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Could you please explain why Mathematica gives the following expression when taking Fourier transform of $\exp(\lambda z)/\lambda$? $$\frac{-\log(-z)+\log(z)}{\sqrt{2\pi}}$$ Why the answer does not depend on $x$?

The source code is as the following:

In[1]:= FourierTransform[Exp[λ z]/λ,λ,x]
Out[1]= (-Log[-z]+Log[z])/Sqrt[2 π]
Mr.Wizard
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Ivan Gankevich
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    I'll venture a guess. The FT does not exist, the FT code fails to catch that, calls Integrate with GenerateConditions->False, and gets the strange result in that way. – Daniel Lichtblau May 28 '16 at 15:32
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    @DanielLichtblau That can't be right; if FourierTransform calls Integrate, there wouldn't be any point of having FourierTransform in the first place. – QuantumDot May 28 '16 at 17:49
  • @QuantumDot that can't be right: FourierTransform is very complex, and it may call Integrate in some specific cases... – AccidentalFourierTransform May 28 '16 at 18:39
  • @AccidentalFourierTransform Integrate is a more general function than FourierTransform. If anything, I would expect Integrate to make calls to FourierTransform for classes of integrals that are of the Fourier form -- not the other way around. – QuantumDot May 28 '16 at 18:52
  • @QuantumDot I would assume that FourierTransform first does a table lookup to see if the function matches any patterns it knows the FT of, and then falls back on doing the integration the old fashioned way. Admittedly, im neither a mathematician or computer scientist, so I'm curious how else would it work? – Jason B. May 28 '16 at 21:03
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    @AccidentalFourierTransform and Jason-with-the-even-longer-moniker are correct. FT sometimes calls Integrate, never the reverse. Integrate itself should probably use more table lookup, like the FT code does, but that's a separate issue. – Daniel Lichtblau May 28 '16 at 21:36
  • @DanielLichtblau Oh ok, I can confirm your guess: Trace[FourierTransform[Exp[λ z]/λ, λ, x], _Integrate, TraceInternal -> True] reveals calls to Integrate. I was so sure of myself that I would've placed a bet that Integrate calls FourierTransform and not the other way around. – QuantumDot May 29 '16 at 12:24

1 Answers1

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Some commenters mentioned that the strange expression that Mathematica gives as a Fourier transform of Exp[\[Lambda] z]/\[Lambda] is a bug. If you are curious what the actual Fourier Transform is, you may find some insights in the related questions (e.g. link).

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Ivan Gankevich
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