Integrals involving products of Bessel functions

I would like to consider the following integral:

$$\displaystyle \int_{\mathbb{R}^d}\left( \int_{\mathbb{R}^d}|\alpha|^{-d/2}|k - \alpha|^{-d/2}J_{d/2}(\rho|\alpha|)J_{d/2}(\rho |k - \alpha|) \ \mathrm{d}\alpha \right) \\ \cdot \left( \int_{\mathbb{R}^d} |\gamma|^{-d/2}|k + \gamma|^{-d/2}J_{d/2}(\rho|\gamma|)J_{d/2}(\rho |k + \gamma|) \ \mathrm{d}\gamma \right) \ \mathrm{d}k,$$

where $J_{\nu}$ denotes the Bessel function of the first kind, $|x|$ denotes the usual Euclidean norm $\|x\|_d$ on $\mathbb{R}^d$, and $\rho$ is large (but in this case it does not depend on any other variables appearing in the integral, so we may treat it as a constant). I would like to show that this whole expression is $O(\rho^a)$ for some power $a$. Originally, these integrals were sums varying over a rational lattice $\Gamma \subset \mathbb{R}^d$. However, due to the oscillatory nature of these functions, it may be more fruitful to consider them as integrals instead. I'll try to explain how this expression arises, and why I care about it.

This integral arises when considering a problem on the distribution of lattice points inside a ball in $\mathbb{R}^d$ with radius $\rho$, which can be thought of as a generalisation of some aspects of the Gauss circle problem. For the full details, I recommend this paper (particularly pages 10-11 and 15-16). The paper considers $\sigma_p$ and proves some asymptotic bounds for $p = 1$ and $p = 2$ which depend on the dimension $d$ and the radius $\rho$. I'm attempting to generalise that work to $p = 4$, and with some extra work, any $p \in \mathbb{N}.$ The derivation of this integral begins here, and uses the estimate at the bottom of this post to arrive at the aforementioned integral. With the help of this answer, I was able to prove that the integral

$$\displaystyle \int_{\mathbb{R^d}} \|x\|^{-d/2}\|\alpha - x\|^{-d/2}J_{d/2}(\rho \|x\|)J_{d/2}(\rho \|\alpha-x\|) \ \mathrm{d}x,$$

converges absolutely in all dimensions $d \geqslant 3,$ after using the asymptotic bound for the Bessel function, $|J_{\nu}(z)| \leqslant C|z|^{-1/2},$ for $z \rightarrow \infty.$ The second innermost integral likely follows similarly. However, this does not help us: if we do bound the Bessel functions in this way (under the assumption that $d \geqslant 3$), we end up with an integral of this form:

$$\displaystyle \int_{\mathbb{R}^d} \frac{1}{|k|^2} \ \mathrm{d}k,$$

which does not converge for $d \geqslant 3$. Morally, it seems that bounding the Bessel function in this way is too reckless, since the positive-negative cancellation that the Bessel functions exhibit may be necessary for convergence.

Numerically, the integral does appear to converge (at least, for $d = 2$ and perhaps $d = 3$). One idea I had was to use the asymptotics of the Bessel function for large argument (or at least, the first few terms of it), but the resulting integral still seems very difficult. Another idea I had was that these integrals look like convolutions, but that does not seem to help.

I don't have much experience in dealing with integrals involving Bessel functions. I would appreciate any help anyone can offer on how to deal with this integral.

I would particularly like to be able to show that the entire expression is $O(\rho^a)$ for some power $a$. I suspect that it will be $O(\rho^{-2 + \epsilon})$ for some $\epsilon > 0$. I have tried several approaches, including expressing both inner integrals as convolutions, finding alternative bounds for the Bessel functions, using the asymptotics of the Bessel functions, integration by parts, but I've not been able to get any results out of any of these attempts.

• The Bessel functions of half-integer order can be expressed in terms of elementary functions, which should simplifiy the problem a bit wolframalpha.com/input/?i=besselj%5B1%2F2,x%5D – tired Jan 23 '17 at 13:17
• Furthermore it seems that due to the heavy oscilallations, the main contributions to the inner two inner integrals might be found by taking into account the points $x=0$ and $x=\pm k$ – tired Jan 23 '17 at 13:29