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MA 201, Mathematics III, July-November 2018,
Fourier Integral and Transforms
Lecture 16
Lecture 16 MA 201, PDE (2018) 1 / 20
Fourier Integral
We already know that Fourier series and orthogonal expansions can be used as
tools
to solve boundary value problems over bounded regions such as intervals, rectangles,
disks and cylinders.
But we can understand that the modelling of certain physical phenomena will
give rise naturally
to boundary value problems over unbounded regions.
For example,
to describe the temperature distribution in a very long insulated wire, we can suppose
that the length of the wire is infinite, which gives rise to a boundary value problem
over an infinite line.
To solve this type of problem we will generalize the notion of Fourier series by
developing Fourier transform.
Lecture 16 MA 201, PDE (2018) 2 / 20
Fourier Integral
Recall that for a piece-wise C1 and continuous function f(t), its Fourier series in
the interval (−L, L) satisfies
f(t) =A0
2+
∞∑
n=1
[
An cosnπt
L+Bn sin
nπt
L
]
where
An =1
L
∫ L
−Lf(t) cos
nπt
Ldt, n = 0, 1, 2, 3, . . . ,
Bn =1
L
∫ L
−Lf(t) sin
nπt
Ldt, n = 1, 2, 3, . . . .
We know
cos
(
nπt
L
)
=1
2[exp(inπt/L) + exp(−inπt/L)],
sin
(
nπt
L
)
=1
2i[exp(inπt/L)− exp(−inπt/L)].
Lecture 16 MA 201, PDE (2018) 3 / 20
Fourier Integral
Putting back these in f(t), An, Bn:
f(t) =A0
2+
∞∑
n=1
[
An
(
1
2[exp(inπt/L) + exp(−inπt/L)]
)
+ Bn
(
1
2i[exp(inπt/L)− exp(−inπt/L)]
)]
We write
Cn =1
2(An − iBn) , Cn =
1
2(An + iBn) = C−n, n = 0, 1, 2, . . .
We obtain Fourier series for f(t) in complex form as
f(t) =
∞∑
n=−∞
Cneinπt/L, (1)
where
Cn =1
2L
∫ L
−Lf(τ)e−inπτ/L dτ, n = 0,±1,±2, . . . (2)
Lecture 16 MA 201, PDE (2018) 4 / 20
Fourier Integral
Substituting (2) into (1):
f(t) =
∞∑
n=−∞
[
1
2π
∫ L
−Lf(τ)e−inπτ/L dτ
]
einπt/L(π
L
)
(3)
Define the frequency of the general term by
σn =nπ
L, (4)
and the difference in frequencies between successive terms by
∆σ = σn+1 − σn =(n+ 1)π
L− nπ
L=
π
L. (5)
Then (3) reduces to
f(t) =
∞∑
n=−∞
[
eiσnt
2π
∫ L
−Lf(τ)e−iσnτ dτ
]
∆σ
=∞∑
n=−∞
F (σn)∆σ, F (σ) =
[
eiσt
2π
∫ L
−Lf(τ)e−iστ dτ
]
. (6)
Lecture 16 MA 201, PDE (2018) 5 / 20
Fourier Integral
If we plot F (σ) against σ, we can clearly see that the sum∞∑
n=1
F (σn)∆σ, ∆σ =π
L(7)
is an approximation to the area under the curve y = F (σ)
Lecture 16 MA 201, PDE (2018) 6 / 20
Fourier Integral
Thus as, L → ∞, (6) can be written as
limL→∞
f(t) = f(t) = limL→∞
∞∑
n=−∞
F (σn)∆σ
=
∫
∞
−∞
F (σ)dσ =
∫
∞
−∞
[
1
2πeiσt
∫
∞
−∞
f(τ)e−iστ dτ
]
dσ, (8)
which is known as the complex form of Fourier integral.
Note that Fourier integral is a valid representation of the non-periodic function, a
function whose period goes to infinity, f(t) provided that
(a) in every finite interval f(t) is defined and is piecewise continuous.
(b) the improper integral∫
∞
−∞|f(t)|dt exists.
