Fourier transform of isotropic Laplace distribution (2D) How would I evaluate the Fourier transform of an isotropic 2D Laplace distribution?
$F(\omega_x,\omega_y)=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \exp(-b \sqrt{x^2+y^2})\exp(-j\omega_x x)\exp(-j\omega_y y)\, \mathrm{d}x\, \mathrm{d}y$
 A: You can use polar coordinates. Define $r=\sqrt{x^2+y^2}$ and $\omega=\sqrt{\omega_x^2+\omega_y^2}$. Then define the angles
$\theta$ and $\alpha$ such that $(x,y)=(r\cos\theta,\,r\sin\theta)$ and $(\omega_x,\omega_y)=(\omega\cos\alpha,\omega\sin\alpha)$. Let us compute 
$$G(b,\omega,\alpha)=\int_0^{2\pi}\mathrm d\theta\int_0^\infty \mathrm e^{-br-\mathrm j\omega r\cos(\theta-\alpha)}\,\mathrm dr=\int_0^{2\pi}\frac{\mathrm d\theta}{b+\mathrm j\omega\cos(\theta-\alpha)}.$$
It becomes clear that $G$ does not depend on $\alpha$ since we can
shift the variable $\theta$ by $\alpha$ without changing the value of the integral. The idea is to replace the cosine by $(\mathrm e^{\mathrm j\theta}+\mathrm e^{-\mathrm j\theta})/2$ and write $\mathrm e^{\mathrm j\theta}$ as $z$. The integral can now be expressed as an integral in the complex plane.
The change of variable also implies that $\mathrm d\theta=-\mathrm j\frac{\mathrm dz}z$. 
$$G(b,\omega)=-\mathrm j\oint_{\cal C}\frac{\mathrm dz}z\frac1{b+\mathrm j\omega\frac{z+z^{-1}}2}=\oint_{\cal C}\frac{-2\mathrm dz}{\omega z^2-2\mathrm jbz+\omega}.$$
${\cal C}$ is the circle of radius 1 centered at the origin.
The poles of $(\omega z^2-\mathrm j b z+\omega)^{-1}$ are 
$$z_\pm=\mathrm j\left(\frac b\omega\pm\sqrt{1+\frac{b^2}{\omega^2}}\right).$$
Since $b$ and $\omega$ are positive, $z_-$ is the only pole in the domain delimited by $\cal C$ in the complex plane.
We conclude that
$$G(b,\omega)=2\pi\mathrm j\frac{-2}{\omega(z_--z_+)}=\frac{2\pi}{\sqrt{b^2+\omega^2}}.$$
To conclude, we remark that $F(\omega)=-\partial G(b,\omega)/\partial b$ and we get
$$F(\omega)=\frac{2\pi\,b}{(b^2+\omega^2)^{3/2}}.$$ 
