4 Fourier Transform(F.T.)integral. Hence evaluate Z1 0 sin cos x d and nd the value of Z1 0 sin d ....

4 Fourier Transform(F.T.)

Fourier integral theorem (without proof) - Fourier transformpair - Sine and Cosine transforms - Properties - Transforms ofsimple functions - Convolution theorem - Parseval’s identity.

4.1 Definition

Let f(x) be a function defined in an interval (a, b). Let k(s, x) be a givenfunction of two variables s and x. The integral transform of f(x) w.r.t

k(s, x) is defined by I[f(x)] = F (s) =

b∫a

f(x)k(s, x)dx (1)

k(s, x) is called the kernel of the integral transform f(x) is called theinverse transform of F (s).The integral transform is said to be finite if both a and b are finite;

otherwise it is called an infinite transform.If there exist a function H(s, x) such that

f(x) =

d∫e

F (s)H(s, x)ds (2)

then (2) is called the inversion formula for (1).

4.2 Fourier integral theorem(without proof)

If f(x) is piecewise continuously differentiable and absolutely integrable

(−∞,∞), then f(x) =1

π

∞∫0

∞∫−∞

f (t) cosλ (t− x)dtdλ (1)

This is known as Fourier integral theorem or Fourier integralformula.

Note : At a point of discontinuity the value of the integral on the left of

245

246 Unit IV - FOURIER TRANSFORM (F.T.)

(1) is1

2[f(x− 0) + f(x+ 0)]

i.e.,f(x− 0) + f(x+ 0)

2=

1

π

∞∫0

∞∫−∞

f(t) cosλ(t− x)dtdλ

4.2.1 Fourier Cosine and Sine Integral

We know that he Fourier integral for f(x) is

f()x =1

π

∞∫0

∞∫−∞


=1

π

∞∫0

∞∫−∞

f(t){cosλt cosλx+ sinλt sinλx}dtdλ

=1

π

∞∫0

∞∫−∞

f(t) cosλt cosλxdtdλ+1

π

∞∫0

∞∫−∞

f(t) sinλt sinλxdtdλ

Case (i) : If f(t) is an even function, then f(t) cosλt is an evenfunction and f(t) sinλt is an odd function.

f(x) =2

π

∞∫0

∞∫0

f(t) cosλt cosλxdtdλ

i.e., f(x) =2

π

∞∫0

cosλx

∞∫0

f(t) cosλtdtdλ

This is known as Fourier cosine integral.

Case (ii) : If f(t) is an odd function, then f(t) cosλt is an oddfunction and f(t) sinλt is an even function.

f(x) =2

π

∞∫0

∞∫0

f(t) sinλt sinλxdtdλ

i.e., f(x) =2

π

∞∫0

sinλx

∞∫0

f(t) sinλtdtdλ

This is known as Fourier sine integral.

MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 247

4.2.2 Complex form of Fourier Integrals :

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

f(x) =1

π

∞∫0

∞∫−∞


f(x) =1

2π

∞∫−∞

∞∫−∞

f(t) cosλ(t− x)dtdλ (1)

(since cosλ(t− x) is an even function of λ)Since sinλ(t− x) is an odd function of λ.

∴

∞∫−∞

sinλ(t− x)dλ =0

1

2π

∞∫−∞

∞∫−∞

f(t) sinλ(t− x)dtdλ =0 (2)

Now (1) + i (2)⇒

f(x) =1

2π

∞∫−∞

∞∫−∞

f(t) {cosλ(t− x) + i sinλ(t− x)} dtdλ

i.e., f(x) =1

2π

∞∫−∞

∞∫−∞

f(t)eiλ(t−x)dtdλ

4.2.3 Examples under Fourier integrals

Example 4.1. Using Fourier integral, show that

∞∫0

ω sinωx

k2 + ω2dω =

π

2e−kx

(x > 0, k > 0).

