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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?

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

1 Answer 1

up vote 12 down vote accepted

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

share|improve this answer
    
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. –  Luap Nalehw Sep 16 '13 at 8:58
    
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 ? –  Luap Nalehw Sep 16 '13 at 9:12
    
@LuapNalehw I think results are exactly the same. For this I gave the simplified form. –  ybeltukov Sep 16 '13 at 9:16
    
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 $(1+i) / 16 i$ –  Luap Nalehw Sep 16 '13 at 9:26
    
Nice work getting two Enlightened badges in the first two weeks of active membership! –  Mr.Wizard Sep 19 '13 at 20:09

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