Compute $\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2$

Question

How would you evaluate $$\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2$$ Wolfram|Alpha says it equals $\;\pi/2-1/2.\;$ If the solution is complicated, ... I can handle complicated.

Answer 1

$\displaystyle\sum_{n=1}^∞ \dfrac{\sin(nx)}{n} =\frac{1}{2i} \sum_{n=1}^∞ \dfrac{(e^{ix})^n}{n}-\dfrac{(e^{-ix})^n}{n}=-\frac{1}{2i}({\log(1-e^{ix})-\log(1-e^{-ix})})$

$= \displaystyle \dfrac{1}{2i} \log\left(\frac{(1-e^{-ix})}{(1-e^{ix})}\right)= \dfrac{1}{2i} \log(-e^{-ix})= \dfrac{\log(e^{iπ})-\log(e^{ix})}{2i} =\dfrac{π-x}{2}. $

Now $\displaystyle\int_0^x \sum_{n=1}^∞\dfrac{ \sin(nx)}{n}dx = \int_0^x \dfrac{π-x}{2}dx $ $\implies \displaystyle -\sum_{n=1}^∞ \left(\dfrac{\cos(nx)}{n^2} -\dfrac{\cos(0)}{n^2}\right)= \dfrac{2πx-x^2}{4}$ $\implies \displaystyle\sum_{n=1}^∞ \dfrac{2\sin^2(\frac{nx}{2})}{n^2} = \dfrac{2πx-x^2}{4} .$

Put $x=2$ the answer will be $\displaystyle\sum_{n=1}^∞ \dfrac{\sin^2(n)}{n^2} =\dfrac{π-1}{2} $.

Edit

In this way we can also solve Basel problem.

$\displaystyle\sum_{n=1}^∞ \dfrac{\sin^2(\frac{nπ}{2})}{n^2} = \dfrac{\frac{π}{2}(π-\frac{π}{2})}{2} = \dfrac{π^2}{8} $

Which means $\displaystyle\dfrac{1}{1^2} +\dfrac{1}{3^2} +\dfrac{1}{5^2} +...= \dfrac{π^2}{8} \implies \dfrac{3\zeta(2)}{4} =\dfrac{π^2}{8} \implies \zeta(2)=\dfrac{π^2}{6}$

Answer 2

The proposed duplicate Evaluating $\sum_{k=1}^{\infty}\left(\frac{\sin(tk)}{k}\right)^2$ does this the hard way, using the Poisson summation formula, which requires finding the Fourier series for a triangular function. (The OP of that question explicitly requested a Poisson summation formula solution.)

It's easier to compute the Fourier series for a rectangular function and use Parseval's theorem, so my solution uses that approach.

Let $f$ be the $2\pi$-periodic function defined as follows: $$f(x) = \begin{cases} 1 & \text{if }|x| < 1 \\ 0 & \text{if }1 \leq |x| \leq \pi \\ \end{cases}$$ Then $f$ has a Fourier series representation $$f(x) = \sum_{n=-\infty}^{\infty}a_n e^{inx}$$ where $$a_n = \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}$$ for each $n \in \mathbb Z$. When $n=0$, this reduces to $$\begin{aligned} a_n &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\ dx \\ &= \frac{1}{2\pi}\int_{-1}^{1}\ dx \\ &= \frac{1}{\pi} \end{aligned}$$ For $n \neq 0$ we have $$\begin{aligned} a_n &= \frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}\ dx \\ &= \frac{1}{2\pi}\int_{-1}^{1}e^{-inx}\ dx \\ &= \frac{1}{-i2\pi n}(e^{-in} - e^{in}) \\ &= \frac{\sin(n)}{\pi n} \end{aligned}$$ Now we can apply Parseval's theorem, which says that if $a_n$ are the Fourier coefficients of $f$ and $b_n$ are the Fourier coefficients of $g$, then $$\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)\overline g(x)\ dx = \sum_{n=-\infty}^{\infty}a_n \overline b_n,$$ where the bar denotes complex conjugation. (In words, the inner product of $f$ and $g$ equals the inner product of their Fourier coefficient sequences.) In particular, for the special case $f=g$, Parseval's theorem gives us $$\frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2\ dx = \sum_{n=-\infty}^{\infty}|a_n|^2$$ For our function $f(x)$, we have $|f(x)|^2 = 1$ for $|x| < 1$ and $|f(x)|^2 = 0$ otherwise, so the integral on the LHS becomes $$\frac{1}{2\pi}\int_{-1}^{1}\ dx = \frac{1}{\pi}$$ And, since our Fourier coefficients are real and even ($a_n = a_{-n}$), the sum on the RHS becomes $$a_0^2 + 2\sum_{n=1}^{\infty}a_n^2 = \frac{1}{\pi^2} + \frac{2}{\pi^2}\sum_{n=1}^{\infty}\left(\frac{\sin(n)}{n}\right)^2 = \frac{1}{\pi^2} + \frac{2}{\pi^2} S$$ where $S$ is your desired sum. Summarizing: $$\frac{1}{\pi} = \frac{1}{\pi^2} + \frac{2}{\pi^2} S$$ Multiplying by $\pi^2$ gives us $$\pi = 1 + 2S$$ and therefore $\displaystyle S = \frac{\pi - 1}{2}$ as desired.