Proof of $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$ I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$
What do you think?
It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this post.
 A: Apply Parseval-Plancherel to $\chi_{[-1,1]}$.
EDIT
If we consider the Fourier transform as given by 
$$f\mapsto \hat{f}(\xi)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\xi}dx$$
then 
$$\int_{-\infty}^\infty |f(x)|^{2}dx=\int_{-\infty}^\infty |\hat{f}(\xi)|^{2}d\xi$$
for $f\in L^2(\mathbb{R})$.
For $f(x)=\chi_{[-1,1]}(x)$, the characteristic function on $I=[-1,1]$ (that is $f(x)=1$ for $x\in I$ and $f(x)=0$ otherwise), we have
$$\hat{f}(x)=\int_{-\infty}^\infty f(x)e^{-2\pi i x\xi}dx =\int_{-1}^1 e^{-2\pi i x\xi}dx=\frac{e^{-2\pi i \xi}-e^{2\pi i \xi}}{{-2\pi i \xi}}=\frac{\sin 2\pi\xi}{\pi\xi}$$
Hence 
$$\int_{-\infty}^\infty \left(\frac{\sin 2\pi\xi}{\pi\xi}\right)^2d\xi=\int_{-1}^1dx = 2$$
by a change of variables, $y=2\pi \xi$,  and using symmetry we arrive at
$$2=\int_{-\infty}^\infty \left(\frac{2\sin y}{y}\right)^2\frac{dy}{2\pi}=\frac{8}{2\pi}\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy$$
or 
$$\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy=\frac{\pi}{2}$$
A: Indeed beforehand, one can show that both integrals are equals without computing their explicit values. Namely we have 
$$ \color{blue}{\int_0^\infty\frac{\sin x}{x} dx = \int_0^\infty\frac{\sin^2u}{u^2} du} $$
From this Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$? We know that , $$\frac{\pi}{2} =\int_0^\infty\frac{\sin x}{x} dx =  \int_0^\infty\frac{\sin 2u}{2u} d(2u) =\int_0^\infty\frac{\sin 2u}{u} du\\ = \underbrace{\left[\frac{\sin^2 u}{u}\right]_0^\infty}_{=0} +\int_0^\infty\frac{\sin^2u}{u^2} du =\color{blue}{\int_0^\infty\frac{\sin^2u}{u^2} du} $$

Given that, $\sin2x = 2\sin x\cos x=(\sin^2x)'$ and $~~\lim\limits _{x\to 0}\frac{\sin^2 x}{x^2} = 1$

A: Well, it's not hard to reduce this integral to $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$: Just integrate by parts in $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$, integrating $\displaystyle {1 \over x^2}$ and differentiating $\displaystyle \sin^2(x)$. You're left with $\displaystyle \int_0^{\infty} {\sin(2x) \over x}\,dx$ which reduces to the $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx$ integral after changing variables from $\displaystyle x$ to $\displaystyle 2x$.
So any elementary proof that $\displaystyle \int_0^{\infty} {\sin(x) \over x}\,dx = {\pi \over 2}$ is effectively also an elementary proof that $\displaystyle \int_0^{\infty} {\sin^2(x) \over x^2}\,dx$ is also $\displaystyle {\pi \over 2}$.
A: Integrating by parts:
$$V_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin^2x}{x^2} \mathrm dx = U_{2n} + U_{2n+1}$$
where:
$$U_n=\int_{n\pi}^{(n+1)\pi} \frac{\sin x}{x} \mathrm dx$$
Thus:
$$\sum U_n = \sum V_n$$
A: An alternative approach that employs a combination of both Feynman's Trick and Laplace Transforms
First, let 
$$I(t) = \int_{0}^{\infty} \frac{\sin^2(xt)}{x^2} \:dx$$
Note that $I = I(1)$
Taking the Laplace Transform:
\begin{align}
\mathscr{L}\left[I(t)\right] &= \int_{0}^{\infty} \frac{\mathscr{L}\left[\sin^2(xt)\right]}{x^2} \:dx = \int_{0}^{\infty} \frac{2x^2}{4x^2s + s^3}\frac{1}{x^2}\:dx \\
&= \frac{1}{2s} \int_{0}^{\infty} \frac{1}{x^2 + \frac{s^2}{4}}\:dx = \frac{1}{2s}\left[ \frac{2}{s}\arctan\left(\frac{2x}{s}\right)\right]_{0}^{\infty} = \frac{1}{s^2}\frac{\pi}{2}
\end{align}
We now take the Inverse Laplace Transform
$$ I(t) =  \mathscr{L}^{-1}\left[ \frac{1}{s^2}\frac{\pi}{2}\right]  = \frac{\pi t}{2}$$
And so, 
$$I = I(1) = \int_{0}^{\infty} \frac{\sin^2(x)}{x^2} \:dx = \frac{\pi\cdot 1}{2} = \frac{\pi}{2}$$
A: Another way (with usage of complex analysis) which I want to add here is the following: 

Let $\Gamma$ be the closed curve in the sketch above. Then the integral $\int_{\Gamma}^{} \! \frac{e^{iz}}{z} \, dz$ is zero. (Cauchy integral theorem).
