Convergence of an integral $\int_0^\infty\frac{dt}{1+t^\alpha\sin^2(t)}$

Question

For what $\alpha\in\Bbb{R}$ does $\displaystyle\int_0^\infty\frac{dt}{1+t^\alpha\sin^2(t)}$ converge ?

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The $0$ bound doesn't seem to be much of a problem, but I don't see how to deal with the $\infty$ bound.

Answer 1

Converges if $\alpha > 1$ and diverges otherwise. If $\alpha < 0$, then the denominator is asymptotic to $1$. If $\alpha > 0$: Note that the denominator is less than $1 + t^\alpha$ which is asymptotic to $t^\alpha$, so the integral diverges when $\alpha \le 1$. If $\alpha > 1$, use the $t^n$ test with $n$ being $\dfrac{1+\alpha}{2}$.