Computing $\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}dx$ I was asked to evaluate the following sum:
$$\sum_{k=1}^{\infty}\frac{1}{k}\int_{\pi k}^{\infty}\frac{\sin(x)}{x}dx$$
I'm trying to use $$\int_{0}^{\infty} \frac{\sin(x)}{x}dx=\frac{\pi}{2}$$ However, it doesn't seem to work. Any help is greatly appreciated. 
 A: $$S:=\sum_{k=1}^\infty\frac1k\int_{k\pi}^\infty\frac{\sin x}{x}~dx\underset{x=kt}{=}\sum_{k=1}^\infty\frac1k\int_\pi^\infty\frac{\sin kt}{t}~dt=\int_\pi^\infty\frac{s(t)}{t}~dt,$$ where $s(t)=\sum_{k=1}^\infty\frac{\sin kt}{k}$ converges uniformly on compact subsets of $\mathbb{R}\setminus 2\pi\mathbb{Z}$ (hence the $\Sigma\smallint\mapsto\smallint\Sigma$ above is valid if $\infty$ is replaced by $2n\pi$ with a positive integer $n$; further, $$0\leqslant\int_{2n\pi}^\infty\frac{\sin kt}{t}~dt\underset{\text{IBP}}{\phantom{[}=\phantom{]}}\frac1k\int_{2n\pi}^\infty\frac{1-\cos kt}{t^2}~dt\leqslant\frac2k\int_{2n\pi}^\infty\frac{dt}{t^2}=\frac{1}{nk\pi},$$ and similarly $\int_{2n\pi}^\infty\frac{s(t)}{t}~dt=\mathcal{O}(n^{-1})$, so the $\Sigma\smallint\mapsto\smallint\Sigma$ is justified completely).
Using $s(t+2\pi)=s(t)$, we have $$S=\sum_{n=1}^\infty\int_{(2n-1)\pi}^{(2n+1)\pi}\frac{s(t)}{t}~dt\underset{t=2n\pi\pm2x}{=}\sum_{n=1}^{\infty}\int_0^{\pi/2}\left(\frac{s(2x)}{n\pi+x}+\frac{s(-2x)}{n\pi-x}\right)dx.$$ The known fact that $s(\pm t)=\pm(\pi-t)/2$ for $0<t<\pi$, and an expansion of $\cot x$, give $$S=-\int_0^{\pi/2}\sum_{n=1}^\infty\frac{x(\pi-2x)}{n^2\pi^2-x^2}~dx=-\frac{1}{2}\int_0^{\pi/2}(\pi-2x)\left(\frac1x-\cot x\right)dx=-\int_0^{\pi/2}\log\frac{x}{\sin x}~dx$$ (after integration by parts). Using $\int_0^{\pi/2}\log\sin x~dx=-(\pi/2)\log 2$, we get finally $$\bbox[5px,border:2px solid]{S=-\frac{\pi}{2}(\log\pi-1).}$$
