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Consider the family of functions $f_n=e^{-x^{2n}}$, where $x$ is a real number. I am interested in the fourier transform $\hat{f_n}(t)=\int_{-\infty}^{\infty}f_n(x)e^{2\pi itx}dx$. While the exact expression for $\hat{f}_n$ would be difficult to obtain, except when $n=1$, are there results that provide the asymptotic behaviour of $\hat{f}_n(t)$, for any fixed $n$ (say $n=2$), but as $t\rightarrow \infty$? In particular, does $\hat{f_n}(t)$ decay as $e^{-t^{\alpha(n)}}$, for some $\alpha(n)>0$?

anurag anshu
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    In the limit, $\hat{f}_n(t) \to \frac{\sin(2\pi t)}{\pi t}$, so it decays slower than exponentially. – Ninad Munshi Jul 12 '20 at 01:06
  • @NinadMunshi Do you mean as n goes to infinity? I meant `asymptotic as a function of t' for every n. I think it was not clear in the question. I will clarify it. – anurag anshu Jul 12 '20 at 02:04
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    It's good you clarified that. That answers your question because it's impossible. If the limit decays as a polynomial, it seems unlikely that for large $n$ the function would decay as a exponential of a power, since those are all faster than the largest possible polynomial decay. But if you want decays for smaller $n$ like $2$ or $3$, numerical computation may be your best bet, but I will think about strategies involving Fourier transform properties. – Ninad Munshi Jul 12 '20 at 02:07
  • @NinadMunshi Right, that does answer the second question. The decay cannot stay super-polynomial for all finite n. Whether the decay stays superpolynomial for small n (say n=2,3) is of interest to me. – anurag anshu Jul 12 '20 at 03:47
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    See this answer. For $n = 2$, $$\hat {f_2}(t) = \exp \left( -\frac {3 (\pi t)^{4/3}} {2^{7/3}} \right) \cos \left( \frac \pi 6 - \frac {3^{3/2} (\pi t)^{4/3}} {2^{7/3}} \right) \left( \frac {(32 \pi)^{1/6}} {\sqrt 3 \hspace {1.5px} t^{1/3}} + o {\left( \frac 1 {t^{1/3}} ! \right)} ! \right)$$ as $t \to \infty$. – Maxim Jul 27 '20 at 22:07

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This can be solved by steepest descent analysis of the integral.

To set things us, it's convenient to first change variables. Exchange $t$ for a new parameter $\lambda$, defined by $$ \lambda = \left(\frac{\pi t}{n}\right)^\frac{2n}{2n-1} \to \infty. $$ Then change variables in the integral to $z$, defined by $$ x = \lambda^\frac{1}{2n} z. $$ The Fourier integral then becomes $$ \lambda^\frac{1}{2n} \int_{-\infty}^\infty e^{\lambda s(z)} \, dz, \qquad s(z)= 2i n z-z^{2n}. $$ The saddle-points are the stationary points $s'(z)=0$ of $s$, lying on the unit circle at points $$ z = \exp\left(\frac{2k-\frac{3}{2}}{2n-1} i \pi\right),\qquad k=1,2,\ldots,2n-1. $$

Now to work out which of these saddle-point is relevant, a diagram is worth $10^3$ words:

A plot of the real part of s, with paths of steepest descent marked.

This plot shows (for $n=3$) the real part of $s(z)$, with saddle-points marked in black. Blue colours indicate regions where the real part is more negative. The red curves are contours of constant imaginary part of $s(z)$: the paths of steepest descent follow such curves. With this picture, you can convince yourself that the contour along the real axis should be deformed to pass through all the saddle-points in the upper half-plane, namely $k=1,2,\ldots,n$.

Next, we should see which saddle-point is most important, giving the largest real part of $s(z)$. On the saddle-points, we have $z^{2n-1}=i$, so $s(z)=(2n-1)iz$: the largest real part of $s(z)$ comes from the smallest imaginary part of $z$, which is $k=1$ and $k=n$.

Finally, we expand $s(z)$ to quadratic order at each of these two saddle-points, and do the Gaussian integral to get the leading order asymptotics. For $n>1$, we get $$ \int_{-\infty}^\infty e^{\lambda s(z)} dz \sim 2\sqrt{\frac{\pi }{\lambda n (2 n-1)}} \Re\left[\exp \left(-i \frac{\pi}{2}\frac{(n-1)}{2n-1}+\lambda e^{\frac{i n \pi }{2 n-1}} (2 n-1)\right)\right]. $$ This is twice the real part of the $k=1$ saddle (with the $k=n$ saddle giving the complex conjugate). For $n=1$, the result is less by a factor of two, because there's a unique saddle (and we get the exact result).

The original Fourier transform therefore decays like $\exp\left(-(2n-1)\sin\left(\frac{\pi}{2(2n-1)}\right)\lambda\right)$, times a rapid oscillation. In terms of the original variable $t$, this is $\exp\left(-\# t^\frac{2n}{2n-1}\right)$.

Holographer
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