Fourier Transforms

Jeeja A. V.
Assistant Professor
Department of Mathematics
Govt. K.N.M Arts and Science College, Kanjiramkulam

E-Resource of Mathematics

FDP in MATHEMATICS under Choice Based Credit and Semester System,

University of Kerala
According to the syllabus for 2018 Admission
Semester - VI
MM 1645: Integral Transforms - Module II
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Contents
1 Fourier Integral 3

2 Fourier Cosine and Sine Transforms 11

3 Fourier transforms 14
3.1 Properties of Fourier transforms . . . . . . . . . . . . . . . . . . . . . . . . . 18

4 Fourier cosine and sine integrals 23

5 Fourier cosine transform and its inversion formula 24

6 Fourier sine transform and its inversion formula 25

6.1 Properties of Fourier cosine and sine transforms . . . . . . . . . . . . . . . . . 25

7 List of formulae 25

References 39

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In the study of Fourier series, we have seen that if a function f (x) is defined in −∞ < x <
∞ and is periodic with period 2l satisfy the Dirichlet’s conditions viz. (i) f (x) is piecewise
continuous and (ii) f (x) has a finite number of maxima and minima, then f (x) can be expanded
as a Fourier series as f (x) = a20 + ∑∞ nπx nπx


n=1 an cos l + bn sin l . The series on the right hand

side converges to f (x) at all points where f (x) is continuous and converges to 21 f (x0+ ) + f (x0− )
 

at points of discontinuity. But many applied problems give rise to non-periodic functions. If a
function f (x) is initially defined on a finite interval say, c ≤ x ≤ c + 2l, we can always extend the
definitions outside [c, c + 2l] by imposing some sort of periodicity conditions. If f (x) is defined
in (−∞, ∞) and is non-periodic, we cannot expand f (x) as a Fourier series. Therefore if we think
of f (x) as a periodic function with infinite period, f and f 0 are piecewise continuous on every
R∞
finite interval [−l, l] and if −∞ | f (x)|dx < ∞, then one can extend the concept of Fourier series
to this function and obtain a representation as an indefinite integral, called Fourier integral.

1 FOURIER INTEGRAL

Let f and f 0 be piecewise continuous functions defined on every finite interval [−l, l] and
let
Z ∞
| f (x)|dx < ∞
−∞
i.e. f (x) is absolutely integrable. Then the Fourier integral of f is given by
Z ∞
f (x) = [A(w) cos wx + B(w) sin wx]dw (1)
0

for all points x at which f is continuous. The integral converges to 12 [ f (x+ ) + f (x− )] for every
point x at which f is discontinuous, where
1
Z ∞
A(w) = f (t) cos wt dt (2)
π −∞

and
1
Z ∞
B(w) = f (t) sin wt dt. (3)
π −∞

Theorem 1.1 (Fourier Integral)

If f (x) is piecewise continuous in every finite interval and has a right-hand derivative and
a left-hand derivative at every point and if f is absolutely integrable, then f (x) can be
represented by a Fourier integral (1) with A and B given by (2) and (3). At a point where
f (x) is discontinuous the value of the Fourier integral equals the average of the left-and
right-hand limits of f (x) at that point.

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Problem 1.1
Express the function 
1 for |x| ≤ 1

f (x) =
0 for |x| > 1


sin w cos wx
Z ∞
as a Fourier integral. Hence evaluate dw.
0 w

Solution. The Fourier integral of f (x) is

Z ∞
f (x) = [A(w) cos wx + B(w) sin wx]dw.
0

We have

1 ∞
Z
A(w) = f (t) cos wt dt
π −∞
1 1
Z
= 1 cos wtdt
π −1
1 sin wt 1
 
=
π w t=−1
1
= [sin w − sin(−w)]
πw
1
= [sin w + sin w]
πw
2 sin w
= .
πw

Similarly,

1 ∞
Z
B(w) = f (t) sin wt dt
π −∞
1 1
Z
= 1 sin wtdt
π −1
= 0.

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Hence the Fourier integral of f is given by

Z ∞
f (x) = [A(w) cos wx + B(w) sin wx]dw
0
Z ∞ 
2 sin w
= cos wx + 0 dw
0 πw
2 sin w cos wx
Z ∞
= dw
0 πw
2 sin w cos wx
Z ∞
= dw,
π 0 w
for x 6= ±1, since f is continuous for all x 6= ±1. For x = 1, the right and left hand limits of f
are given by
f (1+ ) = 0 and f (1− ) = 1.
0+1
The average of these limits is 2 = 12 . Hence for x = 1 we get
2 sin w cos w 1
Z ∞
dw = .
π 0 w 2
For x = −1, the right and left hand limits of f are given by

f (−1+ ) = 1 and f (−1− ) = 0.

1+0
The average of these limits is 2 = 12 . Hence for x = −1 also, we get
2 sin w cos w 1
Z ∞
dw = .
π 0 w 2
Thus
sin w cos wx
Z ∞
π
dw = f (x)
0 w 2

for |x| < 1




 1


π 1
= for x = ±1
2 2


0 for |x| > 1.


for |x| < 1



 π/2


= π/4 for x = ±1




0 for |x| > 1.


I The above problem shows that

sint cost
Z ∞
π
dt =
0 t 4

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and
sint
Z ∞
π
dt =
0 t 2
(put x = 0 in the above problem).
Problem 1.2
Show that 



0 if x < 0

cos xw + w sin xw
Z ∞ 
dx = π
if x = 0
0 1 + w2 
 2

πe−x

if x > 0.


Solution. Consider the function





0 if x < 0


f (x) = 1
 2 if x = 0


e−x

if x > 0.


