UNIT-III

Transform Techniques

MIT SOE
 Preliminaries Basic Formulae

Fourier series:

Fourier series of a periodic function f ( x) in the interval  c, c  2l  is given by

a0   n x n x 
f ( x)    n 1  an cos  bn sin 
2  l l 

Where ,

1 c  2l 1 c  2l n x 1 c  2l n x
a0   f   n c  
x dx , a  f x cos dx , bn   f  
x sin dx
l c l l l c l

MIT SOE
Fourier Integral representation or Fourier Integral Theorem

(i) If f ( x) satisfies the Dirichlet’s conditions and


(ii) If f ( x) is absolutely integrable i.e.  f ( x) dx converges, then


f  x =
1
2 
 

  u 

u 
f  u  e  i  u  x  du d  

If we write 
f (u)eiu du = F ( ) , then (1) can be written as

1 
f  x =  F    e i x d 
 

The function F ( ) is called Fourier Transform of f  x  . We can write F     F  f  x 

The function f  x  is called Inverse Fourier Transform of F ( ) .

MIT SOE
 Fourier Transform and Inverse Fourier Transform:

Fourier Transform of the function f  x  in the interval  ,  is defined by


F ( )   
f (u )e iu du

And the inverse Fourier transform of F ( ) is given by

1 
f ( x) 
2  
F (  ) e i x d 

Even Function:
A function f  x  is said to be even if f   x   f  x   x . Graph of even function is symmetric about y-

f  x  dx  2 f  x  dx .
a a
axis. For even function, 
a 0
Odd Function:
A function f  x  is said to be odd if f   x    f  x   x . Graph of even function is symmetric about

f  x  dx  0 .
a
opposite quadrant. For odd function, 
a

MIT SOE
Fourier Transform for Even Function:

If f  x  is even function in the interval  ,   , then Fourier Transform is given by


F ( )   0
f (u ) cos u du

And the inverse Fourier Transform of F ( ) is given by

2 
f ( x) 
  0
F ( ) cos  x d 

MIT SOE
Fourier Transform for Odd Function:

If f  x  is odd function in the interval  ,   , then from (2)


F     f (u ) sin u du
0

And the inverse Fourier Transform of F ( ) is given by

2 
f ( x) 
  0
F ( ) sin  x d 

MIT SOE
Fourier Cosine Transform:

If f  x  is defined in the interval  0,   , then Fourier Cosine Transform of f  x  is defined by


Fc      f (u ) cos u du
0

And the inverse Fourier Cosine Transform of Fc ( ) is given by

2 
f ( x) 
 0
Fc ( ) cos  x d 

MIT SOE
Fourier Sine Transform:

If f  x  is defined in the interval  0,   , then Fourier Sine Transform of f  x  is defined by


Fs      f (u )sin u du
0

And the inverse Fourier Sine Transform of Fs ( ) is given by

2 

 
f ( x)  Fs ( ) sin  x d 
0

MIT SOE
Some Formulae:

(i) If f ( x)  f ( x)   f ( x) , then function f  x  is neither even nor odd.

(ii) eax is neither even nor an odd function since eax  eax  e ax .

 du 
(iii) Integration by parts  uvdx  u  vdx     vdx  dx , where u and v are functions of x.
 dx 

Generalized Integration by parts  uvdx  u 'v  u"v1  u '''v2  u""v3  ....  .. , where dashes  '

denote derivatives and suffixes   denote the integrations.

|

mn
(iv) B  m, n   .
mn

MIT SOE
 m n
(iv) B  m, n   .
Some Formulae: mn
(v) n  1  n n and n  1  n ! if n is positive integers, 1 2   .

Differentiation Under Integral Sign (DUIS): If I     f  x,   dx , then

b
(vi)
a

dI   d 
f  x,   dx    f  x,   dx
b b

d

d  a a  
 sin ax   2 If a is positive
(vii) 0 x
dx  
   2 If a is negative
.

 a  b
(viii)  0
e  ax cos bx dx 
a 2  b2
and 
0
e  ax sin bx dx 
a 2  b2
.

