Bessel Function Series Closed Form $\sum_{m=1} \frac{1}{2m}J_{2m}(2mx)$ or $\sum_{m=1}J^2_{m\pm 1}(mx)$ or $\sum_{m=1}\frac{J^2_m(mx)}{m}$ I'm trying to find a closed form for the following sum,
$$\sum_{m=1}^\infty\frac{1}{2m}J_{2m}(2mx),$$
where $0<x<1$. I've been using standard references such as Gradshteyn and Ryzhik, Abramovitz and Stegun, Prudnikov et al. which have proven fruitful for almost all cases, except for this sum which I'm now struggling with for a closed form.
Any help would be greatly appreciated!
UPDATE 1:
This arises in the context of finding a closed form for the series
$$
\sum_{m=1}^\infty\left(J^2_{m-1}(mx)-J^2_{m+1}(mx)\right),
$$
I've already found that
$$
\sum_{m=1}^\infty\left(J^2_{m-1}(mx)+J^2_{m+1}(mx)\right)=\frac{1}{\sqrt{1-x^2}},
$$ so by using this, equivalent series to find a closed form of would be
$$
\sum_{m=1}^\infty J^2_{m\pm1}(mx).
$$
Thanks in advance!
UPDATE 2:
Another equivalent infinite series is by using the recurrence relations after completing the square on
$$
\sum_{m=1}^\infty\left(J^2_{m-1}(mx)-J^2_{m+1}(mx)\right),
$$ which gives
$$
\frac{4}{x}\sum_{m=1}J_m(mx)J'_m(mx).
$$
But this can be realised instead as a derivative with respect to $x$ as follows
$$
\frac{2}{x}\frac{d}{dx}\sum_{m=1}\frac{J^2_m(mx)}{m}.
$$
Then, a fourth equivalent series that would answer my question would be
$$
\sum_{m=1}\frac{J^2_m(mx)}{m}.
$$
This can be related to the original question by noting the relation
$$
J^2_m(mx)=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}J_{2m}(2mx\cos\theta)d\theta.
$$ Fubini allows us to bring the infinite sum inside the integral and this is where the original sum involving not a squared Bessel but even order terms only.
Effectively, finding a closed form to any of these series would aid me greatly!
UPDATE 3 (Edited):
After reading the comment below, we use the following two facts
$$
J_n(a)=\frac{1}{\pi}\int_0^\pi dt\,\cos(tn-a\sin t),
$$
and
$$
\sum_{n=1}\frac{\cos(nx)}{n}=-\frac{1}{2}\ln(2(1-\cos x))=-\ln(2\sin(x/2)).
$$ References can be find in either DLMF or Gradshteyn and Ryzhik.
The sum reads
$$
\frac{2}{x}\frac{d}{dx}\sum_{m=1}^\infty\frac{J^2_{m}(mx)}{m}=\frac{2}{x}\frac{d}{dx}\frac{2}{\pi}\int_0^{\frac{\pi}{2}}\sum_{m=1}^\infty\frac{J_{2m}(2mx\cos\theta)}{m}d\theta.
$$ Let's define $z=x\cos\theta<1$.
Then by the expansions, we have
$$
\begin{align*}
&=\frac{4}{\pi^2 x}\frac{d}{dx}\int_0^{\frac{\pi}{2}}d\theta\int_0^\pi dt\sum_{m=1}^\infty \frac{\cos(2m(t-z\sin t))}{m},\\
&=-\frac{4}{\pi^2 x}\frac{d}{dx}\int_0^{\frac{\pi}{2}}d\theta\int_0^\pi dt\ln\left(\sin(t-z\sin t)\right),\\
&=\frac{4}{\pi^2 x}\int_0^{\frac{\pi}{2}}d\theta\int_0^\pi dt\cos\theta\sin t \,\text{cot}(t-x\cos\theta\sin t).
\end{align*}
$$

We use this Bessel J identity to find an integral representation:
$$\sum_{n=1}^\infty \frac{J_{2n}(2nz)}{2n}=\int_0^\pi \sum_{n=1}^\infty \frac{\cos((2t-2z\sin(t))n)}{2\pi n}$$
to get:
$$\sum_{n=1}^\infty \frac{J_{2n}(2nz)}{2n}=-\frac1{2\pi}\int_0^\pi\ln(2\sin\left(t-z \sin(t)\right))dt$$
shown here. Maybe this integral problem will be easier to find a closed form. There are other possible integral representations that may be added later.
