Sum over Bessel functions: what is $\sum_{n=1}^\infty J_n(u)J_n(v)\sin(n\alpha)$? By Neumann's addition theorem, we know that the following identity holds (including for complex $\alpha$):
$$J_0(u)J_0(v)+2\sum_{n=1}^\infty J_n(u)J_n(v) \cos(n\alpha) = J_0 \left( \sqrt{u^2+v^2-2uv \cos\alpha} \right)$$
What is the corresponding identity for
$$\sum_{n=1}^\infty J_n(u)J_n(v)\sin(n\alpha)=\text{ ?}$$
I know that the sum vanishes when it runs over all $n$ from $-\infty$ to $+\infty$ (https://dlmf.nist.gov/10.23) but I haven't been able to derive an answer for the sum over positive $n$ only. Any help is appreciated!
EDIT:
I managed to obtain an integral representation of the sum using the Hilbert transform on the circle:
$$2\sum_{n=1}^\infty J_n(u)J_n(v)\sin(n\alpha)=\frac{1}{2\pi}\int J_0 \left(\sqrt{u^2+v^2-2uv\cos\psi}\right)\cot\left(\frac{\alpha-\psi}{2} \right) \, d\psi+iJ_0 \left(\sqrt{u^2+v^2-2uv\cos\alpha}\right)$$
Does anyone know how to evaluate the above integral in closed form?
 A: Comment
"Neumann's addition theorem"
A special case of Gradshteyn & Ryzhik 8.530.2 is
$$
{{ J}_{0}\left(m\sqrt {{r}^{2}+{\rho}^{2}-2\,r\rho\,\cos \left( 
\phi \right) }\right)}=\sum _{k=-\infty }^{\infty }{{ J}_{k}\left(m
\rho\right)}{{ J}_{k}\left(mr\right)}{{\rm e}^{i k\phi}}
$$
Put $m=1$, then the real part is as the OP and the imaginary part is $0$.
But here $m$ as any complex number.  Maybe we can get more by varying $m$.  Say, put $m=i$ to get formulas with Bessel functions $I_k$.  Or differentiate with respect to $m$.  Or expand both sides in powers of $m$.
Gradshteyn, I. S.; Ryzhik, I. M.; Zwillinger, Daniel (ed.); Moll, Victor (ed.), Table of integrals, series, and products. Translated from the Russian. Translation edited and with a preface by Victor Moll and Daniel Zwillinger, Amsterdam: Elsevier/Academic Press (ISBN 978-0-12-384933-5/hbk; 978-0-12-384934-2/ebook). xlv, 1133 p. (2015). ZBL1300.65001.
A: I don't think a "closed form" expression exists, I tried several approaches. I guess the best one can do is to use a modified version of the OP's integral representation.
Denote
$$
w(\psi) = \sqrt{u^2+v^2-2uv\cos\psi},\tag{1}
$$
then
\begin{align}
S &= 2\sum_{n=1}^\infty J_n(u)J_n(v)\sin(n\alpha) \tag{2a}\\
&= \frac{\sin\alpha}{\pi} \int_0^\pi \mathrm d\psi
\frac{J_0[w(\psi)] - J_0[w(\alpha)]}{\cos\psi-\cos\alpha}.\tag{2b}
\end{align}
Note that the subtraction of $J_0[w(\alpha)]$ removes the simple pole at $\psi=\alpha$ (if $\alpha\in\mathbb R$), such that this integral is along the real axis, while the integral in the OP's edit was shifted infinitesimally to the upper complex plane, leading to the imaginery correction term. The identity (2) also holds for complex $u,v,\alpha$.
Changing the integration variable from $\psi$ to $\omega=w(\psi)$ is possible, but leads to additional square-root denominators, e.g.,
\begin{align}
S = \frac{4u\sin\alpha}{v\pi} \int_{u-v}^{u+v} \mathrm d\omega
\frac{J_0[\omega] - J_0[w(\alpha)]}{\sqrt{[1-(u+\omega)^2][(u-\omega)^2-1]}\,[\omega^2-w(\alpha)^2]}\tag{3}
\end{align}
if $u\geq v > 0$.
