FOURIER TRANSFORMS

FOURIER INTEGRAL THEOREM OR FORMULA

If f(t) is a given function defined in (-∞,∞) and satisfies Dirichlet’s

conditions then

f(x) = ∫ ∫ f(t) cos λ (t-x) dtd λ-------------------------(1)

which is known as Fourier Integral of f(x).

NOTE :

Eqn (1) is true at a point of continuity. At a point of discontinuity

the value of the integral on LHS of (1) is 1/2 [f(x+0) + f(x-0)]

FOURIER SINE AND COSINE INTEGRALS

F (x) = ∫ sinλx ∫ f(t) sinλt dt dλ

Which is known as Fourier sine integral?

f(x) = ∫ cosλx ∫ f(t) cosλt dt dλ

Which is known as Fourier cosine integral?

COMPLEX FROM OF FOURIER INTEGRAL

If f(x) is piecewise continuously differentiable and absolutely

integrable on the entire real line if ∫ |f(x) |dx is bounded then

1

π

∞ ∞

-∞ o

2

π

∞ ∞

o o

2

π

∞ ∞

o o

∞

-∞

f(x) = ∫ ∫ f(t) ei (t-x)λ dt dλ

Problems:

1). Express the function f(x) =

as a Fourier integral. Hence evaluate

and find the value of

Solution :

Fourier Integral formula for f(x) is

f(x) =

=

Given f(t) = 1 for -1 < t < 1

= 0 for -∞ < t < -1

and 1< t < ∞

sub (2) in (1), we get

1

2π

∞ ∞

-∞

-∞

1 for | x | < 1

0 for | x | > 1

ie. -1 < x < 1

ie. -∞< x < -1

ie. 1 < x < ∞

∫ sinλ cosλx

λ dλ

∞

o

∫ sinλ

λ dλ

∞

o

∫ f(t) cosλ (t-x) dt dλ ∞

o

1 π

∫ ∞

-∞

∫ f(t) cosλ (t-x) dt + ∞

o

1 π

∫ -1

-∞

f(t) cosλ (t-x) dt ∫ 1

-1

f(t) cosλ (t-x) dt dλ ∫ ∞

1 + (1)

(2)

f(x) =

=

=

=

=

=

f(x) =

which is the required Fourier integral of the given function

To evaluate

(3)

∫ cosλ (t-x) dt dλ ∞

o

1 π

∫

∫ sin λ (t-x)

λ

∞

o

1 π

dλ

1

1

-1

-1

∫ sin λ (1-x)

λ

∞

o

1 π

dλ sin λ (1+x)

λ +

∫ sin λ (1-x) ∞

o

1 π

dλ sin λ (1-x) +

λ

∫ sin λ cosλ x - cosλ sinλx ∞

o

1 π dλ

+ sinλ cosλx + cosλ sinλx

λ

∫ 2 sinλ cosλx ∞

o

1 π dλ

λ

∫ sinλ cosλx ∞

o dλ

λ

2 π

(3)

∫ sinλ cosλx ∞

o λ

dλ

∫ sinλ cosλx ∞

o λ

dλ = π

2 f(x)

∫ sinλ cosλx ∞

o λ

dλ = π

2 1 if | x | < 1

0 if | x | > 1

(4)

To find the value of

Put x = 0, is (4), we get

=

∞

∫sinλ dλ = π 0 λ 2

Hence the result.

2. Find the Fourier intergral of the function

Verify the representation directly at the point x = 0

Soln :

Fourier integral of f(x) is

f(x) =

=

Given f(t) = 0, -∞ < t < 0

= e-t, 0 < t < ∞

Sub (2) in (1), we get

∫ sinλ ∞

o λ dλ

∫ sinλ cos0

o λ dλ

∞

π

2 (1) , if x = 0

0 , x< 0

1/2, x= 0

e-x

, x>0

f(x) = ie. -∞ < x < 0

ie. 0 < x < ∞

∫ f(t) cosλ (t-x) dt dλ

dλ

∞

o

1 π

∫ ∞

-∞

∫ ∞

o

1 π

∞

dλ f(t) cosλ (t-x) dt + ∫ f(t) cosλ (t-x) dt ∫ o

-∞

o

(1)

(2)

∞ ∞ -t

f(x) = 1 ∫ 0 + ∫ e cos λ(t-x) dt dλ π 0 0

∞ -t ∞

= 1 ∫ e [- cos (λt-λx) + λ sin (λt-λx) dλ π 0 1+λ2

0

∞

= 1 ∫ 0 - 1 [- cos λx + λ sin λx] dλ π 0 1+λ2

∞

f (x) = 1 ∫ 1 (cos λx + λ sin λx ) dλ (3)

π 0 1+λ2

which is the required Fourier integral of the given function.

To verify the value of f(x) at x= 0

Put x = 0 in (3) , we get

∞

f (0) = 1 ∫ dλ π 0 1+λ2

∞

= 1 [ tan-1λ ]

π 0

= 1 π π 2

f(0) = 1

2

Hence verified.

3. Express f(x) = 1 for 0<x<π

0 for x>π as a Fourier Sine integral

and hence evaluate � ��������

sinλt dλ

Soln:

Fourier sine integral for f(x) is

f(x) = � � � ��

�� � ���� �� �� �� ��

=� � � ��

�� �� ���� �� �� �� � � ���� ��

� �����d�

Given f(t) = 1 ��� 0 � � � �

0 for t �

sub (2) in (1) we get,

f(x) =

=

=

f(x) =

which is the required Fourier integral for f(x)

To evaluate

Note : At x = π , which is the point of discontinuity, the value of the

above integral becomes

∫ ∞

o

2

^ sin�x ∫

π

o sin�tdt dλ

∫ ∞

o

2

^ sin�x

π

o

-cos� t

λ

dλ

∫ ∞

o

2

^ sin�x -cos λπ

λ +

cos 0

λ

dλ

∫ ∞

o

2

^ sin�x 1-cos λπ

λ dλ (3)

∫ ∞

o sin�x (1-cos λπ )

λ

dλ

∫ ∞

o sin�x (1-cos λπ)

λ

dλ = ^/2 f(x) (3)

∫ ∞

o sin�x (1-cos λπ)

λ

dλ = ^/2 1 for 0 � x ≤ π

0 for x > π

� �!"����#�������

�� = ��f(π)

= �� $%�����%��#�� &

= �� $��� &

= �'

ASSIGNMENT PROBLEMS

1). Using Fourier sine integral for f(x) = e-ax

, (a > 0)

show that

2). Find the Fourier cosine integral for f(x) = e-ax

. Hence deduce the

value of the integral

3). Using Fourier integral show that

HINT : Consider f(x) =

then apply Fourier sine integral formula because answer is in the

form of sine function.

∫

∫ ∞

o

λsinλx

λ2 +a

2

dλ = /2 e-ax

^

∫ cosλx

1+λ2

∞

o dλ

∫ sin πλ sinxλ

1-λ2

∞

o

dλ = (π/2) sinx, 0 < x < π 0 x > π

(π/2) sinx, 0 ≤ x≤ π and

0 x > π

4). With the help of Fourier integral show that

HINT : Same as previous problem but apply Fourier cosine integral

formula.