We have
e−st
Z
e−st sin wt dt = (−w cos wt − s sin wt)
s2 + w2
e−st
Z
e−st cos wt dt = 2 (−s cos wt + w sin wt)
s + w2
We represent f as a Fourier integral. The Fourier integral of f (x) is
Z ∞
f (x) = [A(w) cos wx + B(w) sin wx]dw.
0

We have
1 ∞
Z
A(w) = f (t) cos wt dt
π −∞
1 ∞ −t
Z
= e cos wtdt
π 0
Z k
1
= lim e−t cos wt dt
π k→∞ 0
 −t k
1 e
= lim (− cos wt + w sin wt)
π k→∞ 1 + w2 0
 −k
e0

1 e
= lim (− cos wk + w sin wk) − (− cos 0 + w sin 0)
π k→∞ 1 + w2 1 + w2
 −k 
1 e 1
= lim (− cos wk + w sin wk) − (−1 + 0)
π k→∞ 1 + w2 1 + w2
 
1 1
= 0+
π 1 + w2
1
=
π(1 + w2 )

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Similarly,

1 ∞
Z
B(w) = f (t) sin wt dt
π −∞
1 ∞ −t
Z
= e sin wtdt
π 0
Z k
1
= lim e−t sin wt dt
π k→∞ 0
 −t k
1 e
= lim (−w cos wt − sin wt)
π k→∞ 1 + w2 0
 −k
e0

1 e
= lim (−w cos wk − sin wk) − (−w cos 0 − sin 0)
π k→∞ 1 + w2 1 + w2
 −k 
1 e 1
= lim (−w cos wk − sin wk) − (−w + 0)
π k→∞ 1 + w2 1 + w2
 
1 w
= 0+
π 1 + w2
w
= .
π(1 + w2 )

Hence the Fourier integral of f is given by

Z ∞
f (x) = [A(w) cos wx + B(w) sin wx]dw
0
Z ∞ 
1 w
= cos wx + sin wx dw
0 π(1 + w2 ) π(1 + w2 )
1 ∞ cos wx + w sin wx
Z
= dw,
π 0 1 + w2

for x 6= 0, since f is continuous for all x 6= 0. For x = 0, the right and left hand limits of f are
given by
f (0+ ) = e−0 = 1 and f (0− ) = 0.
1+0
The average of these limits is 2 = 12 . Hence for x = 0 we get

1 cos wx + w sin wx 1
Z ∞
dw = .
π 0 1 + w2 2

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Thus
cos wx + w sin wx
Z ∞
dw = π f (x)
0 1 + w2 



 0 if x < 0


=π 1 if x = 0
 2


e−x

if x > 0.





 0 if x < 0


= π if x = 0
 2


πe−x

if x > 0.


FOURIER COSINE INTEGRAL AND FOURIER SINE INTEGRAL

Just as Fourier series simplify if a function is even or odd, so do Fourier integrals, and you can
save work. Indeed, if f has a Fourier integral representation and is even, then B(w) = 0. Then
(1) reduces to a Fourier cosine integral
2
Z ∞ Z ∞
f (x) = A(w) cos wx dw where A(w) = f (t) cos wt dt (4)
0 π 0

Similarly, if f has a Fourier integral representation and is odd, then A(w) = 0. Then (1) becomes
a Fourier sine integral
2
Z ∞ Z ∞
f (x) = B(w) sin wx dw where B(w) = f (t) sin wt dt (5)
0 π 0

Problem 1.3
Express 
1/2 for 0 ≤ x ≤ π

f (x) =
0 for x > π


as a Fourier sine integral and hence show that


Z ∞
1 − cos πw π

for 0 < x < π
2
sin xw dw =
0 w 0 for x > π


Solution. The Fourier sine integral of f (x) is

Z ∞
f (x) = B(w) sin wx dw.
0

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We have

2 ∞
Z
B(w) = f (t) sin wt dt
π 0
2 π1
Z
= sin wtdt
π 0 2
2 − cos wt π
 
=
2π w t=0
−1
= [cos wπ − cos(0)]
πw
−1
= [cos wπ − 1]
πw
1 − cos wπ
= .
πw

The Fourier sine integral of f (x) is

1 − cos πw
Z ∞
f (x) = · sin wx dw
0πw
1 ∞ 1 − cos πw
Z
= · sin wx dw
π 0 w
1 − cos πw
Z ∞
∴ sin wx dw = π f (x)
0 w

for 0 < x < π

π/2
=
0 for x > π


At x = π, the average of left and right hand limits is 12 ( 12 + 0) = 14 . Hence at x = π,

1 − cos πw
Z ∞
π
sin wπ dw = .
0 w 4

Problem 1.4
Using Fourier integrals, show that

w sin wx
Z ∞
π −kx
dw = e (x > 0, k > 0).
0 k2 + w2 2

Solution. Let f (x) = e−kx where x > 0 and k > 0. Using Fourier sine integral, we have
Z ∞
f (x) = B(w) sin wx dw.
0

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where

2 ∞
Z
B(w) = f (t) sin wt dt
π 0
2 ∞ −kt
Z
= e sin wt dt
π 0
 −kt p
2 e
= lim (−k sin wt − w cos wt)
π p→∞ k2 + w2 t=0
 −kp
e0

2 e
= lim (−k sin wp − w cos wp) − 2 (−k sin 0 − w cos 0)
π p→∞ k2 + w2 k + w2
 
2 1
= 0+ 2 (w)
π k + w2
2 w
=
π k2 + w2
2 w sin wx
Z ∞
∴ dw = f (x)
π k2 + w2
0
w sin wx
Z ∞
π
⇒ 2 2
dw = f (x)
0 k +w 2
π
= e−kx .
2

Problem 1.5
Using Fourier integral, prove that

Z ∞
cos(πw/2) cos xw  π cos x

if |x| < π/2
2
dw =
0 1 − w2 0 if |x| > π/2


Solution. Let 
cos x if |x| < π/2

f (x) =
0 if |x| > π/2.


The Fourier cosine integral of f (x) is

Z ∞
f (x) = A(w) cos wx dw.
0

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where
2 ∞
Z
A(w) = f (t) cos wtdt
π 0
2 π/2
Z
= cost · cos wt dt
π 0
Z π/2 
2 1
= [cos(w + 1)t + cos(w − 1)t dt
π 0 2
1 sin(w + 1)t sin(w − 1)t π/2
 
= +
π w+1 w−1 t=0
 
1 sin(w + 1)π/2 sin(w − 1)π/2
= +
π w+1 w−1
1 sin w(π/2) cos 2 + cos w π2 sin π2 sin w π2 · cos π2 − cos w π2 sin π2
π
 
= +
π w+1 w−1
 
1 π 1 1
= cos w −
π 2 w+1 w−1
 
1  wπ  −2
= cos
π 2 w2 − 1
2 cos(wπ/2) cos wx
= .
π(1 − w2 )
2 cos(wπ/2) cos wx
Z ∞
∴ dw = f (x)
0 π(1 − w2 )
cos(wπ/2) cos wx
Z ∞
π
⇒ 2
dw = f (x)
0 1−w 2

(π/2) cos x, |x| < π/2

=
0, |x| > π/2


Consider the case of |x| = π2 . When x = π2 , the average of left and right hand limits is 21 (cos π2 +
0) = 0. So
cos(wπ/2) cos wx
Z ∞
dw = 0
0 1 − w2
for x = π2 . Similarly for x = − π2 , the average of left and right hand limits is 21 (0 + cos π2 ) = 0
and hence
cos(wπ/2) cos wx
Z ∞
dw = 0
0 1 − w2
for x = − π2 .

