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

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.

• This looks a bit like a sinc after a change of variable and that has finite integral. It isn't super easy to do, though. – Cameron Williams Jun 21 '18 at 17:03
• You say "exists". Are you simply trying to figure out whether it converges or not? – Michael Hardy Jun 21 '18 at 17:10
• I can't help wondering if $(\sin x)^2$ rather than $\sin (x^2)$ might have been intended. I don't know of anything special about the behavior of the sine function at $\pi^2. \qquad$ – Michael Hardy Jun 21 '18 at 17:19
• The first thing I think of is Dirichlet test. You might give it a try. Or try to somehow use the fact that the integral of $\sin(x^2)$ over the real line converges. – Shashi Jun 21 '18 at 17:20

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 $\frac{1}{\sqrt x\sqrt{x-\pi^2}}$ monotonically decreases to $0$, the Abel-Dirichlet test guarantees that the integral conveges.