1. CHAPTER 4
FOURIER TRANSFORMS
INTEGRAL TRANSFORM
b
The integral transform of a function f (x ) is defined by ∫ f ( x).k (s , x)dx where
a
k(s , x) is a known function of s and x and it is called the kernel of the transform.
When k(s , x) is a sine or cosine function, we get transforms called Fourier sine or
cosine transforms.
FOURIER INTEGRAL THEOREM
If f (x) is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then
∞ ∞
1
π ∫ −∫
f ( x) = f (t ) cos λ (t − x) dt dλ
0 ∞
At a point of discontinuity the value of the integral on the left of above equation is
1
{ f ( x + 0) − f ( x − 0)}.
2
EXAMPLES
1 for x ≤ 1

1. Express the function f ( x) =  as a Fourier Integral. Hence evaluate
0 for x > 1

∞ ∞
sin λ cos λx sin λ
∫ λ
0
dλ and find the value of ∫
0
λ
dλ .
Solution:
We know that the Fourier Integral formula for f (x) is
∞ ∞
1
f ( x) =
π ∫ ∫ f (t ) cos λ (t − x) dt dλ
0 −∞
……………….(1)
Here f (t ) = 1 for t ≤ 1 i.e., f(t) = 1 in -1 < t < 1
f (t ) = 0 for t > 1
f (t ) = 0 in − ∞ < t < −1 and 1 < t < ∞
∞ 1
1
∴ Equation (1) ⇒ f ( x) =
π ∫ ∫ cos λ (t − x) dt dλ
0 −1
∞ 1
1  sin λ (t − x) 
π ∫
=   dλ
0
λ  −1
∞
1 sin λ (1 − x) − sin λ (−1 − x)
π∫
= dλ
0
λ
∞
1 sin λ (1 − x) + sin λ (1 + x )
= ∫ dλ
π 0 λ
1

2. ∞
2 sin λ cos λx
π∫
∴ f ( x) = dλ .………………(2)
0
λ
[Using sin (A+B) + sin (A-B) = 2 sin A cos B]
This is Fourier Integral of the given function. From (2) we get
∞
sin λ cos λx π
∫
0
λ
dλ =
2
f ( x) ……………….(3

1 for x ≤ 1
But f ( x) =  ………………..(4)
0 for x > 1

Substituting (4) in (3) we get
π
∞
sin λ cos λx  for x ≤ 1
∫ λ dλ =  2
0 0 for x > 1

∞
sin λ π
Putting x = 0 we get ∫0
λ
dλ =
2
2. Find the Fourier Integral of the function
0 x<0
1

f ( x) =  x=0
2
e − x x > 0

Verify the representation directly at the point x = 0.
Solution:
The Fourier integral of f (x) is
∞ ∞
1
f ( x) =
π ∫ ∫ f (t ) cos λ (t − x) dt dλ
0 −∞
……………….(1)
1  
∞ 0 ∞
= ∫ −∫∞
π 0
f (t ) cos λ (t − x )dt + ∫ f (t ) cos λ (t − x)dt d λ
0 
1  
∞ 0 ∞
= ∫  ∫ 0. cos λ (t − x)dt + ∫ e −t cos λ (t − x)dt dλ
π 0 − ∞ 0 
∞ ∞
1  e −t
= ∫ 2 [ − cos( λt − λx ) + λ sin(λt − λx)]  dλ

π 0 λ +1 0
∞
1 cos λx + λ sin λx
π∫
f (x) = dλ ……….………(2)
0 λ2 + 1
2

3. Putting x = 0 in (2), we get
∞
1
f (0) = ∫ 2
1
π 0 λ +1
1
d λ = tan −1 ( λ ) 0
π
∞
[ ]
1
[
= tan −1 ( ∞ ) − tan −1 (0)
π
]
1 π  1
=  =
π 2 2
1
The value of the given function at x = 0 is . Hence verified.
2
FOURIER SINE AND COSINE INTEGRALS
The integral of the form
2∞ ∞
f ( x) = ∫ sin λx ∫ f (t ) sin λt dt dλ
π0 0
is known as Fourier sine integral.
The integral of the form
∞ ∞
2
f ( x) = ∫ cos λx ∫ f (t ) cos λt dt dλ
π 0 0
is known as Fourier cosine integral.
PROBLEMS
1. Using Fourier integral formula, prove that
2(b 2 − a 2 ) ∞ u sin xu
∫ (u 2 + a 2 )(u 2 + b 2 ) du (a, b > 0)
− ax − bx
e −e =
π 0
Solution:
The presence of sin xu in the integral suggests that the Fourier sine integral formula
has been used.
Fourier sine integral representation is given by
∞ ∞
2
f ( x) = ∫ sin ux ∫ f (t ) sin ut dt du
π 0 0
∞
∞ 
sin ux du  ∫ ( e − at − e −bt ) sin ut dt 
2
e − ax − e −bx = ∫
π 0 0 
3

