Proving an integral is finite
Solution 1:
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.