Show that $\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2 < 2$ 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!
 A: Hint: $$\lim_{x\to0}\frac{\sin x}{x}=1.$$
A: 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:
$$ I \leq \int_{0}^{1}\frac{x^2}{x^2}\,dx + \int_{1}^{+\infty}\frac{1}{x^2}\,dx = 2.$$
A: 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
$$\int_0^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{\pi}{2} < 2$$
A: 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.
A: 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 $\lim_{x\to 0}\frac{sin^2 x}{x^2} = 1$
A: 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.
