I have the following integral:
$$\displaystyle \int_{\mathbb{R}^2} \left( \int_{\mathbb{R}^2} \frac{J_{1}(\rho |\alpha|)J_{1}(\rho|k- \alpha|)}{|\alpha||k-\alpha|} \ \mathrm{d}\alpha \right)^2 \ \mathrm{d}k,$$
with $\alpha, k \in \mathbb{R}^2$, $\rho$ constant, $J_{\nu}$ the Bessel function of the first kind, $|\cdot|$ the Euclidean norm on $\mathbb{R}^2$.
I want to make the substitution $\alpha = s$, $k = s + t$. If the square wasn't there, this would be easy, and we'd just end up with
$$\displaystyle \int_{\mathbb{R}^2} \int_{\mathbb{R}^2} \frac{J_{1}(\rho |s|)J_{1}(\rho|t|)}{|s||t|} \ \mathrm{d}s \ \mathrm{d}t,$$
since the determinant of the Jacobian corresponding to this change of coordinates is $1$. But the square makes things difficult. What does the integral at the top look like after making that change of coordinates?
IMHO, you should consider apart the transformation of the inside integral.
What follows is a hint, far from a full answer.
2 remarks:
1) $f(\omega)=\pi\dfrac{\sqrt{3}}{2}r^{3/2}\dfrac{J_1(2 \pi \omega r)}{2 \pi \omega r}$ is the Fourier transform of a unit disk with radius $r$.
(http://isi.ssl.berkeley.edu/~tatebe/whitepapers/FT%20of%20Uniform%20Disk.pdf).
2) The inside integral is, up to a multiplicative constant, the convolution of two such transforms, otherwise said, it is a square of convolution.
Thus, the integral can be written, up to a multiplicative constant, as the integral of the Fourier transform of a the area common to two disks with radius $r$, this function being explicitly computable.