When evaluating this expression, Mathematica simplifies the result:

FourierTransform[FourierTransform[u[t], t, s], s, t]

This simplifies to u[-t], as expected.

But let's say I want this expression using different Fourier parameters.

  FourierTransform[u[t], t, s, FourierParameters -> {1, - 1}],
  s, t, FourierParameters -> {1, -1}
] // FullSimplify

When evaluating this, Mathematica leaves the result unaltered, with or without FullSimplify. The result should be $2\pi\ u(-t)$. Is it possible to get this result, or a similar result using these parameters? I would like to test these kinds of expressions with Mathematica, but without the right parameters it is kind of useless.


I would suggest reporting the unevaluated FourierTransform[] with explicitly-set FourierParameters to support.

In the meantime, you can do this instead:

SetOptions[FourierTransform, FourierParameters -> {1, -1}];
FourierTransform[FourierTransform[u[t], t, s], s, t]
   2 π u[-t]
  • 2
    $\begingroup$ I sent a report with this problem. The suggested fix is actually what I want since it saves you from typing FourierParameters -> ... over and over. I have never heard of SetOptions but it seems a really useful function. $\endgroup$ – user3502079 Nov 4 '17 at 19:32

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