Proof of $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$

Question

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.

Answer 1

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

Answer 2

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 $[-1,1]$).

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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).

Answer 3

More generally, there is a result due to Wolstenholme (I can't find a link) that says $$ \int_0^\infty \left( \frac {\sin x}{x} \right)^n dx = \frac{1}{(n-1)!} \frac{\pi}{2^n} \left\lbrace n^{n-1} - { n \choose 1 } (n-2)^{n-1} + { n \choose 2 } (n-4)^{n-1} - \cdots \right\rbrace .$$

Answer 4

$$ \begin{align} \int_0^\infty\frac{\sin^2 x}{x^2}\mathrm dx &= \int_0^\infty\frac{\frac12(1-\cos2x)}{x^2}\mathrm dx \\ &= \int_0^\infty \frac{1-\cos x}{x^2}\mathrm dx \\ &= \frac12\int_{-\infty}^\infty \frac{1-\cos x}{x^2}\mathrm dx \\ &= \frac12\int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm ix}}{x^2}\mathrm dx \\ &= \frac12\int_{-\infty}^\infty\Re\frac{1-\mathrm e^{\mathrm ix}+\mathrm ix/(1+x^2)}{x^2}\mathrm dx \\ &= \frac12\Re\int_{-\infty}^\infty \frac{1-\mathrm e^{\mathrm ix}+\mathrm ix/(1+x^2)}{x^2}\mathrm dx\;. \end{align} $$

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

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

Answer 5

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 $$\pi=\int_\infty^\infty\frac{\sin^2y}{y^2}\,dy.$$

Answer 6

\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}

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

Answer 7

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

Answer 8

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 $$\int_{0}^\infty \left(\frac{\sin y}{y}\right)^2 dy=\frac{\pi}{2}$$

Answer 9

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 $~~\lim\limits _{x\to 0}\frac{\sin^2 x}{x^2} = 1$

Answer 10

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

Answer 11

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,

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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, $${\int_0^\infty\frac{\sin^2x}{x^2} dx}=\frac{\pi}{2}$$