Inverse Fourier transform of sign function

607 Views Asked by At

I am a bit puzzled by the following Mathematica outputs:

{FourierTransform[Sign[x], x, y],  InverseFourierTransform[Sign[x], x, y]}

$$\frac{i \sqrt{2/\pi}}y, \quad -\frac{i \sqrt{2/\pi}}y$$ as expected. But

{FourierTransform[Sign[x], x, y, FourierParameters -> {1, -1}],  
 InverseFourierTransform[Sign[x], x, y,FourierParameters -> {1, -1}]}

$$-\frac{2i}y, \quad \frac i {\pi y} + \delta(y) - 2\pi \delta(y) $$

I wonder if there could be some merit to this nonconventional output in some contexts. For example, my original objective is to calculate the 2D inverse Fourier transform of $\text{sgn}\,(x_1 - x_2)$. This can be done in two ways:

  • After change of variable $x_2' = x_1 - x_2$, we can separate integrals as the 1D direct Fourier of sgn multiplied by 1D inverse Fourier of 1, giving $ -\frac i{\pi y_2}\delta(y_1+y_2)$;
  • Or, by the convolution theorem, we get the inverse Fourier of sgn multiplied by the inverse Fourier of $e^{i x_1 y_2 }$, giving $\text{sgn}^\vee(y_1) \delta(y_1 + y_2)$. Based on Mathematica's second answer, the result after sifting is $$ -\frac i{\pi y_2}\delta(y_1+y_2) + (1-2\pi)\delta(y_1)\delta(y_2) $$ which is a little different.

Finally, Mathematica gives a third answer:

InverseFourierTransform[ Sign[x1 - x2], {x1, x2}, {y1, y2}, FourierParameters -> {1, -1}] // FullSimplify

$$ \frac{i \delta \left(y_1-y_2\right)}{2 \pi y_1}-\frac{i \delta \left(y_2-y_1\right)}{y_1}+\frac{i \delta \left(y_1+y_2\right)}{y_1}+\frac{i \delta \left(y_1+y_2\right)}{2 \pi y_1} +(1-2 \pi) \delta \left(y_1\right) \delta \left(y_2\right) $$

Which one is correct? Does anyone know what's going on here?