What is the closed form of $\sum _{n=1}^{\infty }{\frac { {{\rm J_{0}}\left(2\,n\right)} ^{2}}{{n}^{2}}}$?

Question

Using Maple I am obtaining

$$\sum _{n=1}^{\infty }{\frac { {{\rm J_0}\left(2\,n\right)} ^{2}}{{n}^{2}}} = 0.09845497463 $$

Please check it. Many thanks.

Answer 1

Yes, there is a closed form. Expand $\arcsin(t) \chi_{[-1,1]}(t)$ as a Fourier series on $[-\pi/2, \pi/2]$ (where $\chi_{[-1,1]}(t)$ is the indicator function of the interval $[-1,1]$.

$$ \arcsin(t) \chi_{[-1,1]} = \sum_{n=1}^\infty \dfrac{J_0(2n) - \cos(2n)}{n} \sin(2nt) $$

Now it seems that $$ \sum_{n=1}^\infty \dfrac{\cos(2n)}{n} \sin(2nt) = \cases{ -t - \pi/2 & if $-\pi/2 < t < -1$\cr -t & if $-1 < t < 1$\cr -t + \pi/2 & if $1 < t < \pi/2$\cr }$$

If $$g(t) = \cases{-t - \pi/2 & if $-\pi/2 < t < -1$\cr -t + \arcsin(t) & if $-1 < t < 1$\cr -t + \pi/2 & if $1 < t < \pi/2$\cr }$$ Parseval's theorem says

$$ \eqalign{\sum_{n=1}^\infty \dfrac{J_0(2n)^2}{n^2} &=\dfrac{2}{\pi} \int_{-\pi/2}^{\pi/2} g(t)^2\; dt\cr &= \dfrac{\pi^2}{6} + 1 - \dfrac{8}{\pi}} $$

EDIT:

Somewhat more generally, if $0 \le s \le \pi/2$ we have

$$\arcsin(t/s) \chi_{[-s,s]}(t) = \sum_{n=1}^\infty \dfrac{J_0(2ns) - \cos(2ns)}{n} \sin(2\pi t)$$

and $$\sum_{n=1}^\infty \dfrac{\cos(2ns)}{n} \sin(2nt) = \cases{-t - \pi/2 & if $-\pi/2 < t < -s$\cr -t & if $-s < t < s$\cr -t + \pi/2 & if $s < t < \pi/2$\cr}$$ so that Parseval says $$ \sum_{n=1}^\infty \dfrac{J_0(2ns)^2}{n^2} = \dfrac{\pi^2}{6} - 8 \dfrac{s}{\pi} + s^2 $$