Proving $\sum _{k=1}^{\infty } J_k(k z)^2=\frac{1}{2 \sqrt{1-z^2}}-\frac{1}{2}$

Question

How to prove $$\sum _{k=1}^{\infty } J_k(k z){}^2=\frac{1}{2 \sqrt{1-z^2}}-\frac{1}{2},\ |z|<1$$ The supplementary one $\sum _{k=1}^{\infty } J_k(k z)=\frac{z}{2 (1-z)}$ might be useful. Any help will be appreciated.

Answer 1

We can use the integral representation for the product of Bessel functions DLMF \begin{equation} J_{\mu}\left(x\right)J_{\nu}\left(x\right)=\frac{2}{\pi}\int_{0}^{\pi/2}J_{\mu+\nu}\left(2x\cos\theta\right)\cos\left((\mu-\nu)\theta\right)\mathrm{d}\theta \end{equation} with $\mu=\nu=k$ and $x=kz$: \begin{equation} J_{k}^2\left(kz\right)=\frac{2}{\pi}\int_{0}^{\pi/2}J_{2k}\left(2kz\cos\theta\right)\mathrm{d}\theta \end{equation} From the quoted series, and using parity properties of the Bessel function, \begin{equation} \sum _{k=1}^{\infty } J_k(k x)=\frac{x}{2 (1-x)} \end{equation} gives \begin{align} 2\sum _{k=1}^{\infty } J_{2k}(2k x)&=\sum _{k=1}^{\infty } J_k(k x)+\sum _{k=1}^{\infty } J_k(-k x)\\ &=\frac{x}{2 (1-x)}-\frac{x}{2 (1+x)}\\ &=\frac{x^2}{1-x^2} \end{align} Then \begin{align} \sum_{k\ge1}J_{k}^2\left(kz\right)&=\frac{2}{\pi}\int_{0}^{\pi/2}\sum_{k\ge1}J_{2k}\left(2kz\cos\theta\right)\mathrm{d}\theta\\ &=\frac{1}{\pi}\int_{0}^{\pi/2}\frac{z^2\cos^2\theta}{1-z^2\cos^2\theta}\mathrm{d}\theta\\ &=\frac{1}{\pi}\left[-\frac{\pi}{2}+\int_{0}^{\pi/2}\frac{1}{1-z^2\cos^2\theta}\mathrm{d}\theta\right]\\ &=-\frac12+\frac1\pi\int_0^\infty \frac{dt}{t^2+1-z^2} \text{ ...by changing }t=\tan\theta\\ &=\frac{1}{2\sqrt{1-z^2}}-\frac{1}{2} \end{align} as expected.