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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.

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15 Answers 15

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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 $\displaystyle {\pi \over 2}$.

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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).

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  • $\begingroup$ @TCL: probably there is a proof using contour integration and a proof using differentiation under the integral sign. Those are generally the options. $\endgroup$ Dec 7, 2010 at 6:57
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    $\begingroup$ You might want to correct your integration variables, Andrey. $\endgroup$ Dec 7, 2010 at 8:33
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    $\begingroup$ @Raskolnikov: Stand corrected, thanks! $\endgroup$ Dec 7, 2010 at 9:03
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    $\begingroup$ For lazy people like me... $\endgroup$ Dec 7, 2010 at 13:45
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    $\begingroup$ @J.M. The link seems broken. $\endgroup$ Sep 27, 2013 at 10:33
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More generally, there is a result due to Wolstenholme (I can't find a link) that says $$ \int_0^\infty \left( \frac {\sin x}{x} \right)^n dx = \frac{1}{(n-1)!} \frac{\pi}{2^n} \left\lbrace n^{n-1} - { n \choose 1 } (n-2)^{n-1} + { n \choose 2 } (n-4)^{n-1} - \cdots \right\rbrace .$$

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  • $\begingroup$ This one points to Wolstenholme's rather antique book as a reference. $\endgroup$ Dec 7, 2010 at 13:48
  • $\begingroup$ @J.M.: Thanks for the link. It's interesting that his evaluation is by the trapezoidal rule. I've not yet tried to prove it, but by instinct I would head straight for integration by parts. $\endgroup$ Dec 7, 2010 at 14:23
  • $\begingroup$ I get $0$ for $n=1$ to $4$, where the result should be $\pi/2$ -- are you missing a term $\pi/2$? $\endgroup$
    – joriki
    Feb 20, 2013 at 10:20
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$$ \begin{align} \int_0^\infty\frac{\sin^2 x}{x^2}\mathrm dx &= \int_0^\infty\frac{\frac12(1-\cos2x)}{x^2}\mathrm dx \\ &= \int_0^\infty \frac{1-\cos x}{x^2}\mathrm dx \\ &= \frac12\int_{-\infty}^\infty \frac{1-\cos x}{x^2}\mathrm dx \\ &= \frac12\int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm ix}}{x^2}\mathrm dx \\ &= \frac12\int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm ix}+\mathrm ix/(1+x^2)}{x^2}\mathrm dx \\ &= \frac12\Re\int_{-\infty}^\infty \frac{1-\mathrm e^{\mathrm ix}+\mathrm ix/(1+x^2)}{x^2}\mathrm dx\;. \end{align} $$

Now close the contour in the upper half plane, enclosing the pole at $x=\mathrm i$ with residue $1/(2\mathrm i)$, yielding

$$\int_0^\infty\frac{\sin^2 x}{x^2}\mathrm dx=\frac12\cdot2\pi\mathrm i\cdot\frac1{2\mathrm i}=\frac\pi2\;.$$

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  • $\begingroup$ Hi, could you elaborate between 4-th to 5-th step? $\endgroup$ May 13, 2013 at 17:58
  • $\begingroup$ @Mula: Sorry, I don't understand what you mean. I can elaborate on each of the steps, but what does it mean to elaborate between two of the steps? Perhaps you could quote the equation that you don't understand? Is it the addition of $\mathrm ix/(1+x^2)$ in the numerator of the integrand that you're asking about? $\endgroup$
    – joriki
    May 14, 2013 at 13:38
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    $\begingroup$ particularly this step $ \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 $ how did you get $\frac{ix }{1+x^2}$ and what about the pole at $z=0$ $\endgroup$ May 14, 2013 at 17:25
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    $\begingroup$ @Mula: The numerator is linear in $x$ before that step, so to move $\Re$ outside the integral I need to cancel the linear term. Ideally I'd want to just add $\mathrm ix$ in the numerator (which I can since the real part of that is $0$), but then the integral would diverge at infinity, so I add $\mathrm ix/(1+x^2)$ instead, which also cancels the linear term at $x=0$ and also has real part $0$ but doesn't cause the integral to diverge at infinity. There's no pole at $x=0$; the added term is chosen so as to cancel the pole before $\Re$ is moved outside the integral. $\endgroup$
    – joriki
    May 14, 2013 at 17:36
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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 $$\pi=\int_\infty^\infty\frac{\sin^2y}{y^2}\,dy.$$