FOURIER TRANSFORMS

INFINITE (OR) COMPLEX FOURIER TRANSFORM AND ITS

INVERSION FORMULA

F[S]= F[f(x)] = 1 � ����(!�)��

is called complex Fourier transform of f(x).

f(x) =

is called inversion formula for the complex Fourier transform of f(x)

INFINITE FOURIER SINE AND COSINE TRANSFORMS

FOURIER SINE TRANSFORM :

Fs [f(x)] = �

√�� � ���� �� ���

∫ cosxλ

1+λ2

∞

o dλ = /2 e

-x, x > 0 ^

Infinite Fourier Transform Finite Fourier Transform

2π -∞

∫ ∞

-∞ F [f(x)] e

ds

-isx

1

√2π

∞

is called the Fourier sine Transform of f(x)

f(x) = �

√�� � +s -f�x�0sinsx ds

is called the inversion formula for the Fourier Sine Transform of f(x).

FOURIER COSINE TRANSFORM

c[f(x)] =�

√�� � ���� 4� ���

is called the Fourier Cosine Transform of f(x)

f(x) = �

√�� � +s -f�x�0cossx ds

is called the inversion formula for the Fourier Cosine Transform of f(x).

CONVOLUTION OF TWO FUNCTIONS :

The convolution of two functions f(x) and g(x) over the interval

(-∞,∞) is defined as f *g =�

√�� � ��7�8�7 9 ���7 : ;���

CONVOLUTION THEOREM FOR FOURIER TRANSFORMS

STATEMENT :

The Fourier Transform of the convolution of f(x) and g(x) is the

product of their Fourier Transforms.

F[f * g] = F[s]. G [s]

Proof :

F[f * g] = �

√�� � �� < 8�(!�)��

= �

√�� � - �√��

# � ��7�8�7 9 ���� 0(!�) �7

[by (1)]

(1)

�

√�� � ��7�# $ �

�� � 8�� 9 7�(!�)��# & �7

(by changing the order of integration)

= �√�� � ��7� $ �

√�� � 8�� 9 7�(!�)(!�=(#!�=��# &

# �7

= �

√�� � ��7�(!�= $ �√�� � 8�� 9 7�(!�)#=����

# &# �7

= �

√�� � ��7�(!�= $ �√�� � 8�� 9 7�(!�)#=����� 9 7

# �&# �7

Since d(x-u)=dx-du

=dx-0

=dx

= �

√�� � ��7�(!�=# $ �

√�� � 8���(!�>��# & �7 where x-u=t

= �

√�� � ��7�(!�=�7# $ �

√�� � 8���(!�>��# &

PARSEVAL’S IDENTITY FOR FOURIER TRANSFORMS:

STATEMENT:

If the Fourier Transforms of f(x) and g(x) are F[ s] and

G[s]respectively the

=

f g(t)e 1

√2^

= F[s] G[s]

Hence the theorem

|f(x)|2 dx = |F[s] |

2ds ∫

∞

-∞

∫ ∞

-∞

Proof

We know that

F[f * g] = F[s] . G[s] (by convolution theorem)

F#� [F[s] . G[s]] = f * g

�√�� � +- 0

# @-A0(#!�)� = �√�� ��7�8�� 9 7�(#!�)�

Putting x = 0, we get

sub (2) & (4) in (1), we get

Hence the theorem

FINITE FOURIER SINE AND COSINE TRANSFORMS

Fourier sine Transform

=

∫ -∞

F[s] . G[s] ds = ∫ -∞

∞ f(u) g(-u) du ∫ -∞

∞ (1)

Let g(-u) = f(u) (2)

i.e., g(u) = f(-u) (3)

B G[s] = F[g(u)]

= F[f(-u)]

= F[s]

ie., G[s] = +- 0 (4)

F[s] F[s] ds = ∫ ∞

f(u) f(u) du ∫ ∞

-∞ -∞

|F[s]|2 ds = ∫

∞ |f(u)|

2du ∫

∞

-∞ -∞

Fs [f(x)] = f(x). sin nπx dx ∫ ℓ

l o ℓ

where n is an integer is called the finite Fourier Sine Transform of

in the interval o<x< ℓ

is called the inversion formula for the finite Fourier sine Transform of

f(x)

FOURIER COSINE TRANSFORM :

where n is an integer, is called the finite Fourier Cosine Transform of f(x)

in the interval o<x< ℓ

ℓ

is called the inversion formula for the finite Fourier Cosine Transform of

f(x).

Parseval’s identity

Note

(i).

(ii)

(iii).

(iv)

f(x) = 2/ ℓ

Fs [f(x)] sin nπx dx n=1 ℓ

∑

∞

Fc [f(x)] = f(x). cos nπx dx ∫ ℓ

o ℓ

f(x) = �ℓ Fc [o] +

�ℓ

∑

∑"E� F C [f(x)] cos nπx

Fc[s] Gc[s] ds = ∫ ∞

f(x) g(x) dx ∫ ∞

o o

Fs[s] Gs[s] ds = ∫ ∞

f(x) g(x) dx ∫ ∞

o o

(Fc[s])2 ds = ∫

∞ f(x))

2 dx ∫

∞

o o

(Fs[s])2 ds = ∫

∞ (f(x))

2 dx ∫

∞

o o

PROPERTIES OF FOURIER TRANSFORMS

1. Linearity Property

If F[s] and G[s] are Fourier Transforms of f(x) and g(x)

respectively then F(af(x) + bg(x)] = aF[s] + bG[s] where ‘a’ & ‘b’ are

constants.

Proof :

We have F[S] = F[f(x)] = �

√�� � ����(!�)��#

�√�� � 8���(!�)��

#

= �

√�� � -F�����G8���0(!�)��#

= F- �√�� � ����(!�)��0 �

# G- �√�� � 8���(!�)��0

#

Hence the proof.

