Breadcrumbs Section. Click here to navigate to respective pages.

Chapter

Chapter

# INTEGRAL TRANSFORMS

DOI link for INTEGRAL TRANSFORMS

INTEGRAL TRANSFORMS book

# INTEGRAL TRANSFORMS

DOI link for INTEGRAL TRANSFORMS

INTEGRAL TRANSFORMS book

Click here to navigate to parent product.

## ABSTRACT

The function f(x), -L::; x::; L, can be represented by a Fourier series as follows:

I 00 nn nn 2[f(x +O)+f(x-0)] = a0 + ~)an cos(L x)+ bn sin(L x)]

n =I where:

and

ao = 2 1L Jf(~)d~

I I J " I J nn 2[f(x+0)+f(x-0)]= 2L f(~) d~+ L.J L f(~)cos(L(~-x))d~ -L n = 1 -L

and

then the integrals can be rewritten as: +L Cf) +L

Define the integral to equal F(un), i.e.: +L

n = 1 0 Since the function is absolutely integrable, then the first term vanishes because:

+L Lim - 1-J f(x)dx ~ 0 and F(u) converges L~oo 2L

(7.1)

Iff(x) = f(-x) for -Cf:J < x < Cf:J or, iff(x) = 0 for the range -Cf:J < x < 0 where one can choose f(x) = f(-x) for the range -Cf:J < x < 0, then the second integral in eq. (7.1) vanishes and the integral representation can be rewritten as:

Define the Fourier cosine transform as: 00

Fe (u) = J f(~) cos(u~) d~ 0

(7.2)

then, the inverse Fourier cosine transforms becomes: 00

Iff(x) = -f(-x) in -oo < x < oo or iff(x) = 0 in the range -oo < x < 0, where one can choose f(x) = -f( -x) in the range -oo < x < 0, then the first integral of eq. (7.1) vanishes and:

F5 (u) = J f(~)sin(u~)d~ 0

then the inverse Fourier sine transform becomes: 00

2 J . f(x) =; F5 (u)sm(ux)du x~O 0

where F5(u) is an odd function ofu and sin (ux) is the kernel ofthe Fourier sine transform.