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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!

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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 8 '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 8 '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 8 '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 8 '13 at 14:21

3 Answers 3

up vote 4 down vote accepted

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

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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 8 '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 8 '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 8 '13 at 14:39
    
You can prove that $\displaystyle\lim_{x\rightarrow 0} \frac{\sin x}{x} = 0$ by using L'Hospitals rule. –  SyntacticSugar Jun 8 '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 8 '13 at 22:16

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

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

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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.

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

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