2.

116 Chapter 2 Fourier Series and Fourier Integral

7. Find the half-range sine series of f (x) = sin2 x in 0 < x < p.

È 8 • sin(2n - 1)x ˘
Í Ans. : - Â
p n =1 (2n - 1)(2n + 1)(2n + 3) ˙˚
Î

8. Find the half-range sine series of

2x p
f ( x) = 0£x£
3 3
p -x p
= £x £p
3 3
È 2 • 1 np ˘
Í Ans. : p  2 sin 3 sin nx ˙
Î n =1 n ˚

9. Find the half-range sine series of

f (x) = x 0 £ x <1
=1 1£ x < 2
=3–x 2£x£3

È 6 •
È1 Ê np 2np ˆ ˘ np x ˘
Í Ans. : 2 Â Ín
n =1 Î
2 Á
Ë
sin
3
+ sin ˜
3 ¯˚
˙ sin
3 ˚
˙
Î p

2.8 FourIer IntegrAl

Let f(x) be a function which is piecewise continuous in every finite interval in (–•, •)
and absolutely integrable in (–•, •).
We know that the Fourier series of the function f(x) in any interval (–l, l) is given by
•
np x • np x
f ( x ) = a0 + Â an cos + Â bn sin ...(2.5)
n =1 l n =1 l

1 l
2l Ú- l
where a0 = f ( t ) dt

1 l np t
an = Ú f (t ) cos dt
l -l l
1 l np t
bn = Ú f (t ) sin dt
l -l l
 2.8 Fourier Integral 2.117

Substituting the values of a0, an, and bn in Eq. (2.5),

•
1 l 1 np t np x
f ( x) = Ú f (t ) dt + Â f (t ) cos cos dt
2l - l n =1 l l l
•
1 l np t np x
+Â Ú f (t )sin sin dt
l -l l l
n =1
•
1 l 1 l È np t np x np t np x ˘
= Ú f (t ) dt + Â Ú f (t ) Ícos cos + sin sin dt
2l - l n =1 l -l
Î l l l l ˙˚
1 l 1 • l np
= Ú f (t ) dt + Â Ú f (t ) cos ( t - x ) dt
2l - l l n =1 - l l
np p np p
Putting wn = and Dw n = w n +1 - w n = (n + 1) - = ,
l l l l
•
1 l Dw n l
f ( x) =
2l Ú - l
f ( t ) dt +
p
 Ú-l f (t ) cos w n (t - x) dt
n =1
•
 ÎÍÈÚ-l f (t ) cos w n (t - x) dt ˚˙˘ Dw n
1 l 1 l

2l Ú - l
= f ( t ) dt +
p n =1 ...(2.6)
1 p
As l Æ •, Æ 0 and Dw n = Æ 0, the infinite series in Eq. (2.6) becomes an
l l
integral from 0 to •.
1 •È •
f (t ) cos w (t - x ) dt ˘˙ dw [Q l Æ •, Dw n Æ dw ]
p Ú0 ÍÎ Ú-•
f ( x) = ...(2.7)
˚
Equation (2.7) is called the Fourier integral of f (x).
Expanding cosw(t – x) in Eq. (2.7),
1 •È • ˘
f ( x) =
p Ú0 ÍÎÚ-• f (t )(cos w t cos w x + sin w t sin w x) dt ˙˚ dw
1 • • 1 • •
=
p Ú0 Ú-• f (t ) cos w tdt cos w x dw + p Ú0 Ú-• f (t )sin w t dt sin w x dw
• •
= Ú A(w ) cos w x dw + Ú B(w )sin w x dw ...(2.8)
0 0

1 •
where A(w ) =
p Ú-• f (t ) cos w t dt
1 •
and B(w ) =
p Ú-• f (t )sin w t dt
2.118 Chapter 2 Fourier Series and Fourier Integral