MIT SOE
 1, 𝑥 > 0
Example: Find the Fourier transform F(λ) of 𝑓 𝑥 =ቊ
0, 𝑥 ≤ 0
Solution: We have to find F(λ),which is the Fourier transform of given function 𝑓 𝑥
First we check whether given function f(x) is even or odd

Consider,
1, −𝑥 > 0
𝑓 −𝑥 = ቊ
0, −𝑥 ≤ 0
1, 𝑥 < 0
=ቊ
0, 𝑥 ≥ 0

1, 𝑥 > 0
≠𝑓 𝑥 =ቊ
0, 𝑥 ≤ 0

i.e. 𝑓 −𝑥 ≠ 𝑓 𝑥 , Also 𝑓 −𝑥 ≠ −𝑓 𝑥

∴ 𝑓 𝑥 is neither even function nor odd function

MIT SOE
 ∞

∴ 𝐹 λ = න 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢
−∞

0 ∞
= ‫׬‬−∞ 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢 + ‫׬‬0 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢

0 ∞
= ‫׬‬−∞ 0 𝑒 −𝑖λ𝑢 𝑑𝑢 + ‫׬‬0 (1) 𝑒 −𝑖λ𝑢 𝑑𝑢

∞
= 0 + ‫׬‬0 𝑒 −𝑖λ𝑢 𝑑𝑢

∞
𝑒 −𝑖λ𝑢
=
−𝑖λ 0

𝑒 −∞ 𝑒0 1 1 1
= − = 0 − = i.e. 𝐹 λ =
−𝑖λ −𝑖λ −𝑖λ 𝑖λ 𝑖λ

1, 𝑥 > 0 1
∴ Fourier transform of 𝑓 𝑥 = ቊ is 𝐹 λ =
0, 𝑥 ≤ 0 𝑖λ

MIT SOE
 𝑒 −𝑥 , 𝑥 > 0
Example: Find the Fourier transform F(λ) of 𝑓 𝑥 = ቊ
0, 𝑥 ≤ 0
Solution: We have to find F(λ),which is the Fourier transform of given function 𝑓 𝑥

First we check whether given function f(x) is even or odd

Consider,
𝑒 −(−𝑥) , −𝑥 > 0
𝑓 −𝑥 = ቊ
0, −𝑥 ≤ 0
𝑒𝑥, 𝑥 < 0
=ቊ
0, 𝑥 ≥ 0

𝑒 −𝑥 , 𝑥 > 0
≠𝑓 𝑥 =ቊ
0, 𝑥 ≤ 0

i.e. 𝑓 −𝑥 ≠ 𝑓 𝑥 , Also 𝑓 −𝑥 ≠ −𝑓 𝑥

∴ 𝑓 𝑥 is neither even function nor odd function

MIT SOE
 ∞

∴ 𝐹 λ = න 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢
−∞

0 ∞
= ‫׬‬−∞ 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢 + ‫׬‬0 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢

0 ∞
= ‫׬‬−∞ 0 𝑒 −𝑖λ𝑢 𝑑𝑢 + ‫׬‬0 (𝑒 −𝑢 ) 𝑒 −𝑖λ𝑢 𝑑𝑢

∞
= 0 + ‫׬‬0 𝑒 −(1+𝑖λ)𝑢 𝑑𝑢

∞
𝑒 −(1+𝑖λ)𝑢
=
−(1 + 𝑖λ) 0

𝑒 −∞ 𝑒0 1 1
= − = 0 − =
−(1+𝑖λ) −(1+𝑖λ) −(1+𝑖λ) 1+𝑖λ

𝑒 −𝑥 , 𝑥 > 0 1
∴ Fourier transform of 𝑓 𝑥 = ቊ is 𝐹 λ =
0, 𝑥 ≤ 0 1+𝑖λ

MIT SOE
 𝑥, 𝑥 > 0
Example: Find the Fourier transform F(λ) of 𝑓 𝑥 =ቊ
0, 𝑥 ≤ 0
Solution: We have to find F(λ),which is the Fourier transform of given function 𝑓 𝑥
First we check whether given function f(x) is even or odd