Change of Scale property

If F[s] is the complex Fourier Transform of f(x) then F[f(ax)]

= F [s/a], a ≠ 0

= and G[S] = F[g(x)]

BF[af(x) + bg(x)]

= a F[f(x)] + b[F[g(x)]

= a F[s] + b G[s]

1

a

Proof :

We have F[s] = F[f(x)] = �

√��

F[f(ax)] = �

√��

B F[f(ax)] = �

√�� � (!�H> IJ K# ������ L>

I

= �I $ �√�� � (!�� I⁄ �>

# ��������&

Cor :-

If Fs[s] and Fc[s] are the Fourier sine and cosine transforms of f(x)

respectively, then

(i) Fs [f(ax)] = (1/a) Fs [s/a]

and (ii) Fc[f(ax)] = (1/a) Fc [s/a]

Proof :-

We have Fs[s] = Fs[f(ax)] = N�� � ���� �� ���

Fs[f(ax)] = N��

∫ (!�) f(x) dx

-∞

∞

∫ f(ax)dx -∞

eisx

∞

Put ax = t x = -∞ then t = -∞

adx = dt x = ∞ then t = ∞

1

a =

F[s/a]

∫ f(ax) sinsx dx ∞

o

Put ax = t x = 0 ⇒ t = 0

adx = dt x = ∞ ⇒ t =∞

dx = dt/a

B Fs[f(ax)] =

Hence the proof (i)

Similarly we prove (ii) also

3). Shifting Property

Statement:-

If F[s] is the complex Fourier Transform of f(x) then

F[f(x-a)] = eisa

F[s]

Proof :

+-��� 9 F� : �√�� � (!�)������

#

�

√��

Hencetheproof

∫ f(t) sin(s/a)t dt/a ∞

o 2

^

∫ f(t) sin(s/a)t dt ∞

o 2

^ = 1/a

∞ = 1/a Fs [s/a]

F[s]=F[f(x)] = �

√�� � (!�)������#

Put x-a = t x = -∞ then t = ∞

dx = dt x = ∞ then t = ∞

∫ eis(t+a) f(t) dt -∞

F[f(x-a)] = �

√��

∞

∫ eist

f(t) dt

-∞

∞

= eisa

= eisa

F[f(t]

= eisa F[s]

Modulation Theorem

Statement

If F[s] is the complex fourier transform of f(x), then F[f(x) cosax] =

1/2 {F[s+a] + F[s-a]}

Proof:-

== 1√2�

= �� - P �√��

Cor :-

If Fs[s] and Fc[s] are Fourier sine and cosine transforms of f(x)

respectively, then

(i) Fs [f(x) cosax] = [Fs[s+a] + Fs[s-a]]

(ii) Fs [f(x) sinax] = [Fs[s+a] + Fs[s-a]]

(iii) Fs [f(x) sinax] = [Fc[s-a] + Fc[s+a]]

We have F[s] = F[f(x)] =�

√�� ∫ eisx

f(x) dx ∞

-∞

F[f(x) cosax] = �√�� ∫ e

isx f(x) cosax dx

-∞

∞

∫ eisx f(x) -∞

∞ eisx

+ e-iax

2

dx =

�√��

∫ eisx eiax f(x)dx +

-∞

∞ ∫ eisx e-iax f(x)dx

-∞

∫ ei(s+a)x

f(x)dx +

-∞

∞

∫ ei(s-a)x

f(x)dx

-∞

∞

F[f(x) cosax] = [F[s+a] + F[s-a]]

1

2

1

2

1

2

1

2

Sin cos B = Sin (A+B) + Sin (A-B)

2

CosA SinB = Sin (A+B) - Sin (A-B)

2

Proof :-

Hence the proof (i)

IIIly we prove for (ii) of (iii)

Property :-

If F[s] is the complex fourier transform of f(x) then

F[xn] f(x)] = (-i)

n

Proof :-

First to prove this result for n = 1

ie., F[x f(x)] = t(i)

Now :-

Fs [f(x) cosax] = ∫ f(x) cosax) sin sx dx ∞

o 2

^

= ∫ f(x) o

2

^

∞ sin(ax+sx) – (sin(ax-sx) dx

2

= ∫ f(x) o

2

^ sin(s+a) x +(sin(s-a)x dx

2

∞

= ∫ f(x) o

2

^ sin(s+a)x dx +

∞

∫ f(x) o

2

^ sin(s-a)x dx

∞ 1

2

= 1

2 Fs[s+a] + Fs [s-a]

CosA cosB = Cos(A+B) + Cos(A-B)

2

SinA SinB = Cos (A-B) – Cos (A+B)

2

dnF[s]

dsn

dnF[s]

dsn

d

ds F[s] =

d

ds ∫ eisx

f(x)dx 1

√2^ -∞

∞

∫ eisx

f(x)dx 1

√2^ -∞

∞ =

δ

δs

∫ eisx

xix f(x)dx 1

√2^ -∞

∞ =

~ This is true for n = 1

Next to prove this result of n = z

(1)

ie., F[x2f(x)] = (-i)

2 d

2F[s]

ds

Now

d2F[s]

ds

d

ds

d

ds = F[s]

d

ds = (i) F[xf(x) (by (1))

d

ds = (i) ∫ e

isx xix f(x)dx

1

√2^ -∞

∞

= (i) ∫ eisx xf(x)dx 1

√2^ -∞

∞ δ

δs

= (i) ∫ eisx xix. x f(x)dx 1

√2^ -∞

∞

= (i) ∫ eisx

x2 f(x)dx

1

√2^ -∞

∞ d

2

ds2 F[s]

2

= (i)2 F[x

2f(x)] (2)

This is true for n = 2 also

~ In general

Cor :-

If Fc[s] and F3 [s] are fouier sine and cosine transforms of f(x)

respectively then

(i) Fc[x.f(x)] = Fs[f(x)]

(ii) Fc[x.f(x)] = Fc[f(x)]

Proof :-

(i) We have Fs[f(x)] =

~ Fs[f(x)] =

(ii). We have Fc [f(x) =

~ Fc [f(x) =

= (i)2 F[x

2f(x)] = (-1)

2 F(s)

d2

ds2

F[x2f(x)] = (-1)

2 F[s]

d2

ds2

d

ds

d

ds

∫ f(x) sinsxdx 1

√2^ -∞

∞

∫ f(x) sinsx dx 1

√2^ -∞

∞ δ

δs

∫ f(x) xcosx dx 1

√2^ -∞

∞ =

∫ xf(x) cossxdx 1

√2^ -∞

∞ =

= Fc [x f(x)]

⇒ Fc [x f(x)] = Fs [f(x)]

d

ds

∫ f(x) cossxdx √2

^ -∞

∞

d

ds ∫ f(x) cossx dx √2

^ -∞

∞ δ

δs

∫ f(x) -xsinsx dx √2

^ -∞

∞

= ∫ xf(x) sinsx dx √2

^ -∞ = -

∞

= Fs [x f(x)]

= Fs [x f(x)] = - Fc [f(x)] d

ds

Propety :

If F(s) is the complex fourier trasform of f(x) then

Proof :-

First to prove the result for n = i

i.e.,

Now,

dn f(x)

dxn

F = (-is)n F(s)

df(x)

dx F = (-is) F(s)

df(x)

dx F

∫ eisx f'(x) dx

1

√2^ -∞

∞ =

df(x)

dx ~ = f(x)

∫ eisx

d(f(x)) 1

√2^ -∞

∞ =

∫ f(x) (is)eisx

dx 1

√2^ -∞

∞ = e

isx (f(x)) -

-∞

∞

∫ eisx f(x) dx 1

√2^ -∞

∞ = 0 - is

(Assuming f(x) = 0 as x → + ∞)

= (-is) eisx f(x) dx

1

√2^ ∫ -∞

∞

= (-is) F[s] (1)

