Does a Fourier transform exist for this (smooth) f.?

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SUMMARY

The Fourier transform of the function \( e^{-x^2}\cos(e^{x^2}) \) does not have a known closed form, as confirmed by discussions among users familiar with advanced mathematical concepts. The function is continuously differentiable across all real numbers and approaches zero at infinity, although its derivative becomes unbounded. Given that Mathematica lacks an algorithm to compute this transform, it is unlikely that a simple solution exists.

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rachmaninoff
[tex]e^{-x^2}\cos \left( e^{x^2} \right)[/tex]

Mathematica doesn't have an algorithm for it, does a closed form exist for the Fourier transform? It's continuously differentiable on all intervals in R, and it converges to zero at the infinities (the derivative blows up there).
 
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rachmaninoff said:
[tex]e^{-x^2}\cos \left( e^{x^2} \right)[/tex]

Mathematica doesn't have an algorithm for it, does a closed form exist for the Fourier transform? It's continuously differentiable on all intervals in R, and it converges to zero at the infinities (the derivative blows up there).
Well after looking at it for a few minutes on paper, I'd say I certainly doubt you'll get anything nice out of that. Also if Mathematica can't do it, I doubt I'll find an answer as well.
 

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