Solution : Let f(x) = e−kx.(Here we use sine integral formula since the presence of the term sinwx

in LHS)The Fourier sine integral of f(x) is given by

f(x) =2

π

∞∫0

sinλx

∞∫0

f(t) sinλtdtdλ


Given f(x) = e−kx ⇒ f(t) = e−kt

∴ f(x) =2

π

∞∫0

sinλx

∞∫0

e−kt sinλtdtdλ

=2

π

∞∫0

sinλx

[e−kt

k2 + λ2(−k sinλt− λ cos λt)

]∞0

dλ

f(x) =2

π

∞∫0

sinλxλ

k2 + λ2dλ

=2

π

∞∫0

λ sinλx

k2 + λ2dλ

∞∫0

λ sinλx

k2 + λ2dλ =

π

2f(x)

∞∫0

λ sinλx

k2 + λ2dλ =

π

2e−kx

∞∫0

ω sinωx

k2 + ω2dλ =

π

2e−kx ( put λ by ω )

Example 4.2. Express the function f(x) =

{1 for |x| ≤ 10 for |x| > 1

as a Fourier

integral. Hence evaluate

∞∫0

sinλ cosλx

λdλ and find the value of

∞∫0

sinλ

λdλ.

Solution : The Fourier integral of f(x) is given by


f(x) =1

π

∞∫0

∞∫−∞

f(t) cosλ (t− x) dtdλ

=1

π

∞∫0

1∫−1

cosλ (t− x) dtdλ

=1

π

∞∫0

[sinλ (t− x)

λ

]1−1

dλ

=1

π

∞∫0

sinλ (1− x) + sinλ (1 + x)

λdλ

f(x) =2

π

∞∫0

sinλ cosλx

λdλ

∞∫0

sinλ cosλx

λdλ =

π

2f(x) (1)

=

{ 1

2for |x| < 1

0 for |x| > 1

At x = 1, which is a point of discontinuity of f(x), the value of the aboveintegral

=π

2

[f (1− 0) + f (1 + 0)

2

]=π

2

[1 + 0

2

]=π

4

Hence

∞∫0

sinλ cosλx

λdλ =

π2 for |x| < 1π4 for x = 10 for |x| > 1

Now, putting x = 0 we get

∞∫0

sinλ

λdλ =

π

2

Example 4.3. Find Fourier cosine integral of the function e−ax. Hence


find the value of the integral

∞∫0

cosλx

1 + λ2dλ

Solution : Let f(x) = e−ax.The Fourier cosine integral of f(x) is given by

f(x) =2

π

∞∫0

cosλx

∞∫0

f(t) cosλtdtdλ

Given f(x) = e−ax ⇒ f(t) = e−at

∴ f(x) =2

π

∞∫0

cosλx

∞∫0

e−at cosλtdtdλ

=2

π

∞∫0

cosλx

[e−at

a2 + λ2(−a cosλt+ λ sinλt)

]∞0

dλ

=2

π

∞∫0

cosλxa

a2 + λ2dλ

f(x) =2a

π

∞∫0

cosλx

a2 + λ2dλ

∞∫0

cosλx

a2 + λ2dλ =

π

2af(x)

∞∫0

cosλx

a2 + λ2dλ =

π

2ae−ax

[f(x) = e−ax

]Putting a = 1, we get

∞∫0

cosλx

1 + λ2dλ =

π

2e−x

Example 4.4. Express f(x) =

{ 1

2for 0 ≤ x ≤ π

0 for x > πas a Fourier sine

integral and hence show that∞∫0

1− cos πλ

λsinλxdλ =

{ π

2for 0 < x < π

0 for x > π


Solution : Given

f(x) =

{ 1

2for 0 ≤ x ≤ π

0 for x > π⇒ f(t) =

{ 1

2for 0 ≤ t ≤ π

0 for t > πThe Fourier sine integral of f(x) is given by

f(x) =2

π

∞∫0

sinλx

∞∫0

f(t) sinλtdtdλ

=2

π

∞∫0

sinλx

π∫0

1

2sinλtdtdλ

=1

π

∞∫0

sinλx

[− cosλt

λ

]π0

dλ

=1

π

∞∫0

sinλx

[1− cosλπ

λ

]dλ

1

π

∞∫0

[1− cosλπ

λ

]sinλxdλ =f(x)

∞∫0

[1− cosλπ

λ

]sinλxdλ =πf(x)

=π

{ 1

2for 0 < x < π

0 for x > π

=

{ π

2for 0 < x < π

0 for x > π

4.3 Fourier transform pair[complex Fourier transform / InfiniteFourier transform]

Fourier transform: The Fourier transform of f(x) is given by

F [f (x)] =1√2π

∞∫−∞

f (x) eisxdx (1)

= a function of s

= F (s) = f (s)

Inverse Fourier transform:(Used in deduction part)


The Inverse Fourier transform of F (s) is given by

f (x) =1√2π

∞∫−∞

F [f (x)] e−isxds (2)

The above equations (1) and (2) are jointly called as Fourier Transformpair.

