# Thread: Fourier Integral

1. ## Fourier Integral

Hi! There was a first part to this question to find the fourier series repesentation, I can do that fine but then I cant seem to get around this integral. I am studying for exams tomorrow so I'm going through past papers and this question comes up a lot so if anyone could help it would be really helpful! Thanks

The Fourier transforms of the function f(x) are given by
∞ ∞
A(λ) = ∫ f(x) cosλx dx, B(λ) = ∫ f(x) sin λx dx
-∞ -∞

with the Fourier Integral representation of f(x) given by

f(x) = 1/pi ∫ A(λ) cosλx dx + B(λ)sin λx dx
0

The non-periodic function f is defined by

f(x) =1, −1 < x < 1
0, otherwise.
Obtain a Fourier integral representation for f(x).

2. ## Re: Fourier Integral

Think I finally got it.. Anyone able to tell me if it's right?

A(λ) = ∫(-∞ to ∞) f(x) cos(λx) dx
.......= ∫(-∞ to -1) f(x) cos(λx) dx + ∫(-1 to 1) f(x) cos(λx) dx + ∫(1 to ∞) f(x) cos(λx) dx
.......= ∫(-∞ to -1) 0 * cos(λx) dx + ∫(-1 to 1) 1 * cos(λx) dx + ∫(1 to ∞) 0 * cos(λx) dx
.......= ∫(x = -1 to 1) cos(λx) dx
.......= (1/λ) sin(λx) {for x = -1 to 1}
.......= 2 sin(λ)/λ.

Similarly,
B(λ) = ∫(-∞ to ∞) f(x) sin(λx) dx
.......= ∫(x = -1 to 1) sin(λx) dx
.......= (-1/λ) cos(λx) {for x = -1 to 1}
.......= 0.

Hence, f(x) = (1/π) ∫(x = 0 to ∞) [2 sin(λ)/λ] cos(λx) dx.