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Sum with trilogarithm

Let {\rm Li}_3 denote the trilogarithm function. Prove that

\displaystyle \sum_{n=1}^{\infty} {\rm Li}_3 \left ( e^{-2n \pi} \right ) = \frac{7 \pi^3}{360} - \frac{\zeta(3)}{2}

where \zeta denotes the Riemann zeta function.

(Seraphim Tsipelis)

Solution [by Ramya Datta]

Let us recall the Fourier expansion

\displaystyle \pi\coth (\pi z) = \frac{1}{z} + 2z\sum\limits_{j=1}^{\infty} \frac{1}{z^2+j^2}

Hence

\begin{aligned} \sum\limits_{n=1}^{\infty} \text{Li}_3\left(e^{-2n\pi}\right) &= \sum\limits_{n=1}^{\infty}\sum\limits_{k=1}^{\infty} \frac{e^{-2nk\pi}}{k^3}\\ &= \sum\limits_{k=1}^{\infty} \frac{1}{k^3\left(e^{2k\pi} - 1\right)}\\ &= \frac{1}{2}\sum\limits_{k=1}^{\infty} \frac{1}{k^3}\left(\coth (k\pi) - 1\right)\\ &= \frac{1}{2\pi}\sum\limits_{k=1}^{\infty} \frac{1}{k^3}\left(\frac{1}{k}+2k\sum\limits_{j=1}^{\infty} \frac{1}{k^2+j^2}\right) - \\ &\quad \quad - \frac{1}{2}\zeta(3)\\ &= \frac{1}{2\pi}\left(\zeta(4) + 2\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty} \frac{1}{k^2(k^2+j^2)}\right)  - \frac{1}{2}\zeta(3)\\ &= \frac{1}{2\pi}\zeta(4) +  \frac{1}{2\pi}\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty}\left(\frac{1}{k^2(k^2+j^2)}+\frac{1}{j^2(k^2+j^2)}\right) -\\ &\quad \quad - \frac{1}{2}\zeta(3)\\ &= \frac{1}{2\pi}\zeta(4) +\\ &\quad \quad +\frac{1}{2\pi}\sum\limits_{k=1}^{\infty}\sum\limits_{j=1}^{\infty}\frac{1}{k^2j^2} - \frac{1}{2}\zeta(3) \\ &= \frac{1}{2\pi}\zeta(4) + \frac{1}{2\pi}\zeta(2)^2 - \frac{1}{2}\zeta(3) \\ &= \frac{7\pi^3}{360} - \frac{1}{2}\zeta(3) \end{aligned}

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