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For a problem I am trying to solve, I am required to calculate the Fourier transform of $e^{i f(x)}$. However, it so happens that I do not have access to $f(x)$ but only to its Fourier transform coefficients $\mathcal{F}f(\omega)$. Are you aware of any way to define $\mathcal{F}(e^{i f(x)})$ in terms of $\mathcal{F}f$?

Thank you all in advance!

David Romero
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  • Hint: does $e^{i f(x)}$ satisfy conditions to have Fourier transform? – Yalikesifulei Sep 12 '22 at 18:07
  • I think so. I would think that $e^{if(x)}$ is a sort of oscillatory system since $e^{if(x)}{=}\cos(f(x)) + i \sin(f(x))$. – David Romero Sep 13 '22 at 06:48
  • Hint: Fourier transform is typically defined for functions that absolutely integrable over real line. Does it hold for $e^{i f(x)}$? – Yalikesifulei Sep 13 '22 at 09:19
  • From [this question](https://math.stackexchange.com/questions/171462/fourier-transform-of-function-composition) I would argue that it is possible. What I have tried so far is representing $e^{i f(x)}$ as a series $\sum_{n=0}^{\inf}\frac{((i f(x))^{n}}{n!}$, but this leads me to a series of self-convolutions on the Fourier domain, which is impractical to work with. – David Romero Sep 13 '22 at 13:48

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