4.3.1 Examples under Fourier Transform Pair

Example 4.5. Find the Fourier transform of f(x) =

{1 in |x| < a0 in |x| > a

Solution : The given function can be written as

f(x) =

{1 if − a < x < a0 otherwise[

∵ |x| < a⇒ −a < x < a|x| > a⇒ −∞ < x < −a & a < x <∞

]∴ The Fourier transform of f(x) is given by

F (s) = F [f(x)] =1√2π

∞∫−∞

f(x)eisxdx

=1√2π

a∫−a

eisxdx

=1√2π

a∫−a

(cos sx+ i sin sx)dx

=1√2π

a∫−a

cos sxdx+ i

a∫−a

sin sxdx

=

2√2π

a∫0

cos sxdx [∵ sin sxis an odd function]

=

√2

π

[sin sx

s

]a0

=

√2

π

[sin as

s

]Example 4.6. Find the Fourier transform of the function


f(t) =

0, −∞ ≤ t ≤ 0

sin t, 0 ≤ t ≤ π0, π ≤ t ≤ ∞

Solution : The Fourier transform of f(t) is given by

F (s) = F [f(t)] =1√2π

∞∫−∞

f(t)eistdt

=1√2π

π∫0

sin teistdt

=1√2π

[eist

(is)2 + 1(is sin t− cos t)

]π0

=1√2π

[eist

1− s2(is sin t− cos t)

]π0

=1√2π

[eisπ

1− s2(0 + 1)− 1

1− s2(0− 1)

]=

1√2π

(1 + eiπs

1− s2

)


{x if |x| < a0 if |x| > a


f(x) =

{x if − a < x < a0 otherwise


F (s) = F [f(x)] =1√2π

∞∫−∞

f(x)eisxdx

=1√2π

a∫−a

xeisxdx

=1√2π

a∫−a

x(cos sx+ i sin sx)dx

=1√2π

a∫−a

x cos sxdx+ i

a∫−a

x sin sxdx

=

2i√2π

a∫0

x sin sxdx

[∵ x cos sx is odd function & x sin sx is even fn.]

=i

√2

π

[x

(− cos sx

s

)− 1

(− sin sx

s2

)]a0

=i

√2

π

[−x cos sx

s+

sin sx

s2

]a0

=i

√2

π

[−a cos as

s+

sin as

s2

]F (s) =i

√2

π

[sin as

s2− a cos as

s

]


{x2 if |x| < 10 if |x| > 1


f(x) =

{x2 if − 1 < x < 10 otherwise


F (s) = F [f(x)] =1√2π

∞∫−∞

f(x)eisxdx

=1√2π

1∫−1

x2(cossx + isinsx)dx

=1√2π.2

1∫0

x2 cos sxdx

[∵ x2 sin sx is odd function

]=

√2

π

[x2(sin sx

s

)− 2x

(− cos sx

s2

)+ 2

(− sin sx

s3

)]10

=

√2

π

[x2 sin sx

s+

2xcossx

s2− 2 sin sx

s3

]10

=

√2

π

[sin s

s+

2coss

s2− 2 sin s

s3

]10

Example 4.9. * Find the Fourier transform of f(x) is defined by

f(x) =

0, x < a1, a < x < b0, x > b

.

{Sol: F [f (x)] = F (s) =

1

is√2π

[eisb − eisa

]}

Example 4.10. Find the Fourier transform of

f(x) =

{1− |x| for |x| < 1

0 otherwise. Hence show that

∞∫0

sin2 x

x2dx =

π

2


f(x) =

{1− |x| for − 1 < x < 1

0 otherwise.


F (s) = F [f(x)] =1√2π

∞∫−∞

f(x)eisxdx

=1√2π

1∫−1

(1− |x|)(cos sx + isin sx)dx

=1√2π

1∫−1

(1− |x|) cos sxdx+ i

1∫−1

(1− |x|) sin sxdx

=2√2π

1∫0

(1− |x|) cos sxdx

[∵ (1− |x|) sin sx is odd]

=

√2

π

1∫0

(1− x) cos sxdx

=

√2

π

[(1− x)

sin sx

s− (−1)

