How to compute $\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi\sqrt 3}$?

Question

I came across an old question asking to show that $$\displaystyle\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi}=\frac{\ln 2}{8}.\tag{1}$$ Although I have managed to prove this formula, my proof uses various theta functional relations and looks like an overkill. On the other hand, it suggests a few more identities, for example $$\displaystyle\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi\sqrt3}=\frac{\ln \left(8-4\sqrt{3}\right)}{4}.\tag{2}$$

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Question: can one prove (1) and (2) in a more elementary way? Ideally, the proof should make clear further generalizations.

Answer 1

Don't forget that sinh is an odd function, so we can use $$ n \, sinh(n \pi) = -n \, sinh(-n \pi) $$ to reduce the summation to $$ \sum_{n\in2\Bbb Z+1}\frac{1}{n\,sinh(n\pi)} = \sum_{n\in2\Bbb N+1}\frac{1}{n\,sinh(n\pi)} $$ which is more directly manipulated (without having to play with summation indices) if written anyway as $$ \sum_{n=1}^{\infty}\frac{1}{(2n+1)\,sinh((2n+1)\pi)} $$

Not really sure how to simplify to your end, from here, but perhaps this is helpful anyway.

-Cheers

Answer 2

The identities: $$\sum_{n\geq 1}\frac{(-1)^n}{n^2+m^2} = -\frac{1}{2m^2}+\frac{\pi}{2m\sinh(m\pi)},\tag{A}$$ $$ \frac{1}{m^2+n^2}=\int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-mx}\,dx,\tag{B}$$ $$ \sum_{n\geq 1}(-1)^n\frac{\sin(nx)}{x}=-\frac{x}{2}+\pi\left\lfloor\frac{x+\pi}{2\pi}\right\rfloor\tag{C} $$ give a wide range of possibilities to evaluate our series. For instance, $(A)$ gives: $$\begin{eqnarray*}\sum_{k\geq 0}\frac{1}{(2k+1)\sinh(\pi(2k+1))}&=&\frac{1}{\pi}\sum_{k\geq 0}\frac{1}{(2k+1)^2}+\frac{2}{\pi}\sum_{k\geq 0}\sum_{n\geq 1}\frac{(-1)^n}{n^2+(2k+1)^2}\\&=&-\frac{\pi}{8}+\frac{2}{\pi}\sum_{h\geq 1}\frac{r_2(h)\cdot\eta(h)}{h}\tag{1}\end{eqnarray*}$$ where $\eta(h)$ equals $-1$ if $h\equiv 2\pmod{4}$, $1$ if $h\equiv 1\pmod{4}$, zero otherwise, and: $$ r_2(h) = \#\{(n,k)\in\mathbb{N}^2: h=n^2+(2k+1)^2\}.\tag{2} $$ Now it is well-known that $a^2+b^2$ is the only reduced binary quadratic form of discriminant $-4$, hence: $$\begin{eqnarray*} \#\{(a,b)\in\mathbb{Z}:a^2+b^2=n\} &=& 4\left(\chi_4 * 1\right)(n)\\&=& 4\left(d_{1(4)}(n)-d_{3(4)}(n)\right) \tag{3} \end{eqnarray*}$$ so that we can evaluate the RHS of $(1)$ through Dirichlet convolution.

Since we have class number one also in the case $a^2+3b^2$, the situation is almost the same for the other series.

The Mellin transform gives another chance. See, for instance, this related problem.

Answer 3

Note: This is not a complete answer, but it includes some observations that can possibly be developed further.

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Contour Integral Representation of ${\rm Li}_\nu(e^{-x})$

By Mellin's inversion theorem, the polylogarithm admits the contour integral representation $${\rm Li}_\nu(e^{-x})=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\Gamma(s)\zeta(s+\nu)x^{-s}\ {\rm d}s$$

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The Sum $S(\mu)$ as a Contour Integral

Switching the order of summation, we may write $$S(\mu)\stackrel{\text{def}}=:\sum_{n\ge1}\frac{1}{n\sinh(n\mu)}=2\sum_{n\ge1}\frac{1}{n}\sum_{k\ge0}e^{-(2k+1)n\mu}=2\sum_{k\ge0}{\rm Li}_1\left(e^{-(2k+1)\mu}\right)$$ Therefore, the sum can be expressed as a contour integral $$S(\mu)=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}Q(s,\mu)\ {\rm d}s$$ where \begin{align} Q(s,\mu) &=2\mu^{-s}(1-2^{-s})\Gamma(s)\zeta(s)\zeta(s+1)\\ \end{align}

