4 Fourier Transform(F.T.)
Fourier integral theorem (without proof) - Fourier transformpair - Sine and Cosine transforms - Properties - Transforms ofsimple functions - Convolution theorem - Parseval’s identity.
4.1 Definition
Let f(x) be a function defined in an interval (a, b). Let k(s, x) be a givenfunction of two variables s and x. The integral transform of f(x) w.r.t
k(s, x) is defined by I[f(x)] = F (s) =
b∫a
f(x)k(s, x)dx (1)
k(s, x) is called the kernel of the integral transform f(x) is called theinverse transform of F (s).The integral transform is said to be finite if both a and b are finite;
otherwise it is called an infinite transform.If there exist a function H(s, x) such that
f(x) =
d∫e
F (s)H(s, x)ds (2)
then (2) is called the inversion formula for (1).
4.2 Fourier integral theorem(without proof)
If f(x) is piecewise continuously differentiable and absolutely integrable
(−∞,∞), then f(x) =1
π
∞∫0
∞∫−∞
f (t) cosλ (t− x)dtdλ (1)
This is known as Fourier integral theorem or Fourier integralformula.
Note : At a point of discontinuity the value of the integral on the left of
245
246 Unit IV - FOURIER TRANSFORM (F.T.)
(1) is1
2[f(x− 0) + f(x+ 0)]
i.e.,f(x− 0) + f(x+ 0)
2=
1
π
∞∫0
∞∫−∞
f(t) cosλ(t− x)dtdλ
4.2.1 Fourier Cosine and Sine Integral
We know that he Fourier integral for f(x) is
f()x =1
π
∞∫0
∞∫−∞
f(t) cosλ(t− x)dtdλ
=1
π
∞∫0
∞∫−∞
f(t){cosλt cosλx+ sinλt sinλx}dtdλ
=1
π
∞∫0
∞∫−∞
f(t) cosλt cosλxdtdλ+1
π
∞∫0
∞∫−∞
f(t) sinλt sinλxdtdλ
Case (i) : If f(t) is an even function, then f(t) cosλt is an evenfunction and f(t) sinλt is an odd function.
f(x) =2
π
∞∫0
∞∫0
f(t) cosλt cosλxdtdλ
i.e., f(x) =2
π
∞∫0
cosλx
∞∫0
f(t) cosλtdtdλ
This is known as Fourier cosine integral.
Case (ii) : If f(t) is an odd function, then f(t) cosλt is an oddfunction and f(t) sinλt is an even function.
f(x) =2
π
∞∫0
∞∫0
f(t) sinλt sinλxdtdλ
i.e., f(x) =2
π
∞∫0
sinλx
∞∫0
f(t) sinλtdtdλ
This is known as Fourier sine integral.
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 247
4.2.2 Complex form of Fourier Integrals :
The Fourier integral for the function f(x) is given by
f(x) =1
π
∞∫0
∞∫−∞
f(t) cosλ(t− x)dtdλ
f(x) =1
2π
∞∫−∞
∞∫−∞
f(t) cosλ(t− x)dtdλ (1)
(since cosλ(t− x) is an even function of λ)Since sinλ(t− x) is an odd function of λ.
∴
∞∫−∞
sinλ(t− x)dλ =0
1
2π
∞∫−∞
∞∫−∞
f(t) sinλ(t− x)dtdλ =0 (2)
Now (1) + i (2)⇒
f(x) =1
2π
∞∫−∞
∞∫−∞
f(t) {cosλ(t− x) + i sinλ(t− x)} dtdλ
i.e., f(x) =1
2π
∞∫−∞
∞∫−∞
f(t)eiλ(t−x)dtdλ
4.2.3 Examples under Fourier integrals
Example 4.1. Using Fourier integral, show that
∞∫0
ω sinωx
k2 + ω2dω =
π
2e−kx
(x > 0, k > 0).
Solution : Let f(x) = e−kx.(Here we use sine integral formula since the presence of the term sinwx
in LHS)The Fourier sine integral of f(x) is given by
f(x) =2
π
∞∫0
sinλx
∞∫0
f(t) sinλtdtdλ
248 Unit IV - FOURIER TRANSFORM (F.T.)
Given f(x) = e−kx ⇒ f(t) = e−kt
∴ f(x) =2
π
∞∫0
sinλx
∞∫0
e−kt sinλtdtdλ
=2
π
∞∫0
sinλx
[e−kt
k2 + λ2(−k sinλt− λ cos λt)
]∞0
dλ
f(x) =2
π
∞∫0
sinλxλ
k2 + λ2dλ
=2
π
∞∫0
λ sinλx
k2 + λ2dλ
∞∫0
λ sinλx
k2 + λ2dλ =
π
2f(x)
∞∫0
λ sinλx
k2 + λ2dλ =
π
2e−kx
∞∫0
ω sinωx
k2 + ω2dλ =
π
2e−kx ( put λ by ω )
Example 4.2. Express the function f(x) =
{1 for |x| ≤ 10 for |x| > 1
as a Fourier
integral. Hence evaluate
∞∫0
sinλ cosλx
λdλ and find the value of
∞∫0
sinλ
λdλ.
