$J_0(\gamma)=\sum_{m=-\infty}^\infty(J_m(a)J_m(\beta)e^{im\theta})$ and a.β,γ be triangle edges

Question

How can I proove his relation for a,β,γ be triangle edges and $\gamma^2=a^2+\beta^2-2a\beta cos\theta $ $$J_0(\gamma)=\sum_{m=-\infty}^\infty(J_m(a)J_m(\beta)e^{im\theta})$$

Answer 1

Use the following integral representation : $$J_m(\alpha)=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i(\alpha\sin\tau-m\tau)}$$

Substitute that into the RHS to obtain

$$RHS=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i\alpha\sin\tau}\sum_{m=-\infty}^{\infty}J_{m}(\beta)e^{im(\theta-\tau)}$$

which by virtue of the generating function of the Bessel functions can be written as follows: $$\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i\alpha\sin\tau+i\beta\sin(\theta-\tau)}=\frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i(\alpha-\beta\cos\theta)\sin\tau+i\beta\sin\theta\cos\tau}\\= \frac{1}{2\pi}\int_{-\pi}^{\pi}d\tau~e^{i \gamma\sin(\tau+\phi_0)}$$ where in the last line we defined an angle $\phi_0$ such that $$\gamma\cos\phi_0=\alpha-\beta\cos\theta~,~\gamma\sin\phi_0=\beta\sin\theta$$ Since the function $e^{i\gamma\sin\tau}$ is periodic with period $2\pi$, it is true that $$\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i\gamma\sin(\tau+\phi_0)}d\tau=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i\gamma\sin\tau}d\tau=J_0(\gamma)$$ which concludes the proof.