Improper integral $\int_\pi^\infty\frac{\sin(x^2)}{\sqrt{x^2-\pi ^2}} \, dx$

Question

I've got a problem with checking if the following improper integral exists: $$\int_\pi^\infty \frac{\sin(x^2)}{\sqrt{x^2-\pi ^2}}\,dx$$ $u=x^2$ substitution will probably not work, things started to look worse after that. Seems like it needs to be bounded by something else, but can't figure out any reasonable boundary. Any help will be highly appreciated.

Answer 1

Enforcing the substitution $x\mapsto\sqrt x$ reveals

$$\int_\pi^\infty \frac{\sin(x^2)}{\sqrt{x^2-\pi^2}}\,dx\overbrace{=}^{x\mapsto\sqrt x}=\int_{\pi^2}^\infty \frac{\sin(x)}{2\sqrt x\sqrt{x-\pi^2}}\,dx$$

Since $\int_0^1\frac1{\sqrt x}\,dx$ is integrable, the singularity at the lower limit poses no convergence issue.

And since for any $L$, $\left|\int_{\pi^2}^L \sin(x)\,dx\right|\le 2$ and $\frac{1}{\sqrt x\sqrt{x^2-\pi^2}}$ monotonically decreases to $0$, the Abel-Dirichlet test guarantees that the integral conveges.