# dedekind ramanujan identities

I get the following representation using laplace transform the remainder part it dificult to check numerically could you get some numerically result? Sorry for my bad latex i have correct the formula before $$\sum _{k=1}^{\infty } \frac{1}{k \left(e^{2 \pi k x}-1\right)}=\left(-\frac{\pi x}{12}+\frac{\pi }{12 x}+\frac{\log (x)}{2}\right)-\sum _{n=1}^{\infty } \left(\frac{\pi }{\sqrt{\frac{1}{n^2}} x}+\log \left(\frac{x \left(\pi n \text{csch}\left(\frac{\pi n}{x}\right)\right)}{(2 \pi n) x}\right)\right)$$ you can check that $$\frac{1}{1-e^{-\frac{2 \pi n}{x}}}=\frac{1}{2} e^{\frac{\pi n}{x}} \text{csch}\left(\frac{\pi n}{x}\right)=\log \left(\frac{1}{2} e^{\frac{\pi n}{x}} \text{csch}\left(\frac{\pi n}{x}\right)\right)=\frac{\pi }{\sqrt{\frac{1}{n^2}} x}+\log \left(\frac{x \left(\pi n \text{csch}\left(\frac{\pi n}{x}\right)\right)}{(2 \pi n) x}\right)$$ Paramanand Singh wa right

• Your first sum on left appears to diverge because the n-th term does not tend to $0$. – Paramanand Singh Sep 28 '16 at 8:47
• @ParamanandSingh he plays with wolfram and formulas he doesn't understand – reuns Sep 29 '16 at 7:18
• @user1952009: maybe you are right. that's why i never get to see the complete derivation from his side. – Paramanand Singh Sep 29 '16 at 9:55
• Wolfram can not made the remainder term you need more algebra..anyway thanks for your time – user167276 Sep 29 '16 at 10:44

Let $$q = e^{-\pi / x}$$ and consider the sum \begin{align} S &= \sum_{n = 1}^{\infty}\log\left(\frac{2\pi^{2}n^{2}\operatorname{cosech}(\pi n/x)}{x^{2}}\right) - \frac{\pi n}{x}\notag\\ &= \sum_{n = 1}^{\infty}\log\left(\frac{2\pi^{2}n^{2}e^{-\pi n/x}}{x^{2}\sinh(\pi n/x)}\right)\notag\\ &= \sum_{n = 1}^{\infty}\log\left(\frac{4\pi^{2}n^{2}e^{-\pi n/x}}{x^{2}\{e^{\pi n / x} - e^{-\pi n / x}\}}\right)\notag\\ &= \sum_{n = 1}^{\infty}\log\frac{4\pi^{2}n^{2}q^{2n}}{x^{2}(1 - q^{2n})}\notag\\ \end{align} Now we can see that as $n \to \infty$ we have $n^{2}q^{2n} \to 0$ and hence the $\log$ term tends to $-\infty$ and the series therefore does not converge. Perhaps there is a typo somewhere.
On the other hand the sum of RHS is well defined and is equal to $$\sum_{n = 1}^{\infty}\frac{q^{2n}}{n(1 - q^{2n})} = a(q^{2})$$ and from this post we see that $$a(q^{2}) = -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K}{\pi}\right) - \frac{\pi K'}{12K}$$ Note that $\pi K'/K = -\log q = \pi/x$ and hence we get the term $-\pi/12x$ on LHS.
Note further that if $q' = e^{-\pi x}$ then $$a(q'^{2}) = -\frac{\log(kk')}{6} - \frac{\log 2}{6} - \frac{1}{2}\log\left(\frac{K'}{\pi}\right) - \frac{\pi K}{12K'}$$ and on subtraction we see that $$a(q^{2}) - a(q'^{2}) = \frac{1}{2}\log(K'/K) + \frac{\pi K}{12K'} - \frac{\pi K'}{12K} = -\frac{1}{2}\log x - \frac{\pi }{12 x} + \frac{\pi x}{12}$$ So your identity should look like the following $$\boxed{\sum_{n = 1}^{\infty}\frac{1}{n(e^{2\pi n x} - 1)} + \frac{\pi x}{12} - \frac{\pi }{12x} - \frac{1}{2}\log x = \sum_{n = 1}^{\infty}\frac{1}{n(e^{2\pi n/x} - 1)}}$$ and thus you need to replace the sum $$S = \sum_{n = 1}^{\infty}\log\left(\frac{2\pi^{2}n^{2}\operatorname{cosech}(\pi n/x)}{x^{2}}\right) - \frac{\pi n}{x}$$ with $$S = \sum_{n = 1}^{\infty}\frac{1}{n(e^{2\pi n x} - 1)}$$
• @antonioasis: please see my updated answer. I don't think your sum with $\log$ is correct because it diverges. – Paramanand Singh Sep 28 '16 at 9:19
• @antonioasis: It would be nice if you could also show how you arrived at such log sums. If the work is too lengthy to type in $\mathrm\LaTeX$ here then you can perhaps give a link to some online (scanned) copy of your work. – Paramanand Singh Sep 28 '16 at 9:23