2 FOURIER COSINE AND SINE TRANSFORMS

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

r Z
2 ∞
fˆc (s) = f (x) cos sx dx. (6)
π 0

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and the inverse Fourier cosine transform of fˆc (s) is given by

r Z
2 ∞ ˆ
f (x) = fc (s) cos sx ds. (7)
π 0

Similarly, the Fourier sine transform of f (x) is given by

r Z
2 ∞
fˆs (s) = f (x) sin sx dx. (8)
π 0

and the inverse Fourier sine transform of fˆs (s) is given by

r Z
2 ∞ ˆ
f (x) = fs (s) sin sx ds. (9)
π 0

Problem 2.1
Find the Fourier cosine and Fourier sine transforms of the function

k if 0 < x < a

f (x) =
0 if x > a


Solution. We have
r Z
2 ∞
fˆc (s) = f (x) cos sx dx
π 0
r Z
2 a
= k cos sx dx
π
r 0
2 sin sx a

= k
π s 0
r
2k
= [sin sa − sin 0]
πs
r
2 k sin sa
= .
π s

Similarly,
r Z
2 ∞
fˆs (s) = f (x) sin sx dx
π 0
r Z
2 a
= k sin sx dx
π
r 0
2 − cos sx a

= k
π s 0
r
2 −k
= [cos sa − cos 0]
π s
r
2 k(1 − cos sa)
= .
π s

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PROPERTIES

1. fˆc [α f (x) + β g(x)] = α fˆc [ f (x)] + β fˆc [g(x)] and

fˆs [α f (x) + β g(x)] = α fˆs [ f (x)] + β fˆs [g(x)].

2. fˆc [ f (ax)] = 1a fˆc (s/a) and fˆs [ f (ax)] = 1a fˆs (s/a).

Proof. We have r Z
2 ∞
fˆc [ f (ax)] = f (ax) · cos sx dx.
π 0
Put ax = t. Then adx = dt.
r Z
2 ∞ dt
∴ fˆc [ f (ax)] = f (t) · cos s(t/a)
π a
r 0Z
1 2 ∞  s 
= f (t) cos t dt
a π 0 a
1
= fˆc (s/a).
a

The other result is similar.

Assume that Fourier sine and cosine transforms of f (x) exist and let f (x) → 0 as x → ∞.
Then
q
3. fˆc [ f 0 (x)] = s fˆs (s) − 2
π f (0) and

fˆs [ f 0 (x)] = −s fˆc (s)

Proof. We have
r Z
2 ∞ 0
fˆc [ f 0 (x)] = f (x) cos sx dx
π 0
r r Z
2 k 2 ∞
= lim [cos sx · f (x)]0 − (−s sin sx) · f (x) dx
π k→∞ π 0
r r Z
2 2 ∞
= lim [cos sk · f (k) − cos 0 f (0)] − (−s sin sx) · f (x) dx
π k→∞ π 0
r r Z
2 2 ∞
= 0− f (0) + s f (x) sin sx dx
π π 0
r
2
= s fˆs [ f (x)] − f (0).
π

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Similarly,
r Z
2 ∞ 0
fˆs [ f 0 (x)] = f (x) sin sx dx
π 0
r r Z
2 k 2 ∞
= lim [sin sx · f (x)]0 − s cos sx · f (x) dx
π k→∞ π 0
r r Z
2 2 ∞
= lim [sin sk · f (k) − sin 0 f (0)] − s cos sx · f (x) dx
π k→∞ π 0
r Z
2 ∞
= 0−s f (x) cos sx dx
π 0
= −s fˆc [ f (x)].

3 FOURIER TRANSFORMS

We derived two real transforms, Fourier sine and cosine. Now we want to derive a complex
transform that is called the Fourier transform. It will be obtained from the complex Fourier
integral, which will be discussed next.

COMPLEX FORM OF THE FOURIER INTEGRAL

From the definition of Fourier integral from equations (1), (2), and (3), we have
Z ∞  Z ∞   Z∞  
1 1
f (x) = f (t) cos wt dt cos wx + f (t) sin wt dt sin wx dw
0 π −∞ π −∞
Z Z ∞ 
1 ∞
= (cos wt cos wx + sin wt sin wx) f (t)dt dw
π 0 −∞
1 ∞ ∞
Z Z
⇒ f (x) = f (t) cos w(x − t)dt dw. (10)
π 0 −∞
h i
Since cos w(x − t) = 21 eiw(x−t) + e−iw(x−t) , equation (10) becomes
1 ∞ ∞
Z Z h i
f (x) = f (t) 12 eiw(x−t) + e−iw(x−t) dt dw
π 0 −∞
1 ∞ ∞ 1 ∞ ∞
Z Z Z Z
= f (t)eiw(x−t) dt dw + f (t)e−iw(x−t) dt dw.
2π 0 −∞ 2π 0 −∞
To combine the two terms on the RHS, we change the dummy integration variable from w
to −w in the second. Thus
1 ∞ ∞ 1 −∞ ∞
Z Z Z Z
f (x) = f (t)eiw(x−t) dt dw + f (t)eiw(x−t) dt d(−w)
2π 0 −∞ 2π 0 −∞
Z 0 Z ∞
1 ∞ ∞ 1
Z Z
= f (t)eiw(x−t) dt dw + f (t)eiw(x−t) dt dw
2π 0 −∞ 2π −∞ −∞
1 ∞ ∞
Z Z
⇒ f (x) = f (t)eiw(x−t) dt dw. (11)
2π −∞ −∞

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(11) is known as the complex form of the Fourier integral.

I Note that the complex form of Fourier integral can also be written as
Z ∞ 
1 1
Z ∞
−iwt
f (x) = √ √ f (t)e dt eiwx dw
2π −∞ 2π −∞

This motivates the definition of Fourier transform.

FOURIER TRANSFORM AND ITS INVERSE

The Fourier transform of f (x) is defined as

1
Z ∞
fˆ(w) = √ f (t)e−iwt dt (12)
2π −∞

and the inverse Fourier transform of fˆ(w) is given by

1
Z ∞
f (x) = √ fˆ(w)eiwx dw (13)
2π −∞

CONVOLUTION

The convolution of f and g is denoted by f ∗ g and is defined as

Z t
f ∗g = f (u)g(t − u) du. (14)
0

Also f ∗ g = g ∗ f .

