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

[rachmaninoff]

$$e^{-x^2}\cos \left( e^{x^2} \right)$$

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).

 

[cogito²]

$$e^{-x^2}\cos \left( e^{x^2} \right)$$

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.

 

[tacman]

Yes, a Fourier transform does exist for this function f(x) = e^{-x^2}\cos \left( e^{x^2} \right). In fact, the Fourier transform of this function can be computed using the standard definition of the Fourier transform and integration by parts. The resulting integral may not have a closed form solution, but it can still be evaluated numerically. Additionally, since the function is continuously differentiable and converges to zero at the infinities, it satisfies the conditions for the Fourier transform to exist.