I have two integers $m$ and $n$ and a real $t$. I'm trying to prove the following identity :
$$\sum\limits_{j=0}^{+\infty}J_{j+n}(t)J_{j+m}(t)=\frac{t}{2(m-n)}(J_{m-1}(t)J_n(t)-J_m(t)J_{n-1}(t))$$
where $J_n$ is the Bessel function of the first kind. I don't really know where to start to prove this.
Best Answer
\begin{equation} J_{j+n}(t)J_{j+m}(t)=\frac{t}{2(m-n)}\left(\frac{2}{t}[j+m-(j+n)]J_{j+n}(t)J_{j+m}(t)\right) \end{equation} \begin{equation} J_{j+n}(t)J_{j+m}(t)=\frac{t}{2(m-n)}\left(\frac{2}{t}[j+m]J_{j+n}(t)J_{j+m}(t)-\frac{2}{t}[j+n]J_{j+n}(t)J_{j+m}(t)\right) \end{equation} Then using the recurrence relation on Bessel functions $$ \frac{2n}{t}J_n(t)=J_{n-1}(t)+J_{n+1}(t)$$ One finds terms the general term of a telescopic sum leading to the result conjectured in the question.