4. ∞ ∞
2  e − at − bt

= ∫ sin ux du  2 { − a sin ut − u cos ut} − 2e 2 { − b sin ut − u cos ut} 
π 0 a + u
2
b +u 0
∞
2  u u 
= ∫ sin ux du  a 2 + u 2 − b 2 + u 2 
π 0  
∞
2(b 2 − a 2 ) u sin ux
=
π ∫ (u 2 + a 2 )(u 2 + b 2 ) du
0
2. Using Fourier integral formula, prove that
2
∞
(λ 2
+ 2 ) cos xλ
∫ dλ
−x
e cos x =
π 0 λ2 + 4
Solution:
The presence of cos xλ in the integral suggests that the Fourier cosine integral
formula for e − x cos x has been used.
Fourier cosine integral representation is given by
∞ ∞
2
π∫
f ( x) = cos λx ∫ f (t ) cos λt dt dλ
0 0
2∞ ∞ − t 
∴e −x
cos x = ∫ cos xλ dλ  ∫ e cos t cos λt dt 
π 0 0 
2
∞
1 ∞ 
= ∫ cos xλ dλ  ∫ e −t { cos(λ + 1)t + cos(λ − 1)t } dt 
π 0 2 0 
∞

=
2
π 0 
1 −t
[
∫ cos xλ dλ  (λ + 1) 2 + 1 e { − cos(λ + 1)t + (λ + 1) sin(λ + 1)t} ] ∞
0

+
1
(λ − 1) + 1
2
[
e −t { − cos(λ − 1)t + (λ − 1) sin(λ − 1)t } 0
∞
] )
∞
1  1 1 
= ∫ +  cos xλ dλ
π 0  (λ + 1) + 1 (λ − 1) + 1
2 2
∞
2 (λ2 + 2) cos xλ
= ∫ dλ.
π 0 λ2 + 4
COMPLEX FORM OF FOURIER INTEGRALS
4

5. The integral of the form
∞ ∞
1
∫ e − iλx ∫ f (t ) e
iλt
f ( x) = dt d λ
2π −∞ −∞
is known as Complex form of Fourier Integral.
FOURIER TRANSFORMS
COMPLEX FOURIER TRANSFORMS
∞
1
The function F [ f ( x)] = ∫∞ f (t ).e dt is called the Complex Fourier transform
ist
2π −
of f (x ) .
INVERSION FORMULA FOR THE COMPLEX FOURIER TRANSFORM
∞
1
∫∞F [ f ( x)].e ds is called the inversion formula for the
−isx
The function f ( x) =
2π −
Complex Fourier transform of F [ f ( x)] and it is denoted by F −1 [ F ( f ( x))].
FOURIER SINE TRANSFORMS
∞
2
The function FS [ f ( x )] = ∫ f (t ).sin st dt is called the Fourier Sine Transform of
π 0
the function f (x ) .
∞
2
The function f ( x) =
π ∫ F [ f ( x)]. sin sx ds is called the inversion formula for the
0
S
Fourier sine transform and it is denoted by FS
−1
[ FS ( f ( x))].
FOURIER COSINE TRANSFORMS
∞
2
The function FC [ f ( x)] =
π ∫
f (t ). cos st dt is called the Fourier Cosine
0
Transform of f (x) .
∞
2
The function f ( x) =
π ∫ F [ f ( x)]. cos sx ds
0
C is called the inversion formula for the
Fourier Cosine Transform and it is denoted by FC
−1
[ FC ( f ( x))].
PROBLEMS
1. Find the Fourier Transform of
1 − x 2 in x ≤ 1

f ( x) = 
0
 in x > 1
5

6. ∞
sin s − s cos s s 3π
Hence prove that ∫
0 s 3
cos ds =
2 16
.
Solution:
We know that the Fourier transform of f (x) is given by
∞
1
F [ f ( x )] = ∫ f ( x).e
isx
dx
2π −∞
−1 1 ∞
1 1 1
= ∫ f ( x).e dx + ∫ f ( x).e dx + ∫ f ( x).e
isx isx isx
dx
2π −∞ 2π −1 2π 1
−1 1 ∞
1 1 1
= ∫ 0.e dx + ∫ (1 − x ).e dx + ∫ 0.e
isx 2 isx isx
dx
2π −∞ 2π −1 2π 1
1
1
= ∫ (1 − x
2
).e isx dx
2π −1
1
1  e isx e isx e isx 
=  (1 − x 2 ) − ( −2 x ) 2 2 − 2 3 3 
2π  is i s i s  −1
1  − 2 is 2 is − 2 −is 2 e −is 
=  2 e + 3e + 2 e − 
2π  s is s i s3 
1  − 2 is 2 
= s (e + e −is ) + 3 (e is − e −is )
2π 
2
is 
1 − 4 4  1 4 
=  s 2 cos s + s 3 sin s  =  s 3 (sin s − s cos s )
2π   2π  
By using inverse Fourier Transform we get
∞
1 1 4
f ( x) =
2π
.
2π
∫
−∞ s 3
(sin s − s cos s ).e −isx ds
∞
1 4
=
2π ∫s
−∞
3
(sin s − s cos s ).(cos sx − i sin sx ) ds
∞
1 4
=
2π ∫
−∞ s3
(sin s − s cos s ) cos sx ds
∞
1 4
−
2π ∫
−∞ s3
(sin s − s cos s ) i sin sx ds
6

