- #1

albega

- 75

- 0

## Homework Statement

With the Fourier transform of f(x) defined as F(k)=1/√(2π)∫

_{-∞}

^{∞}dxf(x)e

^{-ikx}and a convolution of g(x) and h(x) defined by f(x)=[g*h](x)=∫

_{-∞}

^{∞}h(x-y)g(y)dy show that the Fourier transform of f(x) equals √(2π)H(k)G(k).

## Homework Equations

In problem

## The Attempt at a Solution

So I understand the general idea, but I'm stuck on a few little issues. I'm going to drop the integral limits from now on.

F(k)=1/√(2π)∫dx{∫h(x-y)g(y)dy}e

^{-ikx}(we can put e

^{-ikx}into the integral over y, allowing the next step)

Now the idea is to let z=x-y (we want to get h(z) which will lead us to the form of a Fourier transform for h). However all proofs seems to say this means dx=dz to give

F(k)=1/√(2π)∫dz{∫h(z)g(y)e

^{-iky}e

^{-ikz}dy}

The question is, why isn't dz=d(x-y)=dx-dy used (given no sources actually discuss this I gather I am being very silly). I also seem

Then

F(k)=√(2π)[1/√(2π)∫h(z)e

^{-ikz}dz][1/√(2π)∫g(y)e

^{-iky}dy]

=√(2π)H(k)G(k)

If anybody could explain that it would be great, thanks :)