Next to prove this result for n = 2

First to prove the result for n = i

i.e.,

Now,

This is true for n = 2

In general

Cor :-

If Fs[s] and Fc[s] are fourier sine and, cosine transforms

respectively then

d2f(x)

dx F

= (-is)2 f [s]

d2(x)

dx F

∫ f''(x) e

isx dx

1

√2^

∞ =

-∞

~ f(x) = f''(x) d

2

dx

= eisx

d [f'(x)] 1

√2^ ∫ -∞

∞

= [eisx f'(x)] - f'(x) eisx (is) dx 1

√2^ -∞

∞

∫ ∞

-∞

= o-(is) f'(x) eisx dx 1

√2^ -∞

∞ ∫

= (-is) f'(x) eisx

dx 1

√2^ -∞

∞ ∫

= (-is) F [f'(x)]

= (-is) F [ f(x)] d

dx

= (-is) (-is) F [s] (by (1))

= (-is) F [s]

dn

dxn

F

= (-is)n F[s] f(x

(i) Fs [f'(x)] = -s Fc[s] (or) (i) find Fs

(ii) Fs [f'(x)] = - f(o) + SFs [s]

Proof

(Assuming f(x) → 0 as x → + ∞)

(Assuming f(x) → 0 as x → + ∞)

~Fx [f'(x) = - f(o) + S Fs [S]

Hence the proof (ii)

df(x)

dx 1

√2^

(i) Fs [f'(x)] = f'(x) sinsx dx 1

√2^ ∫ -∞

∞

= sinsx d(f(x)) 1

√2^ ∫ -∞

∞

= (sinsx f(x)) - f(x) cossx - sdx 1

√2^ ∫ -∞

∞ ∫

-∞

∞

o

∞

= f(x) cossxdx 1

√2^ ∫ ∞

o

= -S f(x) cossxdx 1

√2^ ∫ ∞

o

= -S Fc f(x)

Hence Fs [f'(x)] = -S Fc [f(x)]

(or)

Fourier sine transform of is –S Fc (f(x)) df

dx

(ii) Fc [f'(x)] = f'(x) cosxdx 1

√2^ ∫ -∞

∞

1

√2^ ∫ -∞

∞ = cosx d(f(x))

1

√2^ ∫ -∞

∞ = [(cosx f(x)) + f(x) ssinsx dx]

o

∞

o

∞ ∫

1

√2^ ∫ -∞

∞ = [-f(o) + s f(x) sinsxdx]

o

∞ ∫

1

√2^

Property : 7

If F[S] is the complex Fourier transform of f(x)

then F[f(-x)] = F[-S]

PROOF

Put –x = y When x = -∞ then y = ∞

-dx = dy Whenx = ∞ then y = -∞

~ F[f(-x)] = e-isy

f(y) (-dy)

Hence the proof

Property (8)

If F[S] is the complex Fourier Transform of f(x) then

F [f(x)] = F [-s]

F[f(-x)] = eisx f(-x) dx 1

√2^ ∫

-∞

∞

1

√2^ ∫

-∞

∞

1

√2^ ∫

∞

-∞

= e i(-s) f(y) dy

1

√2^ ∫

∞

-∞ = e i(-s)y f(y) dy

~ - = ∫ -∞

∞ ∫

-∞

∞

= F [-s]

~ F[f(-x) = F [-s]

Proof

We have F[s] = F [f(x)]

Taking complex conjugate on bothsides, we get

Hence the proof.

Property : (9)

If F[s] is the complex fourier transform of f(x) then F[f(-x)] = F[s]

Proof :-

We have F[s] = F[f(x)]

Taking complex cojugate on bothsides, we get

Put x = -y x = ∞ then y = -∞

dx = -dy x = -∞ then y = ∞

~

1

√2^ ∫

∞

-∞

= f(x)eisx

dx

~ F[-s] 1

√2^ ∫

∞

-∞

= f(x)eisx

dx

F[-s] 1

√2^ ∫

∞

-∞

= f(x)eisx

dx

F[-s] = F [f(x)]

1

√2^ ∫

∞

-∞ = f(x)e

isx dx

F[-s] 1

√2^ ∫ ∞

= f(x)eisx

dx -∞

F[-s] 1

√2^ ∫

∞

-∞

= f(-y)eisy

(-dy)

1

√2^ ∫

∞

-∞

= - f(-y)eisy

dy

1

√2^ ∫

∞ = f(-y)e

isy dy

1

√2^ ∫

∞ = f(-x)e

isy dy

1

√2^ ∫

∞ = f(-x)e

isy dx

-∞

-∞

-∞ [Change the variable y into x

Hence the proof.

Problems :-

1). Find the fourier transform of f(x) = 0 for |x| > a

Hence deduce that (i)

Proof :-

Fourier transform of f(x) is

F[f(x)]

Given f(x) = 1 fof –a < x < a

= 0 otherwise

Sub (2) in (1), we get

F[f(x)]

-∞

o ∫

sint

t dt = ^/2

-∞

o ∫

sint

t dt = ^/2 (ii) 2

1

√2^ ∫

∞ = f(-x)e

isy dx

-∞

1

√2^ ∫

∞ = + + f(-x)e

isx ) dx

-a (1)

-a

a ∫

a ∫ a

(2)

1

√2^ ∫

-a = e

isx dx

a

1

√2^ ∫

-a

a = (cosx + isinsx) dx ~ e

iθ = cosθ + isinθ

= cosdx +i sinsxdx 1

√2^ ∫

-a

a ∫

-a

a

~ F[f(x)] =

To deduce (i) dt = ^/2

Using fourier inversion formula, we get

= 2 cosdx + θ 1

√2^ ∫

-a

a

[~ Cossx is even ~ = 2

sinsx is odd ~ = 0]

∫ -a

a ∫

-a

a

∫ -a

a

= cossxdx 1

√2^ ∫

-a

a

= sinsx

s

1

√2^ o

a

= sinQs

s

1

√2^

sinas

s

2

^

∫ o

∞ sint

s

f(x) = e-isx

ds 1

√2^ ∫

-∞

∞ 2

^

sinas

s

sinas

s 1

^ = [cossx – isinsx] ds ∫

∞

-∞

sinas

s

1

^ = cossx ds- (isinsx] ds ∫ ∞

-∞ ∫

∞

-∞

sinas

s

sinas

s 1

^ = 2 cossx ds – 0 ∫

∞

-∞

sinas

s [~ cossx is even,

∫ ∞

-∞ ∫

∞

o

Put a = 1

Put x = 0

Change the variable s to t, we get

(ii) Using parseval’s identity

= 2

and sinsx is odd, sinas

s

∫ ∞

-∞ = 0]

sinas

s ~ f(x) = cossx ds ∫

∞

o

2

^

sinas

s f(x) = cossx ds

2

^ ∫

∞

o

sinas

s f(0) = ds

2

^ ∫

∞

o

sins

s ⇒ ds = f(o) ∫ ∞

o ^

2

sint

t dt = f(o) ∫

∞

o

^

2

^

2 = (1) (~ At x = 0, f(x) = 1)

sint

t ⇒ dt = ^

2

[f(x)]2 dx = (F[s])

