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I`m trying to show that this integral is converges and $<2$ $$\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2dx < 2$$ What I did is to show this expression:
$$\int^{1}_{0}\left(\frac{\sin(x)}{x}\right)^2dx + \int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx$$ Second expression :
$$\int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx < \int^{\infty}_{1}\left(\frac{1}{x^2}\right)^2dx = \lim\limits_{b\to 0} {-\frac{1}{x}}|^b_0 = 1 $$ Now for the first expression I need to find any explanation why its $<1$ and I will prove it.
I would like to get some advice for the first expression. thanks!

Ofir Attia
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    You've probably seen the diagram in this link. One has $0<{\sin x\over x}\le 1$ for $0<x\le1$. – David Mitra Jun 08 '13 at 14:08
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    For the integral from $0$ to $1$, show that $\sin x \le x$ if $x\ge 0$. For proof, let $f(x)=x-\sin x$. Then $f(0)=0$ and since $f'(x)=1-\cos x\ge 0$, the function is increasing, so $x\ge \sin x$. – André Nicolas Jun 08 '13 at 14:09
  • @AndréNicolas got it, I could not think about it alone. I think its enough to show that its less then $1$, write it as answer if you mind. – Ofir Attia Jun 08 '13 at 14:16
  • The important thing is that you now know how to do it. There is already a useful hint given as an answer. – André Nicolas Jun 08 '13 at 14:21
  • $\displaystyle\exists\ \xi \in \left(0,\left\vert x\right\vert\right) \ni {\sin\left(x\right) \over x} = \cos\left(\xi\right)$. – Felix Marin Dec 20 '20 at 23:33

6 Answers6

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Well, this likely isn't what you had in mind, but you could just evaluate the integral. In this case, Parseval-Plancherel's theorem works:

$$\int_{-\infty}^{\infty} dx\, |f(x)|^2 = \frac{1}{2 \pi}\int_{-\infty}^{\infty} dk\, |\hat{f}(k)|^2$$

where $\hat{f}$ is the Fourier transform of $f$. For $f(x)=\sin{x}/x$, we have

$$\int_{-\infty}^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{1}{2 \pi} \int_{-1}^1 dk \, \pi^2 = \pi$$

so that

$$\int_0^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{\pi}{2} < 2$$

Ron Gordon
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Hint: $$\lim_{x\to0}\frac{\sin x}{x}=1.$$

Harald Hanche-Olsen
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  • I know that, but I dont know how to show it, I can just say that? and its enough? there is a connection to continuous fractional? – Ofir Attia Jun 08 '13 at 14:03
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    @OfirAttia Use the Maclaurin expansion of $\sin{x}$; that is, $\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$. – Librecoin Jun 08 '13 at 14:12
  • You should be able to use it, I think. How best to prove it depends a lot on where you are in your learning process. But it is a fundamental identity, often used in the proof of the derivatives of $\sin x$ and $\cos x$. So depending on the Maclaurin expansion, as @Tharsis suggests, might be considered circular. Anyhow, it is enough because now the function which is $(\sin x)/x$ for $x\ne0$ and $1$ for $x=0$ is continuous, and therefore integrable. – Harald Hanche-Olsen Jun 08 '13 at 14:39
  • You can prove that $\displaystyle\lim_{x\rightarrow 0} \frac{\sin x}{x} = 0$ by using L'Hospitals rule. – Yousuf Soliman Jun 08 '13 at 21:52
  • @MuadDib42 But many textbooks use that limit to find the derivative of $\sin x$ in the first place. Seems a bit circular, then. – Harald Hanche-Olsen Jun 08 '13 at 22:16
  • @HaraldHanche-Olsen We could also find the derivative of $\sin x$ by using a power series representation. I suppose it is a bit cyclic in a sense though. – Yousuf Soliman Jun 08 '13 at 23:09
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By the Laplace transform (since $\mathcal{L}(\sin^2 x)=\frac{2}{s(4+s^2)}$ and $\mathcal{L}^{-1}\left(\frac{1}{x^2}\right)=s$) $$ I=\int_{0}^{+\infty}\frac{\sin^2 x}{x^2}\,dx = \int_{0}^{+\infty}\frac{2\,ds}{4+s^2}\stackrel{s\mapsto 2t}{=} \int_{0}^{+\infty}\frac{dt}{1+t^2}=\color{red}{\frac{\pi}{2}}$$ and with a more elementary approach, $\left|\sin(x)\right|\leq\min(|x|,1)$ implies: $$ I \leq \int_{0}^{1}\frac{x^2}{x^2}\,dx + \int_{1}^{+\infty}\frac{1}{x^2}\,dx = 2.$$

Jack D'Aurizio
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If you assume that $0 \le \cos(x) \le 1$ for $0 \le x \le \pi/2$, then, since $\sin'(x) =\cos(x) $, for $0 \le x \le \pi/2$ we have $\sin(x) =\int_0^x \cos(t)dt \le\int_0^x dt =x $.

The integral from $1$ to $\infty$ is easily shown to be bounded by 1.

marty cohen
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From this Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$? We know that , $$2>\frac{\pi}{2} =\int_0^\infty\frac{\sin x}{x} dx = \int_0^\infty\frac{\sin 2u}{2u} d(2u) =\int_0^\infty\frac{\sin 2u}{u} du\\ = \underbrace{\left[\frac{\sin^2 u}{u}\right]_0^\infty}_{=0} +\int_0^\infty\frac{\sin^2u}{u^2} du =\color{blue}{\int_0^\infty\frac{\sin^2u}{u^2} du} $$

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

Guy Fsone
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The answer is 0 < 2. Found using the congruity $\sin^2(x)=\frac {1-\cos(2x)}{2}$. And evaluating $\frac{-1-\cos(2x)}{2x}$ from 0 to infinity. This ends up being the limit of $\frac {\cos^2(x)}{x}$ as x goes to 0.

Daniel Margolis
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