Consider,
−𝑥, −𝑥 > 0
𝑓 −𝑥 = ቊ
0, −𝑥 ≤ 0
𝑥, 𝑥 < 0
= −ቊ
0, 𝑥 ≥ 0

𝑥, 𝑥 > 0
≠ −𝑓 𝑥 = − ቊ
0, 𝑥 ≤ 0

i.e. 𝑓 −𝑥 ≠ −𝑓 𝑥 ,Also 𝑓 −𝑥 ≠ 𝑓 𝑥

∴ 𝑓 𝑥 is neither even function nor odd function

MIT SOE
 ∞

∴ 𝐹 λ = න 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢
−∞

0 ∞
= ‫׬‬−∞ 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢 + ‫׬‬0 𝑓 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢

0 ∞
= ‫׬‬−∞ 0 𝑒 −𝑖λ𝑢 𝑑𝑢 + ‫׬‬0 (𝑢) 𝑒 −𝑖λ𝑢 𝑑𝑢

∞
= 0 + ‫׬‬0 𝑢 𝑒 −𝑖λ𝑢 𝑑𝑢

∞ ∞
𝑒 −𝑖λ𝑢 𝑒 −𝑖λ𝑢
= 𝑢 − න (1) 𝑑𝑢
−𝑖λ 0
−𝑖λ
0
𝑒 −∞ 1 ∞ 1 1 1
= ∞ − (0) + ‫׬‬0 𝑒 −𝑖λ𝑢 𝑑𝑢 ∴ 𝐹 λ = =
−𝑖λ 𝑖λ 𝑖λ 𝑖λ −λ2

𝑥, 𝑥 > 0 1
∴ Fourier transform of 𝑓 𝑥 = ቊ is 𝐹 λ =
0, 𝑥 ≤ 0 −λ2

MIT SOE
 1 − 𝑥 2 ,for 𝑥 ≤ 1
Example 2: Find the Fourier transform of 𝑓(𝑥) = . Hence evaluate
0,for 𝑥 > 1
  x cos x  sin x 
0 
 x 3  cos x dx .


1    x  , for  x  1 
 1  x , for x  1
2 2

Solution: Here f   x      f  x  x

 0, for  x  1 
 0, for x  1

i.e. f  x  is even function of x .

 We find Fourier Cosine Transform.

Fourier Cosine Transform is given by


Fc    =  f (u) cos u du
0

1
2 ).𝑐𝑜𝑠 𝜆 𝑢𝑑𝑢 + ‫ ׬‬0. 𝑐𝑜𝑠 𝜆 𝑢𝑑𝑢 ∞
=
1
‫׬‬0 (1 − 2 = ∞ ‫׬‬ ( 1
𝑢 ).𝑐𝑜𝑠 𝜆 𝑢𝑑𝑢 + ‫ ׬‬0. 𝑐𝑜𝑠 𝜆0 𝑢𝑑𝑢 − 𝑢 1
1

  cos u     sin u   u 
1
 2  sin  u  2  sin
= 

1  u 
  
   2u   =2 
 
  2 
       
1 u 3   

 2u
MIT SOE  0
Illustration

Type 1: Examples on Fourier Integral Representation / Fourier Transform


 1, for x  a
Example 1: Find the Fourier transform of f  x   
0, for x  a.



1, for  x  a 
1, for x  a
Solution: Here f   x      f  x  x

 0, for  x  a 
 0, for x  a

i.e. f  x  is even function of x .

 We find Fourier Cosine Transform.

Fourier Cosine Transform is given by

Fc    =  f (u) cos u du
0

a 
=  1.cos u du  
0 a
0.cos u du
 sin u 
a
1
=     sin  a
MIT SOE 0 
   2
 2sin3 
  0
 2cos  
= 0  2cos

2cos
  2sin  3 
2sin  00 0 00
2sin    


== =
0  0  2  2cos
3 3
 0  0  0  
 0
  
 2   cos  2   3   
202   
    sin 0 0 0 
 Fc    =  3   
      
 == = 2 2 cos 3 23   cos   sin  

2 cos sin
 FFcF  
 cos
 sin  sin
c c
Now, we have tofind the Fc integral.
 = So, we 3 use 3Inverse Fourier Transform which is given by

Now, we have to find the integral. So, we use Inverse Fourier Transform which is isgiven
by byby
2   cosFourier  
Now,
Now,wewe have
have totofind find thethe integral.
 integral. So,So, we we useuseInverse  Fourier
Inverse Fourier Transform Transform which is given
which given
Now, we have to 2 find the integral. So, we 2 use  Inverse   sinTransform which is given by
f ( x) = 
 F ( ) cos  x d  
22  C 
 FCC(()2)cos
2
  
