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I have been working on trying to show that

$\displaystyle \sum _{n=1}^{\infty }\left[\frac{2n}{e^{2\pi n}+1}+\frac{2n-1}{e^{\left(2n-1\right)\pi }-1}\right]=\frac{\varpi ^2}{4\pi ^2}-\frac{1}{8}$

where $\displaystyle \varpi =\frac{\Gamma ^2\left(\frac{1}{4}\right)}{2\sqrt{2\pi }}$ is the lemniscate constant.

Note the individual even and odd indexed series resolve to

$\displaystyle \sum _{n=1}^{\infty }\frac{2n}{e^{2\pi n}+1}=\frac{\varpi ^2}{8\pi ^2}-\frac{1}{12}$

and

$\displaystyle \sum _{n=1}^{\infty }\frac{2n-1}{e^{\left(2n-1\right)\pi }-1}=\frac{\varpi ^2}{8\pi ^2}-\frac{1}{24}$

I have seen evaluations of similar series like $\displaystyle \sum _{n=1}^{\infty }\frac{n}{e^{2\pi n}-1}$ using inverse Mellin transform and I have tried to set this up by using the Mellin transforms of $\displaystyle \frac{x}{e^{\pi x}+1}$ and $\displaystyle \frac{x}{e^{\pi x}-1}$, then taking the inversion and summing over even or odd positive integers.

For example in the even indexed series (similar issues arise in the odd one), I get to

$\displaystyle S_{even} = \frac{1}{2\pi ^3i}\int _{c-i\infty }^{c+i\infty }\left(1-2^{-s}\right)\sin \left(\frac{\pi s}{2}\right)\Gamma \left(1+s\right)\Gamma \left(1-s\right)\zeta \left(1+s\right)\zeta \left(1-s\right)ds$

The integrand has simple poles at $s=-1,1$ and a double pole at $s=0$. I tried to set up a contour by closing in the $\displaystyle \left[c-iR,c+iR\right]$ with a semicircle of radius R centered at $c$ for $c > 1$. After calculating residues, I get $\displaystyle -\frac{1}{24}$ which is definitely wrong and I'm not sure if the circle part vanishes as $R\to\infty$.

I have also seen cases where rectangular contours that have a line on the imaginary axis are used and the function is found odd at $s=0$ so that contribution as well as those on horizontal lines vanish. However, this integrand for the even indexed series is not odd at $s=0$ or any obvious point on the real line that I could use.

I don't know how I can resolve this by picking a better contour or any other manipulations to arrive at the desired result. Help is greatly appreciated.

Paramanand Singh
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Camishere 45
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    The sums belong to theory of elliptic integrals and theta functions and are based on closed form evaluation of the Ramanujan function $P(q) = 1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}$. – Paramanand Singh Jul 15 '24 at 04:15
  • I did recognize this specific kind of Lambert series is another connection that could help for an alternate way to arrive at the solution, but I couldn't find out how to get the proper form or like $q$ as a constant without extra factors depending on $n$. – Camishere 45 Jul 15 '24 at 04:31
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    Let $f(q) =\sum \frac{nq^n} {1-q^n}=\frac{1-P(q)}{24}$ and $g(q) =\sum\frac{nq^n} {1+q^n} $ so that $f(q) - g(q) =2f(q^2)$ and $\sum\frac{2nq^{2n}}{1+q^{2n}}=2g(q^2) =2f(q^2) - 4f(q^4)=\frac{2P(q^4)-P(q^2)}{12}-\frac{1}{12}$. This gives you the sum with even index when $q=e^{-\pi} $. The second sum (which again equals $g(q) $) with odd index can be dealt in same manner with slightly less effort. – Paramanand Singh Jul 15 '24 at 05:29
  • @ParamanandSingh I'm running into issues with the odd indexed series because of the lingering $e^{\pi}$ factor in the denominator from $1-e^{-(2n-1)\pi}$ if I multiply top and bottom by $e^{-(2n-1)\pi}$. And if I split the sum up by the difference $2n-1$, I'm not sure if the sum taking the $-1$ can be expressed as a derivative or other expression in terms of $P(q)$ when missing that $n$ in the numerator. – Camishere 45 Jul 15 '24 at 15:03
  • Nevermind I forgot I can get the odd series from the series over all naturals minus the even series. – Camishere 45 Jul 15 '24 at 15:13
  • I have found sources on StackExchange like here where for $q=e^{-\pi \frac{k^{\prime }}{k}}$ we have $P\left(q\right)=\frac{12E\left(k\right)K\left(k\right)+\left(4k^2-8\right)K^2\left(k\right)}{\pi ^2}$ in terms of elliptic integrals, but I'm not sure of how to prove this and couldn't find references in like Ramanujan's Lost notebook or others. – Camishere 45 Jul 15 '24 at 15:44
  • Well if you really want to get to the proofs you will need to study some amount of theory of elliptic integrals and theta functions. I have covered these topics in my blog posts (see archive page and search for "elliptic" and "theta") but it does require some patience to assimilate all the details. – Paramanand Singh Jul 15 '24 at 20:49
  • You may also have a look at this answer of mine which is based on same ideas: https://math.stackexchange.com/a/1695377/72031 – Paramanand Singh Jul 15 '24 at 20:53

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