Fourier transform of $F(x)$ $$F(x)=\int^{\infty}_{-\infty} e^{-i \xi x}f(x)\ dx$$
Fourier transform of $F$:
$$\int^{\infty}_{-\infty}\left[\int^{\infty}_{-\infty} f(\xi)e^{-x^2\xi^2}\ d\xi\right]\ dx$$
Is that the way to proceed? If yes, how do I continue and if No, show me how I should proceed.
 A: It looks like you might want to perform an inverse Fourier transform, if so, then you should note that your current scheme puts you off by a factor of $ 2 \pi$. You should probably also look this up in a trusted resource or failing that wikipedia.
If you are literally looking to do a fourier transform of a fourier transform then you should probably slow down and write things out:
Consider the fourier transform of a function $f(x)$ to be:
\begin{equation}
\tilde{f}(x') = \int_{-\infty} ^\infty e^{-i x' x}f(x)dx
\end{equation}
then the fourier transform of $\tilde{f}(k)$ would naturally be:
\begin{equation}
\tilde{\tilde{f}}(x'')=\int_{-\infty} ^\infty e^{-i x'' x'}\tilde{f}(x')dx'=\int_{-\infty} ^\infty\int_{-\infty} ^\infty e^{-i x' x}e^{-i x'' x'}f(x)dx dx' = \int_{-\infty} ^\infty\int_{-\infty} ^\infty e^{-i (x''+x) x'}f(x)dxdx'
\end{equation}
Someone correct me if I'm wrong but I believe this leads to:
\begin{eqnarray}
\tilde{\tilde{f}}(x'')&=&\int_{-\infty} ^\infty\int_{-\infty} ^\infty e^{-i (x''+x) x'}f(x)dxdx'\\
&=&\int_{-\infty} ^\infty\delta(x''+x) f(x)dx\\
&=&f(-x'')
\end{eqnarray}
Key differences:


*

*When you multiply exponentials you add the exponent, not multiply

*When performing a fourier transform of a fourier transform you need to introduce a new variable, otherwise you get a number, not a function.

A: $=2 \pi [\int^{\infty}_{-\infty} F(x) e^{ix(-\xi)}$ $dx]=2\pi$ $f(-\xi)$
