Fourier transform of $F(x)$

Question

$$F(x)=\int^{\infty}_{-\infty} e^{-i \xi x}f(x)\ dx$$

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.

Answer 1

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:

1. When you multiply exponentials you add the exponent, not multiply
2. When performing a fourier transform of a fourier transform you need to introduce a new variable, otherwise you get a number, not a function.

Answer 2

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