Solution : The Fourier integral of f(x) is given by
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 249
f(x) =1
π
∞∫0
∞∫−∞
f(t) cosλ (t− x) dtdλ
=1
π
∞∫0
1∫−1
cosλ (t− x) dtdλ
=1
π
∞∫0
[sinλ (t− x)
λ
]1−1
dλ
=1
π
∞∫0
sinλ (1− x) + sinλ (1 + x)
λdλ
f(x) =2
π
∞∫0
sinλ cosλx
λdλ
∞∫0
sinλ cosλx
λdλ =
π
2f(x) (1)
=
{ 1
2for |x| < 1
0 for |x| > 1
At x = 1, which is a point of discontinuity of f(x), the value of the aboveintegral
=π
2
[f (1− 0) + f (1 + 0)
2
]=π
2
[1 + 0
2
]=π
4
Hence
∞∫0
sinλ cosλx
λdλ =
π2 for |x| < 1π4 for x = 10 for |x| > 1
Now, putting x = 0 we get
∞∫0
sinλ
λdλ =
π
2
Example 4.3. Find Fourier cosine integral of the function e−ax. Hence
250 Unit IV - FOURIER TRANSFORM (F.T.)
find the value of the integral
∞∫0
cosλx
1 + λ2dλ
Solution : Let f(x) = e−ax.The Fourier cosine integral of f(x) is given by
f(x) =2
π
∞∫0
cosλx
∞∫0
f(t) cosλtdtdλ
Given f(x) = e−ax ⇒ f(t) = e−at
∴ f(x) =2
π
∞∫0
cosλx
∞∫0
e−at cosλtdtdλ
=2
π
∞∫0
cosλx
[e−at
a2 + λ2(−a cosλt+ λ sinλt)
]∞0
dλ
=2
π
∞∫0
cosλxa
a2 + λ2dλ
f(x) =2a
π
∞∫0
cosλx
a2 + λ2dλ
∞∫0
cosλx
a2 + λ2dλ =
π
2af(x)
∞∫0
cosλx
a2 + λ2dλ =
π
2ae−ax
[f(x) = e−ax
]Putting a = 1, we get
∞∫0
cosλx
1 + λ2dλ =
π
2e−x
Example 4.4. Express f(x) =
{ 1
2for 0 ≤ x ≤ π
0 for x > πas a Fourier sine
integral and hence show that∞∫0
1− cos πλ
λsinλxdλ =
{ π
2for 0 < x < π
0 for x > π
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 251
Solution : Given
f(x) =
{ 1
2for 0 ≤ x ≤ π
0 for x > π⇒ f(t) =
{ 1
2for 0 ≤ t ≤ π
0 for t > πThe Fourier sine integral of f(x) is given by
f(x) =2
π
∞∫0
sinλx
∞∫0
f(t) sinλtdtdλ
=2
π
∞∫0
sinλx
π∫0
1
2sinλtdtdλ
=1
π
∞∫0
sinλx
[− cosλt
λ
]π0
dλ
=1
π
∞∫0
sinλx
[1− cosλπ
λ
]dλ
1
π
∞∫0
[1− cosλπ
λ
]sinλxdλ =f(x)
∞∫0
[1− cosλπ
λ
]sinλxdλ =πf(x)
=π
{ 1
2for 0 < x < π
0 for x > π
=
{ π
2for 0 < x < π
0 for x > π
4.3 Fourier transform pair[complex Fourier transform / InfiniteFourier transform]
Fourier transform: The Fourier transform of f(x) is given by
F [f (x)] =1√2π
∞∫−∞
f (x) eisxdx (1)
= a function of s
= F (s) = f (s)
Inverse Fourier transform:(Used in deduction part)
252 Unit IV - FOURIER TRANSFORM (F.T.)
The Inverse Fourier transform of F (s) is given by
f (x) =1√2π
∞∫−∞
F [f (x)] e−isxds (2)
The above equations (1) and (2) are jointly called as Fourier Transformpair.