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Closing the Contour

We instead consider the integral of $\dfrac{Q(s,\mu)}{2\pi i}$ over a closed contour $C$. In this case, it is a rectangle with vertices $\frac{3}{2}-i\infty$, $\frac{3}{2}+i\infty$, $-\frac{3}{2}+i\infty$, $-\frac{3}{2}-i\infty$. Note that

1. The integral over $\mathrm{Re}(s)=\frac{3}{2}$ is $S(\mu)$.
2. The integral over the horizontal sides vanishes since $\zeta(s)$ grows moderately but $\Gamma(s)$ decays exponentially as $\mathrm{Im}(s)\to-\infty$.
3. Applying Riemann's functional equation twice, the integral over $\mathrm{Re}(s)=-\frac{3}{2}$ is shown to be:

\begin{align} -\frac{1}{2\pi i}\int^{-\frac{3}{2}+i\infty}_{-\frac{3}{2}-i\infty}Q(\mu,s)\ {\rm d}s &=-\frac{1}{2\pi i}\int^{-\frac{3}{2}+i\infty}_{-\frac{3}{2}-i\infty}2\left(\frac{\mu}{4\pi^2}\right)^{-s}(1-2^{-s})\Gamma(-s)\zeta(-s)\zeta(1-s)\ {\rm d}s\\ &=\frac{1}{2\pi i}\int^{\frac{3}{2}+i\infty}_{\frac{3}{2}-i\infty}2\left(\frac{2\pi^2}{\mu}\right)^{-s}(1-2^{-s})\Gamma(s)\zeta(s)\zeta(s+1)\ {\rm d}s\\ \end{align} which is equal to $\displaystyle S\left(\frac{2\pi^2}{\mu}\right)$.

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Functional Equation for $S(\mu)$

By the residue theorem, \begin{align} \frac{1}{2\pi i}\int_C Q(s,\mu)\ {\rm d}s=\sum^1_{k=-1}{\rm Res}(Q(s,\mu);z=k)=\frac{\pi^2}{6\mu}-\ln{2}+\frac{\mu}{12} \end{align} Thus $S(\mu)$ satisfies the functional equation $$S(\pi\mu\sqrt{2})+S\left(\frac{\pi\sqrt{2}}{\mu}\right)=\frac{\pi}{6\sqrt{2}}\left(\mu+\frac{1}{\mu}\right)-\ln{2}$$ For instance, this equation yields the identity $$\sum_{n\ge1}\frac{1}{n\sinh(n\pi\sqrt{2})}=\frac{\pi}{6\sqrt{2}}-\frac{\ln{2}}{2}$$

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Conjectured form of $S(\mu)$

We are actually interested in the quantity $S\left(\pi\right)-\dfrac{1}{2}S(2\pi)$. It is likely that $S(\pi\mu\sqrt{2})$ is of the form $$S(\pi\sqrt{2}\mu)\stackrel{?}=\frac{\pi}{6\sqrt{2}}\mu-\frac{\ln{2}}{2}+f(\mu)\tag1$$ such that $f(\mu)=-f(\mu^{-1})$, $f(0)$ blows up, and $f\left(\dfrac{1}{\sqrt{2}}\right)\stackrel{?}=\dfrac{\ln{2}}{4}$. If this is truly the case, then we indeed get \begin{align} \sum_{n\ \text{odd}}\frac{1}{n\sinh(n\pi)} &=S(\pi)-\frac{1}{2}S(2\pi)\\\ &=\left(\frac{\pi}{12}-\frac{\ln{2}}{2}+\frac{\ln{2}}{4}\right)-\frac{1}{2}\left(\frac{\pi}{6}-\frac{\ln{2}}{2}-\frac{\ln{2}}{4}\right)\\ &=\frac{\ln{2}}{8} \end{align} which of course implies that $\displaystyle\sum_{n\ge1}\frac{1}{n\sinh(n\pi)}=\frac{\pi}{12}-\frac{\ln{2}}{4}$. Numerically, this seems to be true.

Unfortunately, I have no idea how to prove or disprove $(1)$ and find the correct $f(\mu)$.