Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
    
Thanks for the link. –  TCL Dec 7 '10 at 16:18
    
@Alizter Please don't replace $dx$ with your preferred typesetting. Many people write $dx$ for a reason. [I know there's an ISO norm, but we can ignore the bureaucrats.] –  Daniel Fischer Aug 29 at 22:56
    
@DanielFischer I wasn't doing this because I prefer it but because it was a norm. Now I am confused. –  Alizter Aug 29 at 22:57
    
@Alizter Many people (among them I) consider $\mathrm{d}x$ ugly. Really ugly. We choose to not adhere to a norm (which cannot be binding anyway) we deem bad. –  Daniel Fischer Aug 29 at 23:00

9 Answers 9

up vote 18 down vote accepted

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).

share|improve this answer
    
+1. Wonder if there is another more elementary proof. –  TCL Dec 7 '10 at 6:34
1  
You might want to correct your integration variables, Andrey. –  Raskolnikov Dec 7 '10 at 8:33
1  
@Raskolnikov: Stand corrected, thanks! –  Andrey Rekalo Dec 7 '10 at 9:03
2  
For lazy people like me... –  J. M. Dec 7 '10 at 13:45
1  
@J.M. The link seems broken. –  Vishal 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 $\displaystyle {\pi \over 2}$.

share|improve this answer

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

share|improve this answer
    
This one points to Wolstenholme's rather antique book as a reference. –  J. M. 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

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

share|improve this answer
    
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
1  
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
    
thank you very much sire ... awesome trick!! (+1) –  Mula Ko Saag May 14 '13 at 17:45

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

share|improve this answer

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

share|improve this answer

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

share|improve this answer

\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}

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

share|improve this answer

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

share|improve this answer
    
Since $\frac{sinx}{x}$ has a limit of $1$ as $x$ goes to $0$, why can't we just say the residue is $0$? –  Joshua yesterday

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.