period is 2L which can be represented by its Fourier series as follows:

DEPARTMENT OF MATHEMATICS, CVRCE

01

( ) cos sin (1), where L n n n n nn

nf x a a w x b w x w

L

0

1( ) (2)

2

L

L

L

a f x dxL

1

( )cos ; 1,2,... (3)L

n L n

L

a f x w xdx nL

1

( )sin ; 1,2,... (4)L

n L n

L

b f x w xdx nL

FROM FOURIER SERIES TO FOURIER INTEGRAL

From (1), (2), (3), and (4), we get

1

cos ( )cos1 1

( ) ( ) 52

sin ( )sin

L

n L nLL

L L LnL

n L n

L

w x f v w vdv

f x f v dvL L

w x f v w vdv

1

1Set n n

n nw w w

L L L

1

cos ( )cos1 1

( ) ( ) 62

sin ( )sin

L

n L nLL

L L LnL

n L n

L

w x w f v w vdv

f x f v dvL

w x w f v w vdv

FROM FOURIER SERIES TO FOURIER INTEGRAL

The representation (6) is valid for any fixed L, arbitrary large but finite. We now let L and assume that the resultant non-periodic function is absolutely integrable on x-axis , i.e. the resulting non-periodic function

lim LLf x f x

FROM FOURIER SERIES TO FOURIER INTEGRAL

0

0

lim limb

a ba

f x dx f x dx f x dx

Set

0

1( ) cos ( ) cos sin ( )sin 7f x wx f v wvdv wx f v wvdv

is absolutely integrable.

1 ( ) cos (8)

1and ( )sin (9)

A w f v wvdv

B w f v wvdv

Applying (8) and (9) in (7), we get

0

1( ) cos sin 10f x A w wx B w wx dw

FROM FOURIER SERIES TO FOURIER INTEGRAL

Representation (10) with A(w) and B(w) given by (8) and (9), respectively, is called a Fourier integral of f(x).

FOURIER INTEGRAL

0

( ) ( ( ) cos ( )sin )

The above equation is true at a point of continuity

At a point of discontinity,the value of the integral

1on the right is equal to [ ( 0) ( 0)]

2

f x A w wx B w wx dx

f x f x

1( ) ( ) cos

where

A w f v wvdv

1

( ) ( )sin

and

B w f v wvdv

EXISTENCE OF FOURIER INTEGRALTheorem

0

0

lim limb

a ba

f x dx f x dx

If a function f(x) is piecewise continuous in every finite interval and has a right-hand derivative and left-hand derivative at every point and if the integral

exists, then f(x) can be represented by a Fourier integral. At a point where f(x) is discontinuous the value of the Fourier integral equals the average of the left and right hand limits of f(x) at that point.

EXAMPLE-1: Find the Fourier integral representation of the function

1 if 1( )

0 if 1

xf x

x

PROBLEMS ON FOURIER INTEGRAL

The Fourier integral of the given function f(x) is

0

( ) ( ) cos ( ) ( )sinf x A w w x B w wx

Solution:

1 1with ( ) ( ) cos and ( ) ( )sinA w f v wvdv B w f v wvdv

1 1

1 1

1( )cos ( ) cos ( ) cosf v wvdv f v wvdv f v wvdv

1

1

10 cos 0wvdv

PROBLEMS ON FOURIER INTEGRAL

1 1

1 1

10cos 1 cos 0coswvdv wvdv wvdv

1

0

2cos wvdv

1

0

2 sin wv

w

2sin w

w

1( ) ( ) cosA w f v wvdv

1( ) ( )sinB w f v wvdv

1 1

1 1

1( )sin ( )sin ( )sinf v wvdv f v wvdv f v wvdv

0 ( )sin is an odd function off v wv v

PROBLEMS ON FOURIER INTEGRAL

1 1

1 1

10 sin 1 sin 0 sinwvdv wvdv wvdv

1

1

10 sin 0wvdv

1

1

1sin wvdv

0

Replacing the values of ( ) and ( ) in (1) we ge the

fourier integral representation of the given function ( ) s

2 sin cos( )

A w B w

f x a

w wxf x dw

w

PROBLEMS ON FOURIER INTEGRAL

DIRICHLET’S DISCONTINUOUS FACTOR

0

1 if 1 2 sin cos( ) 1

0 if 1

x w wxf x dw

x w

From Example – 1 by Fourier integral representation, we find that

At x= 1, the function f(x) is discontinuous. Hence the value of the Fourier integral at 1 is ½ [f(1-0)+f(1+0)] = ½[0+1]=1/2.

0

1 if 0 1

2 sin cos 1Therefore, 1 if 1

20 if 1

x

w wxdw x

wx

DIRICHLET’S DISCONTINUOUS FACTOR

0

if 0 12

sin cosif 1

40 if 1

x

w wxdw x

wx

The above integral is called Dirichlet’s discontinuous factor.

Sine integral

Dirichlet’s discontinuous factor is given by.

0

if 0 12

sin cosif 1

40 if 1

x

w wxdw x

wx

Putting x= 0 in the above expression we get .

