# Fourier transform in polar coordinate

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?

• You can use Fourier transform properties such as a radial function will have a radial Fourier transform. May 22, 2020 at 19:25
• Does that help for my integral? because I have $\int e^{-\frac{r}{b}}e^{iGrcos(\theta)}r \, \mathrm{d}\mathbf{r} \mathrm{d}\mathbf{\theta}$ May 22, 2020 at 19:31
• 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. May 22, 2020 at 19:43
• @Displayname The $r$ in your $dr$ isn't a vector, so it shouldn't be bold.
– J.G.
May 22, 2020 at 20:08

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.