 
2 222  
cos cos
cos
cosxxddxd    0  2 323   coscos
sin  sin
  
3sin     cos 
xdxd 
x d
f ( x) = 00  FC () cos
f f(f(xx()x) ) === 2  xcos
 cos d
0F
FC ( ) cos0x 
0 0 
d 
    3
3  
sin
  cos  x d 
0
   0    
   cos   sin       
0
4  4 sin cos
==  
44 cos
 4   4cos3 cos sin sin
33cos
   cos  x d44sin
cos xdxd    
sin
 cos
 


cos
   xcos  x d
 000  3  3 cos xcos 0  0   3 3    cos xcos
== sincos     4  sin cos
 3cos
  x
cos d  d 
=
sin d  4 sin cos d  x d 
0x d

3

0     0 
  
0  0 
3

     2 2 2

      
 sin    cos  cos  x d   f x   4

sinsin
   sin 333 sin
cos
cos     
f fx x4 4  1  x  x, for
   
11x1 1x,xfor
2,x for
1  1x  1
x , xfor


0    3  
  cos cos cos
cos  d4
xxdxd 

d    f 0,
 x4  for
, for2 1 x 1
00
  
 
3

cos
 cos 4x 44

f x
4    0,0,0, 4 forforfor
x 1 x 1
x  1 1
0
  0, x for x 1

0
Put x  1 , we have
Put
Put xxx11,,1we we have
have
Put
 Put , we
x  have
1 , we have
 sin    cos  
  sin
 sin    cos  cos  d   0
  3 cos 
00sin
   
 cos
3 sin   cos

cos

cos
cosddd0 0
 cos  d   0
0 0   0
3
 3
  3
 
 sin x  x cos x   b f  x  dx  b f  t  dt 
 
sin x xxx3 cos x  cos x dx  0
 a b b b ba b b b b 
 0sin  sin  3 sin
x cos xxxcos
xcos    a f f xxdx    f ft tdtdt   
  fa f ttdt
x x cos x dx
x 0
 0 
0
 x
x0 
x 3 
x 
3coscos xx 
dx

dx cos 
 0
x
0 dx  0 
 
 

 a 
f  dx
xf 
dxxa adx
  dt 

0
3 a a a

MIT SOE
 Exercise

 1, for - 2  x  0
1. Find the Fourier transform of f  x    . Hence show that
 1, for 0  x  2
 1, for - 2  x  0
1. Find the Fourier transform of f  x    . Hence show that
  cos 2 x  1 sin 2 x    
 1, for - 2  x Ans:
1, for 0 x  0 F ( )  cos 2  1 
2
1. 0 Find the dx  
Fourier transform. of f  x     . Hence show  that
x 2    
  cos 2 x  1 sin 2 x   1, for 0  x 2 cos 2  1 
0  cos 2 xx 1 sin 2 xdx   2 .  Ans: F ( )   
    cos 2   1 
0 x
dx   .
2 -3 


Ans: F ( ) 
 
x <1 sin  cos  x
2. Find the Fourier Transform of f ( x)   . So, evaluate  d  and also deduce
0-3 xx >1 <1  
0 sin  cos  x
2. Find the Fourier Transform of f ( x)   . So, evaluate  d  and also deduce
 0 x >1 
 0

sin   
 sin   sin  cos  x  x  1  sin  
the value of 
 
d  .  F ( )  3 
   
 
d   2 ,   
d   .
0 sin    sin  0 sin  cos  x 0 x 1 1 0 sin  2 
the value of  d  .  F ( )  3   
d   2 x , d   .
      0  2 
0 0
  x 1 0
x 2 
3. Find the Fourier Transform of f ( x)  e  Ans: F ( )  1   2 
x  2 
3. Find the Fourier Transform of f ( x)  e  Ans: F (  ) 
1   2 

MIT SOE
Examples on Fourier Sine and Cosine Transform and Inverse Fourier
Transforms:

Example 1: Find the Fourier cosine and sine transforms of f ( x)  e2 x  4e3 x .

