How to prove this identity $\pi=\sum\limits_{k=-\infty}^{\infty}\left(\frac{\sin(k)}{k}\right)^{2}\;$? 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.
 A: 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
$$f(x) = \begin{cases} \pi & |x|<a\\0&|x|>a \end{cases}$$
A: $\newcommand{\+}{^{\dagger}}%
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$$
\!\!\!\!\!\sum_{k=-\infty}^{\infty}{\sin^{2}\pars{k} \over k^{2}}
=
\sum_{k=-\infty}^{\infty}\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\,
\delta\pars{x - k}\,\dd x
=
\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}
\bracks{\sum_{k=-\infty}^{\infty}\delta\pars{x - k}}\,\dd x\tag{1}
$$

The sum over $k$ in the right hand side of $\pars{1}$ is a periodic function of $x$ with period $1$ and it can be rewritten as follows:
$$
\sum_{k=-\infty}^{\infty}\delta\pars{x - k}
=
\sum_{n = -\infty}^{\infty}a_{n}\expo{2n\pi\ic x}\tag{2}
$$
The set of coefficients $\braces{a_{n}\ \ni\ n \in {\mathbb Z}}$ are given by:
$$
1 = \int_{-1/2}^{1/2}\expo{-2m\pi\ic x}\sum_{k=-\infty}^{\infty}
\delta\pars{x - k}\,\dd x
=
\sum_{n = -\infty}^{\infty}a_{n}\int_{-1/2}^{1/2}\expo{2\pars{n - m}\pi\ic x}\,\dd x
=
a_{m}
$$
which yields $\pars{~\mbox{see expression}\ \pars{2}~}$:
$$
\sum_{k=-\infty}^{\infty}\delta\pars{x - k}
=
\sum_{n=-\infty}^{\infty}\expo{2n\pi\ic x}\tag{3}
$$

We replace $\pars{3}$ in the right hand side of $\pars{1}$ to get:
\begin{align}
&\color{#0000ff}{\large\sum_{k=-\infty}^{\infty}{\sin^{2}\pars{k} \over k^{2}}}
=
\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}
\sum_{n=-\infty}^{\infty}\expo{2n\pi\ic x}\,\dd x
\\[3mm]&=
\sum_{n=-\infty}^{\infty}\int_{-\infty}^{\infty}
\pars{\half\int_{-1}^{1}\expo{\ic tx}\,\dd t}
\pars{\half\int_{-1}^{1}\expo{\ic t'x}\,\dd t'}
\expo{2n\pi\ic x}\,\dd x\tag{4}
\\[3mm]&=
{\pi \over 2}\sum_{n=-\infty}^{\infty}
\int_{-1}^{1}\dd t\int_{-1}^{1}\dd t'\int_{-\infty}^{\infty}\expo{\pars{t + t' + 2n\pi}\ic x}\,
{\dd x \over 2\pi}
\\[3mm]&=
{\pi \over 2}\sum_{n=-\infty}^{\infty}
\int_{-1}^{1}\dd t\int_{-1}^{1}\dd t'\delta\pars{t + t' + 2n\pi}
\\[3mm]&=
{\pi \over 2}\sum_{n=-\infty}^{\infty}
\int_{-1}^{1}\Theta\pars{1 - \verts{-t - 2n\pi}}\,\dd t
=
{\pi \over 2}\sum_{n=-\infty}^{\infty}\overbrace{\left.\int_{-1}^{1}\dd t
\right\vert_{-1 - 2n\pi\ <\ t\ <\ 1 - 2n\pi}}^{\ds{=\ 2\,\delta_{n0}}} =
\color{#0000ff}{\Large\pi}
\end{align}
where we use the identity
$\ds{{\sin\pars{x} \over x} = \half\int_{-1}^{1}\expo{\ic tx}\,\dd t}.\quad$ See line $\pars{4}$.
