# 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.

$$S:=\sum_{k=1}^\infty\frac1k\int_{k\pi}^\infty\frac{\sin x}{x}~dx\underset{x=kt}{=}\sum_{k=1}^\infty\frac1k\int_\pi^\infty\frac{\sin kt}{t}~dt=\int_\pi^\infty\frac{s(t)}{t}~dt,$$ 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 $$0, and an expansion of $$\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).}$$
The first observation we make is a simple substitution, with $$x=nu$$ and $$dx=n\;du$$: $$\int_{n\pi}^{\infty}\frac{\sin x}x\;dx = \int_{\pi}^{\infty}\frac{\sin(nu)}{nu}\;n\;du = \int_{\pi}^{\infty}\frac{\sin(nx)}x\;dx$$Using this identity we can rewrite the given integral as $$\sum_{n=1}^{\infty}\frac1n\int_{n\pi}^{\infty}\frac{\sin x}x\;dx = \sum_{n=1}^{\infty}\frac1n\int_{\pi}^{\infty}\frac{\sin(nx)}x\;dx$$Now (I am causing Fubini to roll over in his grave by doing this) we interchange the sum and integral as follows: $$\int_{\pi}^{\infty}\left[\sum_{n=1}^{\infty} \frac{\sin(nx)}n\right]\;\frac{dx}x$$To evaluate the sum inside the brackets, we take things into the complex plane. In the following, by $$\text{Log}$$ and $$\text{Arg}$$ we refer to the functions with branch cuts along the negative real axis. \begin{align*} \sum_{n=1}^{\infty} \frac{\sin(nx)}n &= \sum_{n=1}^{\infty} \frac{e^{inx} - e^{-inx}}{2in} \\ &= \frac1{2i}\sum_{n=1}^{\infty}\left[\frac{e^{inx}}n - \frac{e^{-inx}}n\right] \\ &= \frac1{2i}\left[-\text{Log}(1-e^{ix}) + \text{Log}(1-e^{-ix})\right] \\ &= \frac1{2i}\left[-\text{log}|1-e^{ix}| - i\text{Arg}(1-e^{ix}) + \text{log}|1-e^{-ix}| + i\text{Arg}(1-e^{-ix})\right] \\ \end{align*}Since $$1-e^{-ix}$$ and $$1-e^{ix}$$ are complex conjugates, we have that their absolute values are equal and their arguments are negatives of each other. So this simplifies to $$\frac1{2i}\left[2i\text{Arg}(1-e^{-ix})\right] = \text{Arg}(1-e^{-ix})$$But we're not yet done simplifying. Now we have turned this into a geometry problem: we have a point $$X$$ on the circle centered at point $$C=1$$ with radius $$1$$, whose terminal angle with the negative real axis (negative because we subtracted from 1) is $$x$$. Now, supposing for the moment that $$0\le x< 2\pi$$, and letting $$O$$ be the origin, the arc subtended by $$\angle OCX$$ of course has measure $$x$$. But if we let point $$D=2$$, then we see that arc $$\overarc{XD}$$ has signed measure $$\pi-x$$ (this behaves as expected if $$\pi \le x < 2\pi$$), and thus inscribed angle $$\angle XOD$$ has measure $$\frac{\pi-x}2$$. But this is exactly the argument of $$X$$! So in summary, $$\text{Arg}(1-e^{-ix}) = \frac{\pi - (x\mod 2\pi)}2$$I should mention that things get a little dicey when $$x$$ is exactly a multiple of $$2\pi$$. The sum at those points can easily be seen to be zero, since all the sines are zero, but the actual function is discontinuous at those points, approaching $$\frac{\pi}2$$ from one direction and $$-\frac{\pi}2$$ from the other. Indeed, the argument is undefined ($$\text{Arg}(0)$$) there. Fortunately these isolated points are not enough