Fourier cosine and sine Integrals

When f(x) is an even function,
2 •
A(w ) = Ú f (t ) cos w t dt
p 0
B(w ) = 0
The Fourier integral of an even function f(x) is given by
• ...(2.9)
f ( x ) = Ú A(w ) cos w x dw
0
Equation (2.9) is called the Fourier cosine integral of f(x).
When f(x) is an odd function,
A(w ) = 0
2 •
B(w ) = Ú f (t )sin w t dt
p 0
The Fourier integral of an odd function f(x) is given by
•
f ( x ) = Ú B(w )sin w x dw ...(2.10)
0
Equation (2.10) is called the Fourier sine integral of f(x).

example 1
Using Fourier integral representation, show that
• cos w x + w sin w x
Ú0 dw = 0 x<0
1+ w2
p
= x=0
2
= p e - x x > 0 [Winter 2014; Summer 2015]
Solution
Let f ( x) = 0 x<0
1
= x=0
2
= e- x x>0
The Fourier integral of f(x) is given by
• •
f ( x ) = Ú A(w ) cos w x dw + Ú B(w )sin w x dw
0 0

1 •
p Ú-•
A(w ) = f (t ) cos w t dt

= ÈÍ Ú 0 ◊ cos w t dw + Ú e - t cos w t dw ˘˙
1 0 •

p Î -• 0 ˚
 2.8 Fourier Integral 2.119

•
1 e- t
= (- cos w t + w sin w t
p 1+ w2 0
1
= [Q cos 0 = 1, sin 0 = 0]
p (1 + w 2 )
1 •
p Ú-•
B(w ) = f (t )sin w t dt

= ÈÍ Ú 0 ◊ sin w t dw + Ú e - t sin w t dw ˘˙
1 0 •

p Î -• 0 ˚
•
1 e- t
= (- sin w t - w cos w t )
p 1+w2 0
1
=- (-w )
p (1 + w 2 )
w
=
p (1 + w 2 )
1 • 1 1 • w
Hence, f ( x ) =
p Ú0 1+w 2
cos w x dw +
p Ú0 1+w2
sin w x dw

1 • cos w x + w sin w x
=
p Ú0 1+w2
dw

• cos w x + w sin w x
Ú0 1+w2
dw = p f ( x )

Ï0 x<0
Ô
Ôp
=Ì x=0
Ô2
ÔÓp e - x x>0

example 2
Express the function f ( x ) = 2 | x | < 2
= 0 |x| > 2
as Fourier integral. [Summer 2017, 2016]
Solution
The function f(x) is an even function. The Fourier cosine integral of f(x) is given by
•
f ( x ) = Ú A(w ) cos w x dw
0
2.120 Chapter 2 Fourier Series and Fourier Integral

2 •
A(w ) =
p Ú0 f (t ) cos w t dt

2 2
=
p Ú0 2 ◊ cos w t dt
2
4 sin w t
=
p w 0

4 sin 2w
= [Q sin 0 = 0]
p w
4 • sin 2w cos w x
Hence, f ( x) =
p Ú0 w
dw

• sin 2w cos w x p
Ú0 w
dw = f ( x )
4
Ïp
Ô |x| < 2
= Ì2
ÔÓ 0 | x | > 2 ...(1)

At | x | = 2, i.e., x = ± 2, f(x) is discontinuous.

At x = 2,
1È ˘
f ( x) = Í lim - f ( x ) + lim + f ( x )˙
2 Î x Æ -2 x Æ -2 ˚
1
= [ 2 + 0]
2
=1
At x = –2,
1È ˘
f ( x) = Í lim f ( x ) + lim + f ( x )˙
2 Î x Æ - 2- x Æ -2 ˚
1
= [ 0 + 2]
2
=1
Hence, from Eq. (1),
Ïp
Ô2 |x| < 2
Ô
• sin 2w cos w x Ôp
Ú0 dw = Ì |x| = 2
w Ô4
Ô0 |x| > 2
Ô
Ó
 2.8 Fourier Integral 2.121

example 3
Find the Fourier integral representation of the function
f ( x) = 1 |x| < 1
=0 |x| >1
• sin w cos w x • sin w
Hence, evaluate (i) Ú0 w
dw (ii) Ú0w
dw
[Winter 2016, 2014, 2013]
Solution
The function f(x) is an even function. The Fourier cosine integral of f(x) is given by
•
f ( x ) = Ú A(w ) cos w x dw
0

