2
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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.

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question cannot be answered without additional information. Questions on problems in code must describe the specific problem and include valid code to reproduce it. Any data used for programming examples should be embedded in the question or code to generate the (fake) data must be included." – halirutan
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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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