Series of Bessel function

Evaluate the series:

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

where $J_0$ is the Bessel function of the first kind.

Solution



We are invoking the formulae:

$ \color{gray}{\bullet} \;\;\;\; \displaystyle  J_0 (2n)= \frac{1}{\pi}\int_{0}^{\pi}\cos (2n \sin x)\, {\rm d}x  $
$ \color{gray}{\bullet} \;\;\;\; \displaystyle \sum_{n=1}^{\infty}\frac{\cos nx}{n^2} = \frac{\pi^2}{6}- \frac{\pi x}{2} + \frac{x^2}{4}, \; x \in [-\pi, \pi] $

Note that the series in the second  equation  converges uniformly due to the $M$ test hence we can interchange integral and summation. Therefore:

$$\begin{aligned} \sum_{n=1}^\infty\frac{J_0(2n)}{n^2} & =\frac 1\pi \int_0^\pi \sum_{n=1}^\infty\frac{\cos (2n \sin x)}{n^2}{\rm d}x \\ & =\frac 1\pi \int_0^\pi\left(\frac{\pi^2}{6}-\pi \sin x+\sin^2 x\right){\rm d}x \\& =\frac{\pi^2}{6}-\frac{3}{2} \end{aligned}$$