How to prove this identity $\pi=\sum\limits_{k=-\infty}^{\infty}\left(\frac{\sin(k)}{k}\right)^{2}\;$?

Question

How to prove this identity? $$\pi=\sum_{k=-\infty}^{\infty}\left(\dfrac{\sin(k)}{k}\right)^{2}\;$$ I found the above interesting identity in the book $\bf \pi$ Unleashed.

Does anyone knows how to prove it?

Thanks.

Answer 1

Find a function whose Fourier coefficients are $\sin{k}/k$. Then evaluate the integral of the square of that function.

To wit, let

$$f(x) = \begin{cases} \pi & |x|<1\\0&|x|>1 \end{cases}$$

Then, if

$$f(x) = \sum_{k=-\infty}^{\infty} c_k e^{i k x}$$

then

$$c_k = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: f(x) e^{i k x} = \frac{\sin{k}}{k}$$

By Parseval's Theorem:

$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{k}}{k^2} = \frac{1}{2 \pi} \int_{-\pi}^{\pi} dx \: |f(x)|^2 = \frac{1}{2 \pi} \int_{-1}^{1} dx \: \pi^2 = \pi $$

ADDENDUM

This result is easily generalizable to

$$\sum_{k=-\infty}^{\infty} \frac{\sin^2{a k}}{k^2} = \pi\, a$$

where $a \in[0,\pi)$, using the function

$$f(x) = \begin{cases} \pi & |x|<a\\0&|x|>a \end{cases}$$

Answer 2

This sum may be calculated by computing a Mellin transform of a suitable function and then inverting that to get the sum. First start by rearranging some terms, so that the target sum $S$ becomes $$ S = 1 + 2 \sum_{k\ge 1} \frac{\sin(k)^2}{k^2}.$$ Now introduce $$ f(x) = \sum_{k\ge 1} \frac{\sin(xk)^2}{k^2}$$ so that we are looking for $f(1).$ Rewrite $f(x)$ as follows: $$ f(x) = - \frac{1}{4} \sum_{k\ge 1} \frac{e^{2ixk}-2+e^{-2ixk}}{k^2} = \frac{1}{2}\sum_{k\ge 1} \frac{1}{k^2} - \frac{1}{4} \sum_{k\ge 1} \frac{e^{2ixk}+e^{-2ixk}}{k^2} \\= \frac{\pi^2}{12} - \frac{1}{4} \sum_{k\ge 1} \frac{e^{2ixk}+e^{-2ixk}}{k^2}.$$ Using Mellin transforms, we find $$\mathfrak{M}\left(\sum_{k\ge 1}\frac{e^{2ixk}}{k^2};s\right)= \Gamma(s) \sum_{k\ge 1} \frac{1}{(2ik)^s k^2} = \frac{1}{(2i)^s}\Gamma(s) \zeta(s+2).$$ Similarly, $$\mathfrak{M}\left(\sum_{k\ge 1}\frac{e^{-2ixk}}{k^2};s\right)= \Gamma(s) \sum_{k\ge 1} \frac{1}{(-2ik)^s k^2} = \frac{1}{(-2i)^s}\Gamma(s) \zeta(s+2).$$ Now observe that $$ \frac{1}{(2i)^s} + \frac{1}{(-2i)^s} = e^{-s \log(2i)} + e^{-s \log(-2i)} = e^{-s \log 2 -s i\pi/2} + e^{-s \log 2 + s i\pi/2} \\ = 2^{-s} 2 \cos(s\pi/2)$$ Putting these two together, we obtain $$ \mathfrak{M}\left(\sum_{k\ge 1}\frac{e^{2ixk}+e^{-2ixk}}{k^2};s\right) = g^*(s)= 2 \times 2^{-s} \cos(s\pi/2) \Gamma(s) \zeta(s+2).$$ We will apply Mellin inversion to this term. There is a pole from the zeta and gamma functions at $s=-1$ (which the cosine turns from a double into a single pole) and one from the gamma function at $s=0$ and another one from the gamma function at $s=-2.$ The cosine term cancels the remaining poles of the gamma term at negative odd integers and the zeta term the ones at even integers. We have $$ \operatorname{Res}(g^*(s)x^{-s}; s=0) = \frac{\pi^2}{3},$$ $$ \operatorname{Res}(g^*(s)x^{-s}; s=-1) = -2\pi x,$$ $$ \operatorname{Res}(g^*(s)x^{-s}; s=-2) = 2 x^2.$$ This yields for the Mellin inversion integral that $$ \mathfrak{M}^{-1}(g^*(s);x) = \int_{1-i\infty}^{1+i\infty} g^*(s) x^{-s} ds = 2 x^2 - 2\pi x + \frac{\pi^2}{3}.$$ Returning to $S$ we have shown that $$S = 1 + 2\left(\frac{\pi^2}{12}-\frac{1}{4} \left( 2-2\pi + \frac{\pi^2}{3}\right) \right) = 1 - \frac{1}{2} (2-2\pi) =\pi. $$

