How to show $\sum_{n=-\infty}^\infty J_n J_{n+m} = \delta(m)$? The following is an identity concerning the Bessel functions of the first kind $J_n(x)$ for integers $n$ and $m$:
$$\sum_{n=-\infty}^\infty J_n(x) J_{n+m}(x) = \delta(m)$$
where $\delta(x)$ is the Kronecker delta function.
This can be derived from the Jacobi-Anger identity, but is there a simpler way to derive it, for instance using well-known recurrence relations of the Bessel functions?
 A: If $m = 2k+1$, this sum always contains the two different terms $J_n(x)J_{n+m}(x)$ and $J_{-(n+m)}(x)J_{-n}(x)$. I believe we can rewrite this sum as $$\sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) + J_{-(n+m)}(x)J_{-n}(x))$$ and then by using the identity $J_{-r}(x) = (-1)^{r}J_r(x)$ we can rewrite this as $$\sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) + (-1)^{2n+m} J_{n+m}(x)J_n(x) = \sum_{n=-k}^\infty (J_n(x)J_{n+m}(x) - J_{n+m}(x)J_n(x)) = 0.$$ So that does it for odd $m$, assuming that reordering the terms is legal in this case.
For even $m = 2k \neq 0$ I haven't a clue. You can try this same trick but it doesn't work, because all the $-1$s cancel out completely rather than leaving one behind to cancel out the whole sum. Plus there's a single awkward $J_{-k}(x)J_{k}(x)$ term in the middle. I'm betting the trick is to use the recurrence $${2r \over x}J_r(x) = J_{r-1}(x) + J_{r+1}(x)$$ to unzip the whole thing starting from the awkward term into two similar sums with $m' = m \pm 1$ (which obviously won't work if $m = 0$, as expected), and then use the above to show both parts are zero. But I have no concrete reason to believe that this will actually work, just a gut instinct.
A: Here's a hand-wavey argument using the generating function $e^{\frac{x}{2}(t-1/t)} = \sum_{m=-\infty}^\infty t^m J_m(z)$:
$$\begin{eqnarray}
e^{\frac{x}{2}(t-1/t)+\frac{x}{2}(u-1/u)} &=& e^{\frac{x}{2}(t-1/t)}e^{\frac{x}{2}(u-1/u)}\\\
&=& \left(\sum_{m=-\infty}^\infty t^m J_m(x)\right)\left(\sum_{n=-\infty}^\infty u^n J_n(x)\right)\\\
&=&\sum_{m=-\infty}^\infty\;\sum_{n=-\infty}^\infty t^{m}u^{n}J_m(x)J_n(x)\\\
&=&\sum_{m=-\infty}^\infty\;\sum_{k=-\infty}^\infty t^{m}u^{m+k}J_m(x)J_{m+k}(x)\\\
&=&\sum_{k=-\infty}^\infty u^k\sum_{m=-\infty}^\infty (tu)^{m}J_m(x)J_{m+k}(x)\;.
\end{eqnarray}$$
Now let $t=1/u$ and we get
$$1 = \sum_{k=-\infty}^\infty u^k \left(\sum_{m=-\infty}^\infty J_m(x)J_{m+k}(x)\right)\;,$$
which gives the desired result (since the parenthesized term is independent of $u$).
[Thanks to Joriki for cleaning up the LaTeX.]
