I'm trying to prove that $$\sum_{n=1}^\infty \text{Ci}(\pi n)=\frac{\ln(2)-\gamma}{2}$$

I've tried parametrizing the sum by replacing $\pi$ with $x$ and differentiating, but this creates to a divergent series (whose partial sum is too messy to integrate), so I had no luck with that.

Any hints? Emphasis on hints - please no full solutions. Just point me in the right direction.

  • $\begingroup$ I deleted my comment before I saw yours - I googled and saw that. Thanks! $\endgroup$ – Prince M Feb 1 '18 at 0:30
  • $\begingroup$ What makes you expect closed form? $\endgroup$ – Yuriy S Feb 1 '18 at 0:32
  • $\begingroup$ @YuriyS The following website says that the value is $\frac{\ln(2)-\gamma}{2}$, but I don't know how to derive that: www-elsa.physik.uni-bonn.de/~dieckman/InfProd/… $\endgroup$ – Franklin Pezzuti Dyer Feb 1 '18 at 0:35
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    $\begingroup$ And this is not in your question because? $\endgroup$ – Did Feb 1 '18 at 0:38
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    $\begingroup$ No. Actually, you even misled potential answerers by mentioning constants that you knew are not involved in the result. $\endgroup$ – Did Feb 1 '18 at 0:47


$\text{Ci}(\pi n)=-\int_{\pi n}^{+\infty}\frac{\cos(x)}{x}=(-1)^{n+1}\int_{0}^{+\infty}\frac{\cos x}{x+\pi n}\,dx\color{red}{=}(-1)^{n+1}\int_{0}^{+\infty}\frac{s e^{-\pi n s}}{1+s^2}\,ds$ where $\color{red}{=}$ is a consequence of $\int_{0}^{+\infty}f(x)\,g(x)\,dx = \int_{0}^{+\infty}(\mathcal{L}f)(s)\,(\mathcal{L}^{-1} g)(s)\,ds$.

The identity $\sum_{n\geq 1}(-1)^{n+1}e^{-\pi n s}\,ds =\frac{1}{1+e^{\pi s}}$ is straightforward to check, hence in order to crack the original series it is enough to compute the integral $$ \int_{0}^{+\infty}\frac{s\,ds}{(1+s^2)(1+e^{\pi s})} $$ for instance, by integration by parts. This might be useful.

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    $\begingroup$ @Nilknarf: you're welcome. I enjoy sharing good ol' tricks I really enjoyed when I learned them. $\endgroup$ – Jack D'Aurizio Feb 2 '18 at 0:50

An overkill. Let $\mathfrak{M}\left(*,s\right) $ the Mellin transform. Using the identity $$\mathfrak{M}\left(\underset{k\geq1}{\sum}\lambda_{k}g\left(\mu_{k}x\right),\, s\right)=\underset{k\geq1}{\sum}\frac{\lambda_{k}}{\mu_{k}^{s}}\mathfrak{M}\left(g\left(x\right),s\right) $$ we have $$\mathfrak{M}\left(\sum_{n\geq1}\mathrm{Ci}\left(nx\right),s\right)=-\zeta\left(s\right)\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}$$ for $\mathrm{Re}\left(s\right)>1$, since $$\mathfrak{M}\left(\mathrm{Ci}\left(x\right),s\right)=-\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}.$$ So, inverting, we obtain $$\sum_{n\geq1}\mathrm{Ci}\left(nx\right)=-\frac{1}{2\pi i}\int_{2-i\infty}^{2+i\infty}\zeta\left(s\right)\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}x^{-s}ds$$ now taking $x=\pi$ and shifting the complex line to the left we have, from the residue theorem, that $$\sum_{n\geq1}\mathrm{Ci}\left(n\pi\right)=\mathrm{Res}_{s=0}\left(-\zeta\left(s\right)\Gamma\left(s\right)\frac{\cos\left(\pi s/2\right)}{s}\pi^{-s}\right)$$ which is...


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