Statement of Fourier Integral Theorem

Unit-IV

FOURIER TRANSFORMS

Statement of Fourier integral theorem - Fourier transform pair Fourier sine and cosine transforms - Properties - Transforms of simple functions - Convolution theorem - Parseval's identity.

INTRODUCTION

Integral transforms are used in the solution of partial differential equations. The choice of a particular transform to be used for the solution of a differential equation depends upon the nature of the boundary conditions of the equation and the facility with which the transform F(s) can be converted to give f(x).

1. STATEMENT OF FOURIER INTEGRAL THEOREM

4.1.a Fourier integral theorem (without proof)

If f(x) is piece-wise continuously differentiable and absolutely integrable in (-∞, ∞), then

Equation (1) can be re-written as

This is known as Fourier integral theorem or Fourier integral formula.

(OR)

Let us assume the following conditions on a function f (x)

1. f (x) is piece-wise continuous in any finite interval (a, b)

2. is convergent.

Then the Fourier integral theorem states that

The double integral in the right hand side is known as a Fourier integral expansion of f (x)

(OR)

If f (x) is a function defined in (-l, l) satisfying Dirichlet's conditions, then

The double integral in the right hand side is known as Fourier integral to represent ƒ (x)

Note: We assume the following conditions on f(x)

(i) f(x) is defined as single-valued except at finite points in (-l, l)

(ii) f(x) is periodic outside (-l, l) with period 2l.

(iii) f(x) and f'(x) are sectionally continuous in (-l, l)

(iv) converges

i.e., f(x) is absolutely integrable in (-∞, ∞)

I.(a) Problems based on Fourier integral theorem :

Example 4.1.a(1): Show that f(x) = 1, 0 < x < ∞ cannot be represented by a Fourier integral.

Solution :

Hence, f (x) = 1 cannot be represented by a Fourier integral.

Example 4.1.a(2) :

Solution: We know that, the Fourier integral formula for ƒ (x) is

Aliter :

We know that, the Fourier integral formula for f (x) is

Example 4.1.a(3) : Express the function as a Fourier integral. Hence evaluate and find the value of

Solution: We know that, the Fourier integral formula for ƒ (x) is

Example 4.1.a(4) Find the Fourier integral of the function

Solution: We know that, the Fourie integral formula for f(x) is

Example 4.1.a(5) Find the Fourier integral of the function

Verify the representation directly at the point x = 0.

Solution: We know that, the Fourier integral formula for f(x) is

Verification :

Put x = 0 in (4), we get

Hence, verified.

4.1.b Complex form of the Fourier integrals.

The Fourier integral formula for f(x) is

because cos [λ (t − x)] is an even function of λ.

Also since sin [λ (t − x)] is an odd function of λ.

which is the complex form of the Fourier integral.

FOURIER SINE AND COSINE INTEGRALS

Fourier sine and cosine integrals

Proof: We know that, the Fourier integral theorem is

Note I :

Equation (2) can be re-written as

Note II :

Equation (3) can be re-written as

I.(b) Problems based on Fourier cosine and Fourier sine integrals

Example 4.1.b(1) : Find Fourier cosine integral of the function e-ax. Hence deduce the value of the integral

Solution: We know that, the Fourier Cosine integral of ƒ (x) is given by

Example 4.1.b(2): Using Fourier integral formula, show that

Solution: We know that, the Fourier Cosine integral of f (x) is given by

Here, f (x) = e-x cos x

Example 4.1.b(3) Express f(x) : as a Fourier sine integral and hence evaluate

Solution :

We know that, the Fourier sine integral of ƒ (x) is given by

Note: If x = π is a finite point of discontinuity of f(x), then

Example 4.1.b(4): Using the Fourier integral representation show that

Solution :

We know that, the Fourier sine integral of ƒ (x) is given by

Example 4.1.b(5) Using Fourier integral formula, prove that

Solution: We know that, the Fourier sine integral of ƒ (x) is given by

Example 4.1.b(6): Applying the Fourier sine integral formula to the function

Solution: We know that, the Fourier sine integral of ƒ (x) is given by