calculus - Prove: $int_0^infty sin (x^2) , dx$ converges.

$\sin x^2$ does not converge as $x \to \infty$, yet its integral from $0$ to $\infty$ does.

I'm trying to understand why and would like some help in working towards a formal proof.

Answer

$x\mapsto \sin(x^2)$ is integrable on $[0,1]$, so we have to show that $\lim_{A\to +\infty}\int_1^A\sin(x^2)dx$ exists. Make the substitution $t=x^2$, then $x=\sqrt t$ and $dx=\frac{dt}{2\sqrt t}$. We have $$\int_1^A\sin(x^2)dx=\int_1^{A^2}\frac{\sin t}{2\sqrt t}dt=-\frac{\cos A^2}{2\sqrt A}+\frac{\cos 1}2+\frac 12\int_1^{A^2}\cos t\cdot t^{-3/2}\frac{-1}2dt,$$
and since $\lim_{A\to +\infty}-\frac{\cos A^2}{2\sqrt A}+\frac{\cos 1}2=\frac{\cos 1}2$ and the integral $\int_1^{+\infty}t^{-3/2}dt$ exists (is finite), we conclude that $\int_1^{+\infty}\sin(x^2)dx$ and so does $\int_0^{+\infty}\sin(x^2)dx$.
This integral is computable thanks to the residues theorem.