Closed form of $\sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2}$

Solution 1:

Here is a route.

Recall that

$$ J_0(2n)=\frac 1\pi \int_0^\pi \cos (2n \sin x)\:{\rm d}x \tag1 $$

and that $$\sum_{n=1}^\infty\frac{\cos nt}{n^2}=\frac{\pi^2}{6}-\frac{\pi t}{2}+\frac{t^2}{4},\quad 0\leq t\leq 2\pi. \tag2 $$ Then, due to normal convergence of the series on $[0,2\pi] $, we may write $$ \begin{align} \sum_{n=1}^\infty\frac{J_0(2n)}{n^2} & =\frac 1\pi \int_0^\pi \sum_{n=1}^\infty\frac{\cos (2n \sin x)}{n^2}{\rm d}x \\\\ & =\frac 1\pi \int_0^\pi\left(\frac{\pi^2}{6}-\pi \sin x+\sin^2 x\right){\rm d}x \\\\ & =\frac{\pi^2}{6}-\frac{3}{2}\\\\ & =0.144934066848226436... \end{align} $$ thus

$$ \sum_{n=1}^{\infty} \frac{J_0(2n)}{n^2} = \frac{\pi^2}{6}-\frac{3}{2}. $$

Using a similar technique, one may obtain the following result.

$$ \begin{align} \sum_{n=1}^\infty\frac{J_0(2 n \alpha )}{n^2} &=\frac{\pi^2}{6}-2\alpha+\frac{\alpha^2}{2},\qquad \alpha \in [0,\pi).\tag3 \end{align} $$

Solution 2:

We could seemingly use Olivier Oloa's approach to also evaluate $$\sum_{n=1}^{\infty} {\color{red}{(-1)^{n-1}}} \frac{J_{0}(2n)}{n^{2}} \tag{1}.$$

But I thought it might be worthwhile to mention that $(1)$ can also be evaluated using the residue theorem from complex analysis.

The first thing to recognize is that $J_{0}(z)$ is an entire function.

Now let $a$ be a positive parameter, and consider the contour integral $$\int_{|z|=N+\frac{1}{2}} \frac{\pi \csc(\pi z)J_{0}(2az)}{z^{2}} \, dz. $$

As $\text{Im}(z) \to + \infty$, the magnitude of $\csc(\pi z)$ decays like $2 e^{- \pi \, \text{Im}(z)}$.

And as $\text{Im}(z) \to - \infty$, the magnitude of $\csc(\pi z)$ decays like $2 e^{\pi \, \text{Im}(z)}$.

Combining this with the asymptotic behavior of $J_{0}(z)$ as $|z| \to \infty$, we can conclude that that as $N \to \infty$ discretely, the contour integral will vanish if $0 < a \le \frac{\pi}{2}$.

(And since $J_{0}(0)=1$, the contour integral will also vanish if $a=0$.)

Summing the residues, and using the fact that $\pi \csc(\pi z)$ has simple poles at the integers with residues that alternate between $1$ and $-1$, we get $$0=2 \pi i \left( -2 \sum_{n=1}^{\infty} (-1)^{n-1} \frac{J_{0}(2an)}{n^{2}} + \text{Res} \left[\frac{\pi \csc(\pi z)J_{0}(2az)}{z^{2}},0 \right] \right). $$

But $$ \begin{align}\frac{\pi \csc(\pi z)J_{0}(2az)}{z^{2}} &= \frac{1}{z^{2}} \left(\frac{1}{z} + \zeta(2) z + \mathcal{O}(z^{3})\right) \left(1-a^{2}z^{2} + \mathcal{O}(z^{4}) \right) \tag{2} \\ &=\frac{1}{z^{3}} + \frac{{\color{red}{\zeta(2) - a^{2}}}}{z} + \mathcal{O}(z) . \end{align}$$

Therefore, $$\text{Res} \left[\frac{\pi \csc(\pi z)J_{0}(2az)}{z^{2}},0 \right] = \zeta(2) - a^{2} = \frac{\pi^{2}}{6}-a^{2}, $$

and

$$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{J_{0}(2an)}{n^{2}} = \frac{\pi^{2}}{12} - \frac{a^{2}}{2} \, , \quad 0 \le a \le \frac{\pi}{2}. $$

(Since $J_{0}(x)$ is an even function, $(3)$ actually holds for $ -\frac{\pi}{2} \le a \le \frac{\pi}{2}$.)

$(2)$ https://en.wikipedia.org/wiki/Trigonometric_functions#Series_definitions

EDIT:

More generally, if $\alpha_{k}$ is a real parameter, $$\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\prod_{k=1}^{m} J_{0}(2 \alpha_{k}n) }{n^{2}} = \frac{\pi^{2}}{12} - \frac{\sum_{k=1}^{m}\alpha_{k}^{2}}{2} \, , \quad \sum_{k=1}^{m} \left| \alpha_{k} \right| \le \frac{\pi}{2}. $$