7. The second integral is odd and hence its values is zero.
∞
2 sin s − s cos s
π −∫
∴ f ( x) = cos sx ds
∞ s3
∞
4 sin s − s cos s
= ∫ cos sx ds
π 0 s3
∞
sin s − s cos s π
i.e., ∫
0 s 3
cos sx ds = f ( x)
4
1
Putting x = , we get
2
∞
sin s − s cos s s π  1  π  1  3π
∫0 s 3
cos ds =
2
f   = 1 −  =
4  2  4  4  16
.
∞
sin s − s cos s s 3π
∫
0 s 3
cos ds =
2 16
.
2. Find the Fourier sine transform of e − x , x ≥ 0 (or) e − x , x > 0. Hence evaluate
∞
x sin mx
∫ 1 + x 2 dx.
0
Solution:
The Fourier sine transform of f(x) is given by
∞
2
FS [ f ( x )] = ∫ f ( x).sin sx dx
π 0
−x −x
Here e =e for x > 0
∞
[ ]
FS e − x =
2
π ∫e
−x
. sin sx dx
0
2 s ∞ − ax b 
=  ∫ e sin bx dx = 2 
π s2 +1 0 a + b2 
Using inverse Fourier sine transform we get
∞
∫ F [e ]. sin sx ds
2 −x
f ( x) = s
π 0
∞
2 2 s
=
π ∫
0
. 2
π s +1
. sin sx ds
∞
2 s
= ∫ s 2 + 1 sin sx ds
π 0
7

8. ∞
π s
i.e., f ( x) = ∫ 2 . sin sx ds
2 0 s +1
∞
s. sin sx π
i.e., ∫0 s +1
2
ds = e − x
2
Replacing x by m we get
∞
s. sin ms π
i.e., ∫
0 s +1
2
ds = e − m
2
∞
x. sin mx π
i.e., ∫
0 x +1
2
dx = e −m
2
[since s is dummy variable, we can replace it by x]
e − ax
3. Find the Fourier cosine transform of .
x
Solution:
∞
2
We know that FC [ f ( x )] = ∫ f ( x). cos sx dx
π 0
− ax
e
Here f ( x) = .
x
∞
2 e − ax
∴ FC [ f ( x )] = ∫ x . cos sx dx
π 0
Let FC [ f ( x)] = F ( s )
∞
2 e − ax
Then F (s) =
π ∫ x . cos sx dx
0
………………(1)
Differentiating on both sides w.r.t. ‘s’ we get,
∞
dF ( s ) d 2 e − ax
ds
=
ds π ∫ x . cos sx dx
0
∞
2 ∂  e −ax 
= ∫ ∂s  x . cos sxdx
π0  
∞ ∞
2 e −ax 2 − ax
= ∫ x (− sin sx).x dx = − π ∫ e sin sx dx
π0 0
dF ( s ) 2 s  ∞
b 
∫e
− ax
=− . 2  sin bx dx = 
ds π a + s2  0 a + b2 
2
Integrating w.r.t. ‘s’ we get
8

9. 2 s
F (s) = − .∫ 2 ds
π s + a2
2 1 1
=− . . log ( s 2 + a 2 ) = − . log ( s 2 + a 2 )
π 2 2π
e − ax − e −bx
4. Find the Fourier cosine transform of .
x
Solution:
We know that the Fourier cosine transform of f(x) is
∞
2
FC [ f ( x )] = ∫ f ( x). cos sx dx
π 0
− ax
e − e −bx
Here f ( x) =
x
∞
 e − ax − e −bx  2 e − ax − e − bx
∴ FC 
 x
=
 π ∫ x . cos sx dx
0
∞ ∞
2 e − ax 2 e −bx
=
π ∫ x . cos sx dx − π ∫ x cos sx dx
0 0
 e −ax   e −bx 
= Fc   − Fc  
 x   x 
1 1
=− log ( s 2 + a 2 ) + log ( s 2 + b 2 )
2π 2π
1  s2 + b2 
= log  2
 s + a2 
2π  
e − as
5. Find f (x) , if its sine transform is . Hence deduce that the inverse sine
s
1
transform of .
s
Solution:
We know that the inverse Fourier sine transform of FS [ f (x )] is given by
∞
2
f ( x) =
π ∫ F [ f ( x)]. sin sx ds
0
S
e − as
Here FS [ f ( x )] =
s
9

10. ∞
2 e − as
∴ f ( x) =
π ∫ s . sin sx ds
0
Differentiating w.r.t. ‘x’ on both sides, we get,
d [ f ( x )]
∞
2 e − as ∂
dx
=
π ∫ s . ∂x (sin sx) ds
0
∞ ∞
2 e − as 2 − as
=
π ∫ s . cos sx s ds = π ∫ e . cos sx ds
0 0
2 a  ∞ −ax a 
=  ∫ e cos bx dx = 2 
π a + x2
2
 0 a + b2 
d [ f ( x )] 2 a
=
dx π x + a2
2
2 1 2 1 −1  x 
∴ f ( x) = a
π ∫ x 2 + a 2 dx = a π a tan  a 
 
