Discuss the convergence or divergence of $$\int_0^\infty\frac{x^\beta dx}{1+x^\alpha \sin^2x} , \alpha >\beta>0$$
I can see that it will be convergent in $[0,K], K\in \mathbf{R}$ as its upper bounded by its numerator function. But not sure how to proceed for $(K, \infty) $ as the denominator will repeatedly come closer to $1$ thus overall fraction will be closer to numerator function, which is divergent.
If $\alpha\le1+\beta$, then $$ \frac{x^\beta }{1+x^\alpha \sin^2x}\ge x^{\beta-\alpha} $$ for large $x$ and the integral diverges. What happens if $\alpha>1+\beta$? Near the points where $\sin x$ vanishes, the integrand is large. So we estimate the integral on intervals $[k\,\pi,k\,\pi+\epsilon]$ for $k\in\Bbb N$ and small $\epsilon>0$. \begin{align} \int_{k\pi}^{k\pi+\epsilon}\frac{x^\beta\,dx}{1+x^\alpha \sin^2x}&\ge\epsilon\,\frac{(k\,\pi)^\beta}{1+(k\,\pi+\epsilon)^\alpha\sin^2\epsilon}\\ &\ge C\,\epsilon\,\frac{k^\beta}{1+k^\alpha\,\epsilon^2} \end{align} for some constant $C>0$ independent of $k$ and $\epsilon$. Now choose $\epsilon=k^{-\gamma}$ for an appropriate $\gamma>0$.