2 ds ∫

∞

-∞ ∫

∞

-∞

[f(x)]2 dx = ds ∫

∞

-∞ ∫

∞

-∞

sins

s

2

^

2

dx = ds ∫ a

-a

sinas

s

2

^ ∫

∞

-∞

2

(x) = ds sinas

s

2

^ ∫

∞

-∞

2 a

-a

a + a = ds sinas

s

2

^

2

∫ ∞

-∞

2a = ds sinas

s

2

^

2

∫ ∞

-∞

∞

Put a=1

Change the variable s to t, we get

2). Find the faurier transform of

f(x) = 1-x2, |x| < | x, -1 < x < 1 i.e.,

0, |x| > 1 Hence

deduce that

^ = ds

sinas

s ∫

∞

-∞

2

⇒ ^ = 2 ds

sinas

s ∫

∞

-∞

2

[ ~ is even

~ = s]

sinas

s

2

∫ ∞

-∞ ∫

∞

o

sint

t

^

2 ⇒ = ds

∫

∫ ∞

o

o

2

sint

t

^

2 = dt =

2

xcosx – sinx x

3

∫ ∞

o

cos x/2 dx = -3^

16

(i)

Soln :-

Fourier Transform of f(x) is

Which is the required fourier transform

To evaluate ∫ ∞

o

1

√2^ ∫

∞ = f(x)e

isy dx

-∞

1

√2^ ∫

-1

= (1-x2)e

isx dx

√2^ ∫

-1 = (1-x

2) (cossx + isinsx) dx

-1

1

√2^ ∫

-1

= (1-x2) (cossx dx + (1-x

2) sinsx dx

1 ∫

-1

1

1

√2^ ∫

-1 = 2 (1-x

2) cossx dx

1

u = 1-x2

v = cossx

u' = -2x v1= sinsx/s

u'' = -2 v2= -cossx/s2

v3= -sinsx/s3

= (1-x2) sinsx/3 - 2cossx/s

2 + 2sinsx /s

3

2

^

= (-2coss/s2 + 2sins/s

3

2

^

F[f(x) = 2

2

^

sins-scoss

s3

xcosx - sinx

x3

cos x/2 dx

By inversion formula, we get

Put x = 1/2, we get

(where s = x)

Which is the requrid result

f(x) 1

√2^ ∫

-∞ = F[f(x)] e

-isx ds

∞

1

√2^ ∫

-∞ =

∞ 2

2

^

sins-scoss

s3 e

-isx ds

1

√2^ ∫

-∞ =

∞ sins-scoss

s3 (cossx – isinsx) ds

1

√2^ ∫

-∞ =

sins-scoss

s3 (cossx–i

∞ sins-scoss

s3

sinsx dy

1

√2^ ∫ -∞

f(x) = 2

sins-scoss

s3 cossx ds

∞

∫ -∞

⇒ sins-scoss

s3 cossx ds = f(x)

∞ ^

4

∫ -∞

sins-scoss

s3 cos s/2 ds = f(1/2)

∞ ^

4

^

4 = (1-(1/2)

2)

^

4 =

4-1

4

3^

4 =

∫ -∞

sins-scoss

s3 cos s/2 ds =

∞ -3^

16 ⇒

∫ -∞

xcosx-sinx

s3

cos x/2 dx =

∞ ⇒

-3^

16

3). Show that the fourier transform of

f(x) =

Hence deduce that dt =

using parseval’s identity. Show that dt =

Soln :- Fourier Transform of f(x) is

a2 – x

2, |x| < a is

0, |x| > a >0

2

2

^

sins-scoss

s3

sint-tcost

t3

∫ ∞

o

^

4 sint-tcost

t3

∫ ∞

o

^

4

2

1

√2^ ∫ -∞

F[f(x)] = f(x) eisx

dx ∞

1

√2^ ∫ -a

= (a2-x

2)e

isx dx

a

1

√2^ ∫ -a = (a

2-x

2) (cossx + isinsx) dx

a

1

√2^ ∫ -a

= (a2-x

2) cossx + i(a

2-x

2)sinsx) dx

a

1

√2^ ∫ -a

(a2-x

2) cossx dx

a = 2

1

√2^ ∫ o (a

2-x

2) cossx dx

∞

=

u = a2-x

2 v = cosx

u' = -2x

v1 = sinsx/s

u'' = -2

v2 = -cossx/s2

v3 = -sinsx/s

3

= ca2-x

2) sinsx/s – 2xcossx/s

2 + 2 sinsx/s

3

2

^

a

o

= -2acosas/s2 + 2sinas/s

3

2

^

F[f(x)] = 2 2

^ sinas-ascosas

s3

Using inversion formula, we get

put x = 0, we get

put a = 1, we get

f(x) 1

√2^ ∫ = F[f(x)] e

isy dx

∞

-∞

1

√2^ ∫ = 2

∞

-∞

e-isx

ds

2

^ sinas-ascosas

s3

1

2^ ∫

= ∞

-∞ (cossx – isinsx) ds

sinas-ascosas

s3

1

2^ ∫ = ∞

-∞ cossx –i sinsx ds

sinas-ascosas

s3

sinas-ascosas

s3

∫ ∞

-∞

1

2^ ∫ = 2 ∞

-∞ sinas-ascosas

s3

cossxds

= -∞

sinas-ascosas

s3

cossxds

f(x) 4

^ ∫ ∞

-∞ sinas-ascosas

s3

cossxds = f(x)

⇒ ∫ ∞

^

4

∞ sinas-scoss

s3

ds = (a2) ⇒ ∫ ^

4

sinas-scoss

s3

ds = ∫ ^

4

o

∞

o

sint-tcost

t3

dt = (where s = t) ⇒ ∫ ^

4 o

∞

[~ The firstintegrand is

even of the second

integrand is odd]

Using Paraseval’s dentity we get

[f(x)]2 dx = (f[s])

2 ds ∫

-∞

∞ ∫

-∞

∞

[f(x)]2 dx = ∫

-∞

∞ ∫

-∞

∞ 2

2

^

sinas-ascosas

s3 ds

(a2-x

2) dx = 8/^

∫ -a

a ∫

-∞

∞ sinas-ascosas

s3 ds

2

(a2-x

2) dx = 8/^

∫ ∫

0

∞ sinas-ascosas

s3 ds

2

0

a 2

2

Put a = 1, we get

(1-x2)dx = 16/^

∫ ∫

0

∞ sins-scoss

s3

ds

0

1 2

(1-2x2+x

4) dx = 8/^

∫ ∫

0

∞ sins-scoss

s3

ds

0

1

x-2x3+x

5 dx = 8/^ ∫

0

∞ sins-scoss

s3

ds

3 5

1

o

x- 2+ 1 = 8/^ ∫ 0

∞ sins-scoss

s3

ds

3 5

15-10+3 = 8/^ ∫ 0

sins-scoss

s3

ds

15

18 = 8/^ ∫ 0

sins-scoss

s3

ds

15

^ = ∫ sins-scoss

s3

ds

15

∞

∞

0

∞

∫ sint-tcost

t3

dt =

0

∞ ^

15

⇒

Where s = t

4). Find the fourier transform of f(x)

where f(x) 1-|x| for |x| < 1 Hence

0 for |x| > 1

deduce that

(i).

