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.

Just with elementary tools: see Here,How to prove only by Transformation that: $ \int_0^\infty \cos(x^2) dx = \int_0^\infty \sin(x^2) dx $

In fact we have: \begin{split} \int_0^\infty \sin(x^2) dx=\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx<\infty\end{split}
  See below

Employing the change of variables $2u =x^2$ after integration by parts we get  \begin{split} \int_0^\infty \sin(x^2) dx&=&\frac{1}{\sqrt{2}}\int^\infty_0\frac{\sin(2x)}{\sqrt{x}}\,dx\\& =&\frac{1}{\sqrt{2}} \int^\infty_0\frac{\sin(2x)}{x^{1/2}}\,dx\\&=&\frac{1}{\sqrt{2}}\underbrace{\left[\frac{\sin^2 x}{x^{1/2}}\right]_0^\infty}_{=0} +\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx \\&= &\frac{1}{2\sqrt{2}} \int^\infty_0\frac{\sin^2 x}{x^{3/2}}\,dx\end{split}
Given that $ \sin 2x  =(\sin^2x)'$ and $$\lim_{x\to 0}\frac{\sin x}{x}=1$$
However, $$ \int^\infty_1\frac{\sin^2 x}{x^{3/2}}\,dx\le \int^\infty_1\frac{1}{x^{3/2}}\,dx<\infty$$
since $|\sin x|\le |x|$ we have, 
$$\int^1_0\frac{\sin^2 x}{x^{3/2}}\,dx \le \int^1_0\frac{\ x^2}{x^{3/2}}\,dx = \int^1_0\sqrt x\,dx<\infty$$