A: This is a great place to apply Abel-Plana's formula
$$\sum_{n\geqslant 0} f(n)=\int_0^\infty f(x)\, dx+\frac{f(0)}{2}+i\int_0^\infty\frac{f(ix)-f(-ix)}{e^{2\pi x}-1}\, dx$$

Allow $f(x)=\left(\frac{\sin x}{x}\right)^2$

We note that, with the fact that $f$ is even $$\sum_{n\geqslant 0} \left(\frac{\sin n}{n}\right)^2=\frac{1}{2}\left(\sum_{n\in\Bbb{Z}}\left(\frac{\sin n}{n}\right)^2-1\right)$$
We already know that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2\, dx=\frac{\pi}{2};$$ and $$\frac{\sin 0}{0}:=1$$
Also, we easily determine that:$$\left(\frac{\sin ix}{ix}\right)^2=\left(\frac{\sin (-ix)}{-ix}\right)^2=\frac{\sinh^2 x}{x^2}$$
With these, the result is obvious.
I'm sorry, I know this is a bit overpowering of a formula to use on this question, but I thought it was worth noting.
A: 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
$$
\int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\pi\tag{9}
$$
A: I'll try to tackle the sum for general powers of $\operatorname{sinc}{x}$. It seems that the general formula is
$$
\begin{align}
\sum_{m = -\infty}^\infty& \left(\frac{\sin{(m)}}{m}\right)^n \tag{0}\\
& = \frac{(-1)^{n}\pi}{2^{n}(n-1)!}\sum_{\ell = -\lfloor n/(2\pi)\rfloor}^{\lfloor n/(2\pi)\rfloor}\left(\sum_{k = 0}^n(-1)^k{n\choose k} (2\pi \ell - n+2k)^{n-1}\operatorname{sign}(2\pi \ell-n+2k)\right).
\end{align}
$$
Before I get into that, I'm going to give another solution to the problem posed originally in the question. I'll use the Poisson summation formula, namely,
$$
\sum_{n = -\infty}^\infty f(x+n) = \sum_{\ell = -\infty}^\infty \hat{f}(2\pi \ell) e^{2\pi i \ell x}.
$$
Here $f$ is a function with reasonable regularity and decay properties, and
$$
\hat{f}(t) = \int_{-\infty}^\infty f(x)e^{-itx}\,dx
$$
is the Fourier transform of $f$. The idea is of course to take take $f(x) = \operatorname{sinc}^2{x}$.

I'm going to start by proving directly that
$$
\begin{align}
\int_{-\infty}^\infty (\operatorname{sinc}{x})^2e^{-itx}\,dx = \int_{-\infty}^\infty \left(\frac{\sin{x}}{x}\right)^2e^{-itx}\,dx = \frac{\pi}{4}(|x-2| + |x+2| - 2|x|). \tag{1}
\end{align}
$$
The right hand side is just a concise expression (that happens to be suggestive of the general situation) for the triangle-shaped function that is $\pi(1-|x|/2)$ when $|x| < 2$ and $0$ otherwise. An application of the Poisson summation formula referenced above then gives
$$
\sum_{n = - \infty}^\infty \left(\frac{\sin{(x+n)}}{x+n}\right)^2 = \pi,
$$
for all $x \in \mathbb{R}$. Taking $x = 0$ will give the result you seek.
To prove the identity $(1)$, write
$$
\operatorname{sinc}{x} = \frac{\sin{x}}{x} = \frac{1}{2}\int_{-1}^1 e^{-ixt}\,dt = \hat{g}(x),
$$
where $g = (1/2)\chi_{[-1,1]}$ is one half times the characteristic function of the interval $[-1,1]$. The Fourier transform converts convolution to multiplication, so
$$
\widehat{g*g}(x) = \left(\frac{\sin{x}}{x}\right)^2.
$$
The right-hand side (the square of $\operatorname{sinc}{x}$) is integrable, so an application of the Fourier inversion formula gives us
$$
\int_{-\infty}^\infty \left(\frac{\sin{x}}{x}\right)^2e^{itx}\,dx = 2\pi (g*g)(t) = \frac{\pi}{2}\int_{-\infty}^\infty \chi_{[-1,1]}(t-y)\chi_{[-1,1]}(y)\,dy.
$$
The evaluation of the integral on the right is straightforward (but a little tedious), and it gives us $(1)$.

