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Help me to find a 2D fourier transform:

$$\int dx dy\ \frac{e^{-ik_x x}e^{-ik_y y}}{\sqrt{x^2 + y^2}}.$$

All I've done so far is

$$\int dx dy\ \frac{e^{-ik_x x}e^{-ik_y y}}{\sqrt{x^2 + y^2}} \rightarrow \int_0^{+\infty} rdr \int_0^{2\pi} d\varphi\ \frac{e^{-i |q| r \cos\varphi}}{r},$$

where $|q|=\sqrt{k_x^2 + k_y^2}$. So

$$\int_0^{+\infty} rdr \int_0^{2\pi} d\varphi\ \frac{e^{-i |q| r \cos\varphi}}{r} = \int_0^{2\pi} d\varphi\ \frac{1}{-i|q| \cos\varphi}e^{-i|q|r \cos\varphi}\bigg|_0^{+ \infty}$$

I can't figure out what I'm sposed to do next?

LexRomah
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1 Answers1

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One way of solving this problem is the following: Using Polar coordinates, we have to calculate $$ \int_{0}^{\infty} dr \int_{0}^{2 \pi} d\phi e^{-i|q|r cos{\phi}} $$

Now the inner integral furnishs a representation of $ 2 \pi J_0( |q| r)$ which denotes a zeroth order Bessel function times $2 \pi$

After rescaling $ |q| r=z$ we end up with

$$ \frac{2 \pi}{|q|}\int_{0}^{\infty} dz J_0(z) $$

This integral is known to be $1$, so our final answer is $$ \frac{2 \pi}{|q|} $$

tired
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  • Thank you! Didn't know that $\int_0^{2\pi} d\varphi e^{i|q|r\cos\varphi}$ is $J_0(|q|r)$! But what about $\int_0^{2\pi} d\varphi e^{-i|q|r\cos\varphi}$ (with a sign of minus in exp)? Is it the same $J_0 (|q|r)$? I use the following fourier representation: $$f_\mathbf{k} = \int d\mathbf{r}\ f(\mathbf{r}) e^{-i\mathbf{k}\mathbf{r}}.$$ – LexRomah Mar 11 '15 at 14:15
  • it doesn't matter, but i corrected the typo – tired Mar 11 '15 at 14:24
  • Thanks! $\int_0^{2\pi}d\phi\ e^{\pm i|q| r \cos\phi} \propto J_0(|q| r)$ is a tabulated integral? Or it's just somehow an integral representation? – LexRomah Mar 11 '15 at 14:28
  • it's one of the possible representations of $J_0$ – tired Mar 11 '15 at 14:38
  • Now I get it! Thank you once again! – LexRomah Mar 11 '15 at 14:55
  • you are welcome – tired Mar 11 '15 at 15:07