2  x
∴ f ( x) = tan −1  
π a
1
To find the inverse Fourier sine transform of :
s
Put a = 0, in (1), we get
2 2 π π
f ( x) = tan −1 (∞) = . =
π π 2 2
PROPERTIES
1. Linearity Property
If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then
F [ a f ( x ) + b g ( x)] = a F ( s) + b G ( s )
Proof:
∞
1
F [ a f ( x) + b g ( x)] = ∫∞[ a f ( x) + b g ( x)] e dx
isx
2π −
∞ ∞
1 1
= ∫ a f ( x).e dx + ∫ b g ( x).e
isx isx
dx
2π −∞ 2π −∞
∞ ∞
a b
= ∫ f ( x).e isx dx + ∫ g ( x).e
isx
dx
2π −∞ 2π −∞
= a F ( s ) + b G ( s)
10

11. 2. Change of Scale Property
1 s
If F(s) is the Fourier transform of f (x ) then F [ f (ax)] = F  ,a>0
a a
Proof:
∞
1
F [ f (ax)] = ∫ f (ax).e
isx
dx
2π −∞
Put ax = y
dy
a dx = dy i.e., dx =
a
When x = −∞, y = −∞ and x = ∞, y=∞
∞ y ∞ s
1 dy 1 1 i  y
F [ f (ax)] =
is
2π
∫ f ( y).e
−∞
a
. =
a a 2π
∫ f ( y).e
−∞
a
.dy
1
= F ( s a)
a
3. Shifting Property ( Shifting in x )
If F(s) is the Fourier transform of f (x ) then F [ f ( x − a )] = e ias F ( s )
Proof:
∞
1
F [ f ( x − a)] = ∫ f ( x − a).e
isx
dx
2π −∞
Put x-a = y
dx = dy
When x = −∞, y = −∞ and x = ∞, y=∞
∞ ∞
1 e ias
F [ f ( x − a)] = ∫ f ( y ).e is ( y + a ) . dy = ∫ f ( y).e
isy
.dy
2π −∞ 2π −∞
∞
e ias
= ∫ f ( x).e .dx = e isa F ( s )
isx
2π −∞
4. Shifting in respect of s
If F(s) is the Fourier transform of f (x ) then F e iax f ( x) = F ( s + a ) [ ]
Proof:
∞
Fe[ iax
]
f ( x) =
1
∫e
iax
f ( x) e isx dx
2π −∞
11

12. ∞
1
∫ f ( x).e
i( s+a) x
= dx = F ( s + a)
2π −∞
5. Modulation Theorem
1
If F(s) is the Fourier transform of f (x ) then F [ f ( x) cos ax ] = [ F ( s + a ) + F ( s − a )]
2
Proof:
∞
1
F [ f ( x) cos ax ] = ∫ f ( x). cos ax.e
isx
dx
2π −∞
∞
1  e iax + e −iax 
= ∫
2π −∞
f ( x).e isx 

 2
dx


∞ ∞
1 1 1 1
∫∞ f ( x).e dx + 2 . 2π ∫ f ( x).e
i( s+a ) x i ( s −a ) x
= . dx
2 2π − −∞
1 1 1
= f ( s + a ) + f ( s − a ) = [ f ( s + a ) + f ( s − a )]
2 2 2
1
F [ f ( x) cos ax ] = [ F ( s + a ) + F ( s − a)]
2
COROLLARIES
1
(i ) FC [ f ( x) cos ax ] = [ FC ( s + a) + FC ( s − a)]
2
1
(ii ) FC [ f ( x) sin ax ] = [ FS (a + s) + FS (a − s )]
2
1
(iii ) FS [ f ( x) cos ax ] = [ FS ( s + a ) + FS ( s − a )]
2
1
(iv ) FS [ f ( x) sin ax ] = [ FC ( s − a ) − FC ( s + a)]
2
6. Conjugate Symmetry Property
If F(s) is the Fourier transform of f (x ) then F f ( − x) = F ( s) [ ]
Proof:
∞
1
We know that F ( s ) = ∫ f ( x). e
isx
dx
2π −∞
Taking complex conjugate on both sides we get
∞
1
∫∞ f ( x). e dx
−isx
F (s) =
2π −
12

13. Put x = -y
dx = -dy
When x = −∞, y = ∞ and x = ∞, y = −∞
−∞
1
∴ F (s) = ∫ f (− y) .e (− dy )
isy
2π ∞
−∞
1
=− ∫ f (− y). e
isy
dy
2π ∞
[ ]
∞
1
= ∫ f (− x). e dx = F f (− x)
isx
2π −∞
7. Transform of Derivatives
If F(s) is the Fourier transform of f (x ) and if f (x) is continuous, f ′(x) is piecewise
continuously differentiable, f (x ) and f ′(x) are absolutely integrable in (−∞ , ∞) and
lim [ f ( x)] = 0 , then
x → ±∞
F ( f ′( x ) ) = −is F ( s )
Proof:
By the first three conditions given, F { f (x)} and F { f ′(x)} exist.
∞
1
F { f ′( x)} = ∫ f ′( x) e
isx
dx
2π −∞
∞
=
1
[e isx
] ∞
f ( x) −∞ −
is
∫e
isx
f ( x) dx, on int egrating by parts.
2π 2π −∞
= 0 − isF { f ( x)} , by the given condition.
= −is F ( s).
The theorem can be extended as follows.
If f , f ′, f ′′, , f ( n −1) are continuous, f (n ) is piecewise continuous, f , f ′, f ′′, , f ( n )
are absolutely integrable in (−∞ , ∞) and f , f ′, f ′′,  , f ( n −1) → 0 as x → ±∞ , then
F ( f ( n ) ( x) ) = (−is ) n F ( s )
13