(ii).

Soln :

Fourier transform of f(x) is

∫ sint

t3

dt =

0

∞ ^

15

2

∫ sint

t3

dt =

0

∞ ^

3

4

[f(x)] = f(x)eisx

dx ∫ -∞

∞ 1

√2^

= (1-|x|)eisx

dx ∫ -1

1 1

√2^

= (1-|x|) (cossx+isinsx) dx ∫ -1

1 1

√2^

= [(1-|x|) (cossx+i(1-|x|) sinsx] dx ∫ -1

1 1

√2^

= (1-|x|) (cossxdx] ∫ 0

1 1

√2^ [2

(~ The first integral is even of the second

integrand is odd)

= (1-|x|) (cossxdx] ∫ 0

1 1

√2^

[~ In (0, 1), (1-|x|) = (1-x)]

= [(1-x) sinsx/s – cossx/s2]

1

√2^

1

0

= -coss + 1 1

√2^ s

2 s

2

F[f(x)] =

(i) Using inversion formula, we get

Put x = 0, we get

1-coss

s2

2

^

f(x) = 1-coss e-isx

ds ∫ ∞ 1

√2^ -∞ s

2

2

^

∫ 1

^ -∞ s

2

∞ 1-coss (cossx – isinsx) ds

∫ -∞

s2

1

^ ∞

1-coss cossx – i ds s

2

1-coss sinsx

f(x) = 1-coss cossx ds ∫ ∞

-∞ s

2

1

^ ~ The first integral is even

of the second integral is odd

⇒ 1-coss cossx ds = ^/2 f(x)

s2

∞

-∞ ∫

1-coss ds = ^/2 f(o)

s2

∞

-∞ ∫

2sin2 s/2 ds = ^/2

s2

∞

-∞ ∫

Put S/2 = t s = ∞ ⇒ t = ∞

S = 2t s = 0 ⇒ t = 0

sint dt = ^/2

t

∞

o ∫

(ii) Using Paraseval’s identity, we get

Find the fourier transform of e-a|x|

, a>0 and hence evaluate

dt and hence deduce F[xe-a|x|] z=I √2/^

Soln :-

Fourier transform of f(x) is

(f(x)2 dx = (F[s])

2 ds

∞

-∞

∫ ∞

-∞

∫

(1-|x|2 dx = ds

1

-1

∫ ∞

-∞

∫ 2

^ s2

1-coss

2

(1-|x|2 dx = ds

1

-1

∫ ∞

-∞

∫

s2

2sin2s/2

s

2

2 4

^

(1-x)2 dx = 4 ds

1

o

∫ ∞

o

∫

s

sin S/2 4

2

^

(1-2x+x)2 dx = ds

1

o

∫ ∞

o

∫

s

sin S/2 4

8

^

x-2x2+x

3 dx = 2dt

2 3

8

^

∞

o

∫

s

sin S/2 4

(1-1+1/3) = dt 16

16^

∞

o

∫

2t

sint 4

^

3

∞

o

∫

t

sin t 4

dt

∞

o

∫

t

sin t 4

dt = ^/3 ⇒

(s2+a

2)

2

2as ∞

-∞

∫

a2+t

2

cosxt

F[f(x)] = f(x) eisx

dx 1

√2^

∞

-∞

∫

= e-a|x|

eisx

dx 1

√2^

∞

-∞

∫ = e

-a|x| (cossx isinsx) dx 1

√2^

∞

-∞

∫

= [e-a|x|

cossx + ie–a|x|

sinsx] dx 1

√2^

∞

-∞

∫

∞

(i) using inversion formula, we get

[ ~ e-a|x|

cosxdx is even

of e-a|x|

sinsx is odd]

F[f(x)] =

a

s2+a

2

2

^

~ e-ax

cosxdx = ∞

-∞

∫ a

s2+a

2

f[f(x)] = e-isx

ds 1

√2^

∞

-∞

∫ 2

^

a

s2+a

2

= (cossx-isinsx) ds a

^

∞

-∞

∫ 1

s2+a

2

= ds-i ds a

^

∞

-∞

∫ cossx

s2+a

2

∞

-∞

∫ sinsx

s2+a

2

= 2 ds a

^

∞

-∞

∫ cossx

s2+a

2

[~ The first integrand is even

The second integrand is odd]

f(x) = 2 ds 2a

^

∞

-∞

∫ cossx

s2+a

2

⇒ ds = f(x) ∞

-∞

∫ cossx

s2+a

2

^

2a

^

2

a

= e-a|x|

⇒ ds = e-a|x|

cosxt

a2+t

2

6. Show that the fourier transform of f(x) = e-x2/2

is is self reciprocal.

Hence evaluate F(xe-x2/2

]

Soln :

Fourier Transform of f(x) is

(ii) F [xe-a|x|

= (-i) F [e-a|x|] d

ds

= (-i)

d

ds

a

s2+a

2

2

^

= (-i) a d

ds 1

s2+a

2

2

^

= (-i) a d

ds -2s

(s2+a

2)

2

2

^

F [xe-a|x|

] = i 2

^

-2as

(s2+a

2)

2

F [f(x)] = f(x) eisx

dx ∞

-∞

∫

= e-x2/2

eisx

dx ∞

-∞

∫

= e-x /2

+ isx

dx ∞

-∞

∫

= e-1/2 (x -2isx)

dx

1

√2^

1

√2^

1

√2^

1

√2^

∞

-∞

∫

= e-1/2 (x -2isx + (is) – (is) )

dx 1

√2^

∞

-∞

∫

2

2

2 2 2

= e-1/2 ((x-is) + s)

dx 1

√2^

∞

-∞

∫ 2 2

= e-s /2 e (x-is)

dx 1

√2^

∞

-∞

∫ 2 -1/2

2

= e-s /2 e x-is

dx 1

√2^

2 - 2 ∞

-∞

∫

√2^

x-is

√2 Put = t

x - is = √2 t

dx = √2 dt

~F[f(x)] = e-t

dt e

-s /2

√ ^

2 ∞

-∞

∫ 2

√2

e-s /2

√ ^

2 ∞

-∞

∫ 2

= e-t

dt

e-s /2

√ ^

2 ∞

o

∫ 2

= 2 e-t

dt

2e-s /2

√2^

2 ∞

o

∫ 2

= e-t

dt

2e-s /2

√2^

2 ∞

o

∫ 2

= e-t

dt

^

2a

2e-s /2

√2^

2 ∞

o

∫ 2

= e-t

dt

=

^

2a ~

F[f(x)] = e-s/2

dt 2

(i) F[f(xe-x/2

] = (-i) F[e-x/2

] 2 2 d

ds

= (-i) [e-s/2

] 2

d

ds

= (-i) -e-s/2

x (1/2 x 2s) 2

F[f(xe-x/2

] = ise-s/2

2 2

7. Find the fourier cosine transform of f.

8. Find the fourier sine transform of the function f(x)

Soln :-

Fourier sine transform of f(x) is

f(x) =

x for 0<x<1

2-x for 1<x<2

0 for x>2

Soln :-

Fourier cosine transform of f(x) is

Fc [f(x)] = f(x) cossx dx ∞

o

∫ 2

^

= [ f(x) cossx dx + f(x) cossxdx + f(x) cossxdx 2

^

∞

o

∫ 2

1

∫ ∞

2

∫

= [ xcossx dx + (2-x) cossxdx+0] 2

^

1

o

∫ 2

1

∫

= [(xsinsx/s + cossx/s2) + [(2-x) sinsx/s - cossx)

