The Identity of the Modified Bessel Function of the first kind Recently, I read a letter, containing the following identity:
$$
\sum _{q=-\infty }^{\infty } \frac{(-1)^q I_q\left(\left| \alpha \right| ^2\right) I_q\left(\left| \alpha \right| ^2\right)}{2 q+1}=\frac{\sinh \left(2 \left| \alpha \right| ^2\right)}{2 \left| \alpha \right| ^2}. 
$$
The letter's author referred to the book "Integrals and Series Vol. 2: Special Functions", 
Gordon and Breach Science Publishers, 1st edition, 1992, A.P. Prudnikov, Y.A. Brychkov, and O.I. Marichev, which contains equation (5.8.6.2):
$$
\sum _{k=1}^{\infty } \frac{(-1)^k I_{\mu -k v}(z) I_{k v+\mu }(z)}{k^2-a^2}=\frac{I_v(z){}^2}{2 a^2}-\frac{\pi  \csc (\pi  a) I_{\mu -v a}(z) I_{a v+\mu }(z)}{2 a}. 
$$
My questions are 


*

*How are these two identities related?

*How to derive the first identity?
Thank you!
 A: I don't see a relation between the two. But your first identity can be deduced from the relation
$$
\sum_{n=-\infty}^\infty \frac{J_n(y)^2}{n+x}=\frac\pi{\sin(\pi x)}J_x(y)J_{-x}(y)
\tag{2.3}
$$
established by M. D. Rogers (2005) (his equation numbering). In fact, putting $x=\frac12$ and using $J_{1/2}(y)=\left(\frac2{\pi y}\right)^{1/2}\sin y$ and $J_{-1/2}(y)=\left(\frac2{\pi y}\right)^{1/2}\cos y$, (2.3) becomes
$$
\sum_{n=-\infty}^\infty \frac{J_n(y)^2}{2n+1}=\frac{\sin(2y)}{2y}.
\tag{4.2}
$$
Now putting $y=iz$ and using $J_n(iz)=i^nI_n(z)$ and $\sin(2iz)=i\sinh(2z)$, we get your first identity.

Edit. The relation with your second identity is as follows (courtesy of Lerche's 1966 paper that you and Zurab quote, p.1074): Multiplying each term of the sum (2.3) by $\frac{n-x}{n-x}$, we get that the left-hand side equals
$$
\sum_{n=-\infty}^\infty \frac{nJ_n(y)^2}{n^2-x^2} -
x\!\!\!\sum_{n=-\infty}^\infty \frac{J_n(y)^2}{n^2-x^2}\\
$$
As the first sum is odd and the second is even, this is $\smash[b]{-x\Bigl[\frac{J_0(y)^2}{-x^2}+2\sum\limits_{n=1}^\infty\frac{J_n(y)^2}{n^2-x^2}\Bigr]}$. Rearranging, we see that (2.3) asserts that
$$
\sum_{n=1}^\infty\frac{J_n(y)^2}{n^2-x^2} 
= \frac{J_0(y)^2}{2x^2} - \frac{\pi J_x(y)J_{-x}(y)}{2x\sin(\pi x)}.
$$
Replacing $y$ by $iz$, this is precisely the case $\mu=0$, $\nu=1$ of your second identity, with this typo corrected: the first term on the right-hand side should involve $I_\mu$, not $I_\nu$.
A: In addition to the Ziegler's answer: it seems the identity (2.3) from the Rogers' paper was first obtained by Lerche in 1966. See a review article http://www.hindawi.com/journals/amp/2009/425164/ref/ (A review of procedures for summing Kapteyn series in mathematical physics, by R.C. Tautz and  I. Lerche).
