I am trying to prove that the improper integral $$ I=\int_0^\infty \sin x\sin(x^2)\mathrm{d}x$$ converges.
Here's my work:
It suffices to show that $$\int_\frac{\pi}{2}^\infty \sin (x) \sin(x^2)\mathrm{d}x$$ converges. Using integration by parts,
\begin{align*} \int_\frac{\pi}{2}^\infty \sin (t) \sin(t^2)\mathrm{d}t&=\int_\frac{\pi}{2}^\infty \frac{\sin (t)}{2t}\cdot 2t\sin(t^2)\mathrm{d}t\\ &=\underline{\bigg[ -\frac{\sin (t)}{2t}\cos(t^2)\bigg]_\frac{\pi}{2}^\infty}+{\int_\frac{\pi}{2}^\infty \left(\frac{\sin (t)}{2t}\right)'\cos(t^2)\mathrm{d}t} \end{align*} The underlined part is a constant...
Then I got stuck. I'd like to use "sandwich rule" using the fact that $-1\leq \cos(t^2)\leq 1$, but I can't find a way to apply it properly.
How can I proceed from here? Any correction and/or help would be appreciated. :)