Solution: Fourier cosine Transform is given by


Fc ( )   0
f (u) cos u du


  0
(e2u  4e3u ) cos u du

 
  0
e2u cos u du  4 e3u cos u du
0

2 4.3   a 
 
22   2 32   2  
0
e  ax cos bx dx 
a 2  b 2 

 1 6 
 FC     2  2  2
   4   9 

MIT SOE
 Fourier Sine Transform is given by


FS ( )   0
f (u)sin u du


  0
(e2u  4e3u )sin u du

 
 0
e2u sin u du  4 e3u sin u du
0

 4. 
 
22   2 32   2

 1 4 
 FS       2  2
   4   9 

MIT SOE
 e  ax  x
Example 2: Find the Fourier sine transforms of f ( x)  and hence evaluate  tan 1   sin x dx .
x 0
a

Solution: Fourier sine Transform is given by


FS ( )  0
f (u)sin u du

 e  au
  0 u
sin u du - -- - - -- - - -- - - -- - - - (1)

Applying Differentiation under Integral Sign (DUIS) Rule

d    e au 
d
FS ( )  0

  u
sin  u  du

 e au
 0 u
cos u. u . du

  0
e au cos u du
d a   a 
 FS ( )    e  ax cos bx dx 
a  b 2 
d a2   2 0 2

MIT SOE
 Integrating
a
 d  FS ( )   a2   2 d   A
a 
FS ( )  tan 1    A - -- - - -- - - -- - - -- - - - (2)
a a

Put   0 in (1) and (2), we have

t 00inin(1)
(1)and
and (2), we have
(2), we have
F (0)  tan 1
0  A  0  0 A A  0
(0)  tan
FS (0) 
tan 00AA
11 S
0 0  0 
0A AA A0 0

 11FS   tan 1  

 FSS   tan
tan    a
aa 

Inverse Sine Transform is given by

2 
f  x  FS    sin  x d 
 0

2  

 
0
tan 1   sin  x d 
a
MIT SOE
    e  ax
 tan   sin  x d  
1

0
a 2 x

Put x  1 , we have

  
 0
tan 1   sin  d   e a
a 2

  x 
  0
tan 1   sin x dx  e  a
a 2

MIT SOE
Example 3: Find the Fourier cosine and sine transforms of the function f ( x)  e x and hence show that
 cos mx  m  x sin mx  m

0 1 x 2
dx 
2
e and 0 1 x 2
dx 
2
e .

Solution: Fourier cosine Transform is given by


Fc ( )  0
f (u) cos u du


  0
eu cos u du

1

1 2

1
 FC    
1 2

MIT SOE
Inverse Fourier Cosine Transform is given by

2 
f  x   FC    cos  x d 
 0

2  1

 0 1  2
cos  x d 

 cos  x   x
 d   f   e
x 
0 1  2
2 2

Put x  m , we have
 cosm   m
0 1  2
d  
2
e

 cos mx  m
  0 1 x 2
dx 
2
e

MIT SOE
 Fourier Sine Transform is given by
 
m is givenFourier
by FSine  0 f (uis)singiven
S ( )Transform u duby 
Fourier Sine Transform is given by
0
eu sin u du
   
0
f (u)sin u duF () 
FS ( ) 
S


10  20
0 u
 efu (sin uududu
)sin
f (u)sin u du 
  u ax
 00
e e sin 0

sin e
bx
u
u
du sin
dx   u b du
a 2  b 2 
    ax  b  ax    ax

FS     220 e sin bx dx  b 
1 2

111  2 a 0 b  0
2 e2 

sin bx
edx 
sin
a 2 bx
 b
dx
2  

 FS    

  
1 2  FS   1  2
1 2

2  2  
f  x   FS    sin  x d    sin  x d 
 0  0 1 2
  sin  x   x
 d   f  x   e
0 1 2 2 2

MIT SOE
 Put x  m , we have
  sin m  m
0 1 2
d  
2
e

 m  f  x  dx   f  t  dt 
 b b

x sin mx
 0 1 x 2
dx 
2
e  a a 

MIT SOE
 sin a
Example 4: Find inverse Fourier Cosine transform f  x  , if FC ( )  .

sin a
Solution: We have given that FC ( )  .

