I am interested in the following sum:
$$\small \sum_{(b,c) \in \mathbb{Z}^2 \backslash \{(0,0)\}} \left| \ \iint\limits_{[0,2\pi) \times [0,2\pi)} (x^2 + y^2)^{-1/2}((b-x)^2 + (c-y)^2)^{-1/2} J_{1}\left(\rho \sqrt{x^2 + y^2}\right)J_{1}\left(\rho \sqrt{(b-x)^2 + (c - y)^2}\right)\mathrm{d}x\mathrm{d}y\right|^2.$$
where $\rho > 0$ is a constant, and $J_{\nu}$ denotes the Bessel function of the first kind. This is a special case of a more general sum I'd like to consider, with $d=2$:
$$\sum_{b \in \mathbb{Z}^d \backslash \{0\}} \left|\int_{[0,2\pi)^d}\|m\|_{d}^{-d/2}\|b-m\|_{d}^{-d/2}J_{d/2}(\rho \|m\|_{d})J_{d/2}(\rho \|b - m\|_{d})\mathrm{d}m \right|^2,$$
where $\|\cdot\|_{d}$ denotes the standard Euclidean norm on $\mathbb{R}^d$. I've tried computing this in a few different ways using Mathematica. The first way to get rid of the Bessel functions is to use the bound $|J_{\nu}(z)| \leqslant C_{\nu}|z|^{-1/2}$ for some positive constant $C_{\nu}$ depending on $\nu$. However, this may be dangerous, since by taking absolute values of the Bessel functions, we lose the ability to take advantage of any positive-negative cancellation that occurs. Mathematica doesn't seem to be able to compute the integral for $d = 2$, although one can get a numerical result by replacing $b, c, \rho$ with numbers instead (but since I want to sum over $b,c$, this is not entirely helpful). We can also use the asymptotic expansions of $J_{\nu}$, and in the case of $d=1$, the Bessel function $J_{1/2}$ has a very simple closed form:
$$\displaystyle J_{1/2}(z) = \sqrt{\frac{2}{\pi z}}\sin(z),$$
and in general, there are also finite sum expansions for half-integer values of Bessel functions. But I haven't managed to get any semblance of a result, using any of these methods.
Can anyone find a way of computing this sum for $d = 2$, or perhaps a better method for general $d$?