2 •
A(w ) =
p Ú0 f (t ) cos w t dt

2 1
=
p Ú0 1 ◊ cos w t dt
1
2 sin w t
=
p w 0
2 sin w
= [Q sin 0 = 0]
p w
2 • sin w cos w x
Hence, f ( x) =
p Ú0 w
dw

(i) • sin w cos w x p

Ú0 w
dw = f ( x )
2
Ïp
Ô |x| < 1
= Ì2
...(1)
ÔÓ0 |x| > 1
At |x| = 1, i.e., x = ± 1, f(x) is discontinuous.
At x = 1,
1
f ( x) = È lim f ( x ) + lim f ( x )˘
2 ÍÎ x Æ1- x Æ1+ ˙˚
1
= (1 + 0)
2
1
=
2
2.122 Chapter 2 Fourier Series and Fourier Integral

At x = –1,
1
f ( x) = È lim f ( x ) + lim f ( x )˘
2 ÍÎ x Æ-1- x Æ-1+ ˙˚
1
=
(0 + 1)
2
1
=
2
Hence, from Eq. (1),
Ïp
Ô2 |x| < 1
Ô
• sin w cos w x Ôp
Ú0 dw = Ì |x| = 1
w Ô4
Ô0 |x| > 1
Ô
Ó
(ii) Putting x = 0 in Eq. (1),
• sin w p p
Ú0 w
dw = f (0) =
2 2
[Q f (0) = 1]

example 4
Find the Fourier integral representation of the function
2
f ( x) = 1 - x |x| £1
=0 |x| >1
Solution
f(x) is an even function. The Fourier cosine integral of f (x) is given by
•
f ( x ) = Ú A(w ) cos w x dw
0

2 •
p Ú0
A(w ) = f (t ) cos w t dt

2 •
= Ú (1 - t 2 ) cos w t dt
p 0
1
2 Ê sin w t ˆ Ê cos w t ˆ Ê sin w t ˆ
= (1 - t )2 Á - ( -2t ) Á - + ( -2) Á -
p Ë w ˜¯ Ë w 2 ˜ ¯ Ë w 3 ˜¯ 0

2 Ê 2 cos w 2 sin w ˆ
= - + [Q sin 0 = 0]
p ÁË w2 w3 ¯
˜

4 Ê sin w - w cos w ˆ
= Á ˜¯
pË w3
 2.8 Fourier Integral 2.123

4 • sin w - w cos w
Hence, f ( x) =
p Ú0 w3
cos w x dw

example 5
Find the Fourier cosine integral of f (x) = e–kx, where x > 0, k > 0.
[Winter 2016]
Solution
The Fourier cosine integral of f(x) is given by
•
f ( x ) = Ú A(w ) cos w x dw
0

2 •
A(w ) =
p Ú0 f (t ) cos w t dt

2 • - kt
=
p Ú0 e cos w t dt
•
2 e - kt
= (- k cos w t + w sin w t
p k2 + w2 0
2Ê k ˆ
= Á 2 [Q cos 0 = 1, sin 0 = 0]
p Ë k + w 2 ˜¯
2k • 1
Hence, f ( x) =
p Ú0 k +w2
2
cos w x dw

• cos w x p
Ú0 2
w +k 2
dw =
2k
f ( x)

p - kx
= e x > 0, k>0
2k

example 6
p -x
Find the Fourier cosine integral of f(x) = e , x ≥ 0. [Winter 2015]
2
Solution
The Fourier cosine integral of f(x) is given by
•
f(x) = Ú A(w ) cos w x dw
0
•
2
A(w) =
p Ú f (t ) cos w t dt
0
2.124 Chapter 2 Fourier Series and Fourier Integral