0

sin(1)

2

wdw

w

The integral (1) is the limit of the integral as u 0

sinu wdw

w

The integral is called the sine integral and it is denoted by Si(u)

0

sinu wdw

w

FOURIER COSINE INTEGRAL

If ( ) is an even function, thenf x

0

1 2( ) ( ) cos ( )cos

Since ( ) cos is an even function of

A w f v wvdv f v wvdv

f v wv v

1and ( ) ( )sin 0

Since ( )sin is an odd function of

B w f v wvdv

f v wv v

0

0

Therefore, the Fourier integral representation of ( ) is given by

( ) cos sin

cos

f x

f x A w wx B w wx dw

A w wxdw

FOURIER COSINE INTEGRAL

0

0

Therefore, the Fourier integral representation of

an even function ( ) is given by

( ) cos ,

2where ( ) ( ) cos

f x

f x A w wxdw

A w f v wvdv

The Fourier integral of an even function is also

known as Fourier cosine integral.

FOURIER SINE INTEGRAL

If ( ) is an odd function, thenf x

1( ) ( ) cos 0

Since ( ) cos is an odd function of

A w f v wvdv

f v wv v

0

1 2and ( ) ( )sin ( )sin

Since ( )sin is an even function of

B w f v wvdv f v wvdv

f v wv v

0

0

Therefore, the Fourier integral representation of ( ) is given by

( ) cos sin

cos

f x

f x A w wx B w wx dw

B w wxdw

FOURIER SINE INTEGRAL

0

0

Therefore, the Fourier integral representation of

an odd function ( ) is given by

( ) sin ,

2where ( ) ( )sin

f x

f x B w wxdw

B w f v wvdv

The Fourier integral of an odd function is also known as Fourier sine integral.

2Example

Find the Fourier cosine and sine integral of

( ) , , 0kxf x e where x k

PROBLEMS INVOLVING FOURIER COSINE AND SINE INTEGRAL

The Fourier cosine integral of the given function f(x) is

0 0

2( ) ( ) cos with ( ) ( ) cosf x A w wxdw A w f v wvdv

Solution:

0

2( ) ( ) cosA w f v wvdv

0

2coskve wvdv

PROBLEMS INVOLVING FOURIER COSINE AND SINE INTEGRAL

2 2

0

2 ( cos sin )kve k wv w wv

k w

2 2

20

k

k w

2 2

2k

k w

2 20

Therefore, the fourier cosine integral representation of ( ) is

2 cos( )

f x

k wxf x dw

k w

PROBLEMS INVOLVING FOURIER COSINE AND SINE INTEGRALThe Fourier sine integral of the given function f(x)

is

0 0

2( ) ( )sin with ( ) ( )sinf x B w wxdw B w f v wvdv

0

2So, we have ( ) sinkvB w e wvdv

2 2

0

2 ( sin cos )kve k wv w wv

k w

2 2

20

w

k w

2 2

2

( )

w

k w

Hence the Fourier sine integral of the given function is

2 20

2 sin( )

w wxf x dw

k w

PROBLEMS INVOLVING FOURIER COSINE AND SINE INTEGRAL

LAPLACE INTEGRALS

From Example – 2 by Fourier cosine integral representation, we find that

2 20

2 cos

( )kx k wx

f x e dwk w

2 20

cos

( ) 2kxwx

dw ek w k

(A)

LAPLACE INTEGRALS

From Example – 2 by Fourier sine integral representation, we find that

2 20

2 sin

( )kx wx

f x e dww k w

2 20

sin

( ) 2kxwx

dw ew k w

The integrals (A) and (B) are called as Laplace integrals.

(B)

20

3. Using Fourier integral prove that

0 0

cos sin0

1 2

0x

if x

xw w xwdw if x

w

e if x

SOME MORE PROBLEMS

0 0

Let 02

0x

if x

f x if x

e if x

Solution:

Fourier integral representation of ( ) is given byf x

0

( ) ( ) cos ( )sin 1f x A w wx B w wx dw

1 1

with ( ) ( ) cos and ( ) ( )sinA w f v wvdv B w f v wvdv

0

0

1So, ( ) ( ) cos + ( )cos A w f v wvdv f v wvdv

0

0

1(0)cos + ( )cos vwv dv e wv dv

SOME MORE PROBLEMS

0

0

1(0) + ( )cos vdv e wv dv

0

( ) cosve wv dv

2 2

0

cos sin1

vewv w wv

w

2 2

10 1

1 w

2

1

1 w

SOME MORE PROBLEMS

0

0

1 ( ) ( )sin + ( )sin B w f v wvdv f v wvdv

0

0

1(0)sin + ( )sin vwv dv e wv dv

0

0

1(0) + ( )sin vdv e wv dv

0

( )sinve wv dv

SOME MORE PROBLEMS

2 2

0

sin cos1

vewv w wv

w

2 2

10

1w

w

21

w

w

SOME MORE PROBLEMS

2 20

So (1) ( ) cos sin1

1 1f x wx wx dw

w

w w

20

cos

1

sin( )

wx wxf x

wdw

w

20

cos sinTherefo

0 0

02

0

re, we ge 1

t

x

wx wxdw f x

if x

if x

e if x

w

w

SOME MORE PROBLEMS

0

0 12

sin cos4.Pr 1

40 1

if x

w wxove that dw if x

wif x

0 12

Let 140 1

if x

f x if x

if x

Solution:

