# Prove $\sum_{n=1}^\infty \text{Ci}(\pi n)=\frac{\ln(2)-\gamma}{2}$

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.

• I deleted my comment before I saw yours - I googled and saw that. Thanks! – Prince M Feb 1 '18 at 0:30
• What makes you expect closed form? – Yuriy S Feb 1 '18 at 0:32
• @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/… – Franklin Pezzuti Dyer Feb 1 '18 at 0:35
• And this is not in your question because? – Did Feb 1 '18 at 0:38
• No. Actually, you even misled potential answerers by mentioning constants that you knew are not involved in the result. – 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.
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...