Lecture 16 MA 201, PDE (2018) 7 / 20
Fourier Integral
Now proceed to obtain the trigonometric representation of the Fourier integral:
Equation (8) can, after writing e−iσ(τ−t) = cos σ(τ − t)− i sin σ(τ − t), be expressed as
f(t) =1
2π
∫
∞
−∞
f(τ)
(∫
∞
−∞
e−iσ(τ−t)dσ
)
dτ
=1
2π
∫
∞
−∞
f(τ)
(∫
∞
−∞
cos σ(τ − t)dσ
)
dτ
(Since sinσ(τ − t) is an odd function of σ)
Also note that cosσ(τ − t) is an even function of σ, therefore
f(t) =1
2π
∫
∞
−∞
f(τ)
(∫
∞
−∞
e−iσ(τ−t)dσ
)
dτ
=1
π
∫
∞
−∞
f(τ)
(∫
∞
0cos σ(τ − t)dσ
)
dτ. (9)
This way the integral has a representation with cosine terms only.
Lecture 16 MA 201, PDE (2018) 8 / 20
Fourier Integral
Fourier cosine and sine integrals:
Expanding cos σ(τ − t) in (9)
f(t) =1
π
∫
∞
0
[∫ 0
−∞
f(τ) cos στ cos σt dτ +
∫
∞
0f(τ) cos στ cos σt dτ
]
dσ
+1
π
∫
∞
0
[∫ 0
−∞
f(τ) sinστ sinσt dτ +
∫
∞
0f(τ) sinστ sinσt dτ
]
dσ. (10)
Changing τ to −τ in the first and third integrals:
f(t) =1
π
∫
∞
0
[∫
∞
0f(−τ) cos στ cos σt dτ +
∫
∞
0f(τ) cos στ cos σt dτ
]
dσ
+1
π
∫
∞
0
[∫
∞
0f(−τ)(− sinστ) sin σt dτ +
∫
∞
0f(τ) sinστ sinσt dτ
]
dσ.
(11)
Remark: Above integral representation does not demand that the function
should be defined in (−∞,∞).
So, we can think of defining Fourier transform of a function defined in (0,∞).
Lecture 16 MA 201, PDE (2018) 9 / 20
Fourier Integral
For a function f defined in (0,∞), we can use even or odd extension to define
Fourier Transform.
For even extension, Fourier transform of f is defined as
f(t) =2
π
∫
∞
0
∫
∞
0f(τ) cos στ cos σt dτ dσ. (12)
Equation (12) is called the Fourier cosine integral of f .
For odd extension
f(t) =2
π
∫
∞
0
∫
∞
0f(τ) sinστ sinσt dτ dσ. (13)
Equation (13) is called the Fourier sine integral of f .
With these definitions of Fourier integrals,
we are in a position to define Fourier transforms including sine and cosine transforms.
Lecture 16 MA 201, PDE (2018) 10 / 20
Complex Fourier Transform
Recalling equation (8):
f(t) =
∫
∞
−∞
[
1
2πeiσt
∫
∞
−∞
f(τ)e−iστ dτ
]
dσ,
we are now in a position to define the complex Fourier transform pair.
The Fourier transform of a function f ∈ L1(R) is defined by
F{f(t)} = g(σ) =1√2π
∫
∞
−∞
e−iστf(τ) dτ. (14)
The inverse Fourier transform is defined as
f(t) = F−1{g(σ)} =1√2π
∫
∞
−∞
g(σ)eiσt dσ. (15)
Here g(σ) is in the frequency domain and f(t) in the time domain.
Lecture 16 MA 201, PDE (2018) 11 / 20
Fourier cosine transform
Recalling equation (12):
f(t) =2
π
∫
∞
0
∫
∞
0f(τ) cos στ cos σt dτ dσ,
we can define the Fourier cosine transform of f as
Fc{f(t)} = gc(σ) =
√
2
π
∫
∞
0f(τ) cos στ dτ. (16)
The inverse Fourier cosine transform is defined as
f(t) = F−1c {gc(σ)} =
√
2
π
∫
∞
0gc(σ) cos σt dσ. (17)
Lecture 16 MA 201, PDE (2018) 12 / 20
Fourier sine transform
Recalling equation (13):
f(t) =2
π
∫
∞
0
∫
∞
0f(τ) sinστ sinσt dτ dσ,
we can define the Fourier sine transform of f as
Fs{f(t)} = gs(σ) =
√
2
π
∫
∞
0f(τ) sinστ dτ. (18)
The inverse Fourier sine transform is defined as
f(t) = F−1s {gs(σ)} =
√
2
π
∫
∞
0gs(σ) sin σt dσ. (19)
Lecture 16 MA 201, PDE (2018) 13 / 20
Some Properties of Fourier transform
Linear Property:
If F{f1(t)} = g1(σ),F{f2(t)} = g2(σ), then
F{c1f1(t)± c2f2(t)} = c1F{f1(t)} ± c2F{f2(t)} = c1g1(σ)± c2g2(σ),
where c1 and c2 are constants.
