Proving an integral is finite I have the following integral:
$$\displaystyle \int_{\mathbb{R}^2} \left( \int_{\mathbb{R}^2} \frac{J_{1}(|\alpha|)J_{1}(|k- \alpha|)}{|\alpha||k-\alpha|} \ \mathrm{d}\alpha \right)^2 \ \mathrm{d}k,$$
where both $\alpha$ and $k$ are vectors in $\mathbb{R}^2$, with $k \neq 0$, and $J_{\nu}$ denotes the Bessel function of the first kind. I'm having some trouble with the best way to approach this integral. If we focus on the inner integral first, then using the fact that for sufficiently large, positive $z$ we have $|J_{\nu}(z)| \leqslant C|z|^{-1/2},$ then the inner integral can be reduced to
$$\displaystyle \int_{\mathbb{R}^2} |\alpha|^{-3/2}|k-\alpha|^{-3/2} \ \mathrm{d}\alpha.$$
However, as can be seen in this answer, this integral is $O(|k|^{-1})$, which, after squaring, is clearly not integrable over all $|k| \geqslant 1$ after switching to polar co-ordinates (obviously not including $0$ in the lower limit of the outer integral). We would need an estimate of at least $O(|k|^{-1 - \epsilon})$ for any $\epsilon > 0$ to guarantee convergence of the outer integral.
One idea might be to try to bring the outer integral inside (though one would need to justify interchanging the order of integration). Using the asymptotics for the Bessel functions gives a product of cosines, and then one can use polar co-ordinates (taking $r = |\alpha|$). This would cancel out the $|\alpha|$ in the denominator, but then the $|k-\alpha|$ terms get very messy, which seems to make things worse. The Bessel functions appear to cause the most trouble. Does anyone have any ideas on how to proceed?
One idea is to notice (as someone suggested in the comments) that the above is the $L^2$ norm of a convolution, and also that (up to a constant) we have
$$\displaystyle f(\xi) = \frac{J_{d/2}(|\xi|)}{|\xi|^{d/2}} = \mathcal{F}(\chi)(\xi),$$
where $\chi$ is the characteristic function of the unit ball in $\mathbb{R}^d$. It can be shown that $f \in L^2(\mathbb{R}^d)$ and even $L^{p}(\mathbb{R}^d)$ for any $p \geqslant 2$. This lets us write the entire expression as
$$\|f \ast f\|_2^2 = \|\mathcal{F}(\chi) \ast \mathcal{F}(\chi)\|_2^2,$$
but unfortunately there is no kind of convolution theorem for functions on $L^2(\mathbb{R}^d)$ that I am aware of, without going into the theory of distributions. Moreover, $f$ does not even belong to $S(\mathbb{R}^d)$ for any $d$, so we cannot say much about the convolution. Thus, the problem is equivalent to asserting the finiteness of the above norm, $\|f \ast f\|_2$. If anyone has any ideas on how else the problem could be approached, then I would be very keen to hear about them.
 A: I think this problem can be tackled by bounding $\left|\frac{J_1(z)}{z}\right|$ with $\frac{1}{2}e^{-z^2/8}$ for $z\in[0,2\pi]$ and with $\sqrt{\frac{2}{\pi z^3}}$ for $z\geq 2\pi$. These bounds come from the Taylor series at the origin and Laplace method. Let $f(z)=\frac{J_1(z)}{z}$ for simplicity. We have:
$$ \int_{\mathbb{R}^2} f(\left|\alpha\right|)\,f(\left|k-\alpha\right|)\,d\mu\\ \leq \frac{1}{2}\int_{\left|k-\alpha\right|\leq 2\pi}f(\left|\alpha\right|)\,e^{-|k-\alpha|^2/8}\,d\mu+\sqrt{\frac{2}{\pi}}\int_{\left|k-\alpha\right|\geq 2\pi}f(\left|\alpha\right|)\frac{d\mu}{|k-\alpha|^{3/2}}$$
and the original integral should be simple to approximate by splitting the integration range in four parts:


*

*$\alpha$ and $k$ being close to each other and close to the origin;

*$\alpha$ and $k$ being close to each other, far from origin;

*$\alpha$ and $k$ being far from each other, both far from the origin;

*$\alpha$ and $k$ being far from each other, one of them being close to the origin.


For instance, let we approximate
$$ I_1 = \int_{\mathbb{R}^2}\left(\int_{\mathbb{R}^2}\exp\left(-\frac{1}{8}(\left|\alpha\right|^2+\left|k-\alpha\right|^2\right)\,d\alpha\right)^2\,dk .$$
By assuming $k=\rho_1(\cos\theta_1,\sin\theta_1)$ and switching to polar coordinates, the innermost integral equals
$$ \int_{0}^{2\pi}\int_{0}^{+\infty}\rho\,\exp\left(-\frac{1}{8}\left(2\rho^2+\rho_1^2-2\rho\rho_1\cos(\theta-\theta_1)\right)\right)\,d\rho\,d\theta $$
that is positive and bounded by
$$ 2\pi\int_{0}^{+\infty}\exp\left(-\frac{1}{8}\left(2\rho^2-\rho_1^2-2\rho\rho_1\right)\right)\,d\rho\leq 16\pi \exp\left(-\frac{(\rho_1-4)^2}{8}\right) $$
that is a Schwartz function in $\rho_1$. It follows that $I_1$ is finite.

On the other hand, most of the mass of the integral is concentrated on
   the region over which
   $\left|\alpha\right|,\left|k\right|,\left|\alpha-k\right|$ are large,
   and by this previous answer (that ultimately boils down to the
   triangle inequality) $I_3$ is not finite. This is how the world
   ends: not with a bang, but with a whimper.

