The following function

$$g(x) = (1 + x^{1/a} )^a $$

should NOT have a Fourier transform, as far as I am aware, for any real values of $a$ since $g(x)$ is not nice in the sense of decays quickly enough to $0$ at infinity. However, doing:

g[x_, a_] := (1 + x^(1/a))^(a);
Plot[g[x, 2], {x, 0, 1000}]
FourierTransform[g[x, 2], x, ω] 

Mathematica somehow comes up with the result:

$$\frac{\left(\frac{1}{2}+\frac{i}{2}\right) (\left| \omega \right| -\omega )}{\omega \left| \omega \right| ^{3/2}}$$

What is going on here?

  • $\begingroup$ I get the same (which doesn't agree with Integrate[g[x, 2] Exp[i x \[Omega]], {x, -Infinity, Infinity}]) on 9.0.1. $\endgroup$ – b.gates.you.know.what Sep 15 '13 at 18:03
  • $\begingroup$ Well, this seems like a bug in V9.0.1 $\endgroup$ – RunnyKine Sep 15 '13 at 18:14
  • $\begingroup$ @RunnyKine I get the same on 8.0.4 and 7.0.1 as well. $\endgroup$ – b.gates.you.know.what Sep 15 '13 at 18:16
  • $\begingroup$ @b.gatessucks. Interesting. Well, then I guess this is one of those sleeping bugs in Mathematica similar to the Eigenvector bug discovered recently. $\endgroup$ – RunnyKine Sep 15 '13 at 18:18
  • 1
    $\begingroup$ is it possible that the generalised fourier transform is being computed? $\endgroup$ – Luap Nalehw Sep 15 '13 at 18:46

It's not a bug, it's a feature

Exact integration returns

1/Sqrt[2 Pi]
  Integrate[(1 + Sqrt[x])^2 Exp[I k x], {x, -Infinity, Infinity}, 
  Assumptions -> {k \[Element] Reals}]

Integrate::idiv: "Integral of E^(I\k\x)\ (1+[Sqrt]x)^2 does not converge on {-Infinity,Infinity}."

However we can multiply by Exp[-b Abs[x]] and then put b -> 0

Limit[1/Sqrt[2 Pi]
   Integrate[(1 + Sqrt[x])^2 E^(I k x)
     E^(-b Abs[x]), {x, -Infinity, Infinity}, 
   Assumptions -> {k \[Element] Reals, b > 0}], b -> 0]

enter image description here

FullSimplify[%, Assumptions -> {k \[Element] Reals}]

enter image description here

| improve this answer | |
  • $\begingroup$ ok - clever thanks. So as a feature it applies a damping factor. This is a quite close to the FourierTransform[] result but not the same I think. In particular the root 2 has somehow dropped out. $\endgroup$ – Luap Nalehw Sep 16 '13 at 8:58
  • $\begingroup$ Sorry. To follow up on this last comment the difference between your direct approach and the FourierTransform[] result is due to a different definition of the transform. But ... the result only applies to positive k ? $\endgroup$ – Luap Nalehw Sep 16 '13 at 9:12
  • $\begingroup$ @LuapNalehw I think results are exactly the same. For this I gave the simplified form. $\endgroup$ – ybeltukov Sep 16 '13 at 9:16
  • $\begingroup$ sorry I didn't see your 1/Sqrt[2 Pi]. You are right. I don't see how the simplified form is working out though. For example, at $k = 2$ I get $(1+i) / 16 i$ $\endgroup$ – Luap Nalehw Sep 16 '13 at 9:26
  • $\begingroup$ Seemingly Mathematica is able to compute Fourier transforms for (tempered) distributions... $\endgroup$ – Henrik Schumacher Jul 4 '17 at 15:31

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