(−cossxs2

)]10

=

√2

π

[(1− x)

sin sx

s− cos sx

s2

]10

=

√2

π

[(0− cos s

s2

)−(0− 1

s2

)]F (s) =

√2

π

(1− cos s

s2

)By inversion formula for Fourier transform


f(x) =1√2π

∞∫−∞

F (s)e−isxds

=1√2π

∞∫−∞

√2

π

(1− cos s

s2

)(cos sx− i sin sx)ds

=1

π

∞∫−∞

[(1− cos s

s2

)cos sx− i

(1− cos s

s2

)sin sx

]ds

f(x) =2

π

∞∫0

(1− cos s

s2

)cos sxds

[∵

(1− cos s

s2

)cos sx is even and

(1− cos s

s2

)sin sx is odd

]∞∫0

(1− cos s

s2

)cos sxds =

π

2f(x)

Put x = 0,∞∫0

(1− cos s

s2

)ds =

π

2.1

∞∫0

2 sin2 s2

s2ds =

π

2

∞∫0

sin2 s2(

s2

)2 ds2 =π

2

Puts

2= t⇒ ds

2= dt

∴

∞∫0

sin2 t

t2dt =

π

2

Example 4.11. Find the Fourier transform of

f(x) =

{1− x2 if |x| < 1

0 if |x| > 1and hence evaluate

∞∫0

(sinx− x cosx

x3

)cos

x

2dx



f(x) =

{1− x2 if − 1 < x < 1

0 otherwise

F (s)=F [f(x)]=1√2π

∞∫−∞

f(x)eisxdx

=1√2π

1∫−1

(1− x2)(cossx + isinsx)dx

=1√2π

1∫−1

[(1− x2) cos sx+ i(1− x2) sin sx

]dx

=2√2π

1∫0

(1− x2) cos sxdx

[∵(1− x2

)cos sx is even and

(1− x2

)sin sx is odd

]=

√2

π

[(1−x2)sin sx

s−(−2x)

(− cos sx

s2

)+(−2)

(− sin sx

s3

)]10

=

√2

π

[(1− x2)

sin sx

s− 2x

cos sx

s2+ (2)

(sin sx

s3

)]=

√2

π

[−2 cos s

s2+

2 sin s

s3

]F (s) =2

√2

π

(sin s− s cos s

s3

)By inversion formula for Fourier transform

f(x) =1√2π

∞∫−∞

F (s)e−isxds

=1√2π

∞∫−∞

2.

√2

π

(sin s− s cos s

s3

)e−isxds

=2

π

∞∫−∞

[(sin s− s cos s

s3

)cos sx− i

(sin s− s cos s

s3

)sin sx

]ds

f(x) =4

π

∞∫0

(sin s− s cos s

s3

)cos sxds

MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 259[∵

(sin s− s cos s

s3

)cos sx is even and

(sin s− s cos s

s3

)sin sx is odd

]∴

∞∫0

(sin s− s cos s

s3

)cos sxds =

π

4f(x)

Put x =1

2∞∫0

(sin s− s cos s

s3

)cos

s

2ds =

π

4

(1− 1

4

)=3π

16

Example 4.12. Find the Fourier inverse transform of e−s2

4

Solution : Solution: The inversion formula for Fourier transform is

f(x) =1√2π

∞∫−∞

F (s)e−isxds

=1√2π

∞∫−∞

e−s2

4 e−isxds

=1√2π

∞∫−∞

e−(

s2

4 +isx)ds

=1√2π

∞∫−∞

e−((

s2

)2+2(s2

)(ix)

)ds

=1√2π

∞∫−∞

e−((

s2

)2+2(s2

)(ix)+(ix)2−(ix)2

)ds

=1√2π

∞∫−∞

e−[(

s2+ix

)2+x2

]ds

=1√2π

∞∫−∞

e−(s2+ix

)2e−x2

ds

=e−x2

√2π

∞∫−∞

e−(s2+ix

)2ds

Puts

2+ ix = t⇒ ds = 2dt


∴ f(x) =e−x2

√2π

∞∫−∞

e−t22dt

=

√2

πe−x2

∞∫−∞

e−t2dt

=

√2

πe−x2

2

∞∫0

e−t2dt

=

√2

πe−x2

2

√π

2

∵ ∞∫0

e−x2

dx =

√π

2

f(x) =

√2e−x2

Note : If the transform of a function f(x) is equal to f(s) [i.e.,F (f(x)) =f(s)] then f(x) is called self reciprocal.

Example 4.13. Find the Fourier transform of e−a2x2

. Hence prove that

e−x2

2 is self reciprocal with respect to the Fourier transform.

