# Fourier transform of imaginary exponential with absolute value argument [closed]

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.

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