4.3.1 Examples under Fourier Transform Pair
Example 4.5. Find the Fourier transform of f(x) =
{1 in |x| < a0 in |x| > a
Solution : The given function can be written as
f(x) =
{1 if − a < x < a0 otherwise[
∵ |x| < a⇒ −a < x < a|x| > a⇒ −∞ < x < −a & a < x <∞
]∴ The Fourier transform of f(x) is given by
F (s) = F [f(x)] =1√2π
∞∫−∞
f(x)eisxdx
=1√2π
a∫−a
eisxdx
=1√2π
a∫−a
(cos sx+ i sin sx)dx
=1√2π
a∫−a
cos sxdx+ i
a∫−a
sin sxdx
=
2√2π
a∫0
cos sxdx [∵ sin sxis an odd function]
=
√2
π
[sin sx
s
]a0
=
√2
π
[sin as
s
]Example 4.6. Find the Fourier transform of the function
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 253
f(t) =
0, −∞ ≤ t ≤ 0
sin t, 0 ≤ t ≤ π0, π ≤ t ≤ ∞
Solution : The Fourier transform of f(t) is given by
F (s) = F [f(t)] =1√2π
∞∫−∞
f(t)eistdt
=1√2π
π∫0
sin teistdt
=1√2π
[eist
(is)2 + 1(is sin t− cos t)
]π0
=1√2π
[eist
1− s2(is sin t− cos t)
]π0
=1√2π
[eisπ
1− s2(0 + 1)− 1
1− s2(0− 1)
]=
1√2π
(1 + eiπs
1− s2
)
Example 4.7. Find the Fourier transform of f(x) =
{x if |x| < a0 if |x| > a
Solution : The given function can be written as
f(x) =
{x if − a < x < a0 otherwise
254 Unit IV - FOURIER TRANSFORM (F.T.)
F (s) = F [f(x)] =1√2π
∞∫−∞
f(x)eisxdx
=1√2π
a∫−a
xeisxdx
=1√2π
a∫−a
x(cos sx+ i sin sx)dx
=1√2π
a∫−a
x cos sxdx+ i
a∫−a
x sin sxdx
=
2i√2π
a∫0
x sin sxdx
[∵ x cos sx is odd function & x sin sx is even fn.]
=i
√2
π
[x
(− cos sx
s
)− 1
(− sin sx
s2
)]a0
=i
√2
π
[−x cos sx
s+
sin sx
s2
]a0
=i
√2
π
[−a cos as
s+
sin as
s2
]F (s) =i
√2
π
[sin as
s2− a cos as
s
]
Example 4.8. Find the Fourier transform of f(x) =
{x2 if |x| < 10 if |x| > 1
Solution : The given function can be written as
f(x) =
{x2 if − 1 < x < 10 otherwise
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 255
F (s) = F [f(x)] =1√2π
∞∫−∞
f(x)eisxdx
=1√2π
1∫−1
x2(cossx + isinsx)dx
=1√2π.2
1∫0
x2 cos sxdx
[∵ x2 sin sx is odd function
]=
√2
π
[x2(sin sx
s
)− 2x
(− cos sx
s2
)+ 2
(− sin sx
s3
)]10
=
√2
π
[x2 sin sx
s+
2xcossx
s2− 2 sin sx
s3
]10
=
√2
π
[sin s
s+
2coss
s2− 2 sin s
s3
]10
Example 4.9. * Find the Fourier transform of f(x) is defined by
f(x) =
0, x < a1, a < x < b0, x > b
.
{Sol: F [f (x)] = F (s) =
1
is√2π
[eisb − eisa
]}
Example 4.10. Find the Fourier transform of
f(x) =
{1− |x| for |x| < 1
0 otherwise. Hence show that
∞∫0
sin2 x
x2dx =
π
2
Solution : The given function can be written as
f(x) =
{1− |x| for − 1 < x < 1
0 otherwise.