2]

s2

2

^

1

o

∫

= (sins + coss ) + 1 -cos2s - sins + coss

s s2 s

2 s

2 s s

2

2

^

= (2coss - cos2s ) + 1

s2 s

2 s

2

2

^

Fc [f(x)] = 2coss – cos2s-1

s2

2

^

sinx, 0<x<a

0, x>a

Fc [f(x)] = f(x) sinsxdx ∞

o

∫ 2

^

9. Find the Fourier sine and cosine transforms of e-2x

. Hence find the

value of the following integrals.

i).

ii).

Soln :-

Fourier sine Transform of f(x) is

= f(x) sinsxdx + f(x) f(x) sinsxdx a

o

∫ 2

^

∞

o

∫

= sinxsinsxdx+o ∞

o

∫ 2

^

= [cos(1-s) x- cos (1+s)x)] dx a

o

∫ 2

^

1

2

a

= sin(1-s)x sin (1+s)x

1-s 1+s

a

o

∫ 1

√2^

Fs [f(x) = sin(1-s)a sin (1+s)a

1-s 1+s

1

√2^

∞

o

∫ dx

(x2+4)

2

∞

o

∫ x

2dx

(x2+4)

2

of Aise find (i) Fs [xe-2x

]

(ii) Fc [xe-2x

]

Fc [f(x)] = f(x) sinsxdx ∞

o

∫ 2

^

= e-2x

sinsxdx ∞

o

∫ 2

^

Fc [f(x)] = S

S2+22

2

^ (1)

IIIly Fourier cosine transform of f(x) is

(i) Using Parseval’s identity, we get

Fc [f(x)] = f(x) cossxdx ∞

o

∫ 2

^

= e-2x

cossxdx ∞

o

∫ 2

^

Fc[F(x)] = 2

s2+s

2

∞

o

∫ 2

^

[Fc (f(x))]2 ds = (f(x))

2dx

∞

o

∫ ∞

o

∫

(2) ds = (e-2x

)2dx

∞

o

∫ ∞

o

∫ 2

^

2

s2+2

2

ds = (e-4x

)2dx

∞

o

∫ 8

^

ds

s2+s

2

∞

o

∫

=

∞

o

e-4x

-4

=

1

4

∞

o

∫ ds

s2+s

2 =

∞

o

∫ dt

s2+s

2

^

32 =

^

32

⇒

(ii) Using Parseval’s identity, we get

10. Using parseval’s identity, prove that

i).

ii).

Soln :-

(ii) We know, Fourier cosine Transform of f(x) = e-ax

is

IIIly if g(x) = e

-bx, then Fc(g(x) =

[Fs (f(x))]2

ds = (f(x))2dx

∞

o

∫ ∞

o

∫

ds = (e-2x

)2dx

∞

o

∫ ∞

o

∫ 2

^

2

s2+2

2

= e-4x

dx ∞

o

∫ 2

^ s

2ds

(s2+

4)

2

∞

o

∫

=

∞

o

e-4x

-4

=

1

4

∞

o

∫ x+dx

(x2+4)

=

^

8

∞

o

∫ sinat

t(a2+t

2)

∞

o

∫ dt

(a2+t

2) (b

2+t

2)

2

^

i.e. Fc[f(x)] = 2

^

dt = ^

8

1-e-a

a2

2

^

2ab(a+b)

a

s2+a

2

a

s2+a

2

2

^

b

s2+a

2

We know,

Fc[f(x)] Fc[g(x)] ds = f(x) g(x) dx ∞

o

∫ ∞

o

∫

2

^

a

s2+a

2

∞

o

∫ b

s2+a

2

, = e-ax

e-bx

dx ∞

o

∫

2ab

(s2+a

2)(s

2+a

2)

∞

o

∫

e-(a+b)x

-(a+b)

ds = e-(a+b)x

dx ∞

o

∫

∞

o

∫

1

a+b

∞

o

∫ ds

(s2+a

2) (s

2+b

2)

^

2ab(a+b)

=

⇒ =

∞

o

∫ ds

(a2+t

2) (b

2+t

2)

^

2ab(a+b) ⇒ =

(i) We know, if f(x) = e-ax

then Fc [f(x)] = 2

^

a

s2+a

2

IIIly if g(x) = 1, 0<x<a then Fc [f(x)] =

0, x>a

2

^

sinas

s

We know, Fc[f(x)] Fc [g(x)] ds = f(x) g(x) dx ∞

o

∫ ∞

o

∫

2

^

a

s2+a

2

∞

o

∫ 2

^

sinas

s ds = e

-ax, dx

∞

o

∫

2

^

∞

o

∫ ds =

a sinas

s(s2+a

2)

e-ax

, dx

-a

a

o

11. Find the Fourier Soxine Transform of e-ax

x

Soln :-

Fourier Cosine Transform of f(x) is

Integrating w.r. to ‘s’. We get

2

^

∞

o

∫ ds = +

sinas

s(s2+a

2)

e-a2

, dx

-a

1

a

∞

o

∫ dx =

sinas

x(x2+a

2)

^

2a2

[1-e-a2

]

Fc[s] = Fc [f(x)] = f(x) cossxdx ∞

o

∫ 2

^

= e-ax cosxdx

x

∞

o

∫ 2

^

Disff w.r.to ‘s’ on both sides, we get

= e-ax cosxdx

x

∞

o

∫ 2

^

Fc[s] = d

ds

d

ds = e-ax cosxdx

x

∞

o

∫ 2

^

= e-ax cosxdx dx

x

∞

o

∫ 2

^

∂

∂s

= e-ax (-xsinsx) dx

x

∞

o

∫ 2

^

= e-ax sinsx dx

∞

o

∫ 2

^

Fc[s] = - d

ds 2

^

s

s2+a

2

12. Find the Fourier Cosine Transfer of

Soln :-

Same as previous problem

13. Find the faourier cosine transform of 3-ax

cosax

Soln :-

Fourier cosine transofrm of f(x) is

Fc[s] = - [Log (s2+a

2) 2

^

e-ax

- e-bx

x

Given f(x) = e

-ax - e

-bx

x

e-ax

x f(x) =

e-bx

x -

Fc [f(x)] = f(x) cossx dx ∞

o

∫ 2

^

= e-ax

cosax cossx dx ∞

o

∫ 2

^

= e-ax

dx

∞

o

∫ 2

^

cos(a+s) x + cos (s-a) x

2

= e-ax

dx

∞

o

∫ 2

^ [cos(a+s) x + cos (s-a) x] ~ e

-ax dx = a

s2+a

2

Fc [f(x)] = 2

^

a

(s+a)2+a

2

+ a

(s-a)2+a

2

14. Find the fourier cosine transform of f(x) = . Hence derive

fourier sine Transform of ϕ(x) =

Soln :-

Fourier cosine transform of f(x) is

Diff w.r. to ‘s’, we get

x

1+x2

x

1+x2

Fc [s] = Fc[f(x)] = f(x) cossx dx ∞

o

∫ 2

^

Fc [s] = Fc[f(x)] = cossx dx ∞

o

∫ 2

^

x

1+x2 (1)