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\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}

$$\color{#ff0000}{\large% \int_{-\infty}^{\infty}{\sin^{2}\left(x\right) \over x^{2}}\,{\rm d}x \color{#000000}{\ =\ } \int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x} $$

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Another option is a Schwinger parametrization. Write $x^{-2}=\int_0^\infty y e^{-xy} dy$ so $$\begin{align}\int_0^\infty x^{-2}\sin^2 x dx&=-\frac{1}{4}\int_0^\infty\int_0^\infty (e^{ix}-e^{-ix})^2 ye^{-xy}dxdy\\&=-\frac{1}{4}\int_0^\infty\int_0^\infty y(e^{-x(y+2i)}+e^{-x(y-2i)}-2e^{-xy})dxdy\\&=-\frac{1}{4}\int_0^\infty\left(\frac{y}{y+2i}+\frac{y}{y-2i}-2\right)dy\\&=2\int_0^\infty\frac{dy}{y^2+4}\\&=\frac{\pi}{2}.\end{align}$$

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    $\begingroup$ This seems to be the solution using the least tools, as only Fubini and integration by parts are required, whereas the rest of solutions relies on Fourier or complex analysis. So props for that, very helpful, as I want to prove this quite early into Fourier analysis. $\endgroup$
    – F.U.A.S.
    Jul 3, 2020 at 12:03
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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 $$\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy=\frac{\pi}{2}$$

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  • $\begingroup$ Could you tell the name of the function $\chi_{[-1,1]}$ ?, thanks. $\endgroup$
    – Zoe Rowa
    Jan 25, 2015 at 14:47
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    $\begingroup$ @zoerowa This is just the function that is 1 on the given set and 0 elsewhere. $\endgroup$ Jan 25, 2015 at 17:49
  • $\begingroup$ isn't that an indicator function? $\endgroup$
    – user459879
    Jun 14, 2019 at 20:36
  • $\begingroup$ en.wikipedia.org/wiki/Indicator_function $\endgroup$
    – user459879
    Jun 14, 2019 at 20:37
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    $\begingroup$ @WesleyStrik Yes, it goes by many names. $\endgroup$ Jun 15, 2019 at 9:42
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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 $~~\lim\limits _{x\to 0}\frac{\sin^2 x}{x^2} = 1$

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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: $$\sum U_n = \sum V_n$$

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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,

$$I = I(1) = \int_{0}^{\infty} \frac{\sin^2(x)}{x^2} \:dx = \frac{\pi\cdot 1}{2} = \frac{\pi}{2}$$

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  • $\begingroup$ This method works extremely well for even powers of $\sin^n(x)/x$ $\endgroup$
    – user150203
    Dec 13, 2018 at 23:34
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Another way (with usage of complex analysis) which I want to add here is the following:

enter image description here

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 $\int_{0}^{\infty} \! \frac{sin(x)}{x} \, dx= \frac{\pi}{2}$

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    $\begingroup$ Since $\frac{sinx}{x}$ has a limit of $1$ as $x$ goes to $0$, why can't we just say the residue is $0$? $\endgroup$
    – user5262
    Dec 16, 2014 at 18:27
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    $\begingroup$ Is this answer misplaced? The OP asked for the computation of $\int_0^\infty \sin^2(x)/x^2\, dx$. $\endgroup$
    – cantorhead
    Mar 24, 2016 at 21:02
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    $\begingroup$ Can someone please explain the step 1 of Marm's answer? $\endgroup$
    – Srini
    Dec 24, 2016 at 6:00
  • $\begingroup$ Why does $\frac{sin(x)}{x}$ become $\frac{e^{iz}}{z}$ ? $\endgroup$
    – Zophikel
    Feb 25, 2018 at 0:01
  • $\begingroup$ @Zophikel $sin(x)=Im(e^{iz})$ $\endgroup$ Mar 19, 2019 at 21:59
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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

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  • $\begingroup$ Can you please explain how you got the limits of $\infty$ and $0$ in this step: Now integrating w.r.t. $a$ gives us: $$\frac{\partial I(t,0)}{\partial t}=-\int^0_{\infty}\frac{t}{a^2+t^2}da$$ $\endgroup$
    – Srini
    Dec 27, 2016 at 6:27
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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, $${\int_0^\infty\frac{\sin^2x}{x^2} dx}=\frac{\pi}{2}$$

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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: $$ \int_{0}^{\infty}\frac{\sin^{2}(x)}{x^{2}}dx=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\sin^{2}(x)}{x^{2}}dx=\frac{\pi}{2} $$

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