11
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
13
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.