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 


1. How are these two identities related?
2. How to derive the first identity?


Thank you!



The Identity of the Modified Bessel Function of the first kind
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)](http://www.ams.org/mathscinet-getitem?mr=2131261) (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](http://dx.doi.org/10.1063/1.1761804) 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 $I\_\nu$.


