What is the closed form of $\sum _{n=1}^{\infty }{\frac { {{\it J}_{0}\left(n\right)} ^2}{{n}^4}}$? Using Maple I am obtaining the numerical approximation
$$0.5902373619$$
Please, let me know what is the closed form.  Many thanks.
 A: Using the wonderful method of Olivier implemented by Maple it is possible to prove that
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)} ^{2}}{{n}^{4}}}=-{\frac {3
}{64}}\,{\alpha}^{4}+{\frac {1}{90}}\,{\pi }^{4}-{\frac {1}{12}}\,{\alpha}^{2}{
\pi }^{2}+{\frac {32}{27}}\,{\frac {{\alpha}^{3}}{\pi }}
$$
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)}  ^{2}}{{n}^{2}}}=\frac{1}{4}\,{
\alpha}^{2}+\frac{1}{6}\,{\pi }^{2}-{\frac {4\alpha}{\pi }}
$$
For this last sum, when $\alpha =2 $ we obtain
$$\sum _{n=1}^{\infty }{\frac {  {{\it J}_{0}\left(\,2\,n\right)}
  ^{2}}{{n}^{2}}}=1+\frac{1}{6}\,{\pi }^{2}-8\,{\pi }^{-1}
$$
that is justly the result derived  here.
Other results:
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)}  ^{2}}{{n}^{6}}}={\frac {1}
{945}}\,{\pi }^{6}+{\frac {1}{64}}\,{\pi }^{2}{\alpha}^{4}-{\frac {1}{
180}}\,{\pi }^{4}{\alpha}^{2}+{\frac {5}{1152}}\,{\alpha}^{6}-{\frac {
512}{3375}}\,{\frac {{\alpha}^{5}}{\pi }}
$$
$$\sum _{n=1}^{\infty }{\frac {  
{{\it J}_{0}\left(\,\alpha\,n\right)}  ^{2}}{{n}^{8}}}=-{\frac {{\pi }^{6}{\alpha}^{2}
}{1890}}\,+{\frac {1}{960}}\,{\pi }^{4}{\alpha}^
{4}-{\frac {35}{147456}}\,{\alpha}^{8}-{\frac {5}{3456}}\,{\pi }^{2}{
\alpha}^{6}+{\frac {1}{9450}}\,{\pi }^{8}+{\frac {4096}{385875}}\,{
\frac {{\alpha}^{7}}{\pi }}
$$
A: Hint.
Observe that
$$
J_0(n)=\frac 1\pi \int_0^\pi \cos (n \sin x)\:{\rm d}x \tag1
$$
and that
$$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi t^3}{12}-\frac{t^4}{48},\quad 0\leq t\leq 2\pi. \tag2 $$
Then, due to normal convergence of the series on $[0,2\pi] $, we may write
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} & =\frac{1}{\pi^2} \int_0^\pi \!\!\int_0^\pi  \sum_{n=1}^\infty\frac{\cos (n \sin x)\cos (n \sin y)}{n^4}{\rm d}x\:{\rm d}y .\tag3
\end{align}
$$
We may plug 
$$
2\cos (n \sin x)\cos (n \sin y)=\cos (n(\sin x+\sin y))+\cos (n(\sin x-\sin y)) \tag4
$$
into $(3)$ and integrate as here.
Hence we obtain

$$ \sum_{n=1}^\infty\frac{J_0^2(n)}{n^4}=\frac{\pi ^4}{90}-\frac{\pi ^2}{12}-\frac{3}{64}+\frac{32}{27 \pi }. \tag5 $$

A numerical value is
$$
\sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} =0.5902373616900361395467798486 \ldots.
$$

Some details.
Identity $(2)$ may be rewritten as
$$\sum_{n=1}^\infty\frac{\cos nt}{n^4}=\frac{\pi^4}{90}-\frac{\pi^2 t^2}{12}+\frac{\pi |t|^3}{12}-\frac{t^4}{48},\quad -\pi \leq t\leq \pi. \tag6 $$
Then using $(3)$, $(4)$ and $(6)$, we get
$$
\begin{align}
\sum_{n=1}^\infty\frac{J_0^2(n)}{n^4} =\frac{1}{2\pi^2}\int_0^\pi \!\!\int_0^\pi  
\left[\left(\frac{\pi^4}{90}-\frac{\pi^2}{12}(\sin x+\sin y)^2+\frac{\pi }{12}(\sin x+\sin y)^3-\frac{1}{48}(\sin x+\sin y)^4\right)+\left(\frac{\pi ^4}{90}-\frac{\pi ^2}{12}(\sin x-\sin y)^2+\frac{\pi }{12}\left|\sin x-\sin y\right|^3-\frac{1}{48}(\sin x-\sin y)^4\right)\right]{\rm d}x\:{\rm d}y .\tag7
\end{align}
$$
We just have to be careful with the computation of the term ($|t|^3=\left({\rm Abs} (t)\right)^3$)
$$
\begin{align}
\frac{1}{2\pi^2}\times \frac{\pi }{12}\times \int_0^\pi \!\!\int_0^\pi \left|\sin x-\sin y\right|^3 {\rm d}x\:{\rm d}y & = \frac{1}{2\pi^2}\times\frac{\pi }{12} \times 4\int_0^{\pi/2} \!\!\int_0^{\pi/2} \left|\sin x-\sin y\right|^3 {\rm d}x\:{\rm d}y \\\\
&=-\frac{13}{36}+\frac{32}{27 \pi }.
\end{align}
$$
