# Fourier transform of product of Bessel functions

I need help finding the Fourier transform of the function

$$\rho(\vec{r}) = \alpha \delta_{\vec{r},0} \left(\lambda\lambda' J_1 (\beta |\vec{r}|)Y_1(\beta |\vec{r}|) - \pi^2 J_0 (\beta |\vec{r}|)Y_0(\beta |\vec{r}|) \right),$$

where $\alpha,\beta \in \mathcal{R}$ and $\lambda,\lambda'=\pm1$. The $J_i(x)$ are the $i$-th Bessel functions of the first, the $Y_i(x)$ are the $i$-th Bessel functions of the second kind. $\delta_{\vec{r},0}$ denotes the Kronecker delta.

The vector $\vec{r} = (x,y)^T$ is discrete because I'm working on a discrete set of points. I'm a physicist, so please don't hesitate to ask for more specific information.

I tried something like

$$\rho(\vec{k}) = \sum_\vec{r} e^{-i\vec{k}\cdot\vec{r}}\rho(\vec{r}),$$

which should in fact not be too difficult because the only vector that contributes to the sum is $\vec{r} = \vec{0}$ due to the Kronecker delta in $\rho(\vec{r})$. The problem is that $\rho(\vec{r})$ diverges at this point. I need some mathematical advice here. Is there anybody who can show me how to compute the Fourier transform of $\rho(\vec{r})$?

• Please confirm that this problem has strict radial symmetry. The reason I ask is that $\delta^2(x,y)$ and $\delta(r)$ are slightly different (for continuous axes), and a 2-dimensional Fourier Transform with radial symmetry is different than a one dimensional Fourier Transform. – Andy Walls 2 days ago
• @AndyWalls: I think we can assume that the problem is radially symmetric. But I wouldn't say no to seeing both Fourier transforms :) – MeMeansMe yesterday