I am interested in computing the following integral, which feels like something that must have been computed before: $$ \int \frac{k^2{\rm d} k}{2\pi^2}j_{\ell}(r k)j_{\ell'}(r' k) $$ From what I understand, there are two adjacent properties of Bessel functions that could be relevant. The first is $$ \int \frac{k^2{\rm d} k}{2\pi^2}j_{\ell}(r k)j_{\ell}(r' k) = \frac{1}{4\pi r^2}\delta(r - r')\,, $$ and the second is $$ \int \frac{{\rm d}z}{z} J_\alpha(z) J_\beta(z) = \frac{2}{\pi}\frac{\sin\left(\frac{\pi}{2}(\alpha - \beta)\right)}{\alpha^2 - \beta^2}\,. $$ Recalling that $$ j_\ell(k r) = \sqrt{\frac{\pi}{2 k r}}J_{\ell + 1/2}(k r)\,, $$ we see $$ \int {\rm d} k j_{\ell}(k r)j_{\ell'}(k r) = \frac{1}{r}\frac{\sin\left(\frac{\pi}{2}(\ell - \ell')\right)}{(\ell(\ell + 1)-\ell'(\ell ' + 1))}\,. $$ The first completeness relation clearly resolves the case of $\ell = \ell'$, while the orthogonality relation doesn't appear to directly help me.
Is there any hope to compute the integral I posed? Are there special cases that have been worked out?
A Conjecture:
The Table of Integrals Series and Products seems to have a closed form for the following integral
$$ \int_0^\infty {\rm d}t J_\nu(\alpha t) J_\mu(\beta t)t^{-\lambda} = \frac{\alpha^\nu \Gamma\left(\frac{\nu + \mu - \lambda + 1}{2}\right)}{2^{\lambda}\beta^{\nu - \lambda + 1}\Gamma\left(\frac{-\nu + \mu + \lambda + 1}{2}\right)\Gamma\left(\nu + 1\right)}F\left(\frac{\nu + \mu - \lambda + 1}{2},\frac{\nu - \mu - \lambda + 1}{2};\nu + 1 ; \frac{\alpha^2}{\beta^2}\right)\,, $$ where $F$ is the hypergeometric function and when $$ {\rm Re}(\nu + \mu - \lambda + 1)> 0, {\rm Re}\lambda > 1, 0 < \alpha < \beta\,. $$ What a mess. Nonetheless, it appears to give the correct answer, even if we bend these limits, taking $\lambda = -1$. We can rewrite this integral in terms of spherical Bessel functions $$ \int_0^\infty {\rm d}t J_\nu(\alpha t) J_\mu(\beta t)t^{-\lambda} = \int_0^\infty {\rm d}t \frac{2 t}{\pi}\sqrt{\alpha\beta}j_{\nu - 1/2}(\alpha t) j_{\mu - 1/2}(\beta t)t^{-\lambda} $$ Taking $\lambda = -1$, we have $$ \int_0^\infty {\rm d}t J_\nu(\alpha t) J_\mu(\beta t)t^{-\lambda} = \frac{2}{\pi}\sqrt{\alpha\beta}\int_0^\infty t^2 {\rm d}tj_{\nu - 1/2}(\alpha t) j_{\mu - 1/2}(\beta t) $$ Thus, taking $\nu = \ell + 1/2$ and $\mu = \ell' + 1/2$, with $\alpha = r$ and $\beta = r'$, we have $$ \int \frac{k^2{\rm d} k}{2\pi^2}j_{\ell}(r k)j_{\ell'}(r' k) =^? \frac{1}{2\pi}\frac{r^{\ell} \Gamma\left(\frac{\ell + \ell' + 3}{2}\right)}{r'^{\ell + 3}\Gamma\left(\frac{\ell'-\ell}{2}\right)\Gamma\left(\ell + \frac{3}{2}\right)}F\left(\frac{\ell + \ell' + 3}{2},\frac{\ell' - \ell + 2}{2};\ell + \frac{3}{2} ; \frac{r^2}{r'^2}\right) $$
The only contribution that this appears to miss is a $\delta$-function. For example, computing explicitly $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_0(k r)j_6(k r') = \Theta(r' - r)\frac{21 \left(33 r^4-30 r^2 r'^2+5 r'^4\right)}{64 \pi r'^7} - \frac{\delta(r - r')}{4\pi r'^2}\,, $$ and $$ \int\frac{k^2{\rm d} k}{2\pi^2}j_1(k r)j_2(k r') = -\frac{-4 r r'^3+12 r^3 r'+\left(r^2-r'^2\right)\left(3 r^2+r'^2\right) \log \left(\frac{(r-r')^2}{(r'+r)^2}\right)}{16 \pi ^2 r^2 r'^3 \left(r^2-r'^2\right) } $$ which matches the hypergeometric expression up to the delta function contribution. Thus, upon considering several other examples, my conjecture is $$ \int \frac{k^2{\rm d} k}{2\pi^2}j_{\ell}(r k)j_{\ell'}(r' k) = \left(\cos^2\left(\frac{\pi}{2}(\ell_< + \ell_>)\right)\Theta(r_> - r_<) + \sin^2\left(\frac{\pi}{2}(\ell_< + \ell_>)\right)\right)\frac{1}{2\pi}\frac{r_<^{\ell_<} \Gamma\left(\frac{\ell_< + \ell_> + 3}{2}\right)}{r_>^{\ell_< + 3}\Gamma\left(\frac{\ell_>-\ell_<}{2}\right)\Gamma\left(\ell_< + \frac{3}{2}\right)}F\left(\frac{\ell_< + \ell_> + 3}{2},\frac{\ell_> - \ell_< + 2}{2};\ell_< + \frac{3}{2} ; \frac{r_<^2}{r_>^2}\right) + \cos\left(\frac{\pi}{2}(\ell_< + \ell_>)\right)\frac{\delta(r_< - r_>)}{4\pi r_>^2}\,. $$ where $\ell_> = \max(\ell,\ell')$ and $\ell_< = \min(\ell,\ell')$, and $r_> = r$ if $\ell_> = \ell$ and $r_> = r'$ if $\ell_> = \ell'$, and $r_< = r$ if $\ell_< = \ell$ and $r_< = r'$ if $\ell_< = \ell'$.
Another integral Another integral in the same family that seems possible is the following $$ \int\frac{k^2{\rm d}k}{2\pi^2}\frac{1}{k^2 - q^2}j_{\ell}(r k)j_{\ell'}(r' k) $$ One specific example that may shed some light on how to compute more general cases is $$ \int\frac{k^2{\rm d}k}{2\pi^2}\frac{1}{k^2 - q^2}j_{0}(r k)j_{2}(r' k) = \frac{q}{4\pi}\Theta(r - r') j_{-1}(q r)j_2(q r') + \frac{q}{4\pi}\Theta(r' - r)\left[j_0(q r)j_{-3}(q r') - \frac{3}{q^3 r'^3}\right] $$ Note that the last term can be rewritten as $$ - \frac{3}{4\pi q^2 r'^3} = \frac{q}{4\pi}(j_2(q r') j_{-1}(q r')-j_{-3}(q r') j_0(q r')) $$