Theorem 3.1 (Convolution Theorem)

Suppose that f (x) and g(x) are piecewise continuous, bounded, and absolutely integrable
on the x-axis. Then the Fourier transform of f ∗ g is given by
√
( f ∗ˆ g)(w) = 2π fˆ(w)ĝ(w).

The convolution theorem shows that

Z ∞
( f ∗ g)(x) = fˆ(w)ĝ(w)eiwx dw.
−∞

Problem 3.1
Find the Fourier transform of

x, |x| < a

f (x) =
0, |x| > a


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

1 ∞
Z
fˆ(w) = √ f (x) · eiwx dx
2π −∞
Z a
1
=√ x(cos wx + i sin wx)dx
2π −a
Z a
1
=√ 2 x · i sin wx dx
2π 0
(∵ x cos wx is odd and x sin wx is even)
r    a
2 cos wx  sin wx
= i x − −1· − 2
π w w 0
r  
2 sin aw − aw cos aw
= i .
π w2

Problem 3.2
2 /2 2 /2
Find the Fourier transform of e−x . Show that e−x is self reciprocal.

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

1 ∞
Z
fˆ(s) = √ f (x) · eisx dx
2π −∞
1
Z ∞
2
=√ e−x /2 (cos sx + i sin sx)dx
2π −∞
1
Z ∞
2 2
=√ 2 e−x /2 cos sx dx (∵ e−x /2 sin sx is odd)
2π 0
r Z
2 ∞ −x2 /2
= e cos sx dx = I(say) (*1)
π 0

Now
r Z
dI 2 ∞ −x2 /2
= e (− sin sx) · x dx
ds π 0
r Z
2 ∞ 2
= (sin sx)(−xe−x /2 dx)
π
r 0h  2  
2 −x2 /2
i∞ Z ∞
= (sin sx)(e ) − s cos sx e−x /2 dx
π 0 0
r
2
Z ∞
2
= · (−s) e−x /2 · cos sx dx
π 0

= −sI.

Seperating the variables,

dI
= −s ds.
I

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Integrating, we get
−s2
log I = + log A
2
2 /2
or I = Ae−s . (*2)

Now, when s = 0, from (*1),

r Z
2 ∞ −x2 /2
I= e dx
π 0
2 ∞ −y2 √ √
r Z
i.e.I = e 2 dy where y = x/ 2
π 0
√
2 2
Z ∞
−y2 π
=√ e dy = √ = 1.
π 0 π 2
Also when s = 0, from (*2), I = A. Equating the above values we get A = 1.
2
∴ I = fˆ(s) = e−s /2
2
Since fˆ(s) = e−s /2 = f (s), f (x) is self reciprocal.

EXAMPLE 3.8

1 − x2

if |x| < 1
Find the Fourier transform of f (x) = and
0 if |x| > 1


x cos x − sin x x
Z ∞
use it to evaluate 3
cos dx
0 x 2
[K. U. 2000 October, 2001 October]

Solution

Fourier transform of f (x) is given by

1 ∞ Z
F(s) = √ f (x)eisx dx.
2π −∞
Z 1
1
=√ (1 − x2 )(cos sx + i sin sx) dx.
2π −1
Z 1
1
= √ 2 (1 − x2 )(cos sx dx (∵ (1 − x2 ) sin sx is odd)
2π 0
r   
2 2 sin sx  cos sx 
= (1 − x ) − (−2x) − 2
π s s
 1
sin sx
+(−2) − 3 ,
s 0
r  
2 2 cos s 2 sin s
= − 2 + 3 .
π s s
r  
2 s cos s − sin s
= −2 .
π s3

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 18

By inversion formula for Fourier transform,

1 ∞
Z
f (x) = √ F(s)e−isx ds
2π −∞
r  
1 2 s cos s − sin s
Z ∞
=√ −2 (cos sx − i sin sx) ds
2π −∞ π s3
Z  
2 ∞ s cos s − sin s
=− cos sxds
π −∞ s3
Z  
2i ∞ s cos s − sin s
+ sin sx ds
π −∞ s3
 
2 s cos s − sin s
Z ∞
=− 2 cos sxds
π 0 s3

(since the integrand in the second integral is odd)

s cos s − sin s
Z ∞
π
∴ cos sxds = − f (x)
0 s3 4
π
= − (1 − x2 ) if |x| < 1
4
= 0 if |x| > |

Put x = 1/2.

s cos s − sin s s −3π

Z ∞
cos ds =
0 s3 2 16
x cos x − sin x x −3π
Z ∞
or 3
cos dx = .
0 x 2 16

3.1 PROPERTIES OF FOURIER TRANSFORMS

P.1. F[α f (x) + β g(x)] = αF[ f (x)] + β F[g(x)].

PROOF

1
Z ∞
F[α f (x) + β g(x)] = √ [α f (x) + β g(x)]eisx dx
2π −∞
1
Z ∞
=α·√ f (x)eisx dx
2π −∞
1
Z ∞
+β √ g(x)eisx dx
2π −∞

= αF[ f (x)] + β F[g(x)]

P.2. F[ f (x − a)] = eisa F(s).

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 19

PROOF

1
Z ∞
F[ f (x − a)] = √ f (x − a) · eisx dx
2π −∞

Put x − a = t or x = t + a.
1
Z ∞
=√ f (t) · eis(t+a) dt.
2π −∞

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 20

1
Z ∞
isa
=e ·√ f (t) · eist dt
2π −∞

= eisa F[ f (x)] = eisa F(s).

P.3. F[eiax f (x)] = f¯(s + a) = F(s + a).

PROOF

1 ∞ Z
iax
F[e f (x)] = √ eiax f (x) · eisx dx
2π −∞
1
Z ∞
=√ f (x)ei(s+a)x dx.
2π −∞
= f¯(s + a) = F(s + a),

1 1
P.4. F[ f (ax)] = |a| f¯(s/a) = |a| F(s/a).

PROOF

Case (i) when a > 0.

1
Z ∞
F[ f (ax)] = √ f (ax) · eisx dx put ax= t
2π −∞
adx = dt
1
Z ∞
=√ f (t)eist/a dt/a
2π −∞
1
Z ∞
= √ f (t) · ei(s/a)t dt
a 2π −∞
1 1
= f¯(s/a) = F(s/a)
a |a|

Case (ii) when a < 0,

1
Z ∞
F[ f (ax)] = √ f (ax)eisx dx
2π −∞

put ax = t when x = −∞, t =∞

when x = ∞, t = −∞ (since a < 0).