Inverse Fourier Cosine Transform is given by
2 
f  x  FC    cos  x d 
 0

1  2sin a cos  x

 
0 
d


 a  x)  sin(a  x)

1 sin(a  x)  sin(a  x)
1 sin(
d d   2sin A cos B  sin( A  B )  sin( A - B )

 0
 0 

2sin A cos B  sin( A  B ) 

1  sin(a  x) 
 
sin(a  x)
  0 0 
  d   d  
 

MIT SOE
 sin ax   2; 2;
  sin ax a  0 

11 22 2, aa  xx00 and
andaa- x- x 0 0 

2, a 0
 
 22      2; a  0 
dx dx
2,
2, aa  xx00 and
andaa- x- x 0 0  0 0 x x  2; a  0  

1, x  a and x  a i.e. | x | a 1, | x | a

   f  x  
0, x  a and x  a i.e. | x | a 0, | x | a

x  a i.e. | x | a 1, | x | a
 f  x  
x  a i.e. | x | a 0, | x | a

MIT SOE
 Exercise

1. Find the Fourier cosine transform of f  x   2e5 x  5e2 x .

  2 2  29 
 Ans: F (  )  10  2 
 (  4)(  25)  
c 2

2. Find the Fourier cosine and sine transform of f  x   xm1 .

 cos  m 2  sin  m 2  
 Ans: FC ( )  m and FS ( )  m
  m
m 
1  2 x
3. Find the inverse Fourier sine transforms if FS ( )  e a   0 .  Ans: f  x   tan 1 
   a
sin a
4. Find the inverse Fourier cosine transforms if FS ( )  .  Ans: f  x   2 


MIT SOE
 Solve integral equations
Illustrations

Example 1: Solve the following integral equation   f  x cos  x dx  e,   0 .
0 f  x   x cos
0    x edx e ,, e0,0..  0 .
Example
Example
Example
Example
1:1: Solve
1:
1: 1:Solve
Example
Example 1:
Solve
Solve
Solve
thethe
Solve
the
the
followingthe
the
following
following
following
following
following integralintegral
integral
integral integral
equation
integral equation
equation
equation
equation
equation


00 

f fx x0fcos
f
 f0 xxcos
cos
x
f 0xcos
cos
fcos
xxcos
xxdx
dx
dx dx
xdx
x
exe dx 
dx
,e,   
e  
e, 0 .
,. 0.0. 0 .
is integral equation 0 f  x  cos  x dx  e ,   0 .
Example
Example 1:1: SolveSolve thethe following
following integral
integral equation
equation   , 
Solution: Given equation
0
0  
Example 1: Given
Solution: Solve equation the following isis is
Solution:
Solution:
Solution:
Solution: Given Given
Given Given
equation equation equation
equation is is
Solution:
Solution: GivenGiven
 equationisisis 
equation
Solution:  Given  f  xequation 
Solution: Given equation
cos  x dx is  e ,   0 - - - - - - - - - - - - - - - (1)
0f  x  cos  x dx
0 f00xf00000fcos

xfffcos
  
xxxxcos
 dx 
xxxedx
, e , 0 ,0 0
e , 0 - - - - - - - - - - ----------- -(1) (1)
eee


cos
xcos
cos
cos
x dx
xdx
dx

xdx
e  
dx
,  
,0e
,   00 ------------------------------------------------------------- -----(1) - -(1) - (1)- - - (1)
- (1)

Fourier f  x Cosine  cos  x dx Transform  e ,  is0 given by - - - - - - - - - - - - - - - (1)
Fourier
FourierFourier
Fourier
0
Fourier Cosine Cosine Cosine
Cosine Transform
Transform
Cosine Transform
Transform
Transformis isgiven given is given
by bygiven
is by
Fourier
Fourier
Fourier Cosine
Cosine
Cosine
 Transform
Transform
Transform isisgiven
given
given bybyby by
Fc     f  u  cosu du e ,   0
Fourier Cosine Transform is given   by - - - - - - - - - - - - - - - (2)
Fc  c F
F FcFFF
0f0u00 fcos
c 
c ccc
   0 u ffffufu
 u uuucos
 
cos
  
uuudu
u  e
du 
 
ee
e  ,   0 - - ----------------------------------------------------------------------(2)
-----(2)
----(2)
--- (2)
(2)
 
