# 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.

• Just for clarification, what happens when $k=0?$ – L. F. Mar 15 '13 at 17:52
• For more fun along these lines, see web.cs.dal.ca/~jborwein/sinc-sums.pdf – user940 Mar 15 '13 at 17:58
• @L.F. at $k=0$, the limit is taken leading to $1$. See Ron's answer below. – nbubis Mar 15 '13 at 18:16
• – Mhenni Benghorbal Mar 19 '13 at 13:42

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

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

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

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

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

• On the Wolfram MathWorld web page for the sinc function it states, "The remarkable fact that the sums of $\text{sinc}(k)$ and $\text{sinc}^{2}(k)$ are equal appears to have first been published in Baillie (1978)." I remember reading that a long time ago and thinking that summing $\text{sinc}^{2}(k)$ must be really difficult if no one realized they were the same until 1978. But the evaluation is mostly straightforward as you showed. Strange. – Random Variable Mar 15 '13 at 18:56
• @RandomVariable: Huh, how odd. Interesting find. (+1) – Ron Gordon Mar 15 '13 at 19:13
• I accidentally plugged $\sum _{k=-\infty }^{\infty } \frac{\sin (k)}{k}$ (without the square) into Mathematica and it also gave $\pi$. I wonder why this identity with the square is the common one. The evaluation of this sum to the first few powers are: $\left\{\pi ,\pi ,\frac{3 \pi }{4},\frac{2 \pi }{3},\frac{115 \pi }{192},\frac{11 \pi }{20}\right\}$ (in case you see these on Jeapardy). The seventh power evaluates to $\frac{129423 \pi -201684 \pi ^2+144060 \pi ^3-54880 \pi ^4+11760 \pi ^5-1344 \pi ^6+64 \pi ^7}{23040}$ and the subsequent powers are all complicated-looking like this. – amr Mar 16 '13 at 20:26
• @amr: see the reference that Random provided. It is kind of interesting that the sum over sinc equals this sum. I'd like to see it derived as something other than a coincidence. Interesting also about the seventh power thing. Even powers of sinc may be summed using Parseval similar to how I outline above. – Ron Gordon Mar 16 '13 at 20:30
• @amr This discussion here was very helpful for me in this topic which you might wish to see: mathematica.stackexchange.com/questions/157444/… – Dr. Wolfgang Hintze Oct 10 '17 at 15:54

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 $$\int_{-\infty}^\infty\frac{\sin(x)}{x}\,\mathrm{d}x=\pi\tag{9}$$

• Interesting approach (+1). How easily does it generalize to other sums of powers of sincs? – Ron Gordon Mar 23 '13 at 8:52
• @RonGordon: I noticed that $(2)$ gives the result for the first power of sinc, and then noticed that these results apply to Riemann sums. I will look into higher powers. – robjohn Mar 23 '13 at 12:15
• Very nifty! I'd upvote you again, but...well, you know. Anyway, that's a really important observation and may also help add more insight into a question I answered yesterday about the link between Fourier transforms and Fourier series. – Ron Gordon Mar 23 '13 at 12:18
• This is that question to which I referred: math.stackexchange.com/questions/329576/… – Ron Gordon Mar 23 '13 at 13:14
• @RonGordon: I generalized this to third powers in this answer. – robjohn Jul 27 '13 at 14:35

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 $$|\ell|. This is where the upper and lower bounds come from in the equation $$(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 $$(2)$$.

• In your equation (1), are you integrating over $x$ or $t$? You integrate over $x$ in the LHS but have a function of $x$ in the RHS. – David Zhang May 27 '16 at 4:04

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 $$x\in (0,\pi).$$

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.

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}{\left\langle #1 \right\rangle}% \newcommand{\braces}{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}{\displaystyle{#1}}% \newcommand{\equalby}{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}{\left\vert #1\right\rangle}% \newcommand{\ol}{\overline{#1}}% \newcommand{\pars}{\left( #1 \right)}% \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}{\underline{#1}}% \newcommand{\verts}{\left\vert\, #1 \,\right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$\!\!\!\!\!\sum_{k=-\infty}^{\infty}{\sin^{2}\pars{k} \over k^{2}} = \sum_{k=-\infty}^{\infty}\int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}}\, \delta\pars{x - k}\,\dd x = \int_{-\infty}^{\infty}{\sin^{2}\pars{x} \over x^{2}} \bracks{\sum_{k=-\infty}^{\infty}\delta\pars{x - k}}\,\dd x\tag{1}$$

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 $\pars{4}$.

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.