Show that $\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2 < 2$ I`m trying to show that this integral is converges and $&lt;2$ 
$$\int^{\infty}_{0}\left(\frac{\sin(x)}{x}\right)^2dx &lt; 2$$
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 :
$$\int^{\infty}_{1}\left(\frac{\sin(x)}{x}\right)^2 dx &lt; \int^{\infty}_{1}\left(\frac{1}{x^2}\right)^2dx = \lim\limits_{b\to 0}  {-\frac{1}{x}}|^b_0 = 1 $$
Now for the first expression I need to find any explanation why its $&lt;1$ and I will prove it.

I would like to get some advice for the first expression. thanks!
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

$$\int_0^{\infty} dx\, \left ( \frac{\sin{x}}{x}\right)^2 = \frac{\pi}{2} &lt; 2$$
The answer is 0 &lt; 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.