Solution : Given f(x) = e−a2x2

.The Fourier transform of f(x) is given by


F (s) = F [f(x)] =1√2π

∞∫−∞

f(x)eisxdx

=1√2π

∞∫−∞

e−a2x2

eisxdx

=1√2π

∞∫−∞

e−(a2x2−isx)dx

=1√2π

∞∫−∞

e−[(ax)2−2(ax)

(is2a

)]dx

=1√2π

∞∫−∞

e−[(ax)2−2(ax)

(is2a

)+(is2a

)2−( is2a

)2]dx

=1√2π

∞∫−∞

e−[(ax− is

2a

)2+ s2

4a2

]dx

=1√2π

∞∫−∞

e−(ax− is

2a

)2e−

s2

4a2 dx

=1√2πe−

s2

4a2

∞∫−∞

e−(ax− is

2a

)2dx

Put ax− is

2a= t⇒ dx =

dt

a

∴ F (s) =1√2πe−

s2

4a2

∞∫−∞

e−t2 dt

a

=1

a√2πe−

s2

4a2

∞∫−∞

e−t2dt


=1

a√2πe−

s2

4a2 2

∞∫0

e−t2dt

=1

a√2πe−

s2

4a2 2

(√π

2

)∴ F (s) =

1

a√2e−

s2

4a2

F [f(x)] =1

a√2e−

s2

4a2

F(e−a2x2

)=

1

a√2e−

s2

4a2

[∵ f(x) = e−a2x2

]Setting a =

1√2

F(e−

x2

2

)=e−

s2

2

∴ f(x) =e−x2

2 is self reciprocal.

Example 4.14. * Find the Fourier Transform of e−x2

2 . (OR) Show that

e−x2

2 is self reciprocal.

Example 4.15. Find the complex Fourier transform of e−a|x|, a > 0.

Solution :

F (s) = F [f(x)] =1√2π

∞∫−∞

f(x)eisxdx

=1√2π

∞∫−∞

e−a|x|(cos sx+ i sin sx)dx

=1√2π

∞∫−∞

(e−a|x| cos sx+ ie−a|x| sin sx)dx

=1√2π

2

∞∫0

e−a|x| cos sxdx[∵ ea|x| sin sx is odd

]


=

√2

π

∞∫0

e−ax cos sxdx

=

√2

π

[e−ax

a2 + s2(−a cos sx+ s sin sx)

]∞0

=

√2

π

[a

s2 + a2

]Example 4.16. Find the Fourier integral representation of f(x) defined

as f(x)=

0, x<01

2, x=0

e−x, x>0

.

F (s)= 1√2π

1 + is

1 + s2, f(x)=

1

π

∞∫0

cos sx+s sin sx

1 + s2ds.

4.4 Parseval’s Identity for Fourier transform:

Suppose F [f(x)] = F (s)&F [g(x)] = G (s), then

1)

∞∫−∞

F [f(x)] .F [g(x)] ds =

∞∫−∞

f(x).g(x)dx If f(x) = g(x),

2)

∞∫−∞

|F [f(x)]|2 ds =∞∫

−∞

|f(x)|2 dx

Example 4.17. Show that the Fourier transform of

f(x) =

{a2 − x2, |x| < a0, |x| > a

as 2

√2

π

(sin as− as cos as

s3

). Hence deduce

(i)

∞∫0

(sin s− s cos s

s3

)ds =

π

4(ii)

∞∫0

(sin s− s cos s

s3

)2

ds =π

15.{

x = 0, a = 1 in (i),Parseval’s with a = 1 in (ii)

}


{a− |x| , |x| < a0, |x| > a

and hence deduce that (i)

∞∫0

(sin t

t

)2

dt =π

2. (ii)

∞∫0

(sin t

t

)4

dt =π

3.