We now compute the several integrals separately:
1)$$ \lim_{R \to \infty} \left| \int_{\Gamma_R}^{} \! \frac{e^{iz}}{z} \, dz \right|  \leq \lim_{R \to \infty}\int_0^{\pi} \! \frac{1}{e^{Rsin(t)}} \, dt =\int_0^{\pi} \! 0 \, dt =0$$
2) $$\lim_{\epsilon \to 0} \int_{\gamma_{\epsilon}}^{} \! \frac{e^{iz}}{z} \, dz   =-i\lim_{\epsilon \to 0}\int_{-\pi}^0 \! e^{i\epsilon cos(t)+\epsilon sin(t)} \, dt =-i\int_{\pi}^0 \! 1 \, dt=-\pi i$$
Hence:
$$0=\int_{\Gamma}^{} \! \frac{e^{iz}}{z} \, dz=\int_{-R}^{-\epsilon} \! \frac{e^{iz}}{z} \, dz+\int_{\gamma_{\epsilon}}^{} \! \frac{e^{iz}}{z} \, dz+\int_{\epsilon}^{R} \! \frac{e^{iz}}{z} \, dz+\int_{\Gamma_R}^{} \! \frac{e^{iz}}{z} \, dz$$
It follows that : $$\int_{-R}^{-\epsilon} \! \frac{e^{iz}}{z} \, dz+\int_{\epsilon}^{R} \! \frac{e^{iz}}{z} \, dz=\pi i$$
By taking the limits $R \rightarrow \infty$ and $\epsilon \rightarrow 0$, we obtain:
$$\int_{-\infty}^{\infty} \! \frac{sin(x)}{x} \, dx =Im \int_{-\infty}^{\infty} \! \frac{e^{iz}}{z} \, dz =Im(i\pi)=\pi$$
Note: $sin(x)$ and $x$ are odd functions, hence $\frac{sin(x)}{x}$ is even. So $\int_{0}^{\infty} \! \frac{sin(x)}{x} \, dx= \frac{\pi}{2}$
A: Let $f(x)=\max\{0,1-|x|\}$. It is easy to calculate the Fourier transform
$$\hat{f}(\xi)=\int_{-\infty}^{\infty}f(x)e^{-ix\xi}dx=\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2.$$
Taking the inverse Fourier transform, we get 
$$\int_{-\infty}^{\infty}\left(\frac{\sin(\xi/2)}{\xi/2}\right)^2e^{ix\xi}d\xi=2\pi f(x),$$ 
and the result follows.
The second integral can be computed in a similar way. Just take $f(x)=\chi_{[-1,1]}(x)$ (the indicator function of the interval $[-1,1]$).

Edit. It might be interesting to note that there are analogous formulas for the sinc 
sums
$$\sum_{n=1}^{\infty}\frac{\sin n}{n}=\sum_{n=1}^{\infty}\left(\frac{\sin n}{n}\right)^2=
\frac{\pi}{2}-\frac{1}{2}.$$
I learned about this from the note "Surprising Sinc Sums and Integrals" by Baillie, Borwein, and Borwein (can be found through a quick web search).
