Post on 03-Mar-2018
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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