14. 8. Derivatives of the Transform
dF ( s )
If F(s) is the Fourier transform of f (x ) then F [ x. f ( x )] = (−i )
ds
Proof:
∞
1
F (s) = ∫ f ( x )e
isx
dx
2π −∞
∞
∫ ds [ f ( x)e ]dx
dF ( s ) 1 d
∴ = isx
ds 2π −∞
∞
i
= ∫ [ x. f ( x)]e dx = iF [ xf ( x)]
isx
2π −∞
dF ( s)
∴ ( −i ) = F [ x. f ( x)]
ds
[ ]
Extending, we get, F x n . f ( x) = (−i ) n
d n F (s)
ds n
DEFINITION
∞
1
∫ f ( x − u ) g (u )du is called the convolution product or simply the convolution
2π −∞
of the functions f (x) and g (x) and is denoted by f ( x) * g ( x ) .
9. Convolution Theorem
If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then the
Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier
transforms.
i.e., F [ f ( x ) * g ( x)] = F ( s ).G ( s)
Proof
∞
1
F [ f ( x ) * g ( x )] = ∫∞ f ( x) * g ( x)e dx
isx
2π −
1  1 ∞
∞
 isx
= ∫∞  2π −∫∞ f ( x − u) g (u )du  e dx
2π −  
1
∞
 1 ∞ 
= ∫∞  2π −∫∞ f ( x − u)e dxdu,
isx
g (u ) 
2π − 
on changing the order of int egration.
14

15. ∞
∫ g (u )[e ]
1
= ius
F ( s) du , by the shifting property.
2π −∞
∞
1
= F ( s). ∫ g (u ).e
ius
du
2π −∞
= F ( s).G ( s )
Inverting, we get
F −1 [ F ( s ).G ( s)] = f ( x) * g ( x )
= F −1 { F ( s )} * F −1 { G ( s )}
10. Parseval’s Identity (or) Energy Theorem
If f (x) is a given function defined in (−∞ , ∞) then it satisfy the identity,
∞ ∞
∫ ∫ F ( s)
2 2
f ( x) dx = ds
−∞ −∞
where F(s) is the Fourier transform of f (x ) .
Proof:
We know that F −1 [ F ( s ).G ( s)] = f ( x) * g ( x )
∞ ∞
1 1
∫ F (s).G(s)e ∫ f (t ) g ( x − t )dt
−isx
ds =
2π −∞ 2π −∞
Putting x = 0, we get
∞ ∞
∫ F (s).G(s)ds = ∫ f (t ) g (−t )dt
−∞ −∞
………………..(1)
Let g ( − t ) = f (t ) .……………….(2)
i.e., g (t ) = f (−t ) ………………..(3)
∴ G ( s ) = F [ f (− x)] = F ( s ) by property (9)
i.e., ∴ G ( s) = F ( s) ………………..(4)
Substituting (2) and (4) in (1) we get
∞ ∞
∫ F (s).F (s) ds =
−∞
∫ f (t ). f (t ) dt
−∞
[ F (s).F (s) = F (s) ]
∞ ∞
∫ ∫
2 2 2
F ( s) ds = f ( x) dx
−∞ −∞
11. If f (x) and g (x) are given functions of x and FC [ f ( x)] and FC [ g ( x)] are their
Fourier cosine transforms and FS [ f ( x)] and FS [ g ( x)] are their Fourier sine transforms then
15

16. ∞ ∞ ∞
(i) ∫
0
f ( x ) g ( x)dx = ∫ FC [ f ( x)].FC [ g ( x)]ds = ∫ FS [ f ( x )].FS [ g ( x)]ds
0 0
∞ ∞ ∞
∫ f ( x) dx = ∫ FC [ f ( x)] ds = ∫ FS [ f ( x )] ds ,
2 2 2
(ii)
0 0 0
which is Parseval’s identity for Fourier cosine and sine transforms.
Proof:
∞ ∞
 2∞ 
(i) ∫ FC [ f ( x )].FC [ g ( x )]ds = ∫ FC [ f ( x)]  ∫ g ( x) cos sx dx  ds
0 0  π 0 
∞
 2∞ 
= ∫ g ( x)  ∫ FC [ f ( x)] cos sx ds  dx,
0  π 0 
Changing the order of integration
∞
= ∫ f ( x) g ( x)dx
0
Similarly we can prove the other part of the result.
(ii) Replacing g ( x) = f * ( x) in (i) and noting that FC [ f ( x)] = FC [ f ( x)] and
FS [ f ( x)] = FS [ f ( x)] , we get
∞ ∞ ∞
∫0
f ( x ). f ( x).dx = ∫ FC [ f ( x)].FC [ f ( x)] ds = ∫ FS [ f ( x )].FS [ f ( x )] ds
0 0
∞ ∞ ∞
∫ f ( x) .dx = ∫ FC [ f ( x)] ds = ∫ FS [ f ( x)] ds
2 2 2
i.e.,
0 0 0
12. If FC [ f ( x)] = FC ( s ) and FS [ f ( x)] = FS ( s ) , then
d
(i) { FC ( s)} = − FS { xf ( x)} and
ds
d
(ii) { FS ( s)} = − FC { xf ( x)}.
ds
Proof:
∞
2
π∫
FC ( s) = f ( x) cos sx dx
0
∞
d
{ FC ( s)} = ∫ f ( x)(− x sin sx)dx
ds 0
∞
= − ∫ {xf ( x)}sin sx dx
0
= − FS {xf ( x)}
Similarly the result (ii) follows.
16