Fc[s] = d

ds 2

^

cossx

1+x2

d

ds

∞

o

∫ dx (2)

2

^ cossx

∞

o

∫ ∂

∂s dx

x

1+x2 =

2

^

∞

o

∫ -xsinsx

1+x2

dx =

(4)

∞

o

∫ sinsx

x(1+x2)

dx Fc[s] = +

(3) d

ds ^

2

2

^

2

^

∞

o

∫ -x

2sinsx

x(1+x2)

dx = -

(~ Multiply of divide by x)

2

^

∞

o

∫ (1-x

2-1) sinsx

x(1+x2)

dx = -

(Add and subtract ‘I’

on the x)

2

^

∞

o

∫ (1-x

2) sinsx

x(1+x2)

dx - = -

∞

o

∫ sinsx

x(1+x2) dx

Again, diff w.r. to ‘s’, we get

2

^

∞

o

∫ sinsx

x

dx - = -

∞

o

∫ sinsx

x(1+x2) dx

2

^ = -

∞

o

∫ sinsx

x(1+x2)

dx ^

2 -

~ dx = ^

2

∞

o

∫ sinsx

x

∞

o

∫

1

x(1+x2)

sdx Fc[s] =

d2

ds2

2

^ ∂

∂s sinsx

∞

o

∫

x cosx

x(1+x2)

dx 2

^ =

∞

o

∫

cosx

(1+x2)

dx 2

^ =

= Fc [s] (by 0)

- Fc [s] = 0 ⇒ d

2 Fc[s]

ds2

(D2-1) Fc[s] = 0

Fc[s] = c, es + c2e

-s (4)

Fc[s] = c1es - c2e

-s (*)

d

ds

When s = 0, (1) ⇒ Fc[s] = ∞

o

∫

dx

(1+x2)

2

^

(5)

(4) ⇒ Fc[s] = C1+C2 (6)

Compare (5) & (6) ⇒ C1 + C2 =

When s = 0, (3) ⇒ F[s] = -

Sub C1 & C2 in (4), we get

∞

o

∫

dx

(1+x2)

2

^

(tan + x) 2

^

∞

o

∫ C1 + C2 =

2

^ =

2 ^

C1 + C2 = 2 ^

d

ds 2 ^ (8)

(9) (4) ⇒ F[s] = C1-C2 d

ds

Compare (3) & (9) ⇒ C1-C2 = - 2 ^ (10)

(9) & (10) ⇒ 2C1 = 0

C1 = 0

C2 =

2 ^

Fc [s] = 0 + e-2

2 ^

Fc [s] = e-2

2 ^

Fourier sine transform of ϕ(x) =

x

(1+x2)

Fs [ϕ(x)] = ∞

o

∫ 2

^ ϕ(x) sinsx dx

= ∞

o

∫ 2

^ sinsx dx

x

(1+x2)

[by (4)] = - Fc[s] d

ds

[by (4)] = - [C1es – C2e

-s]

Fs [ϕ(x)] = 2 ^ e

-s

where C1 = 0

of C2 =

2 ^

Application of Foorier Transforms for Solving Integral Equations

Solve the integral equation f(x) con λxdx = e-1λ

Soln :-

By the definition of Fourier Cosine Transform,

Compare (1) & (2), we get

Using inversion fourmula for the Fourier transofrm, we get

∞

o

∫

Given ∞

o

∫ f(x) con λxdx = e-1λ

i.e., f(x) con λxdx = e-s

∞

o

∫ where λ = s (1)

Fc [f(x)] = f(x) consxdx ∞

o

∫ 2

^

f(x) consxdx = Fc[f(x)] ∞

o

∫ 2 ^

(2)

Fc [f(x)] = e-s

2 ^

Fc [f(x)] = e-s 2

^

[f(x)] = Fc [f(x)] consx ds ∞

o

∫ 2

^

= e-s consx ds

∞

o

∫ 2

^

= 2

^

1

(x2+1)

[f(x)] =

2

^(x2+1)

16. Solve the integral equation f(x) consλxdx =

Soln :-

Comare (1) & (2) , we get

Using inversion formula, we get

1-λ, 0 < λ < 1

0, λ > 1

Hence deduce that ∞

o

∫ sin

2t

t2

dt = ^/2

Given f(x) cosλxdx = ∞

o

∫ 1-λ for 0 < λ < 1

0 for λ > 1

Put A = S, we get

∞

o

∫ 1-s for 0 < s < 1

0 for s > 1

f(x) cossxdx = (1)

We know fourier csine Transform of f(x) is

Fc [f(x)] = f(x) cossxdx 2

^

∞

o

∫

⇒ f(x) cossxdx = Fc([f(x)] ∞

o

∫ ^/2 (2)

^/2 Fc([f(x)] = 1-s for 0 < s < 1

0 for s > 1

Fc [f(x)] = 2

^

1-s for 0 < s < 1

0 for s > 1

f(x) = Fc[f(x)] cossxds 2

^

∞

o

∫

= (1-S) cossxds

2

^

1

o

∫ 2

^

u = 1-s v = cossx

u' = -1 v1= sinsx/x

v2= cossx/x2

(i) To deduce

Put λ = s, get

Put x = 0, we get

= - (1-s) (sinsx

x

2

^

cossx x

2

1

o

∫

= - -cosx

x2

2

^

1 x

2

-cosx

x2

2

^ f(x) = (3)

1

o

∫ sint

t dt = 2

^

Given f(x) cossx dx = 1

o

∫ 1-λ for 0 < λ < 1

0 for λ > 1

1

o

∫ 1-cosx

x2

2

^ cossx dx =

1-s for 0 < s < 1

0 for s > 1

1

o

∫ 1-cosx

x2

2

^ cossx dx =

1-s for 0 < s < 1

0 for s > 1

∞

o

∫ 2sin

2x/2

x2

2

^ dx = 1

∞

o

∫ 2sin x/2

x2

2

dx = ^/4

Put x/2 = t

x = 2t

dx = 2dt

∞

o

∫ ⇒ 2dt = ^/4 sint

2t

∞

o

∫ ^/4 sint

t

1

2 dt =

∞

o

∫ ^/2 sint

t dt =

2

2