# 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.

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).

• @TCL: probably there is a proof using contour integration and a proof using differentiation under the integral sign. Those are generally the options. – Qiaochu Yuan Dec 7 '10 at 6:57
• You might want to correct your integration variables, Andrey. – Raskolnikov Dec 7 '10 at 8:33
• @Raskolnikov: Stand corrected, thanks! – Andrey Rekalo Dec 7 '10 at 9:03
• For lazy people like me... – J. M. is a poor mathematician Dec 7 '10 at 13:45
• @J.M. The link seems broken. – Vishal Gupta Sep 27 '13 at 10:33

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

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

• This one points to Wolstenholme's rather antique book as a reference. – J. M. is a poor mathematician Dec 7 '10 at 13:48
• @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. – Derek Jennings Dec 7 '10 at 14:23
• I get $0$ for $n=1$ to $4$, where the result should be $\pi/2$ -- are you missing a term $\pi/2$? – joriki Feb 20 '13 at 10:20

\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

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

• Hi, could you elaborate between 4-th to 5-th step? – Mula Ko Saag May 13 '13 at 17:58
• @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? – joriki May 14 '13 at 13:38
• 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 $z=0$ – Mula Ko Saag May 14 '13 at 17:25
• @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. – joriki May 14 '13 at 17:36

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

\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}

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

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

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

• Could you tell the name of the function $\chi_{[-1,1]}$ ?, thanks. – Zoe Rowa Jan 25 '15 at 14:47
• @zoerowa This is just the function that is 1 on the given set and 0 elsewhere. – AD. Jan 25 '15 at 17:49
• isn't that an indicator function? – Wesley Strik Jun 14 at 20:36
• en.wikipedia.org/wiki/Indicator_function – Wesley Strik Jun 14 at 20:37
• @WesleyStrik Yes, it goes by many names. – AD. Jun 15 at 9:42

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,

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

• This method works extremely well for even powers of $\sin^n(x)/x$ – user150203 Dec 13 '18 at 23:34

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

• Since $\frac{sinx}{x}$ has a limit of $1$ as $x$ goes to $0$, why can't we just say the residue is $0$? – user5262 Dec 16 '14 at 18:27
• Is this answer misplaced? The OP asked for the computation of $\int_0^\infty \sin^2(x)/x^2\, dx$. – cantorhead Mar 24 '16 at 21:02
• Can someone please explain the step 1 of Marm's answer? – Srini Dec 24 '16 at 6:00
• Why does $\frac{sin(x)}{x}$ become $\frac{e^{iz}}{z}$ ? – Zophikel Feb 25 '18 at 0:01
• @Zophikel $sin(x)=Im(e^{iz})$ – Username Unknown Mar 19 at 21:59

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

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

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

• Can you please explain how you got the limits of $\infty$ and $0$ in this step: Now integrating w.r.t. $a$ gives us: $$\frac{\partial I(t,0)}{\partial t}=-\int^0_{\infty}\frac{t}{a^2+t^2}da$$ – Srini Dec 27 '16 at 6:27

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