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Why does Mathematica return this Fourier transform?

Is my integration wrong? I get a different result regardless of which definition of the Fourier transform I use. The difference is the sign of the ikb term in the exponent.

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POSTED BY: Jay Gourley
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Jay,

If you look at the documentation for FourierParameters I posted in your other thread, you control the exponent with the second parameter.

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In the case that you like the default definition but prefer the sign change in the exponent

FourierParameters-> {0,-1}

should do it.

Regards.
POSTED BY: Neil Singer

I don't know, but we can recover our favourite definition with a change in sign:

myFourierTransform[f_, x_, k_] := FourierTransform[f, x, -k]
POSTED BY: Gianluca Gorni

Is there a version not listed in the documentation that uses the Fourier transform defined in my textbook? It's the inverse of the default definitions.

POSTED BY: Jay Gourley

Thanks, Gianluca. Thumbs up. This community is a great help. I never knew there were so many Fourier transforms. If I had known years ago, I could have found definitions to justify answers marked wrong on my calculus exams.

POSTED BY: Jay Gourley

The documentation states that Mathematica for its FourierTransform uses Exp[I k x] in the integral, rather than Exp[-I k x].
POSTED BY: Gianluca Gorni
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