Theorem I:
If F{f(t)} = g(σ), then
F{f(t− a)} = e−iσag(σ)
Proof:
By definition,
F{f(t− a)} =1√2π
∫
∞
−∞
e−iστf(τ − a)dτ
=1√2π
∫
∞
−∞
e−iσ(ξ+a)f(ξ)dξ, by taking τ − a = ξ
= e−iσag(σ)
�
This is known as shifting property.
Lecture 16 MA 201, PDE (2018) 14 / 20
Some Properties of Fourier transform
Theorem II:
If F{f(t)} = g(σ), then
F{f(at)} =1
ag(σ/a)
Proof:
By definition,
F{f(at)} =1√2π
∫
∞
−∞
e−iστf(aτ)dτ
=1√2π
∫
∞
−∞
e−i(σ/a)ξf(ξ)dξ/a, by taking aτ = ξ
=1
ag(σ/a)
�
This is known as scaling property.
Lecture 16 MA 201, PDE (2018) 15 / 20
Some Properties of Fourier transform
Theorem III:
If F{f(t)} = g(σ), then
F{eiatf(t)} = g(σ − a)
Proof:
By definition,
F{eiatf(t)} =1√2π
∫
∞
−∞
e−iστ eiaτf(τ)dτ
=1√2π
∫
∞
−∞
e−i(σ−a)τ f(τ)dτ
= g(σ − a)
�
This is known as translation property.
Lecture 16 MA 201, PDE (2018) 16 / 20
Some Properties of Fourier transform
Theorem III:
If F{f(t)} = g(σ), f(t) is continuously differentiable and f(t) → 0 as |t| → ∞, then
F{f ′(t)} = iσg(σ).
Proof:
By definition,
F{f ′(t)} =1√2π
∫
∞
−∞
e−iστf ′(τ)dτ
=1√2π
{[f(τ)e−iστ ]∞−∞
+ iσ
∫
∞
−∞
e−iστf(τ)dτ}
= iσg(σ).
�
Theorem IV (Extension of Theorem III):
If f(t) is continuously n-times differentiable and f (k)(t) → 0 as |t| → ∞ for
k = 1, 2, . . . , (n − 1), then the Fourier transform of the n-th derivative of f(t) is given
by
F{f (n)(t)} = (iσ)nF{f(t)} = (iσ)ng(σ).
�
Lecture 16 MA 201, PDE (2018) 17 / 20
Some examples of Fourier integrals
Example
Find the Fourier integral representation of the following non-periodic function
f(t) =
{
eat, t < 0,
e−at, t > 0.a > 0
Solution :-The Fourier transform of f(t):
F{f(t)} = g(σ)
=1√2π
∫
∞
−∞
f(τ)e−iστ dτ
=1√2π
∫ 0
−∞
e(a−iσ)τ dτ +
∫
∞
0e−(a+iσ)τ dτ
=1√2π
2a
a2 + σ2.
Taking the inverse
f(t) = F−1{g(σ)} =1√2π
∫
∞
−∞
g(σ)eiσt dσ
=2a
π
∫
∞
0
cos σt
a2 + σ2dσ.
Lecture 16 MA 201, PDE (2018) 18 / 20
Convolution
A convolution is an integral that expresses the amount of overlap of one function f2 as
it is shifted over another function f1.
It therefore “blends” one function with another.
In other words, the output which produces a third function can be viewed as a
modified version of one of the original functions.
Definition:
The convolution of two functions f1(t) and f2(t), −∞ < t < ∞, is defined as
f1(t)∗f2(t) =1√2π
∫
∞
−∞
f1(τ)f2(t−τ) dτ =1√2π
∫
∞
−∞
f1(t−ξ)f2(ξ) dξ = f2(t)∗f1(t),
provided the integral exists for each t.
This is known as the convolution integral.
Lecture 16 MA 201, PDE (2018) 19 / 20
Convolution: Some algebraic properties
Property I (Commutative):
f1 ∗ f2 = f2 ∗ f1
Property II (Associative):
f1 ∗ (f2 ∗ f3) = (f1 ∗ f2) ∗ f3
Property III (Distributive):
(αf1 + βf2) ∗ f3 = α(f1 ∗ f3) + β(f2 ∗ f3)
Lecture 16 MA 201, PDE (2018) 20 / 20