17. PROBLEMS
a 2 − x 2
 x <a
1. Show that the Fourier transform of f ( x ) =  is
0
 x >a>0
∞
2  sin as − as cos as  sin t − t cos t π
2   . Hence deduce that ∫ dt = . Using Parseval’s
π s 3
 0 t 3
4
∞ 2
 sin t − t cos t  π
identity show that ∫  3  dt = .
0 
t 15
Solution:
We know that
∞
1
F [ f ( x )] = ∫ f ( x).e
isx
dx
2π −∞
1  
−a a ∞
=  ∫ f ( x)e dx + ∫ f ( x)e dx + ∫ f ( x )e dx 
isx isx isx
2π  −∞ −a a 
1  
a a
1
=  0 + ∫ ( a 2 − x 2 )e isx dx + 0 = ∫ (a
2
− x 2 ).e isx dx
2π  − a  2π −a
a
1  2  e isx   e isx   e isx 
=  ( a − x 2 )
 is  − (−2 x) 2 2  − 2 3 3 
 i s  i s 
2π        − a
=
1  − 2a isa
 2 e +e [−isa 2i
] [

+ 3 e −isa − e isa  ]
2π  s s 
1  − 2a 4 
=  2 [2 cos as ] + 3 sin as 
2π  s s 
4  sin as − as cos as 
=
2π 
 s3 

2  sin as − as cos as 
F (s) = 2
π  s3 

2  sin s − s cos s 
When a = 1, F ( s ) = 2 ………………..(A)
π   s3 

Using inverse Fourier Transform, we get
17

18. ∞
1 2 1
f ( x) =
2π
.2.
π ∫s
−∞
3
[sin as − as cos as].e −isx ds
∞
1 2 1
{ sin as − as cos as}{ cos sx − i sin sx} ds
π −∫ s 3
= .2.
2π ∞
∞
2 sin as − as cos as
π −∫
f ( x) = cos sx ds
∞ s3
[The second integral is odd and hence its value is zero]
∞
4 sin as − as cos as
= ∫ cos sx ds
π 0 s3
[since the integrand is an even function of s]
Putting a = 1, we get
∞
4 sin s − s cos s
f ( x) = ∫ cos sx ds
π 0 s3
∞
4 sin t − t cos t
f ( x) = ∫ cos tx dt
π 0 t3
Putting x = 0, in the given function we get
∞
4 sin t − t cos t
π∫
dt = f (0) = 1
0 t3
∞
sin t − t cos t π
∴ ∫
0 t 3
dt =
4
∞ ∞
∫ ∫ F (s)
2 2
Using Parseval’s identity, f ( x) dx = ds [Using (A)]
−∞ −∞
2
 2 
∞  2. (sin s − s cos s)  1
π
∫ 
− ∞ s3
 ds = ∫ (1 − x 2 ) 2 dx
 −1
 
 
∞ 2 1
8  sin s − s cos s 
∫∞  ds = 2 ∫ (1 − x ) dx
2 2
π− s 3
 0
∞ 2
16  sin s − s cos s  8
∫
π 0 s 3  ds = 2.
 15
18

19. ∞ 2
 sin s − s cos s  π
i.e., ∫
0 s 3  ds =
 15
∞ 2
 sin t − t cos t  π
i.e., ∫ t3
0
 dt =
 15
2. Find the Fourier Transform of f (x) if
1 − x , x < 1

f ( x) = 
0,
 x >1
∞ 4
 sin t  π
Hence deduce that ∫   dt =
0
t  3
Solution:
We know that
∞
1
F [ f ( x )] = ∫∞ f ( x).e dx
isx
2π −
1
∫ (1 − x ).e
1
= isx
dx
2π −1
Since x > 1, f ( x) = 0, i.e., in − ∞ < x < −1, and 1 < x < ∞, f ( x ) = 0.
1
∫ [1 − x ] [ cos sx + i sin sx] dx
1
=
2π −1
1
∫ [1 − x ] cos sx dx
1
=
2π −1
The second integral becomes zero since it is an odd function.
1
2
=
2π
∫ (1 − x) cos sx dx
0
[ [1 − x ] cos x is an even function]
1
2   sin sx   − cos sx 
= (1 − x ) s  − (−1) s 2 
π      0
2  − cos s 1 
=  + 2
π  s2 s 
2 1
i.e., F ( s) = . (1 − cos s )
π s2
19