264 Unit IV - FOURIER TRANSFORM (F.T.)F (s) =

√2

π

1

s2

(2 sin2

as

2

), f (x) =

2

π

∞∫−∞

(sin as

2

s

)2

e−isxds,

x = 0, a = 2 in (i),Parseval’s with a = 2 in (ii)

4.5 Sine and Cosine transforms

4.5.1 Fourier Sine transform pair:

Fourier Sine transform:Fourier Sine transform of f(x) is

FS [f(x)] =

√2

π

∞∫0

f (x) sin sxdx

= a function of s = FS (s)

Inverse Fourier Sine transform:Inverse Fourier Sine transform of FS [f(x)] = FS [s] is

f (x) =

√2

π

∞∫0

FS [f(x)] sin sxds

4.5.2 Fourier Cosine transform pair:

Fourier Cosine transform:Fourier Cosine transform of f(x) is

FC [f(x)] =

√2

π

∞∫0

f (x) cos sxdx

= a function of s = FC (s)

Inverse Fourier Cosine transform:Inverse Fourier Sine transform of FC [f(x)] = FC [s] is

f (x) =

√2

π

∞∫0

FC [f(x)] cos sxds

4.5.3 Examples under Fourier Sine & Cosine Transform:

Example 4.19. Find the Fourier Cosine Transform of e−x, x ≥ 0.


Solution : The Fourier Cosine transform of f(x) is given by

Fc(s) = Fc [f(x)] =

√2

π

∞∫0

f(x) cos sxdx

=

√2

π

∞∫0

e−x cos sxdx

=

√2

π

[e−x

1 + s2(− cos sx+ s sin sx)

]∞0

=

√2

π

1

1 + s2

Example 4.20. Find the Fourier Cosine Transform of

f(x) =

{1 if 0 ≤ x ≤ 10 if x > 1

Solution : The Fourier Cosine transform of f(x) is given by

Fc(s) = Fc [f(x)] =

√2

π

∞∫0

f(x) cos sxdx

=

√2

π

1∫0

cos sxdx

=

√2

π

[sin sx

s

]10

=

√2

π

[sin s

s

]Example 4.21. Find the Fourier sine transform of f (x) = e−ax and

deduce

∞∫0

s sin sx

s2 + a2ds =

π

2e−ax.

{FS

[e−ax

]=

√2

π

(s

a2 + s2

)}

Example 4.22. * Find the Fourier cosine transform of f (x) = e−ax and

deduce

∞∫0

cos sx

s2 + a2ds =

π

2

e−ax

a.

{FS

[e−ax

]=

√2

π

(a

a2 + s2

)}

Example 4.23. Find the Fourier sine transform of f (x) =e−ax

xand hence

find

[e−ax − e−bx

x

].

266 Unit IV - FOURIER TRANSFORM (F.T.)FS [e

−ax] =√

2π tan

−1(sa

),

FS

[e−ax − e−bx

x

]=

√2

π

[tan−1

(sa

)− tan−1

(sb

)]

Example 4.24. * Find the Fourier cosine transform of f (x) =e−ax

xand

hence find

[e−ax − e−bx

x

].

FC [e−ax] = − 1√

2πlog(s2 + a2

),

FC

[e−ax − e−bx

x

]=

1√2π

log

(s2 + b2

s2 + a2

)

Example 4.25. Find the Fourier sine transform of xn−1 and deduce1√x

is self reciprocal under sine transform.{FS

[xn−1

]=

√2

π

n

snsin(nπ

2

), FS

[x−1/2

]=

1√s

}Example 4.26. * Find the Fourier cosine transform of xn−1 and deduce1√xis self reciprocal under cosine transform.{

FC

[xn−1

]=

√2

π

n

sncos(nπ

2

), FC

[x−1/2

]=

1√s

}Example 4.27. Find the Fourier cosine transform of x.{FC [x] = −

√2

π

(1

s2

)}

4.6 Convolution theorem

1. F [f (x) ∗ g (x)] = F [f (x)] .F [g (x)] = F (s) .G (s)

2. FS [f (x) ∗ g (x)] = FS [f (x)] .FS [g (x)] = FS (s) .GS (s)

3. FC [f (x) ∗ g (x)] = FC [f (x)] .FS [g (x)] = FC (s) .GC (s)

4.7 Parseval’s identity:

Parseval’s identity for Fourier transform:

a.

∞∫−∞

{F [f (x)] .F [g (x)]} ds =∞∫

−∞

f (x) .g (x)dx [If f(x) 6= g(x)]


b.

∞∫−∞

|F [f (x)]|2 ds =∞∫

−∞

|f (x)|2dx [If f(x) = g(x)]

Parseval’s identity for Fourier Sine transform:

a.

∞∫0

{FS [f (x)] .FS [g (x)]} ds =∞∫0

f (x) .g (x)dx [If f(x) 6= g(x)]

b.

∞∫0

|FS [f (x)]|2 ds =∞∫0

|f (x)|2dx [If f(x) = g(x)]

Parseval’s identity for Fourier Cosine transform:

a.

