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If for a given function $f(x)$, the Fourier transform is $\hat{f}(p)$; Is there a way to find the Fourier transform of $f(x)^{-1}$ in terms of $\hat{f}(p)$?

arsal
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    The title asks about the "inverse function", which is $f^{-1}(x)$. In the post you ask about $f(x)^{-1}$, which is the reciprocal, not the inverse function. So it's hard to know what you mean. Luckily the answer is the same for both: No. – David C. Ullrich Aug 20 '15 at 01:08
  • i did indeed ask about the reciprocal. There was a recent paper(arXiv:1010.5060v1) where the authors have done this for Mellin transform, e.g, if the mellin transform of $f(x)=\sum_n a_n x^n$ is $M[f(x)]=F(u)$ then the mellin transform of $1/f(x)$ is $M[1/f(x)]=F(u)\Gamma(u)\Gamma(n-u)$. That's why i was curious.... – arsal Aug 20 '15 at 04:30
  • @DavidC.Ullrich – arsal Aug 20 '15 at 05:09
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    There are some facts known about it. For example, the Wiener algebra is closed under taking reciprocals. (Fourier series) – Eric Thoma Aug 21 '15 at 01:58
  • A solution to this would solve: https://math.stackexchange.com/questions/4283778/given-product-and-convolution-of-pair-of-functions-can-you-find-original-pair – Sidharth Ghoshal Oct 22 '21 at 17:13
  • What is $n$ in $M[1/f(x)]=F(u)\Gamma(u)\Gamma(n-u)$? Is there really a sum over $n$ in this result? – Steven Clark Jan 30 '24 at 18:53

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