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A Bessel series

Let \mathcal{J}_0 denote the Bessel function of first kind. Evaluate the series:

\displaystyle \mathcal{S}= \sum_{n=1}^{\infty} \frac{\mathcal{J}_0(2n)}{n^2}

Solution

We recall the fact that

(1)   \begin{equation*} \mathcal{J}_0(2n) = \frac{1}{\pi} \int_0^\pi \cos (2n \sin x) \, {\rm d}x \end{equation*}

as well as the Fourier series

(2)   \begin{equation*} \sum_{n=1}^{\infty} \frac{\cos nx}{n^2} = \frac{\pi^2}{6} - \frac{\pi x}{2} + \frac{x^2}{4} \quad , \quad 0 \leq x \leq 2\pi \end{equation*}

Hence,

    \begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{J}_0(2n)}{n^2} &= \sum_{n=1}^{\infty} \frac{1}{\pi n^2} \int_{0}^{\pi} \cos (2 n \sin x) \, {\rm d}x\\ &= \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{align*}

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1 Comment

  1. Some comments and extensions as well:

    Remark 1: Let \alpha \in [0, \pi) then it holds that:

        \begin{align*} \sum_{n=1}^{\infty} \frac{\mathcal{J}_0 \left ( 2n \alpha \right )}{n^2} &= \frac{\pi^2}{6} - 2 \alpha + \frac{\alpha^2}{2} \end{align*}

    Remark 2: Let \alpha \in \left[-\frac{\pi}{2}, \frac{\pi}{2} \right] then it holds that:

        \begin{align*} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}\mathcal{J}_0 \left ( 2n \alpha \right )}{n^2} &= \frac{\pi^2}{12}- \frac{\alpha^2}{2} \end{align*}

    Remark 3: Let \alpha_k be a real parameter such that \sum \limits_{k=1}^{m} \left| \alpha_{k} \right| \leq \frac{\pi}{2} . Then it holds that:

        \begin{align*} \sum_{n=1}^{\infty} (-1)^{n-1} \frac{\prod_{k=1}^{m} \mathcal{J}_{0}(2 \alpha_{k}n) }{n^{2}} &= \frac{\pi^{2}}{12} - \frac{1}{2}\sum_{k=1}^{m}\alpha_{k}^{2} \end{align*}

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