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Given $F(t) = \mathcal{F}\{f(x)\}$ is the Fourier transform of $f$, how can one express $\mathcal{F}\left\{\dfrac{1}{f(x)}\right\}$ in terms of $F(t)$?

EDIT: To be more concrete, I want to compute $\mathcal{F}\left\{\dfrac{1}{f(x)}\dfrac{dg(x)}{dx}\right\}$. So far I have $\mathcal{F}\left\{\dfrac{1}{f(x)}\dfrac{dg(x)}{dx}\right\} = \mathcal{F}\left\{\dfrac{1}{f(x)}\right\}\star\mathcal{F}\left\{\dfrac{dg(x)}{dx}\right\}$, where $\star$ denotes convolution. The point of this computation is to use the resulting expression in a numerical method (spectral Galerkin), so $f$ and $g$ do not have a specific analytical form.

astay13
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  • If you have a more concrete computation to perform it is better you post it. The Fourier transform doesn't behave nicely with respect to multiplication. – Alamos Apr 28 '15 at 04:07
  • @Alamos, see edit. Unfortunately I can't get more concrete than that. – astay13 Apr 28 '15 at 04:23
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    Such questions get asked once in a while... unfortunately, nonlinear transforms like changing $f$ to $1/f$ are completely intractable on the Fourier side. There is no useful answer. –  Apr 28 '15 at 04:31

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