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?
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 $$ \frac{2 \pi}{|q|} $$