•
2 p -t
=
p Ú2e cos w t dt
0

•
-t
= Úe cos w t dt
0
•
e- t
= (- cos w t + w sin w t )
1 + w2 0

1
= [Q cos 0 = 1, sin 0 = 0]
1 + w2
•
1
Hence, f(x) = Ú 1 + w2 cos w x dw
0

•
cosw x
Ú 1+ w2 dw = f(x)
0

•
cosw x p -x
Ú 1+w2 dw =
2
e
0

example 7
p
Find the Fourier cosine integral of the function f ( x ) = cos x | x | <
2
p
=0 |x| >
2
Solution
The Fourier cosine integral of f(x) is given by
•
f ( x ) = Ú A(w ) cos w x dw
0

2 •
A(w ) =
p Ú0 f (t ) cos w t dt
p
2
=
p Ú0
2 cos t cos w t dt
p
2 1
=
p Ú0
2
2
[cos(1 + w )t + cos(1 - w )t ]dt
 2.8 Fourier Integral 2.125

p
1 sin(1 + w )t sin(1 - w )t 2
= +
p 1+w 1-w 0

È p p ˘
sin(1 + w ) sin(1 - w )
1Í 2 2 ˙
= Í + ˙ [Q sin 0 = 0]
p Î 1+w 1-w ˚
È Ê pw ˆ Ê pw ˆ ˘
cos Á
Ë 2 ˜¯
cos Á
1Í Ë 2 ˜¯ ˙
= Í + ˙
p Î 1+w 1-w ˚
Ê pw ˆ
2 cos Á
1 Ë 2 ˜¯
=
p 1-w2
Ê pw ˆ
cos Á
2 • Ë 2 ˜¯
Hence, f ( x) =
p Ú0 1- w2
cos w x dw

example 8
Express the function f ( x ) = 1 0£ x<p
=0 x >p
as a Fourier sine integral and hence, evaluate
• 1 - cos pw
Ú0 w
sin w x dw [Winter 2017]

Solution
The Fourier sine integral of f(x) is given by
•
f ( x ) = Ú B(w )sin w x dw
0

2 •
p Ú0
B(w ) = f (t )sin w t dt

= ÈÍ Ú 1 ◊ sin w t dt + Ú 0 ◊ sin w t dt ˘˙
2 p 0

pÎ 0 p ˚
p
2 cos w t
= -
p w 0
2 Ê - cos pw + 1ˆ
= Á ˜¯ [Q cos 0 = 1]
pË w
2 Ê 1 - cos pw ˆ
= Á ˜¯
pË w
2.126 Chapter 2 Fourier Series and Fourier Integral

2 • 1 - cos pw
p Ú0
Hence, f ( x) = sin w x dw
w
• 1 - cos pw p
Ú0 w sin w x dw = 2 f ( x)
Ïp
Ô 0£ x<p
= Ì2 ...(1)
ÔÓ0 x>p
At x = p, f(x) is discontinuous.
1
f ( x ) = È lim f ( x ) + lim f ( x )˘
2 Î x Æp - x Æp + ˚
1
= (1 + 0)
2
1
=
2
Hence, from Eq. (1),
Ïp
Ô2 0£ x<p
• 1 - cos pw Ô
Ú0 w sin w x dw = ÌÔ p x=p
4
Ô
Ó0 x >p

example 9
Express the function f ( x ) = sin x 0£ x£p
=0 x >p
as a Fourier sine integral and show that
• sin w x sin pw p
Ú0 1 - w 2 dw = 2 sin x 0£ x£p

Solution
The Fourier sine integral of f (x) is given by
•
f ( x ) = Ú B(w ) sin w x dw
0