256 Unit IV - FOURIER TRANSFORM (F.T.)
F (s) = F [f(x)] =1√2π
∞∫−∞
f(x)eisxdx
=1√2π
1∫−1
(1− |x|)(cos sx + isin sx)dx
=1√2π
1∫−1
(1− |x|) cos sxdx+ i
1∫−1
(1− |x|) sin sxdx
=2√2π
1∫0
(1− |x|) cos sxdx
[∵ (1− |x|) sin sx is odd]
=
√2
π
1∫0
(1− x) cos sxdx
=
√2
π
[(1− x)
sin sx
s− (−1)
(−cossxs2
)]10
=
√2
π
[(1− x)
sin sx
s− cos sx
s2
]10
=
√2
π
[(0− cos s
s2
)−(0− 1
s2
)]F (s) =
√2
π
(1− cos s
s2
)By inversion formula for Fourier transform
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 257
f(x) =1√2π
∞∫−∞
F (s)e−isxds
=1√2π
∞∫−∞
√2
π
(1− cos s
s2
)(cos sx− i sin sx)ds
=1
π
∞∫−∞
[(1− cos s
s2
)cos sx− i
(1− cos s
s2
)sin sx
]ds
f(x) =2
π
∞∫0
(1− cos s
s2
)cos sxds
[∵
(1− cos s
s2
)cos sx is even and
(1− cos s
s2
)sin sx is odd
]∞∫0
(1− cos s
s2
)cos sxds =
π
2f(x)
Put x = 0,∞∫0
(1− cos s
s2
)ds =
π
2.1
∞∫0
2 sin2 s2
s2ds =
π
2
∞∫0
sin2 s2(
s2
)2 ds2 =π
2
Puts
2= t⇒ ds
2= dt
∴
∞∫0
sin2 t
t2dt =
π
2
Example 4.11. Find the Fourier transform of
f(x) =
{1− x2 if |x| < 1
0 if |x| > 1and hence evaluate
∞∫0
(sinx− x cosx
x3
)cos
x
2dx
Solution : The given function can be written as
258 Unit IV - FOURIER TRANSFORM (F.T.)
f(x) =
{1− x2 if − 1 < x < 1
0 otherwise
F (s)=F [f(x)]=1√2π
∞∫−∞
f(x)eisxdx
=1√2π
1∫−1
(1− x2)(cossx + isinsx)dx
=1√2π
1∫−1
[(1− x2) cos sx+ i(1− x2) sin sx
]dx
=2√2π
1∫0
(1− x2) cos sxdx
[∵(1− x2
)cos sx is even and
(1− x2
)sin sx is odd
]=
√2
π
[(1−x2)sin sx
s−(−2x)
(− cos sx
s2
)+(−2)
(− sin sx
s3
)]10
=
√2
π
[(1− x2)
sin sx
s− 2x
cos sx
s2+ (2)
(sin sx
s3
)]=
√2
π
[−2 cos s
s2+
2 sin s
s3
]F (s) =2
√2
π
(sin s− s cos s
s3
)By inversion formula for Fourier transform
f(x) =1√2π
∞∫−∞
F (s)e−isxds
=1√2π
∞∫−∞
2.
√2
π
(sin s− s cos s
s3
)e−isxds
=2
π
∞∫−∞
[(sin s− s cos s
s3
)cos sx− i
(sin s− s cos s
s3
)sin sx
]ds
f(x) =4
π
∞∫0
(sin s− s cos s
s3
)cos sxds
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 259[∵
(sin s− s cos s
s3
)cos sx is even and
(sin s− s cos s
s3
)sin sx is odd
]∴
∞∫0
(sin s− s cos s
s3
)cos sxds =
π
4f(x)
Put x =1
2∞∫0
(sin s− s cos s
s3
)cos
s
2ds =
π
4
(1− 1
4
)=3π
16
Example 4.12. Find the Fourier inverse transform of e−s2
4
Solution : Solution: The inversion formula for Fourier transform is
f(x) =1√2π
∞∫−∞
F (s)e−isxds
=1√2π
∞∫−∞
e−s2
4 e−isxds
=1√2π
∞∫−∞
e−(
s2
4 +isx)ds
=1√2π
∞∫−∞
e−((
s2
)2+2(s2
)(ix)
)ds
=1√2π
∞∫−∞
e−((
s2
)2+2(s2
)(ix)+(ix)2−(ix)2
)ds
=1√2π
∞∫−∞
e−[(
s2+ix
)2+x2
]ds
=1√2π
∞∫−∞
e−(s2+ix
)2e−x2
ds
=e−x2
√2π
∞∫−∞
e−(s2+ix
)2ds
Puts
2+ ix = t⇒ ds = 2dt
260 Unit IV - FOURIER TRANSFORM (F.T.)
∴ f(x) =e−x2
√2π
∞∫−∞
e−t22dt
=
√2
πe−x2
∞∫−∞
e−t2dt
=
√2
πe−x2
2
∞∫0
e−t2dt
=
√2
πe−x2
2
√π
2
∵ ∞∫0
e−x2
dx =
√π
2
f(x) =
√2e−x2
Note : If the transform of a function f(x) is equal to f(s) [i.e.,F (f(x)) =f(s)] then f(x) is called self reciprocal.
