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I found the following formula for the Fourier transform on a flat 2-torus, but I don't quite know how to derive it. We have a variable $q=(q_x,q_y) \in [0,2\pi)^2$ and by considering it in polar coordinates we can also define the polar angle $\arg(q).$ The formula then claims:

$$\int_{[0,2\pi)^2} \frac{d^2q}{(2\pi)^2} e^{i(q \cdot r - m\arg(q))} = \frac{i^{|m|} |m| e^{-i m \arg(r)}}{2\pi |r|^2}.$$

Does anybody know how to derive it?

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  • $\begingroup$ Wait, are you asking who discovered it or how to derive it? $\endgroup$
    – mathworker21
    Commented Nov 29 at 12:28
  • $\begingroup$ how to derive it $\endgroup$
    – António Borges Santos
    Commented Nov 29 at 12:33
  • $\begingroup$ $r \in \mathbb R^2$ and $m$ is an integer. $\endgroup$
    – António Borges Santos
    Commented Nov 29 at 13:06

1 Answer 1

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This equality cannot be true in general. Indeed, let $m$ be any fixed nonzero real number, and let $r$ be a varying nonzero vector in $\Bbb R^2$ such that $|r|\to0$. Then the modulus of the left-hand side of your equality is $\le1$, whereas the modulus of the right-hand side of your equality goes to $\infty$.

Mathematica cannot do anything with the integral in question, even in the specific case with $m=1$ and $r=(1,2)$:

enter image description here

So, it is highly unlikely that the integral can be expressed in elementary or special functions.

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  • $\begingroup$ good observation. do you happen to know how to evaluate this Fourier transform? $\endgroup$
    – António Borges Santos
    Commented Nov 29 at 13:06
  • $\begingroup$ @AntónioBorgesSantos : I have now addressed the question in your comment. $\endgroup$
    – Iosif Pinelis
    Commented Nov 29 at 13:54
  • $\begingroup$ I'm curious. Does the "So" in the last line of your answer actually apply? I.e., is Mathematica that reliable that it's inability to simplify strongly indicates lack of a simple expression? $\endgroup$
    – mathworker21
    Commented Nov 29 at 16:12
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    $\begingroup$ @mathworker21 : As I recall, in my experience, there have been at most two cases when Mathematica could not do an integral which was done otherwise. I also know that the Wolfram czar on Integrate has for several decades been Marichev, one of the three co-authors of the most comprehensive collection of integrals. So, I do believe that the "So" applies here. In this particular case, with the integration over a square, rather than over something like a disk, I think the "So" applies even more. $\endgroup$
    – Iosif Pinelis
    Commented Nov 29 at 17:35

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