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