# 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?

• What are the integral bounds in your edit? Please check the formula. Jan 24, 2023 at 8:51
• The integral is over the circle, with $\psi$ ranging from $0$ to $2\pi$. Jan 24, 2023 at 14:26
• (Since the integrand is periodic, it doesn't matter what range you pick as long as you cover the whole circle.) Jan 24, 2023 at 23:28
• There might be a simple formula if one of the Bessel functions is switched to a Neumann function instead.
– Buzz
Jan 25, 2023 at 21:51

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 $$u\geq v > 0$$.

Comment
$${{ J}_{0}\left(m\sqrt {{r}^{2}+{\rho}^{2}-2\,r\rho\,\cos \left( \phi \right) }\right)}=\sum _{k=-\infty }^{\infty }{{ J}_{k}\left(m \rho\right)}{{ J}_{k}\left(mr\right)}{{\rm e}^{i k\phi}}$$
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 $$m$$.