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?