A: More generally, there is a result due to Wolstenholme (I can't find a link) that says
$$ \int_0^\infty \left( \frac {\sin x}{x} \right)^n dx = 
\frac{1}{(n-1)!} \frac{\pi}{2^n} \left\lbrace
n^{n-1} - { n \choose 1 } (n-2)^{n-1} + { n \choose 2 } (n-4)^{n-1} - \cdots
\right\rbrace .$$
A: $$
\begin{align}
\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm dx
&=
\int_0^\infty\frac{\frac12(1-\cos2x)}{x^2}\mathrm dx
\\
&=
\int_0^\infty \frac{1-\cos x}{x^2}\mathrm dx
\\
&=
\frac12\int_{-\infty}^\infty \frac{1-\cos x}{x^2}\mathrm dx
\\
&=
\frac12\int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm ix}}{x^2}\mathrm dx
\\
&=
\frac12\int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm ix}+\mathrm ix/(1+x^2)}{x^2}\mathrm dx
\\
&=
\frac12\Re\int_{-\infty}^\infty \frac{1-\mathrm e^{\mathrm ix}+\mathrm ix/(1+x^2)}{x^2}\mathrm dx\;.
\end{align}
$$
Now close the contour in the upper half plane, enclosing the pole at $x=\mathrm i$ with residue $1/(2\mathrm i)$, yielding
$$\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm dx=\frac12\cdot2\pi\mathrm i\cdot\frac1{2\mathrm i}=\frac\pi2\;.$$
A: From squaring the identity
$$\frac{\sin nx}{\sin x}=\frac{e^{inx}-e^{-inx}}{e^{ix}-e^{-ix}}
=\sum_{k=0}^{n-1}e^{(2k-n+1)ix}$$
and integrating we get
$$n\pi=\int_{-\pi/2}^{\pi/2}\frac{\sin^2 nx}{\sin^2 x}\,dx.$$
Let
$$I_n=\int_{-\pi/2}^{\pi/2}\frac{\sin^2 nx}{nx^2}\,dx
=\int_{-n\pi/2}^{n\pi/2}\frac{\sin^2y}{y^2}\,dy.$$
Then
$$\pi-I_n=\frac{1}{n}
\int_{-\pi/2}^{\pi/2}\sin^2nx(\csc^2x-x^{-2})\,dx.$$
and so
$$|\pi-I_n|\le\frac{1}{n}\int_{-\pi/2}^{\pi/2}|\csc^2x-x^{-2}|\,dx
=O(1/n)$$
as $x\mapsto\csc^2x-x^{-2}$ extends to a continuous
function on $[-\pi/2,\pi/2]$.
Hence $I_n\to\pi$ as $n\to\infty$ and
$$\pi=\int_\infty^\infty\frac{\sin^2y}{y^2}\,dy.$$
A: \begin{align}
{{\rm d} \over {\rm d}\mu}
\int_{-\infty}^{\infty}{\sin^{2}\left(\mu x\right) \over x^{2}}\,{\rm d}x
&=
\int_{-\infty}^{\infty}
{2\sin\left(\mu x\right)\cos\left(\mu x\right)x \over x^{2}}\,{\rm d}x
=
\int_{-\infty}^{\infty}
{\sin\left(2\mu x\right) \over x}\,{\rm d}x
\\[3mm]&=
{\rm sgn}\left(\mu\right)\int_{-\infty}^{\infty}
{\sin\left(x\right) \over x}\,{\rm d}x
\\[5mm]
\int_{0}^{1}{{\rm d} \over {\rm d}\mu}\left[%
\int_{-\infty}^{\infty}{\sin^{2}\left(\mu x\right) \over x^{2}}\,{\rm d}x
\right]\,{\rm d}\mu
&=
\int_{0}^{1}{\rm sgn}\left(\mu\right)\left[\int_{-\infty}^{\infty}
{\sin\left(x\right) \over x}\,{\rm d}x\right]\,{\rm d}\mu
\end{align}
$$\color{#ff0000}{\large%
\int_{-\infty}^{\infty}{\sin^{2}\left(x\right) \over x^{2}}\,{\rm d}x
\color{#000000}{\ =\ }
\int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x}
$$
A: Another option is a Schwinger parametrization. Write $x^{-2}=\int_0^\infty y e^{-xy} dy$ so $$\begin{align}\int_0^\infty x^{-2}\sin^2 x dx&=-\frac{1}{4}\int_0^\infty\int_0^\infty (e^{ix}-e^{-ix})^2 ye^{-xy}dxdy\\&=-\frac{1}{4}\int_0^\infty\int_0^\infty y(e^{-x(y+2i)}+e^{-x(y-2i)}-2e^{-xy})dxdy\\&=-\frac{1}{4}\int_0^\infty\left(\frac{y}{y+2i}+\frac{y}{y-2i}-2\right)dy\\&=2\int_0^\infty\frac{dy}{y^2+4}\\&=\frac{\pi}{2}.\end{align}$$
A: Here is a slightly different way using Feynman integration.
