2

While answering this question I found through Wolfram Alpha that

$$\sum_{k=1}^\infty (-1)^{k+1}\frac{ \text{csch}(\pi k)}{k} = \frac{1}{12}(\ln(64) - \pi).$$

Sadly Wolfram Alpha just spits the summation value out without any justification or references.

Is there an explanation or reference for this identity?

orlp
  • 10,508
  • 4
    As I said in the other question, I.J.Zucker - THE SUMMATION OF SERIES OF HYPERBOLIC FUNCTIONS, 1979. Expand $\frac{\text{csch}(\pi k)}{k}$ as an Eisenstein series and carefully rearrange the terms you get that way. – Jack D'Aurizio Jun 20 '17 at 21:33
  • Am I wrong? I got approx. $0.085$ for the series, but $(1/2)(\ln 64 - \pi) \approx 0.509$. – GEdgar Jun 20 '17 at 21:41
  • Mathematica 11.1 give: $\sum _{k=1}^{\infty } \frac{(-1)^{k+1} \text{csch}(\pi k)}{k}=\frac{1}{12} (-\pi +\log (64))$ – Mariusz Iwaniuk Jun 20 '17 at 21:44
  • I am going to write a full answer, just give me a few moments. – Jack D'Aurizio Jun 20 '17 at 21:49
  • @JackD'Aurizio I had already looked at that paper, but missed it the first time. After a more close inspection the identity in question is found in Table 1(iv), $\text{XIII}{-1}(1) = 2D{-1}(1)$, however $D$ is defined as a $q$-series (I believe), which I'm woefully unfamiliar with. – orlp Jun 20 '17 at 21:53
  • @GEdgar The issue was that I typo'd $\frac{1}{2}$ instead of $\frac{1}{12}$. – orlp Jun 20 '17 at 22:44

1 Answers1

4

The meromorphic function $f(z)=\frac{1}{z\sinh(\pi z)}$ has a double pole at the origin and simple poles at $z\in\{\pm i,\pm 2i,\pm 3i,\ldots\}$. By computing the residues at such points it follows that: $$ \frac{\text{csch}(\pi z)}{z} = \frac{1}{\pi z^2}+\frac{i}{\pi}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m}\left(\frac{1}{z-mi}-\frac{1}{z+mi}\right)\tag{1} $$ and now we need to evaluate both sides at $z=k$, multiply by $(-1)^{k+1}$ and sum over $k\geq 1$.
The contribute provided by the double pole at the origin is simple to compute: $$ \sum_{k\geq 1}\frac{(-1)^{k+1}}{\pi k^2} = \frac{\eta(2)}{\pi} = \frac{\pi}{12}\tag{2} $$ It remains to evaluate: $$ \frac{2}{\pi}\sum_{k\geq 1}(-1)^{k+1}\sum_{m\geq 1}\frac{(-1)^{m+1}}{m^2+k^2}\tag{3}$$ and that is possible through the identities $$ \frac{1}{m^2+k^2} = \int_{0}^{+\infty}\frac{\sin(mx)}{m}e^{-kx}\,dx = \int_{0}^{+\infty}\cos(mx)\frac{e^{-kx}}{k}\,dx.\tag{4} $$ and Fourier series, or by recognizing in $(3)$ a Jacobi theta function. The final outcome is $$ \sum_{k\geq 1}(-1)^{k+1}\frac{\text{csch}(\pi k)}{k}=\color{red}{\frac{\log 2}{2}-\frac{\pi}{12}}.\tag{5}$$

Jack D'Aurizio
  • 353,855