Example 4.13. Find the Fourier transform of e−a2x2
. Hence prove that
e−x2
2 is self reciprocal with respect to the Fourier transform.
Solution : Given f(x) = e−a2x2
.The Fourier transform of f(x) is given by
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 261
F (s) = F [f(x)] =1√2π
∞∫−∞
f(x)eisxdx
=1√2π
∞∫−∞
e−a2x2
eisxdx
=1√2π
∞∫−∞
e−(a2x2−isx)dx
=1√2π
∞∫−∞
e−[(ax)2−2(ax)
(is2a
)]dx
=1√2π
∞∫−∞
e−[(ax)2−2(ax)
(is2a
)+(is2a
)2−( is2a
)2]dx
=1√2π
∞∫−∞
e−[(ax− is
2a
)2+ s2
4a2
]dx
=1√2π
∞∫−∞
e−(ax− is
2a
)2e−
s2
4a2 dx
=1√2πe−
s2
4a2
∞∫−∞
e−(ax− is
2a
)2dx
Put ax− is
2a= t⇒ dx =
dt
a
∴ F (s) =1√2πe−
s2
4a2
∞∫−∞
e−t2 dt
a
=1
a√2πe−
s2
4a2
∞∫−∞
e−t2dt
262 Unit IV - FOURIER TRANSFORM (F.T.)
=1
a√2πe−
s2
4a2 2
∞∫0
e−t2dt
=1
a√2πe−
s2
4a2 2
(√π
2
)∴ F (s) =
1
a√2e−
s2
4a2
F [f(x)] =1
a√2e−
s2
4a2
F(e−a2x2
)=
1
a√2e−
s2
4a2
[∵ f(x) = e−a2x2
]Setting a =
1√2
F(e−
x2
2
)=e−
s2
2
∴ f(x) =e−x2
2 is self reciprocal.
Example 4.14. * Find the Fourier Transform of e−x2
2 . (OR) Show that
e−x2
2 is self reciprocal.
Example 4.15. Find the complex Fourier transform of e−a|x|, a > 0.
Solution :
F (s) = F [f(x)] =1√2π
∞∫−∞
f(x)eisxdx
=1√2π
∞∫−∞
e−a|x|(cos sx+ i sin sx)dx
=1√2π
∞∫−∞
(e−a|x| cos sx+ ie−a|x| sin sx)dx
=1√2π
2
∞∫0
e−a|x| cos sxdx[∵ ea|x| sin sx is odd
]
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 263
=
√2
π
∞∫0
e−ax cos sxdx
=
√2
π
[e−ax
a2 + s2(−a cos sx+ s sin sx)
]∞0
=
√2
π
[a
s2 + a2
]Example 4.16. Find the Fourier integral representation of f(x) defined
as f(x)=
0, x<01
2, x=0
e−x, x>0
.
F (s)= 1√2π
1 + is
1 + s2, f(x)=
1
π
∞∫0
cos sx+s sin sx
1 + s2ds.
4.4 Parseval’s Identity for Fourier transform:
Suppose F [f(x)] = F (s)&F [g(x)] = G (s), then
1)
∞∫−∞
F [f(x)] .F [g(x)] ds =
∞∫−∞
f(x).g(x)dx If f(x) = g(x),
2)
∞∫−∞
|F [f(x)]|2 ds =∞∫
−∞
|f(x)|2 dx
Example 4.17. Show that the Fourier transform of
f(x) =
{a2 − x2, |x| < a0, |x| > a
as 2
√2
π
(sin as− as cos as
s3
). Hence deduce
(i)
∞∫0
(sin s− s cos s
s3
)ds =
π
4(ii)
∞∫0
(sin s− s cos s
s3
)2
ds =π
15.{
x = 0, a = 1 in (i),Parseval’s with a = 1 in (ii)
}
Example 4.18. Find the Fourier transform of f(x) =
{a− |x| , |x| < a0, |x| > a
and hence deduce that (i)
∞∫0
(sin t
t
)2
dt =π
2. (ii)
∞∫0
(sin t
t
)4
dt =π
3.
