series involving $\log(\tanh(\pi k/2))$ II

Question

continuation of the question above series involving $\log\tanh(\pi k/2)$) it is posible to prove that $$\sum_{k=1}^\infty \log (\tanh (k x))=\sum _{n=1}^\infty \left(\frac{\tanh \left(\frac{\pi ^2 n}{2 x}\right)}{2 n}-\frac{\sqrt{\frac{1}{n^2}}}{2}\right)+\frac{\pi^2}{8x}+\frac{1}{2} \log \left(\frac{x}{2 \pi }\right)=-\sum _{n=1}^\infty \frac{1}{n \left(e^{2 \pi n}+1\right)}+\frac{\pi ^2}{8 x}+\frac{1}{2} \log \left(\frac{x}{2 \pi }\right))$$

Answer 1

There is clearly a typo in the last equation. The sum $$\sum_{n = 1}^{\infty}\frac{1}{n(e^{2\pi n} + 1)}$$ should be replaced by $$\sum_{n = 1}^{\infty}\frac{1}{n(e^{n\pi^{2}/x} + 1)}$$

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As shown in this answer we have $$\sum_{n = 1}^{\infty}\log\tanh (nx) = \log\vartheta_{4}(e^{-2x})\tag{1}$$ where $\vartheta_{4}(q)$ is a Jacobi theta function given by $$\vartheta_{4}(q) = \prod_{n = 1}^{\infty}(1 - q^{2n})(1 - q^{2n - 1})^{2} = \prod_{n = 1}^{\infty}\frac{(1 - q^{n})^{2}}{1 - q^{2n}} = \prod_{n = 1}^{\infty}\frac{1 - q^{n}}{1 + q^{n}} = \sqrt{\frac{2k'K}{\pi}}\tag{2}$$ In what follows we have $q = e^{-2x}$. We next calculate the value of a sum defined by $$F(q) = \sum_{n = 1}^{\infty}\log(1 - q^{n})\tag{3}$$ Note that the Dedekind eta function defined by the equation $$\eta(q) = q^{1/24}\prod_{n = 1}^{\infty}(1 - q^{n})\tag{4}$$ satisfies the relation $$\eta(q) = 2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{5}$$ where $q = e^{-\pi K'/K}$. Taking logs we get $$\frac{\log q}{24} + F(q) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right)$$ or $$F(q) = \frac{\log k}{12} + \frac{\log k'}{3} + \frac{\log 2}{3} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{24K}\tag{6}$$ Similarly from the relation $$\eta(q^{2}) = 2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}$$ we get $$F(q^{2}) = \frac{\log(kk')}{6} + \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{12K}\tag{7}$$ Similarly we get $$F(q^{4}) = \frac{\log(k^{4}k')}{12} - \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\pi K'}{6K}\tag{8}$$ Next we calculate the value of sum $$S = -\sum_{n = 1}^{\infty}\frac{1}{n(e^{2ny } + 1)}$$ We have \begin{align} S &= -\sum_{n = 1}^{\infty}\frac{1}{n(e^{2ny } + 1)}\notag\\ &= -\sum_{n = 1}^{\infty}\frac{q'^{n}}{n(1 + q'^{n})}\text{ (where }q' = e^{-2y})\notag\\ &= -\sum_{n = 1}^{\infty}\frac{1}{n}\sum_{k = 1}^{\infty}(-1)^{k - 1}q'^{nk}\notag\\ &= \sum_{k = 1}^{\infty}(-1)^{k}\sum_{n = 1}^{\infty}\frac{q'^{kn}}{n}\notag\\ &= \sum_{k = 1}^{\infty}(-1)^{k + 1}\log(1 - q'^{k})\notag\\ &= \sum_{k \text{ odd}}\log(1 - q'^{k}) - \sum_{k\text{ even}}\log(1 - q'^{k})\notag\\ &= \sum_{k = 1}^{\infty}\log(1 - q'^{k}) - 2\sum_{k = 1}^{\infty}\log(1 - q'^{2k})\notag\\ &= F(q') - 2F(q'^{2})\notag\\ &= F(q_{1}^{2}) - 2F(q_{1}^{4})\text{ (where }q_{1} = e^{-y})\notag\\ &= \frac{\log(ll')}{6} + \frac{\log 2}{6} + \frac{1}{2}\log\left(\frac{L}{\pi}\right) + \frac{\pi L'}{12L}\notag\\ &\,\,\,\, - \left(\frac{\log(l^{4}l')}{6} - \frac{\log 2}{3} + \log\left(\frac{L}{\pi}\right) + \frac{\pi L'}{3L}\right)\notag\\ &= -\frac{\log l}{2} + \frac{\log 2}{2} - \frac{1}{2}\log\left(\frac{L}{\pi}\right) - \frac{\pi L'}{4L}\tag{9} \end{align} where $l,l', L$ correspond to $q_{1} = e^{-y} = e^{-\pi L'/L}$.

