# Finishing off integral of a Bessel function

I have the following $d$-dimensional integral:

$$\displaystyle \int_{|\alpha - k| \leqslant 3|k|}\frac{J_{d/2}(\rho |\alpha|)J_{d/2}(\rho |\alpha - k|)}{|\alpha|^{d/2}|\alpha - k|^{d/2}} \ \mathrm{d}\alpha,$$

where $k \in \mathbb{R}^d$ is such that $\frac{1}{2}k \leqslant |\alpha| \leqslant 2|k|$, $\rho \gg 0$ is a constant, $J_{\nu}$ denotes the Bessel function of the first kind, and $|\cdot|$ denotes the standard Euclidean norm $\|\cdot\|$. What I am trying to do is imitate the argument given in this answer but applying it to my integrand. (The limits here represent the partition $E_1$, defined in the answer linked to above.) After using the bound $|\alpha - k| \leqslant 3|k|$, and using the bound $J_{\nu}(z) \leqslant C_{\nu}|z|^{-1/2}$, we arrive at:

$$\displaystyle |k|^{-d/2}\int_{0}^{3|k|} \int_{|\alpha - k| = r}\frac{J_{d/2}(\rho |\alpha|)J_{d/2}(\rho |\alpha - k|)}{|\alpha - k|^{d/2}} \ \mathrm{d}S \ \mathrm{d}r \\ \leqslant C_d|k|^{-d/2}\int_{0}^{3|k|}r^{\frac{d-2}{2}}J_{d/2}(\rho r) \ \mathrm{d}r.$$

Does anyone know what this integral is, explicitly? For technical reasons, I don't want to use the same bound on the Bessel function to bound the Bessel function in the integrand (the short answer is that I then want to integrate this whole thing over $\mathbb{R}^d$ with respect to $k$, which would result in something divergent).

Does anyone know if this appears in a table of integrals somewhere?

I've just found is equation 5.52 in Gradshteyn-Ryzhik, which says that

$$\displaystyle \int x^{p+1}J_{p}(x) \ \mathrm{d}x = x^{p+1}J_{p+1}(x).$$

I'd also like to know if something can also be said about the integral

$$\displaystyle \int_{|k|}^{\infty} r^{-1}J_{d/2}(\rho r) \ \mathrm{d}r,$$

which arises when considering $E_3$ (in the answer linked to above).