9
$\begingroup$

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!

$\endgroup$
  • 1
    $\begingroup$ You've probably seen the diagram in this link. One has $0<{\sin x\over x}\le 1$ for $0<x\le1$. $\endgroup$ – David Mitra Jun 8 '13 at 14:08
  • 1
    $\begingroup$ 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 $x\ge \sin x$. $\endgroup$ – André Nicolas Jun 8 '13 at 14:09
  • $\begingroup$ @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. $\endgroup$ – Ofir Attia Jun 8 '13 at 14:16
  • $\begingroup$ The important thing is that you now know how to do it. There is already a useful hint given as an answer. $\endgroup$ – André Nicolas Jun 8 '13 at 14:21
5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Ofir Attia Jun 8 '13 at 14:03
  • 2
    $\begingroup$ @OfirAttia Use the Maclaurin expansion of $\sin{x}$; that is, $\sin{x}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$. $\endgroup$ – Librecoin Jun 8 '13 at 14:12
  • $\begingroup$ 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. $\endgroup$ – Harald Hanche-Olsen Jun 8 '13 at 14:39
  • $\begingroup$ You can prove that $\displaystyle\lim_{x\rightarrow 0} \frac{\sin x}{x} = 0$ by using L'Hospitals rule. $\endgroup$ – yousuf soliman Jun 8 '13 at 21:52
  • $\begingroup$ @MuadDib42 But many textbooks use that limit to find the derivative of $\sin x$ in the first place. Seems a bit circular, then. $\endgroup$ – Harald Hanche-Olsen Jun 8 '13 at 22:16
12
$\begingroup$

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

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

$\endgroup$
3
$\begingroup$

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: $$ I \leq \int_{0}^{1}\frac{x^2}{x^2}\,dx + \int_{1}^{+\infty}\frac{1}{x^2}\,dx = 2.$$

$\endgroup$
  • 1
    $\begingroup$ Nice work, as usual, Jack. $\endgroup$ – marty cohen Jul 3 '17 at 23:20
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $\lim_{x\to 0}\frac{sin^2 x}{x^2} = 1$

$\endgroup$
  • $\begingroup$ Please avoid posting duplicate answers. $\endgroup$ – Jack D'Aurizio Nov 7 '17 at 14:21
-2
$\begingroup$

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.

$\endgroup$
  • 5
    $\begingroup$ The integral of a nonnegative, nonzero function can't be zero. you've made a mistake somewhere. $\endgroup$ – icurays1 Jun 8 '13 at 15:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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