20. Using Parseval’s identity
∞ ∞
∫ ∫ F ( s)
2 2
f ( x) dx = ds
−∞ −∞
1 ∞
[1 − x ] dx = 2 ∫ (1 − cos s) ds
2
∫
2
−1
π −∞ s4
1 ∞
2 (1 − cos s) 2
2 ∫ [1 − x ] dx =
π −∫
2
ds
0 ∞ s4
∞
2 2 (1 − cos s ) 2
3 π −∫
= ds
∞ s4
put s = 2t when s = ∞, t = ∞
ds = 2dt when s = −∞, t = −∞
π ∞ (1 − cos 2t ) 2
3 −∫
= .2 dt
∞ 16t 4
π ∞ (1 − cos 2t ) 2
3 −∫
= dt
∞ 8t 4
∞
π (1 − cos 2t ) 2
= 2∫ dt
3 0 8t 4
π ∞ sin 4 t
3 ∫ t4
= dt
0
∞ 4
 sin t  π
i.e., ∫  t  dt = 3
0 
∞
dx
3. Evaluate ∫ (x
0
2
+ a )( x 2 + b 2 )
2 using transforms.
Solution:
2 a
We know that the Fourier cosine transform of f ( x) = e − ax is . 2 .
π s + a2
2 b
Similarly the Fourier cosine transform of f ( x) = e − ax is . 2 .
π s + b2
20

21. ∞ ∞
We know that ∫ FC [ f ( x)].FC [ g ( x)] ds = ∫ f ( x).g ( x) dx
0 0
∞ ∞
2 a 2 b
i.e., ∫
0
. 2 . . 2
π s +a π s +b
2 2
. ds = ∫ e − ax .e −bx dx
0
∞ ∞
2 ab
∫ (s 2 + a 2 )(s 2 + b 2 ) ds = ∫ e
−( a +b ) x
i.e., ds
π 0 0
∞
 e −( a + b ) x  1 1
=  = 0− =
 − ( a + b)  0 − ( a + b) a + b
∞
dx π
i.e., ∫ (x
0
2 2 2 2
=
+ a )( x + b ) 2ab( a + b)
4. Find the Fourier transform of e − a x and hence deduce that
∞
cos xt π −a x
(i) ∫ 2 2 dt = e
0 a +t
2a
[
(ii) F xe
−a x
]=i 2 2as
π (s + a 2 ) 2
2
Solution:
∞
1
F [ f ( x )] = ∫ f ( x).e
isx
dx
2π −∞
1  
0 ∞
= ∫ f ( x)e isx dx + ∫ f ( x)e isx dx 
2π  −∞ 0 
e − ax
 if 0 ≤ x < ∞
Here f ( x ) =  ax
e
 if − ∞ < x < 0
1  ax isx 
0 ∞
= ∫ e. e dx + ∫ e −ax .e isx dx 
2π  −∞ 0 
1  ( a +is ) x 
0 ∞
=  ∫ e. dx + ∫ e −( a −is ) x dx 
2π  −∞ 0 
21

22. 1  e ( a +is ) x   e −( a −is ) x  
0 ∞
=   +  
2π  (a + is)  −∞  − (a − is )  0 
 
1  1 1 
=  a + is + a − is 
2π  
Fe[ −a x
]= 2 a
π s + a2
2
Using inversion formula, we get
∞
1 2 a
f ( x) =
2π
∫
−∞
. 2
π s +a 2
e −isx ds
∞
a cos sx − i sin sx
π −∫
= ds
∞ s2 + a2
∞
a cos x
=
π−∫∞ s 2 + a 2 ds
∞
cos sx π π −a x
∫s
0
2
+a 2
dx =
2a
f ( x) =
2a
.e (or )
∞
cos tx π −a x
∫s
0
2
+a 2
dt =
2a
.e
Putting a = 1, we get,
Fe[ ]=−x 2 1
.
π s2 +1
∞ ∞
cos sx π −x cos tx π −x
and ∫ s 2 + 1 ds = 2 e
0
(or ) ∫t
0
2
+1
dt = e
2
FINITE FOURIER TRANSFORMS
If f (x) is a function defined in the interval (0 , l) then the finite Fourier sine
transform of f (x) in 0 <x < l is defined as
22

23. nπx
l
FS [ f ( x)] = ∫ f ( x). sin dx where ‘n’ is an integer
0
l
The inverse finite Fourier sine transform of FS [ f (x )] is f (x) and is given by
2 ∞ nπx
f ( x) = ∑ FS [ f ( x )] sin
l n =1 l
The finite Fourier cosine transform of f (x ) in 0 < x < l is defined as
nπx
l
FC [ f ( x)] = ∫ f ( x). cos dx where ‘n’ is an integer
0
l
The inverse finite Fourier cosine transform of FC [ f (x)] is f (x) and is given by
1 2 ∞ nπx
f ( x) = FC (0) + ∑ FC [ f ( x )] cos
l l n =1 l
PROBLEMS
1. Find the finite Fourier sine and cosine transforms of f ( x) = x 2 in 0 < x < l.
Solution:
The finite Fourier sine transform is
nπx
l
FS [ f ( x)] = ∫ f ( x). sin dx
0
l
Here f ( x) = x 2
[ ] nπx
l
FS x 2 = ∫ x 2 . sin dx
0
l
l
  nπx   nπx   nπx 
  − cos   − sin   cos 
= x 2 l  − 2 x l  + 2 l 
  nπ   n 2π 2   n 3π 3 
       
 l   l2   l 3  0 
− l3 2l 3 2l 3
= cos nπ + 3 3 cos nπ − 3 3
nπ nπ nπ
=
l3
nπ
2l 3
[
(−1) n +1 + 3 3 (−1) n − 1
nπ
]
The finite Fourier cosine transform is
23

