# Replacing sum by integral

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?

• Do you need to compute (an accurate approximation of) your series or is it enough to show it is convergent? The latter is quite simple. – Jack D'Aurizio Feb 23 '17 at 18:44
• I already know that the series converges, but I would like to know if we can say that $$\displaystyle \frac{\sum_{m \in \mathbb{Z}^d \backslash \{0\}}\frac{J^{2}_{d/2}(\rho|m|)}{|m|^d}}{\int_{|\alpha| \geqslant 1}\frac{J^{2}_{d/2}(\rho|\alpha|)}{|\alpha|^{d}} \ \mathrm{d}\alpha} \sim 1.$$ An accurate approximation of my series (together with the error) would also be something I would like to compute. – user363087 Feb 23 '17 at 18:48

You are allowed to replace $\sum$ with $\int$ for the following reasons:

1. Gauss circle problem in dimension $d$. For any moderately large $R$, the number of lattice points in $\mathbb{Z}^d$ having a modulus between $R-\frac{1}{2}$ and $R+\frac{1}{2}$ is (very) close to the surface area of the $d$-dimensional sphere with radius $R$;
2. Tricomi's inequalities. By the Bessel differential equation, $z\cdot J_{d/2}(z)^2$ is (very) close to a squared cosine wave.

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.

• I would be very interested in seeing how these two facts could be used to show that the integral is near the sum. Do you have a reference I could look at? – Antonio Vargas Feb 23 '17 at 20:46
• @AntonioVargas: this is a nice article on a classical topic: arxiv.org/pdf/math/0410522 – Jack D'Aurizio Feb 23 '17 at 20:51
• May I ask how you computed the very last integral? I had previously asked a similar question about that integrand here. – user363087 Feb 24 '17 at 22:02
• @user363087: a simple substitution and the lemma $\int_{0}^{+\infty}\frac{J_m(x)^2}{x}\,dx = \frac{1}{2m}$. – Jack D'Aurizio Feb 24 '17 at 22:23
• I see, thanks. I did not know that integral had such a nice solution. – user363087 Feb 24 '17 at 22:23