F f - - - - -

cos u du
ducos ,  
 0
,0,e
Fc     000 f0  u  cosu du  e ,   0 ucos
cos
cos u edu u
du , du

00

, 0  0 - - - - -
(2) - - (2)
- - - - - - - - - - - - - - - (2)
Now Inverse 0 Fourier Cosine Transform is given by,
NowNow
NowInverse
Now Now Inverse
Inverse
Inverse
InverseFourier Fourier Fourier
Fourier
Fourier CosineCosine Cosine
Cosine
Cosine Transform
Transform Transform
Transform isisgiven
given by,
isby,
isisgiven given
given by, by,
by,
Now Now Inverse
Inverse Fourier
Fourier Cosine
Cosine Transform
Transform given
is given by,by,
NowInverse
Now Inverse Fourier 2 Fourier  Cosine Cosine Transform Transform is given is
by, given by,
f  x  2 F2222FC0Fcos
f fxffxffxxxx
x  2 

 F C cos  x d 
 cos  x d 
ff xx     2 
  0 00200FCCFCCCcosFF F   xcos
cos
cos dx xddxxddd
   2 00 0 FC    cos  x d   
0 C C
cos x

 2 222222e0ecos ecos xd dxxd d 


 cos
 0 eeeeecos 
 0 
  
02 x
  e 2 cos   x x dxxddd  

  cos
0  cos cos
cos 
 0 e cos  x d 

 
 0 00

2 221 11 0    ax ax  ax a a  a 

  2 12  
 0 e0ecos    
  0e e e  cos
e bx cos dx bx dx a 22 aaa 2 

1  cos bx 
dx a

 a
   1 
ax

2222x  111122  22
 0 0
bx dx 2 b
 ax 2 2


2
xx22   
ax  cos
ax bxdx dx adx b  2ab   2
 1 x 2112 cosax bx a
 e cos bx dx 2 
ax
  
0
  e coscos bx bx a aab b
2 dx 2
  2  2 2 
 1 x x   b 
0

 2112 x 2  bbba
2221
aa 
22


00  ax adx
 0 e cos bx
 f  xffxx  2 22 1 2 x
2 a 2  b 2 
 f  x  f f  x 1 21x1 22 xx222
 
  f f 
 x x 
  111xx x2222
   
  


1 x  
f x 1 x
f  x   MIT SOE

  e3
Example 2: Solve the following integral equation  f x  sin  x dx 3 .
 e3
fxex3dx
 e
 
f  x  sin   x dx .
0
Example
xample 2: the
2: Solve Solve the following
following integralintegral
equation equation sin .
Example 2: Solve
Solution: the following
Given equationintegral
is equation  f  x  sin  x dx 
0 0
.  
0 
Solution:Given Given
olution: equation equation
is is e3
Given equation is
f  x  sin  x dx3
Solution:
 
 0 
ee3
  3
e
- - - - - - - - - - - - - - - (1)
f f
x0x  sin x dx 
0 0 Fourier
sinf xx sin
dx  x dx  - - - - - - -- -- --- -- --- -------(1)
-- -- -- -- -- - -(1)
-- -- (

Sine Transform  given by
is
Fourier
Fourier SineTransform
Sine Transform isis given
givenby by
Fourier Sine Transform is given by
  e3 e3
S f
FS    F
0 0 f u udu sin
u  sin  u du 
 
- - - - - - - - - - - - -- ---- -(2)
- - - - - - - - (2)

Now Inverse Fourier Sine Transform

Now Inverse is given by,is given by,
Now Inverse FourierFourier Sine Transform
Sine Transform is given by,
2 
f  x  F  2 sin
f  x  f 0 x S  0 FS 2  sin
  x d

x d sin
  x d
 
 0
F S
2  e3

 2sinex3d - - - - - - - - - - - - - - - (3)

2 0e3
 sinex3d
0
 2 - - - - - - - - - - - - - - - (3)
 
