department of mathemati cs [ year of establishment – 1997 ] department of mathematics, cvrce
TRANSCRIPT
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DEPARTMENT OF MATHEMATICS, CVRCE
DEPARTMENT OF MATHEMATICS
[YEAR OF ESTABLISHMENT –
1997]
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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]
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MATHEMATICS - II
DEPARTMENT OF MATHEMATICS, CVRCE
FOURIER INTEGRAL [chapter – 10.8]
LECTURE :16
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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
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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
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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
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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
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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
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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
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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).
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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
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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.
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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:
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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
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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
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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
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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
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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.
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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
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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
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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.
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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
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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.
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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:
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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
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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
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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
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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)
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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)
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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:
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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
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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
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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
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2 2
0
sin cos1
vewv w wv
w
2 2
10
1w
w
21
w
w
SOME MORE PROBLEMS
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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
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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
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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
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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
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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
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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
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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
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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:
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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
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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
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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
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Solution:
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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
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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
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2
0
2( )cosv ve e wvdv
2
0 0
2cos cosv ve wvdv e wvdv
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2
0
2
0
cos sin12
2cos sin4
v
v
ewv w wv
w
ewv w wv
w
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2 2
2 1 2
1 4w w
2 20
2 1 2( ) cos
1 4f x xwdw
w w
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2 2
2 1 20 0
1 4w w
Hence the Fourier cosine integral of the given function is
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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
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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
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0
2(1)sin (0)sin
a
a
wvdv wvdv
Hence the Fourier sine integral of the given function is
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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
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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