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Bug introduced in 5 or earlier and persisting through 11.3.0


On Mathematica 11.0.0 the Fourier transform of $x\theta(x)$ gives the expected result:

In: FourierTransform[x*UnitStep[x], x, t]
Out: -(1/(Sqrt[2*Pi]*t^2)) - I*Sqrt[Pi/2]*Derivative[1][DiracDelta][t]

But the Fourier transform of $(x-a)\theta(x)$ misses the derivative of a delta function, and does not reduce to the previous answer when $a\rightarrow 0$:

In: FourierTransform[(x - a)*UnitStep[x], x, t]
Out: -(1/(Sqrt[2*Pi]*t^2)) - (I*a)/(Sqrt[2*Pi]*t) - a*Sqrt[Pi/2]*DiracDelta[t]
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  • $\begingroup$ Related? $\endgroup$ – corey979 Sep 9 '16 at 16:08
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    $\begingroup$ @march -- the difference between the first and second integrands is $a\theta(x)$, which is a smooth function of $a$, without any discontinuity at $a=0$, so taking the limit $a\rightarrow 0$ of the integral over $x$ should be fine. $\endgroup$ – Carlo Beenakker Sep 9 '16 at 17:54
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    $\begingroup$ @march There is clearly an inconsistency if you expect the Fourier transform to be invertible (a fundamental property): InverseFourierTransform[FourierTransform[(x-a) Sign[x],x,t],t,x] doesn't produce the original function back again. So I think it has to be called a bug. $\endgroup$ – Jens Sep 9 '16 at 18:13
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    $\begingroup$ Distribution over + should clearly be valid in this case e.g. (FourierTransform[x*UnitStep[x], x, t] - FourierTransform[a*UnitStep[x], x, t]) - FourierTransform[(x - a)*UnitStep[x], x, t] // FullSimplify should be zero, and it is not. This is a bug $\endgroup$ – mikado Sep 11 '16 at 9:22
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    $\begingroup$ FWIW, the bug persists even if you replace UnitStep[] with HeavisideTheta[], even if the latter function is supposedly the function intended for use with the built-in integral transforms. $\endgroup$ – J. M. will be back soon Oct 4 '16 at 2:06
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I've reported the issue to Wolfram and received this response:

Thank you for taking the time to send in this report. It does appear that the inverse Fourier transform of the Fourier transform of a particular function is not behaving properly. In particular it is not giving the original function back, as it should. I will forward an incident report to our developers regarding this issue.

So this is a bug. The case number is CASE:3710457.

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  • $\begingroup$ I decided to put the bug report case number in the answer instead of a comment, because this will make it easier to find in the future, in case I get additional feedback... $\endgroup$ – Jens Sep 12 '16 at 23:07
  • $\begingroup$ thank you, Jens, nice to catch up with you here! $\endgroup$ – Carlo Beenakker Sep 13 '16 at 9:41

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