I am considering the following sum
$$\displaystyle \sum_{m \in \mathbb{Z}^d \backslash \{0\}}\frac{J^{2}_{d/2}(\rho|m|)}{|m|^d},$$
where $J_{\nu}$ denotes the Bessel function of the first kind, and $\rho > 0$ is independent of $m$ and $d$. I would like to be able to do this:
$$\displaystyle \sum_{m \in \mathbb{Z}^d \backslash \{0\}}\frac{J^{2}_{d/2}(\rho|m|)}{|m|^d} \sim \int_{|\alpha| \geqslant 1}\frac{J^{2}_{d/2}(\rho|\alpha|)}{|\alpha|^{d}} \ \mathrm{d}\alpha,$$
because the integral of that expression is far easier to deal with than the sum. (I am aware that the Bessel function can be bounded in the sum to get something convergent, but I don't want to do this.) However, I would like to know if this is valid. Obviously, it is not necessarily true that the integral bounds the sum from above because the summand is not monotonically decreasing -- it is non-negative, but has lots of oscillations. Can we estimate the sum by the integral in this way? Is there any way to justify this?
You are allowed to replace $\sum$ with $\int$ for the following reasons:
The resulting integral is not difficult to compute. Integrating along shells, $$ \int_{\mathbb{R}^d}\frac{J_{d/2}^2(\rho|\alpha|)}{|\alpha|^d}\,d\alpha = \frac{2\pi^{d/2}}{\Gamma(d/2)}\int_{0}^{+\infty}\frac{J_{d/2}^2(\rho z)}{z}\,dz=\color{red}{\frac{\pi^{d/2}}{(d/2)!}}.$$ This is exactly the volume of the unit ball in dimension $d$.
Not by chance, Voronoi's method for improving the error term in Gauss circle problem for $d=2$ employs Bessel functions, since they arise from the Fourier transform of the characteristic function of a ball.