2D Fourier of Bessel I need help with the fourier transform of Bessel function of first kind on 2 dimensions.
$$G(w_1,w_2) = F[J_0(a\sqrt{x^2+y^2})]$$
where $J_0$ is the bessel function of first kind of order 0, and $a$ is a possibly complex constant.
I think it will be easier in polar coordinates, but couldn't make much progress so far. Any kickstarter will be appreciated.
 A: From Bracewell, the 2D cartesian Fourier Transform
$$F(u,v) = \mathscr{F}\left\{f(x,y)\right\} = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} f(x,y) e^{-2\pi i (xu+yv)} dx dy$$
for functions with circular symmetry, in polar coordinates, can be manipulated into a Hankel Transform of zero order
$$F(u,v) = F(q) = \mathscr{H}_0\left\{f(r)\right\} = 2\pi\int_{0}^{\infty}f(r)J_0(2\pi qr)r\space dr$$
with a strictly reciprocal inverse transform
$$f(x,y) = f(r) = \mathscr{H}_0\left\{F(q)\right\} = 2\pi\int_{0}^{\infty}F(q)J_0(2\pi qr)q\space dq$$
Note: $r = \sqrt{x^2+y^2}$ and $q=\sqrt{u^2+v^2}$
$ $
Looking at the inverse transform of $F(q) = \dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)$, where $\delta()$ represents a 1 dimensional delta function:
$$\begin{align*}f(r) = \mathscr{H}_0\left\{\dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)\right\} &= 2\pi\int_{0}^{\infty}\dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)J_0(2\pi qr)q\space dq\\
\\
&= \dfrac{2\pi}{a}\dfrac{a}{2\pi}J_0\left(2\pi \dfrac{a}{2\pi}r\right)\\
\\
&=J_0(ar)\\
\end{align*}$$
So that gives you your desired transform pair for real $a$:
$$\mathscr{F}\left\{J_0(ar)\right\} = \mathscr{H}_0\left\{J_0(ar)\right\} = \dfrac{1}{a}\delta\left(q-\dfrac{a}{2\pi}\right)$$
The transform is an impulse ring at a radius of $\dfrac{a}{2\pi}$ from the origin.
