I want to show that this integral converges:
$$I := \int_0^\infty \frac{\sin x^2}{\sqrt{x}}dx$$
My attempt is to say that $$ I < \int_0^\infty {\sin x^2}dx := J $$, and this one converges ( Prove: $\int_0^\infty \sin (x^2) \, dx$ converges. the first answer is awesome).
Do you have any other idea to show that $I$ converges? I tried to use substitution and int. by part but in vain :/. I was lucky to find that $J$ converges, but I would never have been able to guess this result alone.