7

Let us define the sums

$$S_k=\sum_{n\geq 1}\frac{1}{\sinh^{2k}(n \pi)}$$

A few days ago, i was able to give a surprisingly simple derivation of the fact that $$ S_1=\sum_{n\geq 1}\frac{1}{\sinh^2(\pi n)}=\frac{1}{6}-\frac{1}{2\pi} $$ as an answer to this nice old question in terms of contour integration. Being curious i immediatly tried to generalize this result to higher orders of $\sinh$, which as it turned out will not work that easily. For example, considering the case of $k=2$ it turns out that

$$\text{Res}\left(\frac{ \cot(\pi z)}{ \sinh^4(\pi z)},z=z_k\right)= \begin{cases} \frac{1}{\pi\sinh^4(\pi k)} \quad\text {if}\,\,k \in \mathbb{Z}/0,\\ -\frac{1}{\pi\sinh^4(\pi k)} -\frac{4}{3\pi\sinh^2(\pi k)} \quad\text {if}\,\,ik \in \mathbb{Z}/0 \end{cases} $$

So integrating over a large quadratic contour gives us absolutly no information about $S_2$, since the residues cancel (For details of the exact procedure please have a look at the post linked above). This is in fact a simple consequence of the transformation properties of $\sinh^{2k}(z)$ along the imaginary axis: $\sinh^{2k}(i y)=(-1)^k\sin^{2k}(y)$

Looking at $S_3$ with the same approach, it turns out that it contains $S_2$ as well as $S_1$ so it is clear that we need this value to make any progress in a generalization at all.

I also want to mention that a Laplace/Mellintransform approach seem to suffer from the same cancellations then what i tried above, so this is maybe also not the way to go...:/

My questions are

1.) Is there any chance to proof the result for $S_2$ (and $S_k$ in general if possible) using an an approach which is close to the one i used to calculate $S_1$ ?

2.) What is a general approach to derive calculate such sums using a minimum of knowledge about the realm of elliptic integrals, Jacobi theta functions etc. (where i'm sure they can be derived but i have far too less knowledge in this field to do this by my own)

PS: It turns out that a closed form is, according to mathematica, given by

$$ S_2=-\frac{11}{90}+\frac{1}{3\pi}+\frac{\bar{\omega}^4}{7680 \pi^4} $$ where $\bar{\omega}$ is the Lemniscate constant.

tired
  • 12,325

1 Answers1

4

($1$) IMHO there is little chance to compute $S_2$ using contour integration due to existence of the lemniscate constant in all $S_n, n\ge 2$ (see below).

($2$) Here is a general approach. The normalized Eisenstein series can be written as $$G_{2k}(i)=\sum_{m,n\in\mathbb R, mn\not=0}\ \frac{1}{(m+ni)^{2k}}=2\zeta(2k)+2\sum_{n>0}\sum_{m\in\mathbb R}\frac{1}{(m+ni)^{2k}}$$ Where the inner sum can be computed through differentiating Mittag-Leffler expansion of $\cot$: $$\sum_{m\in\mathbb R}\frac{1}{(m+ni)^{2k}}=\sum_{j=1}^{k}a_{k,j}\text{csch}^{2j}(n\pi)$$ Thus the following recurrence $$G_{2k}(i)=2\zeta(2k)+2\sum_{j=1}^{k}a_{k,j}S_k$$ with well-known initial value $S_1=\frac{1}{6}-\frac{1}{2\pi}$ (see for instance J. Borwein's monograph Pi and AGM) allows us to compute all $S_n$ in terms of $G_{2k} (k=2,\cdots,n)$, all of which are expressible via products of $G_4(i)=\frac{\Gamma \left(\frac{1}{4}\right)^8}{960 \pi ^2}, G_6(i)=0$ (due to theory of modular forms). Therefore we conclude that $S_n$ lie in the algebra of $\mathbb Q$ generated by $\frac{1}{\pi}, \Gamma\left(\frac14\right)$, for instance $$S_5=\frac{14797}{187110}-\frac{64}{315 \pi }-\frac{\Gamma \left(\frac{1}{4}\right)^{16}}{2580480 \pi ^{12}}-\frac{41 \Gamma \left(\frac{1}{4}\right)^8}{90720 \pi ^6}$$

Infiniticism
  • 8,644
  • 1
  • 22
  • 67