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.

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...
