Convergence/divergence of $\int_{0}^{\infty}{x^{3}\,\mathrm{d}x \over 1 + x^{a} \sin^2{x}}$ I am trying to conclude about the convergence/divergence of
$$
\int_{0}^{\infty}{x^3 \over 1+x^a \sin^{2}\left(\,x\,\right)}\,\mathrm{d}x
\qquad\mbox{for}\quad a \in \mathbb{R}.
$$
First, we notice
$${x^3 \over 1+x^a} \leq{x^3 \over 1+x^a \sin^2{x}}$$
$${x^3 \over 1+x^a} \approx {x^3 \over x^a}={1 \over x^{a-3}}$$
So for $a \leq 4$ the integral diverges by the comparison test.
How do I approach this for $a > 4$ ?. We do seem to run into trouble for
$x = k\pi$.

First note that, for $m>0$ we have
$$\eqalign{\int_{0}^\pi\frac{dx}{1+ m \sin^2x}&=2\int_{0}^{\pi/2}\frac{1}{1+m+\cot^2x}\frac{dx}{\sin^2x}\cr
&=2\int_0^\infty\frac{dt}{1+m+t^2}=\frac{\pi}{\sqrt{1+m}}
}$$
Thus, if
$$a_n=\int_{n\pi}^{(n+1)\pi}\frac{x^3}{1+x^a\sin^2x}dx
=\int_{0}^{\pi}\frac{(n\pi + u)^3}{1+(n\pi +u)^a\sin^2x}dx
$$
Then
$$ \frac{\pi^4n^3}{\sqrt{1+\pi^a\min(n^a,(n+1)^a)}}\le a_n\le \frac{\pi^4(n+1)^3}{\sqrt{1+\pi^a\max(n^a,(n+1)^a)}}$$
It follows that
$$\lim_{n\to\infty} n^{\frac{a}{2}-3} a_n =\pi^{4-a/2}>0$$
So, the series $\sum a_n$, and consequently the considered integral, does converge if and only if $\frac{a}{2}-3>1$ or equivalently $a>8$.$\square$