Remark, Feb 29 2020. It is proved at the following MSE link that the contribution from the left vertical line segment $\sigma \pm i\infty$ vanishes as we left-shift to $\sigma=-\infty$ when $x\in (0,\pi).$

Answer 3

Assume $a\in\left[0,\frac\pi2\right]$.

An integral $$ \begin{align} \int_0^a\frac{\sin(2kx)}{k}\mathrm{d}x &=\int_0^a\frac{2\sin(kx)}{k^2}\mathrm{d}\sin(kx)\\ &=\left.\frac{\sin^2(kx)}{k^2}\right]_0^a\\ &=\frac{\sin^2(ka)}{k^2}\tag{1} \end{align} $$ and a sum $$ \begin{align} \sum_{k=1}^\infty\frac{\sin(2kx)}{k} &=\sum_{k=1}^\infty\frac{e^{i2kx}-e^{-i2kx}}{2ik}\\ &=\frac1{2i}\left(-\log(1-e^{i2x})+\log(1-e^{-i2x})\right)\\ &=\frac1{2i}\log(-e^{-i2x})\\[4pt] &=\frac\pi2-x\quad\text{for }x\in\left(0,\pi\right)\tag{2} \end{align} $$ Putting $(1)$ and $(2)$ together $$ \begin{align} \sum_{k=1}^\infty\frac{\sin^2(ka)}{k^2} &=\int_0^a\left(\frac\pi2-x\right)\,\mathrm{d}x\\ &=\frac\pi2a-\frac{a^2}2\tag{3} \end{align} $$ If we take $\dfrac{\sin(ka)}{ka}=1$ when $k=0$, we get the answer to the question using $a=1$: $$ \sum_{k\in\mathbb{Z}}\left(\frac{\sin(ka)}{ka}\right)^2=\frac\pi a\tag{4} $$

---

Application to a Riemann Sum

If we multiply $(4)$ by $a$ and set $a=\frac1n$, we get $$ \sum_{k\in\mathbb{Z}}\frac{\sin^2(k/n)}{(k/n)^2}\frac1n=\pi\tag{5} $$ $(5)$ is a Riemann sum which shows that $$ \int_{-\infty}^\infty\frac{\sin^2(x)}{x^2}\,\mathrm{d}x=\pi\tag{6} $$

---

First Power of Sinc

We can also use $(2)$ at $x=\frac a2$, assuming $\frac{\sin(ka)}{ka}=1$ when $k=0$, to get $$ \sum_{k\in\mathbb{Z}}\frac{\sin(ka)}{ka}=\frac\pi a\tag{7} $$ Again, multiplying $(7)$ by $a$ and letting $a=\frac1n$, we get $$ \sum_{k\in\mathbb{Z}}\frac{\sin(k/n)}{k/n}\frac1n=\pi\tag{8} $$ and $(8)$ is a Riemann sum which shows that $$ \int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\pi\tag{9} $$