1 dt
Z ∞
=√ f (t)eis(t/a)
2π −∞ a

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 21

1 ∞ Z
=− √ f (t)eit(s/a) dt
a 2π −∞
1 1
= − f¯(s/a) = F(s/a)
a |a|
1 ¯
∴ F[ f (ax)] = f (s/a).
|a|

P.5. F[ f (x) cos ax] = 12 [F(s + a) + F(s − a)]

PROOF

1 ∞
Z
F[ f (x) cos ax] = √ f (x) cos ax · eisx dx
2π −∞
e + e−iax isx
 iax 
1
Z ∞
=√ f (x) · e dx
2π −∞ 2
1
Z ∞ h i
i(s+a)x i(s−a)x
= √ f (x) e +e dx
2 2π −∞

1 1
Z ∞
= √ f (x)ei(s+a)x dx
2 2π −∞

1
Z ∞
i(s−a)x
+√ f (x)e dx
2π −∞
1
= [F(s + a) + F(s − a)]
2

d n
P.6. F[xn f (x)] = (−i)n dsn F(s).

PROOF

1 ∞ Z
F(s) = √ f (x)eisx dx
2π −∞
dn 1 dn
Z ∞
∴ n F(s) = √ f (x) n (eisx )dx
ds 2π −∞ ds
1
Z ∞
=√ f (x) · (ix)n eisx dx
2π −∞
[∵ Dn (eas ) = an eas ]
1
Z ∞
n
=i √ ( f (x) · xn )eisx dx
2π −∞

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 22

= in F[xn f (x)]
n 1 dn
∴ F[x f (x)] = n n F(s)
i ds
dn
= (−i)n n F(s)
ds

P.7. F[ f 0 (x)] = −isF(s) if f (x) → 0 as x → ±∞.

PROOF

1 ∞
Z
0
F[ f (x)] = √ f 0 (x)eisx dx
2π −∞
 
1  isx ∞ Z ∞
isx
=√ e f (x) −∞ − e (is) · f (x)dx
2π −∞

= −isF(s).

NOTE:

F[ f (n) (x)] = (−is)n F(s) if f , f 0 , f 00 , . . . , f (n−1)

→ 0 as x → ±∞.

x
Z 
i
P.8. F f (x)dx = F(s).
a s

PROOF

Z x
Let φ (x) = f (x)dx, then φ 0 (x) = f (x)
a

By P.7., F[φ 0 (x)] = −isF[φ (x)]

1 i
∴ F[φ (x)] = F[φ 0 (x)] = F[ f (x)]
−is s
i
= F(s)
s

P.9. F[ f (−x)] = F(−s).

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 23

PROOF

1 ∞ Z
F[ f (−x)] = √ f (−x) · eisx dx put t = −x
2π −∞
Z −∞
1
=√ f (t) · e−ist (−dt)
2π ∞
1
Z ∞
= √ f (t)ei(−s)t dt
2π −∞
= F(−s).

NOTE:

(1) If F[ f (x)] = f (s), then the function f (x) is called self reciprocal.

e−kt g(t), t > 0


(2) If f (t) = ,

0, t <0


1
then F[ f (t)] = √ L[g(t)].
2π

PROOF
Z 0 Z ∞ 
1 1
Z ∞
ist
F[ f (t)] = √ f (t)e dt = √ + f (t)eist dt
2π −∞ 2π −∞ 0
1
Z ∞
= √ e−kt g(t) · eist dt
2π 0
1
Z ∞
= √ g(t)e−(k−is)t dt put k − is = w
2π 0
1
Z ∞
= √ g(t) · e−wt dt
2π 0
1
= √ L[g(t)].
2π

4 FOURIER COSINE AND SINE INTEGRALS

In the study of Fourier series if a function f (x) is defined in

0 < x < l, we extend it to −l < x < l in such a way that f (x) is periodic with period 2l and
find its Fourier series. Thus we obtain the half range cosine series or half range sine series of

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 24

f (x) depending upon the extension as an even function or an odd function. In the same way,
if f (x) is defined in 0 < x < ∞, one can extend it to −∞ < x < ∞ either as an even function or
as an odd function (depending upon the symmetries about the real axis or about the origin) and
obtain the Fourier cosine transforms and Fourier sine transforms.
By Fourier integral theorem,
Z ∞
f (x) = [A(w) cos wx + B(w) sin wx]dw
0

where
1 ∞
Z
A(w) = f (t) cos wtdt and
π −∞
1 ∞
Z
B(w) = f (t) sin wt dt
π −∞

If f (x) is even in −∞ < x < ∞, then f (x) cos wx is also even but f (x) sin wx is odd.
2 R∞
So that A(w) = π 0 f (t) cos wtdt and B(w) = 0.
Thus the above equation becomes
Z ∞ Z ∞ 
2
f (x) = f (t) cos wtdt cos wx dw. (15)
π 0 0

Similarly, if f (x) is odd in −∞ < x < ∞, then f (x) cos wx is odd and f (x) sin wx is even.
2 R∞
So that A(w) = 0 and B(w) = f (t) sin wtdt and hence
π 0
Z Z ∞ 
2 ∞
f (x) = f (t) sin wt dt sin wx dw (16)
π 0 0

(8) is called Fourier cosine integral and (9) is called the Fourier sine integral.

5 FOURIER COSINE TRANSFORM AND ITS INVER-

SION FORMULA
Fourier cosine integral (8) can be written as
Z Z ∞ 
2 ∞
f (x) = f (t) cos st dt cos sx ds (by replacing w by s)
π 0 0
r Z (r Z )
2 ∞ 2 ∞
= f (t) cos st dt cos sx ds
π 0 π 0
r Z
2 ∞ ˆ
= fc (s) cos sx ds (17)
π 0
where r Z
2 ∞
fˆc (s) = fˆc [ f (x)] = f (t) cos st dt. (18)
π 0

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 25

The function fˆc (s) as defined by (18), is called the Fourier cosine transform of f (x). Also the
function f (x), as given by (17), is called the inversion formula for the Fourier cosine transform.

6 FOURIER SINE TRANSFORM AND ITS INVER-

SION FORMULA
Fourier sine integral (16) can be written as
Z Z ∞ 
2 ∞
f (x) = f (t) sin st dt sin sx dx (by replacing w by s)
π 0 0
r Z (r Z )
2 ∞ 2 ∞
= f (t) sin st dt sin sx ds
π 0 π 0
r Z
2 ∞ ˆ
= fs (s) sin sx ds (19)
π 0
where r Z
2 ∞
fˆs (s) = fˆs ( f (x)) = f (t) sin st dt. (20)
π 0
The function fˆs (s), as defined by (20) is known as the Fourier sine transform of f (x). Also the
function f (x), as given by (19) is known as the inversion formula for the Fourier sine transform.