2 •
B(w ) = Ú f (t )sin w t dt
p 0
= ÈÍ Ú sin t sin w t dt + Ú 0 ◊ sin w t dt ˘˙
2 p •

p Î 0 p ˚
 2.8 Fourier Integral 2.127

2 1
Ú0 2 [cos(w - 1)t - cos(w + 1)t ]dt
p
=
p
p
1 sin(w - 1)t sin(w + 1) t
= -
p w -1 w +1 0
1 È sin(w - 1)p sin(w + 1)p ˘
= Í - [Q sin 0 = 0]
p Î w -1 w + 1 ˙˚
1 È sin pw sin pw ˘
= Í- +
p Î w -1 w + 1 ˙˚
1 Ê 2 sin pw ˆ
= Á- 2
p Ë w - 1 ˜¯
2 Ê sin pw ˆ
=
p ÁË 1 - w 2 ˜¯
2 • sin pw
Hence, f ( x) =
p Ú0 1-w2
sin w x dw , w π1

• sin w x sin pw p
Ú0 1-w 2
dw =
2
f ( x)

Ïp
Ô sin x 0£ x£p
= Ì2
ÔÓ0 x>p

example 10
Find the Fourier sine integral of f(x) = e–bx.
p • w sin w x
Hence, show that e- bx = Ú0 2 dw
2 b +w2
Solution
The Fourier sine integral of f (x) is given by
•
f ( x ) = Ú B(w )sin w x dw
0

2 •
B(w ) =
p Ú0 f (t )sin w t dt

2 • - bt
=
p Ú0 e sin w t dt
•
2 e - bt
= (-b sin w t - w cos w t )
p b2 + w 2 0
2.128 Chapter 2 Fourier Series and Fourier Integral

2Ê w ˆ
= [Q cos 0 = 1, sin 0 = 0]
p ÁË b2 + w 2 ˜¯
2 • w sin w x
p Ú0 b2 + w 2
Hence, f ( x) = dw

• w sin w x p
Ú0 b2 + w 2 dw = 2 f ( x)
p
= e - bx
2

example 11
•
l 3 sin l x p -x
Show that Ú 4
dl = e cos x, where x > 0. [Winter 2015]
0 l +4 2

Solution
p -x
f(x) = e cos x, x > 0
2
The Fourier sine integral of f(x) is given by
•
f(x) = Ú B(l )sin l x dl
0
•
2
B(l) =
p Ú f ( x) sin l x dx
0
•
2 p -x
=
p Ú2e cos x sin l x dx
0
•
-x
= Úe cos x sin l x dx
0
•
1 -x
2 Ú0
= e (2 cos x sin l x ) dx

•
1 -x
e [sin(l + 1) x + sin(l - 1) x ] dx
2 Ú0
=

1 È -x ˘
• •
= Í Ú e sin(l + 1) x dx + Ú e - x sin(l - 1) x dx ˙
2 ÍÎ 0 0 ˙˚
 2.8 Fourier Integral 2.129

•
1È e- x
= Í {- sin (l + 1) x - (l + 1) cos(l + 1) x}
2 ÍÎ 1 + (l + 1)2
0

•˘
e- x ˙ , ( x > 0)
+ {- sin ( l - 1) x - ( l - 1) cos( l - 1) x}
1 + (l - 1)2 ˙
0 ˚

1 È (l + 1) (l - 1) ˘
= Í + ˙
2 ÍÎ 1 + (l + 1) 2
1 + (l - 1)2 ˙˚
1 È l +1 l -1 ˘
= +
2 ÍÎ l 2 + 2 l + 2 l 2 - 2 l + 2 ˙˚

1 È (l + 1) (l 2 - 2 l + 2) + (l - 1)(l 2 + 2 l + 2) ˘
= Í ˙
2 ÍÎ (l 2 + 2 l + 2)(l 2 - 2 l + 2) ˙˚

1 È l 3 - 2l 2 + 2l + l 2 - 2l + 2 + l 3 + 2l 2 + 2l - l 2 - 2l - 2 ˘
= Í ˙
2 ÍÎ l4 + 4 ˙˚