264 Unit IV - FOURIER TRANSFORM (F.T.)F (s) =
√2
π
1
s2
(2 sin2
as
2
), f (x) =
2
π
∞∫−∞
(sin as
2
s
)2
e−isxds,
x = 0, a = 2 in (i),Parseval’s with a = 2 in (ii)
4.5 Sine and Cosine transforms
4.5.1 Fourier Sine transform pair:
Fourier Sine transform:Fourier Sine transform of f(x) is
FS [f(x)] =
√2
π
∞∫0
f (x) sin sxdx
= a function of s = FS (s)
Inverse Fourier Sine transform:Inverse Fourier Sine transform of FS [f(x)] = FS [s] is
f (x) =
√2
π
∞∫0
FS [f(x)] sin sxds
4.5.2 Fourier Cosine transform pair:
Fourier Cosine transform:Fourier Cosine transform of f(x) is
FC [f(x)] =
√2
π
∞∫0
f (x) cos sxdx
= a function of s = FC (s)
Inverse Fourier Cosine transform:Inverse Fourier Sine transform of FC [f(x)] = FC [s] is
f (x) =
√2
π
∞∫0
FC [f(x)] cos sxds
4.5.3 Examples under Fourier Sine & Cosine Transform:
Example 4.19. Find the Fourier Cosine Transform of e−x, x ≥ 0.
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 265
Solution : The Fourier Cosine transform of f(x) is given by
Fc(s) = Fc [f(x)] =
√2
π
∞∫0
f(x) cos sxdx
=
√2
π
∞∫0
e−x cos sxdx
=
√2
π
[e−x
1 + s2(− cos sx+ s sin sx)
]∞0
=
√2
π
1
1 + s2
Example 4.20. Find the Fourier Cosine Transform of
f(x) =
{1 if 0 ≤ x ≤ 10 if x > 1
Solution : The Fourier Cosine transform of f(x) is given by
Fc(s) = Fc [f(x)] =
√2
π
∞∫0
f(x) cos sxdx
=
√2
π
1∫0
cos sxdx
=
√2
π
[sin sx
s
]10
=
√2
π
[sin s
s
]Example 4.21. Find the Fourier sine transform of f (x) = e−ax and
deduce
∞∫0
s sin sx
s2 + a2ds =
π
2e−ax.
{FS
[e−ax
]=
√2
π
(s
a2 + s2
)}
Example 4.22. * Find the Fourier cosine transform of f (x) = e−ax and
deduce
∞∫0
cos sx
s2 + a2ds =
π
2
e−ax
a.
{FS
[e−ax
]=
√2
π
(a
a2 + s2
)}
Example 4.23. Find the Fourier sine transform of f (x) =e−ax
xand hence
find
[e−ax − e−bx
x
].
266 Unit IV - FOURIER TRANSFORM (F.T.)FS [e
−ax] =√
2π tan
−1(sa
),
FS
[e−ax − e−bx
x
]=
√2
π
[tan−1
(sa
)− tan−1
(sb
)]
Example 4.24. * Find the Fourier cosine transform of f (x) =e−ax
xand
hence find
[e−ax − e−bx
x
].
FC [e−ax] = − 1√
2πlog(s2 + a2
),
FC
[e−ax − e−bx
x
]=
1√2π
log
(s2 + b2
s2 + a2
)
Example 4.25. Find the Fourier sine transform of xn−1 and deduce1√x
is self reciprocal under sine transform.{FS
[xn−1
]=
√2
π
n
snsin(nπ
2
), FS
[x−1/2
]=
1√s
}Example 4.26. * Find the Fourier cosine transform of xn−1 and deduce1√xis self reciprocal under cosine transform.{
FC
[xn−1
]=
√2
π
n
sncos(nπ
2
), FC
[x−1/2
]=
1√s
}Example 4.27. Find the Fourier cosine transform of x.{FC [x] = −
√2
π
(1
s2
)}
4.6 Convolution theorem
1. F [f (x) ∗ g (x)] = F [f (x)] .F [g (x)] = F (s) .G (s)
2. FS [f (x) ∗ g (x)] = FS [f (x)] .FS [g (x)] = FS (s) .GS (s)
3. FC [f (x) ∗ g (x)] = FC [f (x)] .FS [g (x)] = FC (s) .GC (s)
4.7 Parseval’s identity:
Parseval’s identity for Fourier transform:
a.
∞∫−∞
{F [f (x)] .F [g (x)]} ds =∞∫
−∞
f (x) .g (x)dx [If f(x) 6= g(x)]
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 267
b.
∞∫−∞
|F [f (x)]|2 ds =∞∫
−∞
|f (x)|2dx [If f(x) = g(x)]
Parseval’s identity for Fourier Sine transform:
a.
∞∫0
{FS [f (x)] .FS [g (x)]} ds =∞∫0
f (x) .g (x)dx [If f(x) 6= g(x)]
b.