∞∫0

{FC [f (x)] .FC [g (x)]} ds =∞∫0

f (x) .g (x)dx [If f(x) 6= g(x)]

b.

∞∫0

|FC [f (x)]|2 ds =∞∫0

|f (x)|2dx [If f(x) = g(x)]

4.7.1 Examples under Parseval’s identity:

Example 4.28. Evaluate

∞∫0

dx

(x2 + a2) (x2 + b2).

{π

2ab (a+ b)

}


∞∫0

x2dx

(x2 + a2) (x2 + b2).

{π

2 (a+ b)

}


∞∫0

dx

(x2 + a2)2.

{ π

4a3

}Example 4.31.

Evaluate

∞∫0

x2dx

(x2 + a2)2.

{ π4a

}

4.8 Properties

4.8.1 Fourier Transform properties:

Property 4.1. Fourier Transform is linear i.e.,


F [af(x) + bg(x)] = aF [f(x)] + bF [g(x)]

Proof :

F [af(x) + bg(x)] =1√2π

∞∫−∞

(af(x) + bg(x)) eisxdx

=a1√2π

∞∫−∞

f(x)eisxdx+ b1√2π

∞∫−∞

g(x)eisxdx

=aF [f(x)] + bF [g(x)]Note : Fourier cosine transform and Fourier sine transform are linear.

i.e.,

Fc[af(x) + bg(x)] =aFc[f(x)] + bFc[g(x)]

Fs[af(x) + bg(x)] = aFs[f(x)] + bFs[g(x)]

Property 4.2. (Shifting theorem) F [f(x− a)] = eiasF (s)

Proof : F [f(x− a)] =1

2π

∞∫−∞

f(x− a)eisxdx

Put x− a =t when x = −∞, t =−∞dx =dt when x = ∞, t = ∞

=1

2π

∞∫−∞

f(t)eis(a+t)dt

=eias√2π

∞∫−∞

f(t)eistdt

∴ F [f(x− a)] =eiasF (s)

Property 4.3. F [eiaxf(x)] = F (s+ a).

Proof :

F [eiaxf(x)] =1√2π

∞∫−∞

eiaxf(x)eisxdx

=1√2π

∞∫−∞

f(x)ei(s+a)xdx

=F (s+ a)

Property 4.4. (Change of scale) F [f(ax)] =1

|a|F(sa

)where a 6= 0.


Proof : Suppose a > 0

F [f(ax)] =1√2π

∞∫−∞

f(ax)eisxdx

Put ax =t when x = infty, t =infty

dx =dt

awhen x = ∞, t =∞

=1√2π

∞∫−∞

f(t)ei(sa

)tdt

a=

1

a

1√2π

∞∫−∞

f(t)ei(sa

)tdt

=1

aF(sa

)Similarly, if a < 0

F [f(ax)] =1√2π

−∞∫∞

f(t)ei(sa

)tdt

a

since when x =−∞, t = −∞when x =∞, t = ∞

=−1

a.

1√2π

∞∫−∞

f(t)ei(sa

)tdt

=−1

aF(sa

)Hence F [f(ax)] =

1

|a|F(sa

)Note : Fc[f(ax)] =

1

aFc

(sa

)Fs[f(ax)] =

1

aFs

(sa

), where a 6= 0.

1. Linearity property:F [af (x) + bg (x)] = aF [f (x)] + bF [g (x)] = aF (s) + bG (s)

2. Shifting theorem:F [f (x− a)] = eisaF [f (x)] = eisaF (s)

3. Shifting theorem in s:F[eiaxf (x)

]= [F (s)]s→s+a = F (s+ a)

4. Change of scale property:

F [f (ax)] =1

aF(sa

), a > 0


F [f (ax)] = −1

aF(sa

), a < 0

F [f (ax)] =1

|a|F(sa

)5. F [xnf (x)] = (−i)n dn

dsn{F [f (x)]} = (−i)n

{dn

dsn[F (s)]

}6. F

[dn

dxnf (x)

]= (−is)n {F [f (x)]} = (−is)n [F (s)]

7. F [f (−x)] = F (−s)

8. F[f (x)

]= F (−s)

9. F[f (−x)

]= F (s)

10.Modulation property:

a. F [f (x) cos ax] =1

2{F (s− a) + F (s+ a)}

b. F [f (x) sin ax] =1

2{F (s− a)− F (s+ a)}

4.8.2 Fourier Sine & Cosine Transform properties:

11. Linearity property:

a. FS [af (x) + bg (x)] = aFS [f (x)] + bFs [g (x)]

b. FC [af (x) + bg (x)] = aFC [f (x)] + bFC [g (x)]