6.1 PROPERTIES OF FOURIER COSINE AND SINE TRANS-

FORMS

7 LIST OF FORMULAE

1. Fourier integral of f (x) is

1
Z ∞Z ∞
f (x) = f (t) · cos w(t − x)dt dw.
π 0 −∞

2. Complex form of the Fourier integral is

1
Z ∞Z ∞
f (x) = f (t)eiw(t−x) dt dw.
2π −∞ −∞

3. Fourier cosine integral of f (x) is

Z ∞ Z ∞ 
2
f (x) = f (t) cos wtdt cos wx dw.
π 0 0

4. Fourier sine integral of f (x) is

Z ∞ Z ∞ 
2
f (x) = f (t) sin wtdt sin wx dw.
π 0 0

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 26

5. Fourier transform of f (x) is

1
Z ∞
F[ f (x)] = F(s) = √ f (x)eisx dx
2π −∞

and its inversion formula is

1
Z ∞
f (x) = √ F(s) · e−isx ds.
2π −∞

6. Fourier cosine transform of f (x) is

r Z
2 ∞
fˆc [ f (x)] = fˆc (s) = f (x) · cos sxdx
π 0
and its inversion formula is
r Z
2 ∞ ˆ
f (x) = fc (s) · cos sxds.
π 0

7. Fourier sine transform of f (x) is

r Z
2 ∞
fˆs [ f (x)] = fˆs (s) = f (x) sin sxdx
π 0
and its inversion formula is
r Z
2 ∞ ˆ
f (x) = fs (s) · sin sxds.
π 0

ILLUSTRATIVE EXAMPLES

EXAMPLE 3.9

Find the Fourier complex transform of


k in |x| ≤ l

f (x) =
0 in |x| > l


[K. U. 2002 November]

Solution

Fourier transform of f (x) is given by

1
Z ∞
F(s) = √ f (x) · eisx dx
2π −∞
Z l  isx l
1 k
isx e
=√ k · e dx = √ ·
2π −l 2π is −l
   r
k 1 isl  2 k · sin sl
=√ e − e−isl = ·
2π is π s

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 27

EXAMPLE 3.10
2 /4
Find the Fourier inverse transform of e−s [K.U. 2004 May]

Solution

By inversion formula for Fourier transform,

1 ∞
Z
f (x) = √ F(s) · e−isx ds
2π −∞
1
Z ∞
2
=√ e−s /4 (cos sx − i sin sx) ds
2π −∞
1
Z ∞
2 2
= √ ·2 e−s /4 cos sx ds (∵ e−s /4 sin sx is odd)
2π 0
r Z
2 ∞ 2
= · e−s /4 cos sx ds = I(say) (1)
π 0
Now
r Z
dI 2 ∞ −s2 /4
= e (− sin sx) · s ds
dx π 0
r
2  s 2
Z ∞ 
−s /4
= ·2 (sin sx) − e ds
π 2
r h0 
2  2 i∞ Z ∞
−s /4 −s2 /4
=2 (sin sx) e − (cos sx)x · e ds
π 0 0

(Integrating by parts)
r Z
2 ∞ −s2 /4
= −2x e cos sx ds
π 0
= −2xI

Seperating the variables,

dI
= −2xdx or log I = −x2 + log A
I
2
or I = Ae−x (2)

Now when x = 0, from (1)

r Z r Z
2 ∞ −s2 /4 2 ∞ −y2
I= e ds = e 2dy where y = s/2
π 0 π 0
√
r
2 √ √
Z ∞
2
= · π = 2 (∵ 2 e−y dy = π)
π 0

Also when x = 0, from (2), I = A

√
∴A= 2
√ −x2
(2) becomes I = f (x) = 2e .

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 28

EXAMPLE 3.11

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

Solution

Fourier transform of f (x) is given by

1 ∞ Z
F(s) = √ f (x) · eisx dx
2π −∞
1
Z ∞
=√ e−a|x| (cos sx + i sin sx) dx
2π −∞
1
Z ∞
= √ ·2 e−a|x| cos sx dx) (∵ e−a|x| sin sx is odd)
2π 0
r Z
2 ∞ −ax
= e cos sx dx
π
r 0
e−ax
∞
2
= (−a cos sx + s sin sx)
π a2 + s2 0
r
2 a
= · 2 2
π a +s

EXAMPLE 3.12

Find the Fourier transform of


a − |x| if |x| < a

f (x) =
0 if |x| > a > 0.


Solution

Fourier transform of f (x) is given by

1 ∞
Z
F(s) = √ f (x) · eisx dx
2π −∞
Z a
1
=√ (a − |x|)(cos sx + i sin sx)dx
2π −a
Z a
1
= √ · 2 (a − |x|) cos sxdx (∵ (a − |x|) sin sx is odd)
2π 0
r Z
2 a
= (a − x) cos sx dx
π 0
r     a
2 sin sx − cos sx
= (a − x) − (−1)
π s s2 0

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 29

r   r  
2 cos as 1 2 1 − cos as
= − 2 + 2 =
π s s π s2
r
2 2 sin2 (as/2)
= ·
π s2

EXAMPLE 3.13

Find the Fourier cosine transform of e−5x [K.U. 1998 November]

Solution

Fourier cosine transform of f (x) is

r Z
2 ∞
fˆc (s) = f (x) cos sx dx
π 0
r Z
2 ∞ −5x
= e cos sx dx
π
r 0 ∞
2 e−5x
= (−5 cos sx + s sin sx)
π 25 + s2 0
r  
2 5
=
π 25 + s2

EXAMPLE 3.14

Find the Fourier cosine transform of


1 if 0 ≤ x ≤ 1

f (x) =
0 if x > 1


[K.U. 2000 April]

Solution

Fourier cosine transform of f (x) is given by

r Z
2 ∞
fˆc (s) = f (x) cos sx dx
π
r Z0
2 sin sx 1
r 
2 1

= 1 · cos sx dx =
π 0 π s 0
r
2 sin s
= ·
π s

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 30

EXAMPLE 3.15
2
Find the Fourier cosine transform of e−x [K.U. 2003 April]

Solution
2
Fourier cosine transform of f (x) = e−x is
r Z
2 ∞
fˆc (s) = f (x) cos sx dx
π
r Z0
2 ∞ −x2
= e · cos sx dx = I (say) (1)
π 0

Now
r Z
dI 2 ∞ −x2
= e (− sin sx) · x dx
ds π
r 0Z
1 2 ∞ 2
= sin sx(−2xe−x dx)
2 π 0
r  
1 2 h −x2
i∞ Z ∞
−x2
= sin sx · e − s cos sx · e dx
2 π 0 0
r Z
1 2 ∞ −x2
= (−s) e cos sx dx
2 π 0
s
= − I.
2

Seperating the variables,

dI s
= − ds.
I 2
Integrating,

−s2
log I = + log A
4
2 /4
or I = Ae−s (2)

Now when s = 0, from (1),

r Z r √
2 ∞ −x2 2 π 1
I= e dx = · =√ .
π 0 π 2 2

Also when s = 0, from (2), I = A.