1 È 2l 3 ˘
= Í ˙
2 ÍÎ l 4 + 4 ˙˚

l3
=
l4 + 4
•
l3
Hence, f(x) = Ú l 4 + 4 sin l x dl
0
• 3
l p -x
Ú l 4 + 4 sin l x dl =
2
e cos x
0

exercIse 2.4
1. Find the Fourier integral representations of the following functions:
(i) f (x) = x |x| < 1 (ii) f (x) = -e ax x<0
- ax
=0 |x| > 1 =e x>0
È • sin w - w cos w ˘
Í Ans. : (i) Ú-• ipw 2
eiw x dw ˙
Í ˙
Í 2 • w
dw ˙˙
ÍÎ (ii)
p Ú0
sin w x 2
a +w 2
˚
2.130 Chapter 2 Fourier Series and Fourier Integral

2. Find the Fourier sine integral of f(x) = e–ax – e–bx.

È 2 • (b 2 - a 2 )w sin w x ˘
Í Ans. : Ú0 2 dw ˙
Î p (a + w )(b + w )
2 2 2
˚
3. Find the Fourier cosine integral of f(x) = e–xcos x.
È 2 • w2 + 2 ˘
Í
Î
Ans. : Ú
p w +4
0 4
cos w x dw ˙
˚
4. Express the function
p
f ( x) = 0<x <p
2
=0 x <p

as the Fourier sine integral and show that

• 1 - cos pw p
Ú0 w
sin w x dw =
2
È -1 1 - cos pw ˘
ÍÎ Ans. : Ú0 w
sin w x dw ˙
˚

Points to remember
Fourier Series in the Interval (0, 2p)
∞ ∞
f ( x) = a0 + ∑ an cos nx + ∑ bn sin nx
n =1 n =1

1 2p

2p Ú0
a0 = f ( x ) dx

1 2p
an = Ú f ( x ) cos nx dx
p 0
1 2p
bn = Ú f ( x )sin nx dx
p 0

Fourier Series in the Interval (c, c + 2l)

•
np x • np x
f ( x ) = a0 + Â an cos + Â bn sin
n =1 l n =1 l
1 c + 2l
a0 = Ú f ( x ) dx
2l c
 Points to Remember 2.131

1 c + 2l np x
an =
l Úc
f ( x ) cos
l
dx

1 c + 2l np x
bn = Ú f ( x )sin dx
l c l
where 2l is the length of the interval.
Fourier Series of Even Function in the Interval (–p, p)
•
f ( x ) = a0 + Â an cos nx
n =1
1 p
a0 =
p Ú0 f ( x ) dx

2 p
an =
p Ú0 f ( x ) cos nx dx

bn = 0

Fourier Series of Even Function in the interval (–l, l)

•
np x
f ( x ) = a0 + Â an cos
n =1 l
1 l
a0 = Ú f ( x ) dx
l 0
2 l np x
an = Ú f ( x ) cos dx
l 0 l
bn = 0

Fourier Series of Odd Function in the Interval (–p, p)

•
f ( x ) = Â bn sin nx
n =1
a0 = 0
an = 0
2 p
bn =
p Ú0 f ( x )sin nx dx

Fourier Series of Odd Function in the Interval (–l, l)

•
np x
f ( x ) = Â bn sin
n =1 l
a0 = 0
an = 0
2 l np x
l Ú0
bn = f ( x )sin dx
l
2.132 Chapter 2 Fourier Series and Fourier Integral

Half-Range Cosine Series in the Interval (0, p)

•
f ( x ) = a0 + Â an cos nx
n =1
1 p
a0 =
p Ú0 f ( x ) dx
2 p
an =
p Ú0 f ( x ) cos nx dx

Half-Range Cosine Series in the Interval (0, l)

•
np x
f ( x ) = a0 + Â an cos
n =1 l
1 l

l Ú0
a0 = f ( x ) dx

2 l np x
an = Ú f ( x ) cos dx
l 0 l

Half-Range Sine Series in the Interval (0, p)

