Let $f:\mathbb R \to \mathbb R$ be a continuous bounded function such that $$|\sin x|\le f(x) \le|\sin x|(1+x^2), \forall x\in\mathbb R.$$
The question is to find all $\alpha$ such that the integral $$\displaystyle\int_0^{+\infty}\frac{(f(x))^{1+\alpha+\alpha^2}}{x^{1+\alpha}}dx$$ converges.
I've got no clue where to begin with this. Trying to compare the integral with $$\displaystyle\int_0^{+\infty}\frac{(|\sin x|(1+x^2))^{1+\alpha+\alpha^2}}{x^{1+\alpha}}dx$$ or with $$\displaystyle\int_0^{+\infty}\frac{(|\sin x|)^{1+\alpha+\alpha^2}}{x^{1+\alpha}}dx$$doesn't really seem like a prosperous way to go. Any hint or answer would be appreciated.
Not a complete solution, but it might help.
Because $f$ is bounded $\frac{(f(x))^{1+\alpha+\alpha^2}}{x^{1+\alpha}} \le \frac {M}{x^{1+\alpha}}$ therefore the integral is convergent for $\alpha \gt 0$.
Now, we have $\frac{(f(x))^{1+\alpha+\alpha^2}}{x^{1+\alpha}} \ge \frac{|\sin x|^{1+\alpha+\alpha^2}}{x^{1+\alpha}}$ and $\displaystyle\int_0^{+\infty} \frac{|\sin x|^{1+\alpha+\alpha^2}}{x^{1+\alpha}}$ is divergent for $ \alpha \le -1$.
The only remaining case is $\alpha \in (-1, 0]$