Convergence of $\int_0^{\infty} \frac{x^n}{(1+x^2)^m}dx$ Suppose we have an integral of the form $$I(n,m):= \int_0^{\infty} \frac{x^n}{(1+x^2)^m}dx$$
Most of my test cases computed seem to indicate if $n\geq 2m-1$, then this integral diverges.
I have a non-rigorous justification for this: just expand the bottom and look at leading terms. If we ignore lower order terms as we move towards $\infty$, the integrand will be of order $x^{n-2m}$. If $n= 2m-1$, then this will be $x^{-1}$, whose integral diverges here, and similarly for $n>2m-1$, whereas for $n<2m-1$, the integrand behave like $x^{-2}$ or greater negative powers, which will converge.
This is all very nice intuitively, but I'm wondering if anyone has a more rigorous, formal approach to this?
Here's a related question, of which this is a generalization.

If $n<2m-1$, we can just directly bound the integrand: 
$$\frac{x^n}{(1+x^2)^m}\leq \frac{x^n}{x^{2m}+1}.$$