SOME MORE PROBLEMS

Fourier cosine integral representation of ( ) is given byf x

0

( ) ( ) cos 1f x A w wxdw

0

2with ( ) ( ) cos A w f v wvdv

1

0 1

2( )cos + ( )cos f v wvdv f v wvdv

1

0 1

2cos + (0)cos

2wv dv wv dv

SOME MORE PROBLEMS

1

0

cos wv dv 101

sin wvw

sin w

w

0

So (1) ( ) cossin w

wf x wx dw

0

ssin co wxd

w

ww

0

si cosTherefore, we

0

n

12

140

g

1

etwx

dw f

if x

if x

if x

w

wx

SOME MORE PROBLEMS

0

01 cos5.Pr sin 2

0

if xwove that wxdw

wif x

0

2( ) ( )sinwhere B w f v wvdv

0

The fourier sine integral of ( ) is given by

( ) ( )sin (1)

f x

f x B w wxdw

SOME MORE PROBLEMS

Solution: 0Let 2

0

if xf x

if x

0

2( )sin ( )sinf v wvdv f v wvdv

00

2 cossin 0

2

wvwvdv

w

1 1 coscos 1

ww

w w

SOME MORE PROBLEMS

0

1 cos( ) sin

wf x wxdw

w

Hence from (1) we obtain

0

01 cossin 2

0

if xwwxdw f x

wif x

SOME MORE PROBLEMS

20

cos6.Pr 0

1 2x xxw

ove that dw e if e if xw

Let ( ) 02

xf x e if x

0

( ) ( ) cos (1)f x A w xwdw

The Fourier cosine integral of the function f(x) is given by

SOME MORE PROBLEMS

Solution:

0 0

2 2Where ( ) ( ) cos cos

2vA w f v wvdv e wvdv

2

0

( cos sin )

1

ve wv w wv

w

2 2

0 ( 1) 1

1 1w w

SOME MORE PROBLEMS

20

cos(1) ( )

1

xwHence from we obtain f x dw

w

20

cos. 0

1 2x xxw

i e dw e if e if xw

SOME MORE PROBLEMS

7.Find Fourier Cosine Integral Representation

1 if 0 1( )

0 if 1

xf x

x

The Fourier cosine integral of f(x) is given by

0

( ) ( ) cosf x A w xwdw

1

0 0

2 2Where ( ) ( ) cos 1.cosA w f v wvdv wvdv

SOME MORE PROBLEMS

Solution:

1

0

2 sin 2sinwv w

w w

Hence from (1) we obtain the required Fourier cosine integral as

0

2 sin cos( )

w xwf x dw

w

SOME MORE PROBLEMS

28. Find the cosine integral of ( ) ( 0)x xf x e e x

2Solution: Given function is ( ) where 0x xf x e e x

The Fourier cosine integral of the given function f(x) is

0

( ) ( ) cos (1)f x A w xwdw

0

2Where ( ) ( ) cosA w f v wvdv

SOME MORE PROBLEMS

2

0

2( )cosv ve e wvdv

2

0 0

2cos cosv ve wvdv e wvdv

SOME MORE PROBLEMS

2

0

2

0

cos sin12

2cos sin4

v

v

ewv w wv

w

ewv w wv

w

2 2

2 1 2

1 4w w

2 20

2 1 2( ) cos

1 4f x xwdw

w w

SOME MORE PROBLEMS

2 2

2 1 20 0

1 4w w

Hence the Fourier cosine integral of the given function is

1 09. Find Fourier Sine Integral of ( )

0

if a af x

if x a

1if 0Solution: Given function is ( )

0 if

a af x

x a

The Fourier sine integral of the given function f(x) is

0

( ) ( )sin (1)f x B w xwdw

SOME MORE PROBLEMS

0

2( ) ( )sinWhere B w f v wvdv

0

2( )sin ( )sin

a

a

f v wvdv f v wvdv

00

2 2 cossin 0

aa wvwvdv

w

2 cos 1 2(1 cos )aw aw

w w

0

2 1 cos( ) sin

awf x xwdw

w

SOME MORE PROBLEMS

0

2(1)sin (0)sin

a

a

wvdv wvdv

Hence the Fourier sine integral of the given function is

Test your knowledge

20

cos cos cos2 2 21. Pr

10

2

wxw x if x

ove that dww

x

3

40

sin.2Pr cos 0

4 2xw xw

Q ove that dw e x if xw

2 0 1.3. sin ( )

0 1

x if xQ Find fourier Co e Integral of f x

if x

2 2 0.4. sin ( )

0

a x if x aQ Find fourier Co e Integral of f x

if x a

sin 0.5. sin ( )

0

x if xQ Find fourier e Integral of f x

if x

0 1.6. sin ( )

0 1

xe if xQ Find fourier e Integral of f x

if x