# Why does Mathematica return this Fourier transform?

Posted 2 months ago
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 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 2 months ago
 The documentation states that Mathematica for its FourierTransform uses Exp[I k x] in the integral, rather than Exp[-I k x].
Posted 2 months ago
 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 2 months ago
 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.
 I don't know, but we can recover our favourite definition with a change in sign: myFourierTransform[f_, x_, k_] := FourierTransform[f, x, -k] 
 Jay,If you look at the documentation for FourierParameters I posted in your other thread, you control the exponent with the second parameter.In the case that you like the default definition but prefer the sign change in the exponent FourierParameters-> {0,-1} should do it.Regards.