The method just used to prove the identity $(1)$ has the benefit of being elementary, and does generalize to higher powers of $\operatorname{sinc}{x}$ (in the sense that $(-1)^n2\pi$ times the $n$-fold convolution of $(1/2)\chi_{[-1,1]}$ with itself is the Fourier transform of $\operatorname{sinc}{x}$ raised to the $n$th power) but unfortunately the computations become quite involved.  Thus, I'll use another method, one that requires some basic knowledge of distribution theory, to prove—for positive integers $n$—that
$$
\begin{align}
\widehat{\operatorname{sinc}^n}{(x)} &= \int_{-\infty}^\infty \left(\frac{\sin{t}}{t}\right)^ne^{-ixt}\,dt \\
&  =  \frac{(-1)^n\pi}{2^{n}(n-1)!}\sum_{k = 0}^n(-1)^k{n\choose k} (x - n+2k)^{n-1}\operatorname{sign}(x-n+2k). \tag{2}
\end{align}
$$
Applying the Poisson summation formula to this identity leads to $(0)$—almost. I still have to explain why the upper and lower bounds of summation in the right hand side of $(0)$ are finite, which I'll do now. Since $(2)$ is equal to a constant multiple of the $n$-fold composition of $\chi_{[-1,1]}$ with itself, it is supported in $[-n,n]$. (This can also be seen directly.) Therefore, in the right hand side of the Poisson summation formula, we need only sum over those indices $\ell$ satisfying $|\ell|<n/2\pi$. This is where the upper and lower bounds come from in the equation $(0)$.
So all that remains is to prove the equation $(2)$. Let me briefly outline the tools we'll need. Let $\mathscr{S} = \mathscr{S}(\mathbb{R})$ denote the Schwartz space on $\mathbb{R}$. Let $u_n : \mathscr{S} \to \mathbb R$ be the distribution defined for $\varphi \in \mathscr{S}$ by
$$
u_n(\varphi) = \lim_{\epsilon \to 0^+}\int_{|x|>\epsilon} \frac{\varphi(x) - \sum_{k=0}^{n-2}\varphi^{(k)}(0)x^k/k!}{x^n}\,dx.
$$
Basically, $u_n$ is the distribution best resembling the function $x^{-n}$. In fact, if $h$ is a smooth function that vanishes to order $n$ at the origin, then the distribution $h\cdot u_n$ is equal to the function $h(x)/x^n$. As one would expect from the relation $\partial^k x^{-1} = (-1)^{k}k! x^{-k-1}$, the $k$th distributional derivative of $u_1$ is given by $\partial^ku_1 = (-1)^kk!\, u_{k+1}$. This is all proved straightforwardly.
It turns out that
$$
\hat{u}_1(t) = -i\pi \operatorname{sign}(t),
$$
where $\operatorname{sign}(t)$ is the usual sign function that returns $0$ when $t = 0$ and $t/|t|$ otherwise. (This can be proved by regarding $\operatorname{sign}(t)$ as a limit $\lim_{k\to \infty} \chi_{(0,k]}(t) - \chi_{[-k,0)}(t)$, taking the distributional limit of the Fourier transforms, and then applying the inverse Fourier transform.) Now, for any distribution $v$, one has
$$
\widehat{\partial^k v} = (it)^k \hat{v}.
$$
Thus, using $u_n = (-1)^{n-1}(n-1)!^{-1}\partial^{n-1} u_1$,
$$
\hat{u}_n(t) = \frac{(-1)^{n-1}}{(n-1)!}\widehat{\partial^n u_1}(t) = \frac{(-1)^{n-1}}{(n-1)!}(it)^{n-1}\hat{u}_1(t) = \frac{(-1)^{n}i^n\pi}{(n-1)!}t^{n-1} \operatorname{sign}(t). \tag{3}
$$
Finally, another property of the Fourier transform dictates that
$$
\widehat{e^{ihx}u_n}(t) = \hat{u}_n(x-h). \tag{4}
$$
We're now prepared to evaluate the Fourier transform of $(\operatorname{sinc}{x})^n$.