∞∫0
|FS [f (x)]|2 ds =∞∫0
|f (x)|2dx [If f(x) = g(x)]
Parseval’s identity for Fourier Cosine transform:
a.
∞∫0
{FC [f (x)] .FC [g (x)]} ds =∞∫0
f (x) .g (x)dx [If f(x) 6= g(x)]
b.
∞∫0
|FC [f (x)]|2 ds =∞∫0
|f (x)|2dx [If f(x) = g(x)]
4.7.1 Examples under Parseval’s identity:
Example 4.28. Evaluate
∞∫0
dx
(x2 + a2) (x2 + b2).
{π
2ab (a+ b)
}
Example 4.29. Evaluate
∞∫0
x2dx
(x2 + a2) (x2 + b2).
{π
2 (a+ b)
}
Example 4.30. Evaluate
∞∫0
dx
(x2 + a2)2.
{ π
4a3
}Example 4.31.
Evaluate
∞∫0
x2dx
(x2 + a2)2.
{ π4a
}
4.8 Properties
4.8.1 Fourier Transform properties:
Property 4.1. Fourier Transform is linear i.e.,
268 Unit IV - FOURIER TRANSFORM (F.T.)
F [af(x) + bg(x)] = aF [f(x)] + bF [g(x)]
Proof :
F [af(x) + bg(x)] =1√2π
∞∫−∞
(af(x) + bg(x)) eisxdx
=a1√2π
∞∫−∞
f(x)eisxdx+ b1√2π
∞∫−∞
g(x)eisxdx
=aF [f(x)] + bF [g(x)]Note : Fourier cosine transform and Fourier sine transform are linear.
i.e.,
Fc[af(x) + bg(x)] =aFc[f(x)] + bFc[g(x)]
Fs[af(x) + bg(x)] = aFs[f(x)] + bFs[g(x)]
Property 4.2. (Shifting theorem) F [f(x− a)] = eiasF (s)
Proof : F [f(x− a)] =1
2π
∞∫−∞
f(x− a)eisxdx
Put x− a =t when x = −∞, t =−∞dx =dt when x = ∞, t = ∞
=1
2π
∞∫−∞
f(t)eis(a+t)dt
=eias√2π
∞∫−∞
f(t)eistdt
∴ F [f(x− a)] =eiasF (s)
Property 4.3. F [eiaxf(x)] = F (s+ a).
Proof :
F [eiaxf(x)] =1√2π
∞∫−∞
eiaxf(x)eisxdx
=1√2π
∞∫−∞
f(x)ei(s+a)xdx
=F (s+ a)
Property 4.4. (Change of scale) F [f(ax)] =1
|a|F(sa
)where a 6= 0.
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 269
Proof : Suppose a > 0
F [f(ax)] =1√2π
∞∫−∞
f(ax)eisxdx
Put ax =t when x = infty, t =infty
dx =dt
awhen x = ∞, t =∞
=1√2π
∞∫−∞
f(t)ei(sa
)tdt
a=
1
a
1√2π
∞∫−∞
f(t)ei(sa
)tdt
=1
aF(sa
)Similarly, if a < 0
F [f(ax)] =1√2π
−∞∫∞
f(t)ei(sa
)tdt
a
since when x =−∞, t = −∞when x =∞, t = ∞
=−1
a.
1√2π
∞∫−∞
f(t)ei(sa
)tdt
=−1
aF(sa
)Hence F [f(ax)] =
1
|a|F(sa
)Note : Fc[f(ax)] =
1
aFc
(sa
)Fs[f(ax)] =
1
aFs
(sa
), where a 6= 0.
1. Linearity property:F [af (x) + bg (x)] = aF [f (x)] + bF [g (x)] = aF (s) + bG (s)
2. Shifting theorem:F [f (x− a)] = eisaF [f (x)] = eisaF (s)
3. Shifting theorem in s:F[eiaxf (x)
]= [F (s)]s→s+a = F (s+ a)