12. Modulation property:

a. FS [f (x) sin ax] =1

2{FC (s− a)− FC (s+ a)}

b. FS [f (x) cos ax] =1

2{FS (s− a) + FS (s+ a)}

c. FC [f (x) sin ax] =1

2{FS (a− s) + FS (s+ a)}

d. FC [f (x) cos ax] =1

2{FC (s− a) + FC (s+ a)}

13. Change of scale property:

a. FS [f (ax)] =1

aFS

(sa

), a > 0


b. FC [f (ax)] =1

aFC

(sa

), a > 0

14. a. FS [f′ (x)] = −sFS (s) ifx→ ∞ ⇒ f (x) = 0

b. FC [f ′ (x)] = −√

2

πf (0) + sFS (s) ifx→ ∞ ⇒ f (x) = 0

15. a. FS [xf (x)] = − d

ds{FC [f (x)]} = − d

ds{FC (s)}

b. FC [xf (x)] =d

ds{FS [f (x)]} = − d

ds{FS (s)}

4.8.3 Examples under Properties:

Example 4.32. Find Fourier sine transform of xe−ax.{2

√2

π

[as

(a2 + s2)2

]}

Example 4.33. Find Fourier cosine transform of xe−ax.{√2

π

[a2 − s2

(a2 + s2)2

]}.

Example 4.34. Find Fourier cosine transform of e−a2x2

and evaluateFourier sine transform of xe−a2x2

.{FC

[e−a2x2

]=e−s2/4a2

√2a

,FS

[xe−a2x2

]=se−s2/4a2

2√2a3

}

Example 4.35. Find f(x) if its sine transform ise−sa

sand hence find

F−1S

(1

s

). {

f (x) =

√2

πtan−1

(xa

), F−1

S

(1

s

)=

√π

2

}

Example 4.36. Find FS

(e−x), FC

(e−x), FC

(1

1 + x2

), FS

(x

1 + x2

).{√

2

π

s

s2 + a2,

√2

π

a

s2 + a2,

√π

2e−s,

√π

2e−s

}


Example 4.37. Solve the integral equation

∞∫0

f (x) cos sxdx = e−s, show

that

∞∫0

cos sx

1 + s2ds =

π

2e−x.

{f (x) =

2

π

(1

1 + x2

), e−x

}

4.9 Assignment II[Fourier Transforms]

1. Find the Fourier integral representation of f(x) defined as

f(x) =

0, x < 012 , x = 0e−x, x > 0

.

2. Find the Fourier transform of the function f(x) defined by

f(x) =

{1− x2; if |x| < 10; if |x| ≥ 1

. Hence prove that

(i)

∞∫0

(sin s− s cos s

s3

)cos(s2

)ds =

3π

16

(ii)

∞∫0

(sin s− s cos s

s3

)2

ds =π

15.

3. Find F.T. of f(x) =

{1− |x| ; if |x| < 1,0; if |x| > 1,

and hence find the value of

∞∫0

(sin t

t

)dt and

∞∫0

(sin t

t

)4

dt

4. Find the Fourier transform of e−a2x2

. Hence prove e−x2/2 is selfreciprocal.

5. Find the Fourier sine and cosine transform of

f(x) =

x; 0 < x < 12− x; 1 < x < 20; x > 2

.

6. Prove that e−x2/2 is self reciprocal under Fourier cosine transform.

7. Find the Fourier sine and cosine transform of xn−1 and hence prove1√xis self reciprocal under Fourier sine and cosine transforms.


8. Find the Fourier sine transform of e−ax and hence evaluate Fouriercosine transforms of xe−ax and e−ax sin ax.

9. Find F.S.T. and F.C.T. of e−ax, a > 0. Hence evaluate

∞∫0

x2

(a2 + x2)2dx

and

∞∫0

dx

(x2 + a2) (x2 + b2)

10. State and prove convolution theorem and Parseval’s identity forFourier transforms.

4 Fourier Transform(F.T.)integral. Hence evaluate Z1 0 sin cos x d and nd the value of Z1 0 sin d ....

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Transcript of 4 Fourier Transform(F.T.)integral. Hence evaluate Z1 0 sin cos x d and nd the value of Z1 0 sin d ....