We consider the function:
$$I(t,a)=\int^\infty_{0}\frac{e^{-ax}\sin^2(xt)}{x^2}dx$$
Differentiating w.r.t. $t$ gives us:
$$\frac{\partial I(t,a)}{\partial t}=\int^\infty_{0}\frac{e^{-ax}\sin(2xt)}{x}dx$$
and now w.r.t $a$ giving us:
$$\frac{\partial^2 I(t,a)}{\partial a\partial t}=-\int^\infty_{0}e^{-ax}\sin(2xt)dx$$
$$=-Im \left(\int^\infty_0 e^{(it-a)x}dx\right)$$
$$=-Im\left( \frac{1}{a-it} \right)$$
$$=-\frac{t}{a^2+t^2}$$
Now integrating w.r.t. $a$ gives us:
$$\frac{\partial I(t,0)}{\partial t}=-\int^0_{\infty}\frac{t}{a^2+t^2}da$$
making the substitution $a=t \tan(\theta)$ this becomes:
$$\frac{\partial I(t,0)}{\partial t}=-\int^0_{\pi/2} da$$
$$=\frac{\pi}{2}$$
Then integrating w.r.t $t$:
$$ I(1,0)=\int^1_0 \frac{\pi}{2} dt$$
$$=\frac{\pi}{2}$$
References
SuperAbound (https://math.stackexchange.com/users/140590/superabound), Integration of Sinc function, URL (version: 2014-08-09): https://math.stackexchange.com/q/891822
A: Fourier Transform of $$\frac{\sin t}{t} $$ is a rectangular pulse $$\pi Rect(w/2) $$ of amplitude equal to $$\pi$$ for $$-1<w< 1$$.
Using Parseval's theorem, $$ {\int_{-\infty}^\infty\frac{\sin^2t}{t^2} dt}={\frac{1}{2\pi}\int_{-\infty}^\infty\left[{\pi Rect(w/2)}\right]^2 dw} = {\frac{1}{2\pi}\int_{-1}^{+1}\left[{\pi}\right]^2 dw} = \pi$$
As sinx/x is an even function of x,
$${\int_0^\infty\frac{\sin^2x}{x^2} dx}=\frac{\pi}{2}$$
A: You can first add a term in the denominator to eliminate the "problematic"
$x^{2}$:
$$
I(\alpha)=\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{x^{2}+\alpha^{2}}dx
$$
We can now solve this integral using the Residue Theorem:
\begin{align*}
I(\alpha) & =\frac{1}{2}Re\left\{ \int_{-\infty}^{\infty}\underbrace{\frac{1-e^{2xi}}{x^{2}+\alpha^{2}}}_{\triangleq f(x)}dx\right\} \\
 & =\frac{1}{2}Re\left\{ 2\pi iRes\left(f,i\alpha\right)\right\} \\
 & =\frac{1}{2}Re\left\{ 2\pi i\frac{1-e^{-2\alpha}}{2\alpha i}\right\} \\
 & =\frac{\pi}{2}\frac{1-e^{-2\alpha}}{\alpha}
\end{align*}
Using the fact that $I(\alpha)$ is continuous in $\alpha$:
$$
\lim_{\alpha\to0}I(\alpha)=\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{x^{2}}dx=\lim_{\alpha\to0}\frac{\pi}{2}\frac{1-e^{-2\alpha}}{\alpha}=\pi
$$
Therefore, the wanted integral is:
$$
\int_{0}^{\infty}\frac{\sin^{2}(x)}{x^{2}}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{x^{2}}dx=\frac{\pi}{2}
$$
