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

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.

For $x>1$ $$\frac{\sin^2x}{x^2}\le \frac{1}{x^2}$$ and For $x<1$ $$ |\sin x|\le |x|\implies \frac{\sin^2 x}{x^2}\le 1.$$

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 $\int_{0}^{\infty} \frac{\sin (ax)}{x} dx=\frac{\pi}{2}$