Now we use the fact that $y = \pi^{2}/2x$ and hence so that $q = e^{-2x}$ and $q_{1} = e^{-y}$ are sort of conjugate in the sense that $l = k', l' = k, L = K', L' = K$. Then we can see that the sum $$S = -\frac{\log k'}{2} + \frac{\log 2}{2} - \frac{1}{2}\log\left(\frac{K'}{\pi}\right) - \frac{\pi K}{4K'}\tag{10}$$

From equations $(1)$ and $(10)$ we can see that $$\log\vartheta_{4}(q) = S + \frac{1}{2}\log(2k') + \frac{1}{2}\log\left(\frac{K}{\pi}\right) - S$$ or $$\log\vartheta_{4}(q) = S + \frac{1}{2}\log(2k') + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + \frac{\log k'}{2} - \frac{\log 2}{2} + \frac{1}{2}\log\left(\frac{K'}{\pi}\right) + \frac{\pi K}{4K'}$$ or $$\log\vartheta_{4}(q) = S + \log k' + \frac{\pi K}{4K'} + \frac{1}{2}\log\left(\frac{KK'}{\pi^{2}}\right)$$ We now use the fact that $$2x = -\log q = \frac{\pi K'}{K}$$ so that $x = \pi K'/2K$. Then we get $$\log\vartheta_{4}(q) = S + \frac{\pi^{2}}{8x} + \frac{1}{2}\log\left(\frac{k'^{2}KK'}{\pi^{2}}\right)$$ I wonder if the last term $(1/2)\log(x/2\pi)$ in question is correct or not.

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Update: I checked after some manipulation that if we use $$\log\vartheta_{4}(q) = -S + \frac{1}{2}\log(2k') + \frac{1}{2}\log\left(\frac{K}{\pi}\right) + S$$ then we get $$\log\vartheta_{4}(q) = -S + \log 2 + \frac{1}{2}\log(K/K') - \frac{\pi K}{4K'} = -S - \frac{1}{2}\log\left(\frac{x}{2\pi}\right) - \frac{\pi^{2}}{8x}$$ and therefore the right formula is \begin{align}\sum_{n = 1}^{\infty}\log \tanh (nx) &= \log\vartheta_{4}(e^{-2x})\notag\\ &= \sum_{n = 1}^{\infty}\frac{1}{2n}\left(1 - \tanh \left(\frac{\pi^{2}n}{2x}\right)\right) - \left\{\frac{\pi^{2}}{8x} + \frac{1}{2}\log\left(\frac{x}{2\pi}\right)\right\}\notag\\ &= \sum_{n = 1}^{\infty}\frac{1}{n(e^{n\pi^{2}/x} + 1)} - \left\{\frac{\pi^{2}}{8x} + \frac{1}{2}\log\left(\frac{x}{2\pi}\right)\right\}\notag \end{align} Thus the original formula given in the question is differing only in sign (apart from the typo mentioned in the beginning of my answer). The formula corresponds not to $\sum\log\tanh(kx)$ but rather to $\sum\log\coth(kx) = -\sum\log\tanh(kx)$.