 0   
sin  x d  sin  x d 
0 
- - - - - - - - - - - - -- --- - (3)
- - -- - -

MIT SOE
 By using DUIS Rule, we have
d 2    e3 
 f  x        d
dx 
sin x
 0 x   

2  e3

 cos  x.  d 
2  e 3 0 

 
0 
cos  x.  d 
2 

2 

  e3 cos  x d 

3 0
 e cos  x d 
 0
2 3 6
   32  ax 

2 3

6
 0 e x 2cos bx9dx
 
x2

 32  x 2  9  x 2  

Integrating

 
d  f  x   
 
6 1
 32  x 2  c

MIT SOE
 6 x  1 1 1  x  
f  x  tan 1    A   a2  x2 dx  tan  
3
6  x  3    1 1 1 ax
a
f  x  tan 1    A  a dx  tan   
3  3x  0 , we have
Put 
2
x 2
a  a 
Put x f 0, we0 =have
0 A A  0

f  0  0 =f0   
x A tan 1  0
2 A  x
    
 3
2  x
 f  x  tan 1  
 3

MIT SOE
   sin  x
Example 3: Using Fourier integral representations prove that e  x  e 2 x   d , x  0
0
 
 2
 1  2
 4 
Solution: Consider f  x   e x  e2 x

 f  u   eu  e2u

Since integral contains sin  x and x  0 x>0

Since integral contains sin  x and x  0 x>0
 Using Fourier Sine Transform
 Using Fourier Sine Transform

Fs ( )   f  u  sin u du
0

 0  eu  e2u  sin u du



 
 0 e sin u du 0 e2u sin u du
u

  3  
 2  2  sin bx  c  d
 ax
= 2  e
  1   4    1  2  4   0
MIT SOE
Now, inverse Fourier Sine Transform, we have

2 
f  x   FS    sin  x d 
 0

2  3

  
0

 1   4
2 2
sin  x d 

 e x  e2 x

6  
  sin  x d 
x 2 x
 e e 
 0 2

1  2  4 

MIT SOE
 Exercise
Exercise
Exercise
1 0  
1 10 01 1

1. Solve 1 0  the 1 integral equation

 f  x    x dx  2 1  
sin
1.
1. Solve  the integralExercise equation 0 f0 fx sin
 0 xdxx dx
x  sin  2 21 12. 2 .
 x  sin  x dx  2 1    2 . 0    2 0 
Solve the integral equation
 0
0 1 02 1
  2  2cos  
cos 2 x  2cos 2 x  1    
1. Solve
 Ans: f  
x  
2  
cos
x
the 
x
2cos

2
2cos
x
2

x
1
 
1
 

integral
  
equation  f  x  sin  x dx  2 1    2 .
 Ans: f  x   
Ans: f x 
      0
0
  x x   2
  x  
  2   x  2  x
2. 
Solve the 2
following  cos x 
integral2cos 2 x 
equation 1 
 f  x  

sin fsin xxsin
 x dx  e 
xedx
, 
 e  0  .
,  .  21Ans:
 0 Ans:
Ans: 2x   

2. Solve

2. SolveAns:
thef 
the x 
following
 
following integral
integral equation  0
equation
0 f  x  dx ,   0 .      2   1  x 2
x
   1  x  
0
  x   
 2  x    1 0    1
.  Ans:  2  e   1 0 .1  10Ans:
  21  x  
2. Solve the  1  x
following 
3.  Solve the following integral

integral equation

equation 0    
ff xx  sin
sin xxdxdx   ,1  
1   
1    2
0 2  2 
3. SolveSolvethe thefollowing
following integral equation
integral 0
equation 0 0 f  x 
 sin
f  
x x dx
sin 

 0
x 
dx  
1 
  2  1 1    
2  1  x 

0   1 0 1  2
 0 0   1   2
2 2cos x  3cos 2 x  1 sin 2 x  
  3. 2 Solve the following integral equation 0 fAns:

x  sinf xx dx
 2 2cos 1 1x3cos  22 x 1 sin 2 x  
  Ans: f  x    
    x  3cos
x 2 x2 1  sin 2 x 
2
2 x 2cos
2   Ans: f  x 0    x 2 x   
  x x2
 2  2cos x  3cos 2 x  1 sin 2 x    2  2cos x  3cos 2 x  1 sin 2 x  
Ans: f  x    
x  3cos 2 x  1 sin
          2 
x  
 
2 x x 2 Ans: f x
   x x  
x x 2  

MIT SOE