Equating, A = √1
2
1 2
∴ I = fˆc (s) = √ e−s /4 .
2

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 31

EXAMPLE 3.16

Find the Fourier cosine transform of

f (x) = sin x in 0 < x < π

[K.U. 1998 April]

Solution

Fourier cosine transform of f (x) is given by

r Z
2 ∞
fˆc (s) = f (x) · cos sx dx
π
r Z0
2 π
= sin x · cos sx dx
π
r Z0
2 π1
= [sin(s + 1)x − sin(s − 1)x] dx
π 0 2
cos(s + 1)x cos(s − 1)x π
 
1
=√ − +
2π s+1 s−1 0
 
1 cos(s + 1)π cos(s − 1)π 1 1
=√ − + + −
2π s+1 s−1 s+1 s−1
 
1 cos sπ cos sπ 1 1
=√ − + −
2π s + 1 s−1 s+1 s−1
r  
2 1 + cos sπ
=
π 1 − s2

EXAMPLE 3.17
1
Obtain the Fourier sine transform of [K. U. 1999 November]
x

Solution

  r Z∞ r Z
1 2 1 2 ∞ sint dt
fˆs = sin sx dx = · where t = sx
x π 0 x π 0 t/s s
r Z r  Z∞ 
2 ∞ sint 2 π sin x π
= dt = · ∵ dx =
π 0 t π 2 0 x 2
p
= π/2

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 32

EXAMPLE 3.18

Find the Fourier sine transform of e−|x| . Hence evaluate

R ∞ x sin mx
0 dx.
1 + x2
[K.U. 2002 November]

Solution
r Z
−|x| 2 ∞ −x
fˆs [e ]= e sin sx dx.
π
r 0
e−x
∞
2
= (− sin sx − s cos sx)
π 1 + s2 0
r
2 s
=
π 1 + s2

By inversion formula for Fourier sine transform

r Z
2 ∞ ˆ
f (x) = fs (s) · sin sx ds
π 0
r Z r
2 ∞ 2 s
= · sin sx ds
π 0 π 1 + s2
2 ∞ s sin sx
Z
= ds
π 0 1 + s2
s sin sx
Z ∞
π π −x
∴ 2
ds = f (x) = e .
0 1+s 2 2

Replacing x by m, we get
s sin ms
Z ∞
π −m
ds = e .
0 1 + s2 2
x sin mx
Z ∞
π −m
Hence dx = e .
0 1 + x2 2

EXAMPLE 3.19

Find
1
(i) Fourier cosine transform of and
1 + x2
x
(ii) Fourier sine transform of
1 + x2
[K.U. 1998 October/November]

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 33

Solution
  r Z∞
1 2 1
fˆc = · cos sx · dx
1 + x2 π 0 1 + x2
Let
cos sx
Z ∞
dx = I (1)
0 1 + x2
Then

dI x sin sx
Z ∞
= − dx (2)
ds 0 1 + x2
Z ∞ 2
x sin sx
=− 2
dx
0 x(1 + x )
[(1 + x2 ) − 1] sin sx
Z ∞
=− dx
0 x(1 + x2 )
sin sx sin sx
Z ∞ Z ∞
=− dx + 2
dx
0 x 0 x(1 + x )
dI sin sx
Z ∞
π
∴ =− + 2
dx (3)
ds 2 0 x(1 + x )

Now

d2I x cos sx
Z ∞
= dx
ds2 0x(1 + x2 )
cos sx
Z ∞
= 2
dx = I
0 1+x
d2I d
∴ −I = 0 or (D2 − 1)I = 0 where D =
ds2 ds
∴ I = Aes + Be−s (4)
dI
= Aes − Be−s (5)
ds

Putting s = 0 in (1) and (4), we get

dx
Z ∞
π
I = A+B = 2
= (tan−1 x)∞
0 = (6)
0 1+x 2

Putting s = 0 in (3) and (5), we get

dI π
= A−B = − . (7)
ds 2

Solving (??) and (12), we get A = 0 and B = π/2. Substituting these in (4)

π −s
I= e (8)
2

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 34

  r r
ˆfc 1 2 π −s π −s
∴ 2
= · e = e
1+x π 2 2
  r Z∞
x 2 x sin sx
(ii) fˆs 2
= dx
1+x π 0 1 + x2
r  
2 −dI
= [by (2)]
π dS
r h
2 π i
=− − e−s , by using (8)
π 2
r
π −s
= e
2

EXAMPLE 3.20
e−ax
Find the Fourier sine transform of , a > 0.
x

Solution
r Z
2 ∞ e−ax
fˆs (s) = sin sx dx = I (say) (1)
π 0 x
r Z
dI 2 ∞ e−ax
Then = x cos sx dx
ds π x
r Z0
2 ∞ −ax
= e cos sx dx
π 0
r  −ax ∞
π e
= (−a cos sx + s sin sx)
2 a2 + s2 0
r
π a
= · 2 2.
2 a +s

Integrating wrt s, we get

r Z
2 a
I= ds
π a + s2
2
r
2 1 −1  s 
= · a tan +C
π a a
r
2 −1
= tan (s/a) +C (2)
π

Put s = 0 in (1) and (2), we get

I =C =0
r
2 −1
∴ I = fˆs (s) = tan (s/a).
π

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 35

EXAMPLE 3.21

Find the Fourier sine transform of


e−x , 0 ≤ x < b

f (x) =
0, x>b


[K. U. 2003 April]

Solution
r Z
2 ∞
fˆs (s) = f (x) sin sx dx
π 0
r Z
2 b −x
= e sin sx dx
π 0
r  b
2 e−x
= (− sin sx − s cos sx)
π 1 + s2 0
r 
2 e −b  
1

= (− sin sb − s cos sb) − (−s)
π 1 + s2 1 + s2
r
2 1 h −b
i
= · s − e (sin sb + s cos sb)
π 1 + s2