4. Change of scale property:
F [f (ax)] =1
aF(sa
), a > 0
270 Unit IV - FOURIER TRANSFORM (F.T.)
F [f (ax)] = −1
aF(sa
), a < 0
F [f (ax)] =1
|a|F(sa
)5. F [xnf (x)] = (−i)n dn
dsn{F [f (x)]} = (−i)n
{dn
dsn[F (s)]
}6. F
[dn
dxnf (x)
]= (−is)n {F [f (x)]} = (−is)n [F (s)]
7. F [f (−x)] = F (−s)
8. F[f (x)
]= F (−s)
9. F[f (−x)
]= F (s)
10.Modulation property:
a. F [f (x) cos ax] =1
2{F (s− a) + F (s+ a)}
b. F [f (x) sin ax] =1
2{F (s− a)− F (s+ a)}
4.8.2 Fourier Sine & Cosine Transform properties:
11. Linearity property:
a. FS [af (x) + bg (x)] = aFS [f (x)] + bFs [g (x)]
b. FC [af (x) + bg (x)] = aFC [f (x)] + bFC [g (x)]
12. Modulation property:
a. FS [f (x) sin ax] =1
2{FC (s− a)− FC (s+ a)}
b. FS [f (x) cos ax] =1
2{FS (s− a) + FS (s+ a)}
c. FC [f (x) sin ax] =1
2{FS (a− s) + FS (s+ a)}
d. FC [f (x) cos ax] =1
2{FC (s− a) + FC (s+ a)}
13. Change of scale property:
a. FS [f (ax)] =1
aFS
(sa
), a > 0
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 271
b. FC [f (ax)] =1
aFC
(sa
), a > 0
14. a. FS [f′ (x)] = −sFS (s) ifx→ ∞ ⇒ f (x) = 0
b. FC [f ′ (x)] = −√
2
πf (0) + sFS (s) ifx→ ∞ ⇒ f (x) = 0
15. a. FS [xf (x)] = − d
ds{FC [f (x)]} = − d
ds{FC (s)}
b. FC [xf (x)] =d
ds{FS [f (x)]} = − d
ds{FS (s)}
4.8.3 Examples under Properties:
Example 4.32. Find Fourier sine transform of xe−ax.{2
√2
π
[as
(a2 + s2)2
]}
Example 4.33. Find Fourier cosine transform of xe−ax.{√2
π
[a2 − s2
(a2 + s2)2
]}.
Example 4.34. Find Fourier cosine transform of e−a2x2
and evaluateFourier sine transform of xe−a2x2
.{FC
[e−a2x2
]=e−s2/4a2
√2a
,FS
[xe−a2x2
]=se−s2/4a2
2√2a3
}
Example 4.35. Find f(x) if its sine transform ise−sa
sand hence find
F−1S
(1
s
). {
f (x) =
√2
πtan−1
(xa
), F−1
S
(1
s
)=
√π
2
}
Example 4.36. Find FS
(e−x), FC
(e−x), FC
(1
1 + x2
), FS
(x
1 + x2
).{√
2
π
s
s2 + a2,
√2
π
a
s2 + a2,
√π
2e−s,
√π
2e−s
}
272 Unit IV - FOURIER TRANSFORM (F.T.)
Example 4.37. Solve the integral equation
∞∫0
f (x) cos sxdx = e−s, show
that
∞∫0
cos sx
1 + s2ds =
π
2e−x.
{f (x) =
2
π
(1
1 + x2
), e−x
}
4.9 Assignment II[Fourier Transforms]
1. Find the Fourier integral representation of f(x) defined as
f(x) =
0, x < 012 , x = 0e−x, x > 0
.
2. Find the Fourier transform of the function f(x) defined by
f(x) =
{1− x2; if |x| < 10; if |x| ≥ 1
. Hence prove that
(i)
∞∫0
(sin s− s cos s
s3
)cos(s2
)ds =
3π
16
(ii)
∞∫0
(sin s− s cos s
s3
)2
ds =π
15.
3. Find F.T. of f(x) =
{1− |x| ; if |x| < 1,0; if |x| > 1,
and hence find the value of
∞∫0
(sin t
t
)dt and
∞∫0
(sin t
t
)4
dt
4. Find the Fourier transform of e−a2x2
. Hence prove e−x2/2 is selfreciprocal.
5. Find the Fourier sine and cosine transform of
f(x) =
x; 0 < x < 12− x; 1 < x < 20; x > 2
.
6. Prove that e−x2/2 is self reciprocal under Fourier cosine transform.
7. Find the Fourier sine and cosine transform of xn−1 and hence prove1√xis self reciprocal under Fourier sine and cosine transforms.
MA6351 Transforms and Partial Differential Equations by K A Niranjan Kumar 273
8. Find the Fourier sine transform of e−ax and hence evaluate Fouriercosine transforms of xe−ax and e−ax sin ax.
9. Find F.S.T. and F.C.T. of e−ax, a > 0. Hence evaluate
∞∫0
x2
(a2 + x2)2dx
and
∞∫0
dx
(x2 + a2) (x2 + b2)
10. State and prove convolution theorem and Parseval’s identity forFourier transforms.