to change the value of the integral. In fact, the next thing we're going to do is split our integral at these points. \begin{align*} \int_{\pi}^{\infty}\left[\sum_{n=1}^{\infty} \frac{\sin(nx)}n\right]\;\frac{dx}x &= \int_{\pi}^{\infty}\left[\frac{\pi - (x\mod 2\pi)}2\right]\;\frac{dx}x \\ &= \frac12\left[\int_{\pi}^{2\pi} \frac{\pi-(x\mod 2\pi)}x\;dx + \sum_{n=1}^{\infty}\int_{2n\pi}^{2(n+1)\pi} \frac{\pi-(x\mod 2\pi)}x\;dx\right] \\ &= \frac12\left[\int_{\pi}^{2\pi} \frac{\pi-x}x\;dx + \sum_{n=1}^{\infty}\int_{2n\pi}^{2(n+1)\pi} \frac{\pi-(x-2n\pi)}x\;dx\right] \\ &= \frac12\left[\int_{\pi}^{2\pi} \left(\frac{\pi}x - 1\right)\;dx + \sum_{n=1}^{\infty}\int_{2n\pi}^{2(n+1)\pi} \left(\frac{(2n+1)\pi}x-1\right)\;dx\right] \\ &= \frac12\left[\left[\pi\log(x)-x\right]\biggr\rvert_{x=\pi}^{2\pi} + \sum_{n=1}^{\infty}\left[(2n+1)\pi\log(x)-x\right]\biggr\rvert_{x=2n\pi}^{2(n+1)\pi}\right] \\ &= \frac12\left[\pi\log(2)-\pi + \sum_{n=1}^{\infty}\left[(2n+1)\pi\log\left(\frac{2(n+1)\pi}{2n\pi}\right) - 2\pi\right]\right] \\ &= \frac{\pi}2\left[\log(2)-1 + \sum_{n=1}^{\infty}\left[(2n+1)\log\left(\frac{n+1}n\right) - 2\right]\right] \\ &= \frac{\pi}2(\log(2)-1) + \pi\sum_{n=1}^{\infty}\left[\left(n+\frac12\right)\log(n+1) - \left(n+\frac12\right)\log(n) - 1\right] \\ \end{align*}To evaluate the sum, we write it as a limit of a finite sum: $$\sum_{n=1}^{\infty}\left[\left(n+\frac12\right)\log(n+1) - \left(n+\frac12\right)\log(n) - 1\right] = \lim_{N\rightarrow\infty}\sum_{n=1}^{N}\left[\left(n+\frac12\right)\log(n+1) - \left(n+\frac12\right)\log(n) - 1\right]$$$$= \lim_{N\rightarrow\infty}\left[\frac32\log2 - \frac32\log1 - 1 + \frac52\log3 - \frac52\log2 - 1 + \cdots + \left(N+\frac12\right)\log(N+1) - \left(N+\frac12\right)\log(N) - 1\right]$$$$= \lim_{N\rightarrow\infty}\left[\left(N+\frac12\right)\log(N+1) - \log(N) - \log(N-1) - \cdots - \log2 - \log1 - N\right]$$$$= \lim_{N\rightarrow\infty}\left[\left(N+\frac12\right)\log(N+1) - \log(N!) - N\right]$$And finally, we finish with Stirling's approximation. \begin{align*} \lim_{N\rightarrow\infty}\left[\left(N+\frac12\right)\log(N+1) - \log(N!) - N\right] &= \lim_{N\rightarrow\infty}\left[\left(N+\frac12\right)\log(N+1) - \log\left(\sqrt{2\pi N}\left(\frac{N}e\right)^N\right) - N\right] \\ &= \lim_{N\rightarrow\infty}\left[\left(N+\frac12\right)\log(N+1) - \frac12\log(2\pi N) - N\log N + N - N\right] \\ &= \lim_{N\rightarrow\infty}\left[\left(N+\frac12\right)\log(N+1) - \frac12\log(2\pi) - \left(N+\frac12\right)\log N\right] \\ &= -\frac12\log(2\pi) + \lim_{N\rightarrow\infty}\left(N+\frac12\right)\log\left(\frac{N+1}N\right) \\ &= -\frac12\log(2\pi) + \lim_{N\rightarrow\infty}\left(N+\frac12\right)\log\left(1+\frac1N\right) \\ &= -\frac12\log(2\pi) + \lim_{N\rightarrow\infty}\frac{N+\frac12}N \\ &= 1 - \frac12\log(2\pi) \end{align*}And so, all together, the integral is $$\frac{\pi}2(\log2-1) + \pi\left[1-\frac12\log(2\pi)\right] = \frac{\pi}2(\log2 - 1 + 2 - \log2 - \log\pi) = \frac{\pi}2(1-\log\pi)$$Yay, that was fun! $$\blacksquare$$