Convergence of the integral $\int_0^\infty \frac{\sin^2x}{x^2}~\mathrm dx.$ 
Determine whether the integral $$\int_0^\infty \frac{\sin^2x}{x^2}~\mathrm dx$$ converges.

I know it converges, since in general we can use complex analysis, but I'd like to know if there is a simpler method that doesn't involve complex numbers. But I cannot come up with a function that I could compare the integral with.
 A: Hint:$$x>1\implies0\le\frac{\sin^2(x)}{x^2}\le\frac1{x^2}\\0<x<1\implies1\le\frac{\sin^2(x)}{x^2}\le\frac1{\cos^2(x)}\le\frac1{\cos^2(1)}$$The second of the two coming from the proof of the derivative of $\sin$ using squeeze theorem.
A: For $x>1$ $$\frac{\sin^2x}{x^2}\le \frac{1}{x^2}$$ and  For $x<1$ 
$$ |\sin x|\le |x|\implies \frac{\sin^2 x}{x^2}\le  1.$$ 

A: Using integration by part $$\int_0^\infty \frac{\sin^2(x)}{x^2}dx=\\
-\frac{1}{x}\sin^2(x) |^b_a-\int_{0}^{\infty}-\frac{1}{x}2\sin(x)\cos(x)dx$$note that $\lim_{x\to 0}-\frac{1}{x}\sin^2(x)\to 0$and also $\lim_{x\to \infty}-\frac{1}{x}\sin^2(x)\to 0$ so 
$$-\frac{1}{x}\sin^2(x) |^b_a-\int_{0}^{\infty}-\frac{1}{x}2\sin(x)\cos(x)dx=\\0-(-)\int_{0}^{\infty}\frac{1}{x}2\sin(x)\cos(x)dx=\\
\int_{0}^{\infty}\frac{\sin(2x)}{x}dx=\\\frac{\pi}{2}$$ It is well known $\int_{0}^{\infty}\frac{\sin(ax)}{x}dx=\frac{\pi}{2}$   
$\bf{Remark}$: let:
$$I(t)=\int_{0}^{\infty} \frac{\sin x}{x} e^{-tx} dx$$
Note:
$$\frac{\partial}{\partial t} \frac{\sin x}{x} e^{-tx}=\frac{\sin x}{x} e^{-tx}(-x)$$
So by differentiation of parameter,we have
$$I'(t)=-\int_{0}^{\infty} e^{-tx} \sin x dx$$
And through integration by parts twice we have:
$$I'(t)=-\frac{1}{t^2+1}$$
Hence,
$$I(t)=\int -\frac{1}{t^2+1} dt$$
$$I(t)=-\arctan (t) +c$$
when $t \to \infty$, $I(t) \to 0$ so :
$$I(t)=\frac{\pi}{2}-\arctan t$$  
Let $t \to 0^+$:
$$\int_{0}^{\infty} \frac{\sin x}{x} dx=\frac{\pi}{2}$$and general form is $\int_{0}^{\infty} \frac{\sin (ax)}{x} dx=\frac{\pi}{2}$