•
f ( x ) = Â bn sin nx
n =1
2 p
bn =
p Ú0 f ( x )sin nx dx

Half-Range Sine Series in the Interval (0, l)

•
np x
f ( x ) = Â bn sin
n =1 l
2 l np x
bn =
l Ú0
f ( x )sin
l
dx

Fourier Integral Theorem

• •
f ( x ) = Ú A(w ) cos w x dw + Ú B(w )sin w x dw
0 0
1 •
A(w ) =
p Ú-• f (t ) cos w t dt
1 •
B(w ) =
p Ú-• f (t )sin w t dt
Fourier Cosine Integral
•
f ( x ) = Ú A(w ) cos w x dw
0
 Multiple Choice Questions 2.133

2 •
A(w ) =
p Ú0 f (t ) cos w t dt

B(w ) = 0

Fourier Sine Integral

•
f ( x ) = Ú B(w )sin w x dw
0
A(w ) = 0
2 •
B(w ) = Ú f (t )sin w t dt
p 0

multiple choice Questions

Select the most appropriate response out of the various alternatives given in each
of the following questions:
Ï-1 -1 < x < 0
1. If f(x) = Ì
Ó 1 0 < x <1
then f(x) is a/an _______ function in (–1 , 1).
(a) even (b) odd (c) constant (d) none of these
Ï- x -p < x < 0
2. If f(x) = Ì
Ó x 0< x<p
then f(x) is a/an _____ function in (–p, p).
(a) even (b) odd (c) constant (d) none of these
Ï 2x
ÔÔ1 + p -p £ x £ 0
3. If f(x) = Ì
Ô1 - 2 x 0£ x£p
ÔÓ p
then f(x) is a/an _____ function in (–p , p).
(a) even (b) odd (c) constant (d) none of these
Ï- x 2
-p < x £ 0
4. The Fourier series expansion of f(x) = ÔÌ contains no
ÔÓ x
2
0£ x£p
______ terms.
(a) sine (b) cosine (c) constant (d) none of these
Ï- x -4 £ x £ 0
5. The Fourier series expansion of f(x) = Ì contains no ______
Ó x 0£ x£4
terms.
(a) sine (b) cosine (c) constant (d) none of these
2.134 Chapter 2 Fourier Series and Fourier Integral

6. If f(x) is an even function in (–p , p), then the graph of f(x) is symmetrical about
the ______.
(a) x-axis (b) y-axis (c) origin (d) none of these
7. If f(x) is an odd function in (–l, l), then the graph of f(x) is symmetrical about the
______
(a) x-axis (b) y-axis (c) origin (d) none of these
8. If f(x) is an even function in the interval (–l, l), then the value of bn is
p
(a) (b) p (c) 1 (d) 0
2
9. If f(x) is an odd function in (–l, l) , then the values of a0 and a1 are
p
(a) 0, 0 (b) p, p (c) ,p (d) 1, 1
2
10. If f(x) = x in (–p, p), then the Fourier coefficient a2 is
(a) p (b) 0 (c) 1 (d) –1
11. If f(x) = cos x in (–p, p), then the Fourier coefficient bn is
(a) 0 (b) p (c) 1 (d) none of these
12. In the Fourier series expansion of f(x) = x sin x in (–p, p), the ____ terms are
absent.
(a) sine (b) cosine (c) constant (d) none of these
13. If f(x) = x cos x in (–p, p), then b1 is
(a) 0 (b) p (c) 1 (d) none of these
14. Which of the following is neither an even function nor an odd function?
(a) x sin x (b) x2 (c) e–x (d) x cos x
15. Fundamental period of sin 2x is
p p
(a) (b) (c) 2p (d) p
4 2
16. Fundamental period of tan 3x is
p p p
(a) (b) (c) p (d)
2 3 4
17. If f(x + nT) = f(x) where n is any integer, then the fundamental period of f(x) is
T
(a) 2T (b) (c) T (d) 3T
2
18. For half-range sine series of f(x) = cos x, 0 £ x £ p and period 2p, Fourier series
•
is represented by  bn sin nx, then Fourier coefficient b1 is
n =1

