For what p $\int_{1}^{\infty} x^p \cdot \sin(\frac{1}{x^2}) dx$ converges?

Question

For what values of $p$ does $\displaystyle\int_{1}^{\infty} x^p \cdot \sin\bigl(\frac{1~}{x^2} \bigr)\, \mathrm{d}x$ converge?

I was able to see that for $p \geq 2$ it diverges because the limit exists and its different than 0.

Then I used $\sin\bigl( \frac{1~}{x^2} \bigr) < \frac1{x^2}$, and saw that the integral converges for $p<1$.

However, I am completely stuck for $1<p<2$, can anyone give some guidance?

The limit does seem to be $0$.

Thanks a lot.

EDIT: The only thing we were taught in class is $\sin t < t$ and nothing beyond that. Answers without using other trigonometric inequalities will be preferred.

Answer 1

Hint: if $x$ is sufficiently small, $\frac{x}{2}\le \sin x$

Answer 2

Recall $\frac{2}{\pi}\theta\leq\sin\theta\leq\theta$ for all $\theta\in[0,\pi/2]$. We will use the $\frac{2}{\pi}\theta\leq\sin\theta$ estimate here, giving $$ x^p\sin(x^{-2})\geq\frac2\pi x^{p-2} $$ for all $x\geq 1$. Since $p-2\geq -1$ if $p\geq 1$, we have $\int_1^\infty x^{p-2}\,\mathrm{d}x=\infty$ and thus $\int_1^\infty x^p\sin(x^{-2})\,\mathrm{d}x$ diverges too.