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Consider the Fourier transform of $\exp(-z^k)$ where $k$ is a positive integer. As the function is analytic, I expect it to have exponential decay at infinity. Is there some known theorem giving a quantitative estimate for that decay?
(Some variant of the Paley–Wiener theorem useful for this case?)

Off course if k=2 we get (up to a constants) the same function, but I don't expect an explicit formula for other values of k.

Pablo De Napoli
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You have to restrict yourself to $k$ even, as for odd $k$ the function is not integrable and thus the Fourier transform does not exists (at least in a conventional sense).

So you are interested in the value of $$ F_n(k)=\int_{-\infty}^\infty e^{i k x - x^{2n}}\,dx $$ for $|k|\to \infty$. The saddle point $x_*$ is given by the solution of $$ \frac{d}{dx} (i k x - x^{2n}) = i k - 2n x^{2n-1} =0\,.$$ We obtain $$ x_*= c_1 k^{\frac{1}{2n-1}}\,.$$ So the dominating behavior of the integral is given by $$ F_n(k) \sim c_2 \exp\left(-c_3 k^{\frac{2n}{2n-1} } \right)$$ where $c_1,c_2,c_3$ are constants which can be determined by applying the method of stationary phase carefully.

Fabian
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  • You are right about your observation! For odd k it should be exp(-|z|^k) (but then the function is not analytic anymore). I also had forgotten about the method of stationary phase! Many thanks! – Pablo De Napoli Apr 10 '19 at 19:20
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    To apply the steepest descent method to this problem, you have to choose a contour going through two complex critical points. You get two complex exponentials, yielding sine/cosine terms. It's not exactly correct to say that $F$ is asymptotically equivalent to this expression $G$: $F/G$ does not tend to $1$, $F$ and $G$ have zeroes at different points. – Maxim Apr 15 '19 at 20:59
  • How to apply the method of stationary phase precisely, can you give me some reference? Thanks a lot. – yufeng lu Mar 10 '21 at 14:38