I'm trying to compute the Fourier transform of this potential: V(r)= e$^{-\frac{r}{b}}$, where b is a constant and r the distance in the x-y plane. The problem being that I'm in 2D so I thought I have to use polar coordinate but I'm blocked so does someone have any idea?

  • $\begingroup$ You can use Fourier transform properties such as a radial function will have a radial Fourier transform. $\endgroup$ May 22, 2020 at 19:25
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    $\begingroup$ Does that help for my integral? because I have $\int e^{-\frac{r}{b}}e^{iGrcos(\theta)}r \, \mathrm{d}\mathbf{r} \mathrm{d}\mathbf{\theta}$ $\endgroup$ May 22, 2020 at 19:31
  • $\begingroup$ This method means avoiding integrals. If you wanted to do the actual integral, just do it. It's not hard, just maybe tedious. Integration by parts then trig identities. $\endgroup$ May 22, 2020 at 19:43
  • $\begingroup$ @Displayname The $r$ in your $dr$ isn't a vector, so it shouldn't be bold. $\endgroup$
    – J.G.
    May 22, 2020 at 20:08

1 Answer 1


I'll sketch the key details, because the rest can be filled in with various strategies to taste. The Fourier transform is$$\int_0^{2\pi}d\theta\int_0^\infty r\exp[-r(1/b-iG\cos\theta)]dr=b^2\int_0^{2\pi}d\theta(1-ibG\cos\theta)^{-2}.$$Since the odd powers of cosine integrate to $0$ while$$\int_0^{2\pi}\cos^{2n}\theta d\theta=4\int_0^{\pi/2}\cos^{2n}\theta d\theta=2\operatorname{B}(n+\tfrac12,\,\tfrac12),$$the FT is$$2b^2\sum_{n\ge0}(2n+1)(ibG)^n\operatorname{B}(n+\tfrac12,\,\tfrac12)=\frac{2\pi b^2}{(1-ibG)^{3/2}}.$$That the modulus is maximal when $G=0$ gives a useful sanity check.


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