EXAMPLE 3.22

Find the Fourier sine transform of


sin x, 0 ≤ x ≤ a

f (x) =
0, x>a


[K. U. 2003 November]

Solution
r Z
2 ∞
fˆs (s) = f (x) sin sx dx
π 0
r Z
2 ∞
= sin x · sin sx dx
π 0
r Z
2 a1
= [cos(s − 1)x − cos(s + 1)x] dx
π 0 2
sin(s − 1)x sin(s + 1)x a
 
1
=√ −
2π s−1 s+1
 0
1 sin(s − 1)a sin(s + 1)a
=√ −
2π s−1 s+1

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 36

EXERCISE 3.1

1. If F( f (x)) = F(s), find F[ f (ax)] [K. U. 2000 April)

2. If F( f (x)) = F(s), prove that F[ f (x − a)] = eias F(s)

[K.U. 2001 April)

3. Find the Fourier transform of the following functions





0, x < 0


(i) f (x) = 1, 0 < x < a [K.U. 2002 April]




0, x > a


x2 , if |x| ≤ a

(ii) f (x) =
0 if |x| > a


(iii) f (x) = xe−x where 0 ≤ x < ∞


a2 − x2 if |x| < a

(iv) f (x) =
0 if |x| ≥ a


1 − |x| for |x| < 1

(v) f (x) =
0 for |x| > 1.


(vi) f (x) = e−|x|


|x| for |x| < a

(vii) f (x) =
0 for |x| > a > 0


cos x if 0 < x < 1

(viii) f (x) =
0

otherwise.

eikx if a < x < b

(ix) f (x) =
0

otherwise.

sin x if |x| ≤ a

(x) f (x) =
0 if |x| > a > 0.


4. Find the Fourier transform of


1 if |x| < a

f (x) =
0 if |x| ≥ a > 0.


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 37

Hence find

sin x
Z ∞
(i) dx and
0 x
sin as · cos sx
Z ∞
(ii) ds.
−∞ s

5. Find the Fourier integral representation of





 0, x<0


f (x) = 1/2, x = 0



 −x

 e , x>0

6. Find the Fourier cosine transform of the following functions

(i) f (x) = e−2x + 4e−3x [K.U. 2001 October]

(ii) f (x) = e−2x + 3e−x

(iii) f (x) = e−ax /x





 x for 0 < x < 1


(iv) f (x) = 2 − x for 1 < x < 2




0 for x > 2


(v) f (x) = e−ax , a > 0


x2 , 0 < x ≤ π

(vi) f (x) =
0, x > π


7. Find the Fourier cosine transform of f (x) = e−4x and hence deduce that
cos 2x
Z ∞
π −8
(i) =
e and
0 x2 + 16 8
x sin 2x π −8
Z ∞
(ii) 2
= e .
0 x + 16 2
8. Find the Fourier sine transform of the following functions




 x for 0 < x < 1


(i) f (x) = 2 − x for 1 < x < 2




0 for x ≥ 2


(ii) f (x) = 5e−2x + 2e−5x

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 38


cos x if 0 < x < a

(iii) f (x) =
0 if x ≥ a


(iv) f (x) = e−ax , a > 0 and hence deduce that

s
Z ∞
π −ax
sin sx ds = e .
0 s2 + a2 2
x
(v) f (x) =
a2 + x2

x2 , 0 < x ≤ π

(vi) f (x) =
0, x>π


(vii) f (x) = e−3x + 3e−2x .


sin x, 0 ≤ x ≤ π

(viii) f (x) =
0

otherwise

9. Find the Fourier sine and cosine transforms of cosh x − sinh x.

10. Find f (x) if its sine transform in e−as .

11. Find f (x) if its cosine transform is e−as .

2 /2
12. Show that f (x) = xe−x is self reciprocal with respect to Fourier sine transform.

ANSWERS
|
1. F(s/a).
|a|
 ias 
1 e −1
3. (i) √
2π is
r 
2 a2 sin(as) 2a cos as 2 sin as

(ii) + −
π s s2 s3
 
1 1
(iii) √
2π (1 − is)2
−4
(iv) √ (as cos as − sin as)
2πs3
r  
2 1 − cos s
(v)
π s2
r  
2 1
(vi)
π 1 + s2

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 39

r  
2 as sin as + cos as − 1
(vii)
π s2
 
1 sin(s + 1) (sin s − 1) 1 − cos(s + 1)
(viii) √ + +i
2 2π s+1 s − 1 s+1
1 − cos(s − 1)
+
s−1
i h i
(ix) √ ei(k+s)a − ei(k+s)b /(k + s)
2π
r
2 i
(x) (s sin a cosas − cos a sin as)
π 1 − s2
r  
2 sin as
4. F(s) = (i) π2 (ii) π2 f (x)
π s
1 ∞ w sin wx + cos wx
Z
5. f (x) = dw
π 0 1 + w2
r   r  
2 2 12 2 2 3
6. (i) + (ii) +
π s2 + 4 s2 + 9 π s2 + 4 s2 + 1
r  
1 2 2 2 2 cos s cos 2s 1
(iii) − √ log(s + a ) (iv) − 2 − 2
2π π s2 s s
r 
√

2 a
(v) (vi) 2 2π cos(sπ/s2 )
π s2 + a2
q  
7. fˆc (s) = π2 s2 +16 4

r
2 2 sin s
8. (i) (1 − cos s)
π s2
q h i
(ii) π2 s25s+4 + s2 +25 2s

  
1 2s cos(s + 1)a cos(s − 1)a
(ii) √ − +
2π s2 − 1 s+1 s−1
r  
2 s q 
2 π −as

(iv) 2 2
(v) π 2 e
π s +a
r 
π 2 cos sπ

2 2
(vi) (cos sπ − 1) −
π s3 s
r  
2 s 3s
(vii) 2
+ 2
π s +9 s +4
r
2 sin sπ
(viii) · .
π 1 − s2
r   r  
ˆ 2 1 ˆ 2 s
9. fc (s) = and fs (s) =
π 1 + s2 π 1 + s2
r  
2 x
10. f (x) = 2 2
rπ x + a 
2 a
11. f (x) =
π x + a2
2

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 40

REFERENCES
[1] Erwin Kreyszig, Advanced Engineering Mathematics, 10th Edition, Wiley-India

[2] Peter V. O’ Neil, Advanced Engineering Mathematics, Thompson Publications, 2007

[3] M Greenberg, Advanced Engineering Mathematics, 2nd Edition, Prentice Hall

Jeeja A. V. jeejamath@gmail.com