# Fourier transforms and their inverse

• Jul 2nd 2012, 07:36 AM
Alexrey
Fourier transforms and their inverse
I'm having a little bit of trouble that I know should be trivial but for some reason I cannot get it. As a simple example consider $f(x)&=&x$. I want to perform a Fourier transform on $f(x)$ and then perform an inverse Fourier transform on this transformed function to get back to the original function. Now, the Fourier transform is given by
$\tilde{f}(\omega)&=&\frac{1}{\sqrt{2\pi}}\int f(x)e^{i\omega x}dx$,
and the inverse Fourier transform is
$f(x)&=&\frac{1}{\sqrt{2\pi}}\int\tilde{f}(\omega)e ^{-i\omega x}d\omega$.
What I did was the following: The Fourier transform of $f(x)&=&x$ is given by
$\tilde{f}(\omega)&=&\frac{1}{\sqrt{2\pi}}\int xe^{i\omega x}dx$,
then performing an inverse Fourier transform on this new function yields
$f(x)=\frac{1}{\sqrt{2\pi}}\int\left[\frac{1}{\sqrt{2\pi}}\int xe^{i\omega x}e^{-i\omega x}dxd\omega\right]$.
Now, the $e$ terms become 1 and I'm assuming $dx$ cancels with $d\omega$, so we are left with
$f(x)=\frac{1}{2\pi}\int\int x$,
which is clearly not the original function I started with. So, does anyone know what rookie error I made to go so horribly wrong?