# Obtaining generating function via Fourier transform

Series coefficient for a function can be obtained via Fourier transform:

$$f^{(s)}(0)=\frac{1}{2\pi}\int_{-\infty}^{+\infty} (- i \omega)^s \int_{-\infty}^{+\infty}f(t)e^{i\omega t}dt \, d\omega$$

What is the inverse operator, how to get the original function $f(x)$ if series coefficient $f^{(s)}(0)$ is known, via Fourier transform?

• Why would you need a Fourier transform to get $f(x)$? Constructing a function from its derivatives at a point is just forming the Taylor series... – Semiclassical Oct 15 '14 at 2:56
• @Semiclassical yes but I wonder whether it is possible using Fourier transform. – Anixx Oct 15 '14 at 3:06