When asked to compute

FourierTransform[Exp[(-I*Abs[x])], x, k, FourierParameters -> {1, 1}]

Mathematica returns

(2 I)/(-1 + k^2)

My question is: how much "legitimate" is such Fourier transform? I can't fint it on any book of tables; furthermore when asked to compute

InverseFourierTransform[(2 I)/(-1 + k^2), k, x, FourierParameters -> {1, 1}]

Mathematica gives

-(1/2) E^(-I x) (-1 + E^(2 I x)) Sign[x]

which does not seem the same as the original.


closed as off-topic by rhermans, halirutan Aug 1 '18 at 12:26

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  • 1
    $\begingroup$ How interesting: FourierTransform[-I Sign[x] Sin[x], x, k, FourierParameters -> {1, 1}] == FourierTransform[Exp[-I Abs[x]], x, k, FourierParameters -> {1, 1}] $\endgroup$ – J. M. is away Apr 12 '18 at 12:57
  • $\begingroup$ Are you teasing me? If so please be more explicit... $\endgroup$ – Massimiliano Malgieri Apr 12 '18 at 17:23

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