I am computing the Fourier transform


for large $k$. Since $e^{-x^4}$ is $C^{\infty}$ on the whole real line, I expect its Fourier transform to decay faster than any power of $k$ at infinity. Indeed, with a saddle point approximation I find that the leading behavior is $\sim e^{-k^{4/3}}\times k^{-1/3}$ for large positive $k$.

Mathematica can do the integral in a closed form:

f[k_] = FourierTransform[Exp[-x^4], x, k]


(* (2 Gamma[5/4] HypergeometricPFQ[{}, {1/2, 3/4}, k^4/256] -
    1/4 k^2 Gamma[3/4] HypergeometricPFQ[{}, {5/4, 3/2}, k^4/256])/Sqrt[2 π] *)

However, expanding this around $+\infty$ with

Series[f[k], {k, Infinity, 1}]

shows a leading behavior of $e^{k^{4/3}}\times k^{-1/3}$. Plotting the function $f(k)$ with Mathematica at large $k$ also shows a divergent result.

I am confused about this. I would expect the Fourier transform to be convergent at $k\to +\infty$, for the reasons explained above.


After fixing a mistake in the saddle point approximation, from the analytic calculation I get $f(k) = 0$ at large $k$. There is a cancellation between the contributions of two saddle points (each of which showed a behavior $\sim e^{-k^{4/3}}\times k^{-1/3}$, in any case).

  • $\begingroup$ When I plot the answer you get for f[k] I see it decays exponentially as k goes to infinity. The asymptotic expansion answer is suspect since there is likely an essential singularity at $k = \infty$, but the expression f[k] looks correct and has the correct (numerical) asymptotic behavior. $\endgroup$
    – ulvi
    Jul 4, 2017 at 23:52
  • $\begingroup$ Until k = 20 I also get a decay when I plot the function. But if I plot up to, e.g., k = 40 the function f[k] seems to oscillate and diverge. I guess that it is just a numerical problem, and Mathematica's closed form answer is correct. $\endgroup$
    – HR_8938
    Jul 5, 2017 at 8:56
  • $\begingroup$ the correct asymptotics is given at mathoverflow.net/q/366816/11260 $\endgroup$ Jul 28, 2020 at 20:39

1 Answer 1


The Landau-O-notation $\operatorname{O}((1/k)^\alpha)$ for $k \to \infty$ in Mathematica has to be taken with a grain of salt. It just tells us that Mathematica stopped the expansion at some point.

In this case it might be better to read it as: Something decays at least as fast as $(1/k)^{\alpha}$ without a statement on the sharpness of this estimate. Do higher order Laurent expansion and Simplify to highlight that. Just try

Series[f[k], {k, Infinity, 10}] // Simplify
Series[f[k], {k, Infinity, 20}] // Simplify


  • $\begingroup$ Expanding to higher order with Series still gives the overall factor of $e^{k^{4/3}}$. This will still diverge very fast at infinity whatever is the $(1/k)^n$ factor from the Laurent expansion. $\endgroup$
    – HR_8938
    Jul 4, 2017 at 15:07
  • 2
    $\begingroup$ Not if you never hit a nonzero coefficient... The function you are expanding might not be meromorphic at $\infty$ so that it might not coincide with its Laurant series... Consider the Taylor expansion of $x \mapsto \exp(-x^{-2})$ at $x = 0$ for real $x$. The Taylor/Laurant expansion is all zero. The function ist not zero. The problem is that $\exp(1/z)$ has an essential sigularity at $z=0$. $\endgroup$ Jul 4, 2017 at 15:23
  • $\begingroup$ Just to understand better your point: do you think that it is correct to say that $f(k)$ should decay at infinity? I thought it should be true because the original function I am Fourier transforming is $C^{\infty}$. $\endgroup$
    – HR_8938
    Jul 4, 2017 at 16:00
  • $\begingroup$ Thinking of it: The result of generated be FourierTransform is weird indeed, as you essentially apply it to the squared of a Gauss bell. The result should be something like a Gauss bell convoluted with itself -- being again essentially a Gauss bell. $\endgroup$ Jul 4, 2017 at 16:00
  • 1
    $\begingroup$ I don't think it's the square of a Gaussian: this would give $e^{-2x^2}$, not $e^{-x^4}$. $\endgroup$
    – HR_8938
    Jul 4, 2017 at 16:02

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