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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2 thoughts on “An series related to the eta Dedekind function”

Tolaso says:

November 25, 2020 at 3:56 pm

Using the same technique with the function

$f_a(z)=\frac{\cot(\pi z)}{\sinh^2(a z \pi)}$

we get that

$\sum_{n=1}^{\infty} \frac{1}{\sinh^2 n \pi a}+\frac{1}{a^2}\sum_{n=1}^{\infty} \frac{1}{\sinh^2 n \pi/a}=\frac{1+a^2}{6 a^2}-\frac{1}{\pi a}$

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Tolaso says:

November 25, 2020 at 3:58 pm

An interesting result is the following

$\sum_{k=1}^{\infty} \frac{k}{e^{2 k \pi} - 1 } = \frac{1}{24} - \frac{1}{8 \pi}$

since

$\begin{align*} \sum_{n=1}^{\infty}\frac{1}{\sinh^2\!\pi n}&=4\sum_{n=1}^{\infty}\frac{e^{-2\pi n}}{(1-e^{-2\pi n})^2}\\ &=4\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}k e^{-2\pi n k} \\ &=4\sum_{k=1}^{\infty}k \sum_{n=1}^{\infty}e^{-2\pi n k} \\ &=4\sum_{k=1}^{\infty}\frac{k}{e^{2\pi k}-1} \\ &=\frac{1}{6}-\frac1{2\pi} \end{align*}$

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