Since $(\sin{x})^n$ vanishes to order $n$ at the origin, we have
$$
(\operatorname{sinc}{x})^n = (\sin{x})^n x^{-n} = (\sin{x})^nu_n
$$
as distributions. If we now expand $(\sin{x})^n$ into powers of $e^{ix}$, we get
$$
(\sin{x})^nu_n = \frac{1}{(2i)^n} \sum_{k = 0}^n(-1)^k{n\choose k}e^{i(n-2k)x}u_n.
$$
Taking the term-wise Fourier transform of the right hand side and inserting the formulas $(3)$ and $(4)$, we arrive at
$$
\begin{align}
\widehat{\operatorname{sinc}^n}\,(t) &= \frac{1}{(2i)^n} \sum_{k = 0}^n(-1)^k{n\choose k}\widehat{e^{i(n-2k)x}u_n}(t) \\
& = \frac{1}{(2i)^n} \sum_{k = 0}^n(-1)^k{n\choose k}\hat{u}_n(t-n+2k)\\
& = \frac{1}{(2i)^n} \sum_{k = 0}^n(-1)^k{n\choose k}\frac{(-1)^{n}i^n\pi}{(n-1)!}(t-n+2k)^{n-1}\operatorname{sign}(t-n+2k).
\end{align}
$$
Simplifying then gives $(2)$.
A: 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 $x\in (0,\pi).$ 
A: There is a simple way to compute the sum. Note
$$ \sum_{k=-\infty}^\infty\frac{\sin^2k}{k^2}=1+2\sum_{k=1}^\infty \frac{\sin^2k}{k^2}=1+\sum_{k=1}^\infty\frac{1-\cos(2k)}{k^2}=1+\frac{\pi^2}{6}-\sum_{k=1}^\infty\frac{\cos(2k)}{k^2}.$$
Letting $x=e^{2\theta i}$ in $\sum_{k=1}^\infty\frac{x^k}{k}=-\ln(1-x)$ gives us
$$ \sum_{k=1}^\infty\frac{1}{k}(\cos(2k\theta)+i\sin(2k\theta))=-\ln(1-\cos(2\theta)-i\sin(2\theta)). $$
So
$$ \sum_{k=1}^\infty\frac{1}{k}\sin(2k\theta)=-\Im\ln(1-\cos(2\theta)-i\sin(2\theta))=-\arctan(-\cot \theta)=\frac{\pi}{2}-\theta. $$
Integrating this derives
$$ -\sum_{k=1}^\infty\frac{1}{2k^2}\cos(2k\theta)=\frac{\pi}{2}\theta-\frac{1}{2}\theta^2+C. $$
Letting $\theta=0$, we have $C=-\frac{\pi^2}{12}$. Thus
$$ \sum_{k=1}^\infty\frac{1}{k^2}\cos(2k\theta)=-\pi \theta+\theta^2+\frac{\pi^2}{6}. $$
Letting $\theta=1$, we have
$$ \sum_{k=1}^\infty\frac{\cos(2k)}{k^2}=-\pi+1+\frac{\pi^2}{6} $$
and hence
$$ \sum_{k=-\infty}^\infty\frac{\sin^2k}{k^2}=1+\frac{\pi^2}{6}-\sum_{k=1}^\infty\frac{\cos(2k)}{k^2}=1+\frac{\pi^2}{6}-(-\pi+1+\frac{\pi^2}{6})=\pi.$$
It is easy to use the same trick to generalize this result to $\sum_{k=-\infty}^\infty\frac{\sin^2(ak)}{k^2}$. I omit the detail.
