I need some help proving the following identity:
$$\sum_{n\in\mathbb{Z}}J_n(x)\,J_{n+m}(x) = 0.$$
I wonder if it is possible to use the generating function identity:
$$ g(x,t) = \exp\left[\frac{x}{2}\left(t-\frac{1}{t}\right)\right]=\sum_{n\in\mathbb{Z}}J_n(x)\,t^n,$$
maybe through the substitution $t\to \frac{1}{t}$.
A little confused. Help appreciated.
Thank you.
From the Jacobi-Anger expansion: $$e^{ix\cos\theta}=\sum_{n\in\mathbb{Z}}i^n J_n(x)\, e^{i n\theta}\tag{1}$$ it follows that: $$e^{i(x\cos\theta-m\theta)}=\sum_{n\in\mathbb{Z}}i^n J_n(x) e^{i (n-m)\theta}=i^m\sum_{n\in\mathbb{Z}}i^n J_{n+m}(x)e^{in\theta}$$ so, mapping $\theta$ into $\frac{\pi}{2}-\theta$ then into $-\theta$: $$e^{ix\sin\theta} = \sum_{n\in\mathbb{Z}}(-1)^n J_n(x)\, e^{-in\theta},\tag{2}$$ $$(-1)^m e^{i(x\sin\theta+m\theta)}=\sum_{n\in\mathbb{Z}}(-1)^n J_{n+m}(x)\,e^{-in\theta}$$ $$(-1)^m e^{-i(x\sin\theta+m\theta)}=\sum_{n\in\mathbb{Z}}(-1)^n J_{n+m}(x)\,e^{in\theta}\tag{3}$$ and by the orthogonality relations for the complex exponential, provided that $m\neq 0$: $$\sum_{n\in\mathbb{Z}}J_n(x)\,J_{n+m}(x)=\frac{(-1)^m}{2\pi}\int_{0}^{2\pi}e^{ix\sin\theta}e^{-i(x\sin\theta+m\theta)}\,d\theta = 0.\tag{4}$$