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Could anyone give an insight on how to prove the following formula?

$$\sum_{n=-\infty}^{+\infty}J_{n}(\alpha)J_{N+n}(\alpha)=\delta_{N0} \, ,$$

where $N$ is an integer. I checked many references but failed to figure out the calculation method. By the way, I found this relation during some numerical calculations about Bessel functions.

Thank you very much in advance.

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  • $\begingroup$ Hi, welcome to MO. Did you mean $\delta_{N,0}$? $\endgroup$
    – Amir Sagiv
    Commented Apr 19, 2017 at 6:44
  • $\begingroup$ Hi, thanks a lot for your answer. What I mean is that the value equals to 1 when $N = 0$, and equals to 0 in other cases. I have got the answer from Francois. :) $\endgroup$
    – Islaspace
    Commented Apr 19, 2017 at 7:06

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This is the case $y+z=0$ of the addition formula $J_N(y+z)=\sum_{n\in\mathbf Z}J_n(y)J_{N-n}(z)$, plus the fact that $J_k(z)=\frac1{2\pi}\int_0^{2\pi}\cos(k\theta-z\sin\theta)\,d\theta=J_{-k}(-z)$.

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