Convergence of the integral $\int _0^\infty \ln^2x\sin(x^2)\,dx$ $$\int _0^\infty (\ln^2x)\sin(x^2)dx$$

Does this converge?

In the answers in our textbook, we've been told that since the inner function limit's in in infinity is not 0 then the integral doesn't converge. I couldn't find a theorem that states this though.
Are there any other good ways to prove that this integral diverges?
There is no real difficulty at $0$, since near $0$ the function $\sin(x^2)$ behaves like $x^2$, so $\lim_{x\to 0^+}\ln^2 x\sin(x^2)=0$. So we examine
$$\int_1^B (\ln^2 x)( \sin(x^2))\,dx.\tag{1}$$
Rewrite as
$$\int_1^B \frac{\ln^2 x}{2x} 2x \sin(x^2)\,dx,$$
and use integration by parts, letting $u=\frac{\ln^2 x}{2x}$ and $dv=2x\sin(x^2)\,dx$. Then  $du=\frac{2\ln x-\ln^2 x}{2x^2}\,dx$ and we can take $v=-\cos(x^2)$. Thus our integral (1) is
$$\left.\left(-\frac{\ln^2 x}{2x}\cos(x^2)\right)\right|_1^B +\int_1^B \frac{2\ln x-\ln^2 x}{2x^2}\cos(x^2)\,dx.$$
The first part gives no problem, indeed it vanishes as $B\to\infty$. The remaining integral has a (finite) limit as $B\to\infty$, because $\cos(x^2)$ is bounded and the $2x^2$ in tthe denominator crushes the $\ln$ terms in the numerator.  

It follows that our original integral converges.
$\sin(x^2)&gt;.5$ on $(\sqrt{\pi/6+2k\pi},\sqrt{5\pi/6+2k\pi}),k\in\mathbb{N}$.  When $x&gt;e$, $\log^2(x)&gt;1$, so we can see that the positive part of the integral is greater than $.5\sum_{k=2}^K \left(\sqrt{5\pi/6+2k\pi}-\sqrt{\pi/6+2k\pi}\right)$ for any $K\in\mathbb{N}\setminus\{1\}$.  Now see that 

$$
\sum_{k=2}^K \left(\sqrt{5\pi/6+2k\pi}-\sqrt{\pi/6+2k\pi}\right)=\sum_{k=2}^K \left(\frac{2\pi/3}{\sqrt{5\pi/6+2k\pi}+\sqrt{\pi/6+2k\pi}}\right)\geq\sum_{k=2}^K \left(\frac{2\pi/3}{\sqrt{5\pi/6+2k\pi}}\right)
$$

which we see is unbounded.  We conclude that the positive part of the integral is unbounded.  Similarly, we can conclude that the negative part of the integral is unbounded.  Therefore it is not integrable.