Fourier transform of $\frac{1}{f}$

Question

There dosnt seem to be any place in which $\mathcal{F}(\frac{1}{f}(x))(n)$ is being computed nor talking about its relation to $\mathcal{F}(f(x))(n)$. Prodcuts looks like they are easy to handle but is there somthing making this harder?

Answer 1

The main problem is that if $f \in L^1,$ (or $L^2$) then $f(x)$ must more-or-less tend to $0$ as $|x| \to \infty,$ which implies that $1/f(x)$ more-or-less tends to $\infty$ so $\frac1f \not\in L^1$ (or $L^2$).

Using distribution we can get away from this. For example, $\mathcal{F}\{x\} = 2\pi i\delta'(\xi)$ and $\mathcal{F}\{\frac1x\} = -i\pi \operatorname{sign}(\xi)$. One has $\delta'*\operatorname{sign}=\delta,$ but generally it's not easy to find the convolutional inverse of a function or distribution.