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An series related to the eta Dedekind function

Prove that

    \[\sum_{n=1}^{\infty} \frac{1}{\sinh^2 n \pi} = \frac{1}{6} - \frac{1}{2\pi }\]

Solution

Let us consider the function

    \[f(z) = \frac{\cot \pi  z}{\sinh^2 \pi z}\]

and integrate it along a quadratic counterclockwise contour \Gamma_N with vertices \displaystyle \frac{N}{2} \left ( \pm 1 \pm i \right ) where N is a big odd natural number. Hence,

    \begin{align*} \oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z &= \int_{N/2}^{-N/2} f \left ( x + \frac{i N}{2} \right )\, \mathrm{d}x + i \int_{N/2}^{-N/2} f \left ( i y + \frac{N}{2} \right ) \, \mathrm{d}y \\ &\quad \quad \quad + \int_{-N/2}^{N/2} f \left ( x - \frac{i N}{2} \right ) \, \mathrm{d}x + i \int_{-N/2}^{N/2} f \left ( iy + \frac{N}{2} \right ) \, \mathrm{d}y \end{align*}

We note that \displaystyle \lim_{N \rightarrow +\infty} f \left (x \pm \frac{i N}{2} \right ) = \frac{\mp i}{\cosh^2 \pi x} and that \displaystyle \lim_{N \rightarrow +\infty} f \left (i y \pm \frac{N}{2} \right ) = 0.

It’s also easy to see that

    \[\int_{-\infty}^{\infty} \frac{\mathrm{d}x}{\cosh^2 \pi x } = \left [ \frac{\tanh \pi x}{\pi} \right ]_{-\infty}^{\infty} = \frac{2}{\pi}\]

Hence, as N \rightarrow +\infty we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = - \frac{4 i}{\pi}\]

By Residue theorem we have that

    \[\oint \limits_{\Gamma_N} f(z) \, \mathrm{d}z = 2 \pi i \sum \mathfrak{Res} \left ( f; z = k \right ) + \mathfrak{Res} \left ( f; z = ik \right )\]

It is straightforward to show that

    \[\begin{matrix} \mathfrak{Res} \left ( f;z=k \right ) = &\mathfrak{Res} \left ( f;z=ik \right ) = & \dfrac{1}{\pi \sinh^2 \pi k} \\\\ \mathfrak{Res} \left ( f;z=0 \right ) & = & -\dfrac{2}{3\pi} \end{matrix}\]

Hence,

    \[\oint \limits_{\Gamma_N} f(z)\, \mathrm{d}z = 8 i\sum_{k=1}^{\infty}\frac{1}{\sinh^2 n \pi}-\frac{4 i}{3}\]

in the limit N \rightarrow +\infty. The result follows.

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