Closed-form for $\sum_{n=1}^{\infty} \left(m n \, \text{arccoth} \, (m n) - 1\right)$

Solution 1:

Denote $$\mathscr{F}(\xi)=\sum_{n=1}^{\infty}n\left ( \operatorname{arcoth}(n\xi) -\frac{1}{n\xi} \right ).$$ Note that $\lim_{\xi\to\infty}\mathscr{F}(\xi)=0$ and $$ \begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\xi}\mathscr{F}(\xi) &=\sum_{n=1}^{\infty}n^2\left ( \frac{1}{n^2\xi^2} -\frac{1}{n^2\xi^2-1} \right ). \end{aligned} $$ Consider for $0\le\omega\le1$, $$\begin{aligned} \mathscr{S}(\omega) &=\sum_{n=1}^{\infty}\cos(2\pi\omega n)\left ( \frac{1}{n^2\xi^2} -\frac{1}{n^2\xi^2-1}\right )\\ &=\frac{1}{\xi^2} \left ( \frac{\pi^2}{6}-\pi^2\omega+\pi^2\omega^2 \right )-\sum_{n = 1}^{\infty} \frac{\cos(2\pi\omega n)}{n^2\xi^2-1}. \end{aligned}$$ Now integrate $$f(z)=\frac{e^{2\pi i \omega z}}{z^2\xi^2-1} \frac{2\pi i}{e^{2\pi iz}-1}.$$ The residue evaluation gives $$ \sum_{n = 1}^{\infty} \frac{\cos(2\pi\omega n)}{n^2\xi^2-1} =\frac{1}{2}-\frac{\pi}{2\xi}\csc\left ( \frac{\pi}{\xi} \right ) \cos\left ( \frac{\pi-2\pi \omega}{\xi} \right ). $$ Then the $\mathscr{S}$ is solvable. Take the differentiation two times and let $\omega$ go to $0$, we find $$ \frac{\mathrm{d}}{\mathrm{d}\xi}\mathscr{F}(\xi) =-\frac{1}{2\xi^2}+\frac{\pi}{2\xi^3}\cot\left ( \frac{\pi}{\xi} \right ). $$ Therefore, $$ \mathscr{F}(\xi)=\frac{(1-\ln2)}{2\xi} -\frac{1}{2\xi} \ln\left ( \sin \left ( \frac\pi\xi\right ) \right ) -\frac{1}{4\pi}\operatorname{Cl}_2\left ( \frac{2\pi}{\xi} \right )+c. $$ The constant is $0$.

In your notations, final expression is $$\sum_{n=1}^{\infty}(mn\operatorname{arcoth}(mn)-1) =\frac{1-\ln2}{2} -\frac{1}{2} \ln\left ( \sin \left ( \frac\pi m\right ) \right ) -\frac{m}{4\pi}\operatorname{Cl}_2\left ( \frac{2\pi}{m} \right ). $$ for any real number such that $m>1$.

Using $$\sum_{n=1}^{\infty}(2f(2n)-f(n)) =\sum_{n=1}^{\infty}(-1)^nf(n).$$ The alternating case is determined.

Solution 2:

For $|x| >1 $, the inverse hyperbolic cotangent function has the Laurent series representation $$\operatorname{arcoth}(x) = \frac{1}{2} \ln \left(\frac{1+ \frac{1}{x}}{1- \frac{1}{x}} \right) = \frac{1}{2} \log \left[ \left(1+ \frac{1}{x} \right) - \log\left(1- \frac{1}{x} \right) \right]= \sum_{k=0}^{\infty}\frac{1}{(2n+1)x^{2n+1}}. $$ And for $0 < |x| < 1$, $\cot (\pi x)$ has the Laurent series representation $$\cot(\pi x) = \frac{1}{\pi x} - \frac{2}{\pi x} \sum_{k=1}^{\infty}\zeta(2k) x^{2k}. $$

Therefore, we have $$ \begin{align} \sum_{n=1}^{\infty} \left(mn \operatorname{arcoth} (mn)-1 \right) &= \sum_{n=1}^{\infty} \left(mn \sum_{k=0}^{\infty} \frac{1}{(2k+1) (mn)^{2k+1}} -1\right) \\ &= \sum_{n=1}^{\infty} \sum_{k=1}^{\infty} \frac{1}{(2k+1)(mn)^{2k+1}} \\ &= \sum_{k=1}^{\infty}\frac{1}{(2k+1)m^{2k}} \sum_{n=1}^{\infty} \frac{1}{n^{2k}} \\ &= \sum_{k=1}^{\infty}\frac{\zeta(2k)}{(2k+1)m^{2k}} \\ &= \frac{m}{2} \int_{0}^{1/m} \, \mathrm dx - \frac{m \pi}{2} \int_{0}^{1/m} x \cot (\pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{m x \log(\sin \pi x)}{2}\Bigg|_{0}^{1/m} + \frac{m}{2} \int_{0}^{1/m} \log(\sin \pi x) \, \mathrm dx \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi} \int_{0}^{2 \pi /m} \log \left(\sin \frac{ u}{2}\right) \, \mathrm du \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} + \frac{m}{4 \pi}\int_{0}^{2 \pi /m} \log \left(2 \sin \frac{u}{2}\right) \, \mathrm du - \frac{\log 2}{2} \\ &= \frac{1}{2} - \frac{\log \left(\sin \frac{\pi}{m} \right)}{2} - \frac{m \operatorname{Cl}_{2} \left(\frac{2 \pi}{m} \right)}{4 \pi } - \frac{\log 2}{2}. \end{align} $$