24. nπx
l
FC [ f ( x)] = ∫ f ( x). cos dx
0
l
Here f ( x) = x 2
[ ] nπx
l
FC x 2 = ∫ x 2 . cos dx
0
l
l
  nπx   nπx   nπx 
  sin   − cos   − sin 
= x 2  l  − 2 x l  + 2 l 
  nπ   n 2π 2   n 3π 3 
  l     
   l2   l3  0
2l 3
= cos nπ
n 2π 2
2l 3
= (−1) n
nπ
2 2
2. Find the finite Fourier sine and cosine transforms of f ( x ) = x in (0 , π ) .
Solution:
The finite Fourier sine transform of f ( x) = x in (0 , π ) is
π
FS [ f ( x)] = ∫ f ( x ). sin nx dx
0
Here f ( x) = x in (0 , π )
π π
  − cos nx   − sin nx 
FS [ x ] = ∫ x. sin nx dx =  x  − 1 2 
0   n   n  0
π π
=− cos nπ = (−1) n +1 .
n n
The finite Fourier cosine transform of f ( x) = x in (0 , π ) is
π
FC [ f ( x)] = ∫ f ( x). cos nx dx
0
Here f ( x) = x in (0 , π )
π π
  sin nx   − cos nx 
FC [ x ] = ∫ x. cos nx dx =  x  − 1 2 
0   n   n  0
=
1
n 2
1 1
[
cos nπ − 2 = 2 (−1) n − 1
n n
]
24

25. 2π (−1) p−1
3. Find f (x) if its finite sine transform is given by , where p is positive
p3
integer and 0 < x < π .
Solution:
We know that the inverse Fourier sine transform is given by
∞
2
f ( x) = ∑ FS [ f ( x )] sin px ………………..(1)
π p =1
2π (−1) p−1
Here FS [ f (x )] = ………………..(2)
p3
Substituting (2) in (1), we get
2 ∞ 2π (−1) p −1
f ( x) = ∑ sin px
π p =1 p3
∞
(−1) p −1
= 4∑ sin px
p =1 p3
 2 pπ 
cos 
 3  find FC [ f ( p )] if 0 < x <1.
−1
4. If
f ( p) =
(2 p + 1) 2
Solution:
∞
nπx
We know that FC
−1
[ f ( p)] = 1 FC (0) + 2 ∑ FC [ f ( x)] cos
l l n =1 l
 2 pπ 
cos 
Here
f ( p) =  3 
(2 p + 1) 2
Let FC [ f ( x)] = f ( p)
∞
nπx
∴ FC
−1
[ f ( p)] = 1 f C ( 0) +
2
∑ f ( p ) cos [ l = 1]
l l n =1 l
 2 pπ 
∞
cos 
= 1 + 2∑  3  . cos nπx
n =1 (2 p + 1) 2
25

26. UNIT-4
PART A
1. State the Fourier integral theorem.
Ans:
If f (x) is a given function defined in (-l , l) and satisfies Dirichlet’s conditions, then
∞ ∞
1
f ( x) =
π ∫ ∫ f (t ) cos λ (t − x) dt dλ
0 −∞
2. State the convolution theorem of the Fourier transform.
Ans:
If F(s) and G(s) are the Fourier transform of f (x ) and g (x) respectively then the
Fourier transform of the convolution of f(x) and g(x) is the product of their Fourier
transforms.
i.e., F [ f ( x ) * g ( x)] = F ( s ).G ( s)
3. Write the Fourier transform pair.
Ans:
F [ f (x)] and F −1 [ F ( S )] are Fourier transform pairs.
4. Find the Fourier sine transform of f ( x) = e − ax (a > 0).
Ans:
∞
2
FS [ f ( x)] =
π ∫
f ( x ). sin sx dx
0
2
∞
2 s   ∞ −ax b 
∫ e . sin sx dx =  ∫ e sin bx dx = 2
− ax
= 
π 0
π  s2 + a2 
   0 a + b2 
2 s 
=
π  s2 + a2 
 
5. If the Fourier transform of f (x ) is F(s) then prove that . F [ f ( x − a )] = e isa F ( s)
Ans:
∞
1
F [ f ( x − a)] = ∫∞ f ( x − a).e dx
isx
2π −
Put x-a = y
dx = dy
When x = −∞, y = −∞ and x = ∞, y = ∞
∞ ∞
1 e ias
F [ f ( x − a)] = ∫ f ( y).e
is ( y + a )
. dy = ∫ f ( y).e
isy
.dy
2π −∞ 2π −∞
∞
e ias
= ∫ f ( x).e .dx = e isa F ( s )
isx
2π −∞
6. State the Fourier transforms of the derivatives of a function.
26