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

The humps for $x\mapsto \sin(x^2)$ go up and down. Each has an area smaller than that of the last. The areas converge to 0 as you progress down the $x$-axis. By the alternating series test, this converges.

$x\mapsto \sin(x^2)$ 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 $$\int_1^A\sin(x^2)dx=\int_1^{A^2}\frac{\sin t}{2\sqrt t}dt=-\frac{\cos A^2}{2\sqrt {A^2}}+\frac{\cos 1}2+\frac 12\int_1^{A^2}\cos t\cdot t^{-3/2}\frac{-1}2dt,$$ and since $\lim_{A\to +\infty}-\frac{\cos A^2}{2\sqrt {A^2}}+\frac{\cos 1}2=\frac{\cos 1}2$ 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.