1 2 2
(a) (b) 0 (c) (d) –
p p p
 Multiple Choice Questions 2.135

19. A function f(x) is said to be periodic of period T if

(a) f(x + T) = f(x) for all x (b) f(x + T) = f(T) for all x
(c) f(–x) = f(x) for all x (d) f(–x) = –f(x) for all x
20. Fourier series representation of a periodic function f(x) with period 2p which
satisfies Dirichlet’s conditions is
•
(a) a0 + Â (an cos nx + bn sin nx)
n =1

•
(b) a0 + Â (an cos np x + bn sin np x)
n =1

•
(c) a0 + Â (an cos nx)(bn sin nx)
n =1

(d) a0 + an cos nx + bn sin nx

21. A function f(x) is said to be even if
(a) f(–x) = f(x) (b) f(–x) = –f(x)
(c) f(x + 2p) = f(x) (d) f(–x) = [f(x)]2
22. A function f(x) is said to be odd if
(a) f(–x) = f(x) (b) f(–x) = –f(x)
(c) f(x + 2p) = f(x) (d) f(–x) = [f(x)]2
23. Which of the following is an odd function?
(a) sinx (b) ex + e–x (c) e| x | (d) p2 – x2
24. Which of the following is an even function?
(a) sinx (b) ex – e–x (c) x cosx (d) cosx
25. For an even function f(x) defined in the interval -p £ x £ p and f ( x + 2p ) = f ( x ),
the Fourier series is
• •
np x
(a) Â bn sin x (b) a0 + Â an cos l
n =1 n =1

• •
np x
(c) a0 + Â an cos nx (d) Â bn sin l
n =1 n =1

26. For an odd function f(x) defined in the interval -p £ x £ p and f(x + 2p) = f(x),
the Fourier series is
• •
np x
(a) Â bn sin nx (b) a0 + Â an cos l
n =1 n =1

• •
np x
(c) a0 + Â an cos nx (d) Â bn sin l
n =1 n =1
2.136 Chapter 2 Fourier Series and Fourier Integral

27. Half-range Fourier cosine series for f(x) defined in the interval (0, p) is
• •
nx
(a) a0 + Â an cos nx (b) an + Â an cos l
n =1 n =1

• •
(c) Â an cos nx (d) an + Â (an cos nx + bn sin nx)
n =1 n =1

28. Half-range Fourier sine series for f(x) defined in the interval (0, p) is
• •
np x
(a) Â bn sin nx (b) Â bn sin l
n =1 n =1

• •
np x
(b) a0 + Â an cos l
(d) a0 + Â (an cos nx + bn sin nx)
n =1 n =1

29. The Fourier series of an odd periodic function contains only

(a) odd harmonics (b) even harmonics
(c) cosine terms (d) sine terms
30. The trigonometric Fourier series of an even function does not have
(a) constant (b) cosine terms (c) sine terms (d) odd harmonic terms
Ï- k -p < x < 0
31. For the function f ( x ) = Ì , the value of a0 in the Fourier series
Ók 0< x<p
expansion will be

(a) k (b) 2k (c) 0 (d) –k

32. The value of a0 in Fourier series expansion of f(x) = x2, –1 < x < 1 is
1 1
(a) (b) 3 (c) (d) 1
3 2

Answers
1. (b) 2. (a) 3. (a) 4. (b) 5. (a) 6. (b) 7. (c) 8. (d)
9. (a) 10. (b) 11. (a) 12. (a) 13. (a) 14. (c) 15. (d) 16. (b)
17. (c) 18. (b) 19. (a) 20. (a) 21. (a) 22. (b) 23. (a) 24. (d)
25. (c) 26. (a) 27. (a) 28. (a) 29. (d) 30. (c) 31. (c) 32. (a)