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.
 A: I am assuming $n$ and $m$ are positive integers. Over $[0,1]$, the integrand is continuous and integrable. Use the limit comparison test to check convergence/divergence of the improper integral over $[1, \infty).$
Note that 
$$\lim_{x \to \infty} \frac{x^n}{(x^2 +1)^m} \frac{1}{x^{n - 2m}} = \lim_{x \to \infty} \frac{1}{(1 + x^{-2})^m} = 1$$
Hence, for sufficiently large $x$
$$ \frac{3}{2}  x^{n-2m}   >\frac{x^n}{(x^2 +1)^m} > \frac1{2}x^{n-2m},$$
and the integral converges if $n < 2m-1$ and diverges if $n \geqslant 2m-1,$ since the integral $\int_1^\infty x^{a} \, dx$ converges for $a < -1$ and diverges for $a \geqslant -1$. 
A: Assuming $n,m\geq 0$, the integrand function is bounded in any right neighbourhood of the origin. The behaviour for large values of $x$ is the same as $\frac{1}{x^{2m-n}}$, hence we have convergence iff $2m>n+1$. Under such assumptions, through the substitution $x=\tan t$, $t=\arcsin u$ and $u=\sqrt{v}$ we have:
$$ I(n,m) = \int_{0}^{\pi/2}\cos^{2m-n-2}(t)\sin^{n}(t)\,dt = \int_{0}^{1}u^{n} (1-u^2)^{\frac{2m-n-3}{2}}\,du$$
so:

$$ I(n,m) = \frac{1}{2}\int_{0}^{1}v^{\frac{n-1}{2}}(1-v)^{\frac{2m-n-3}{2}}\,dv = \color{red}{\frac{\Gamma\left(\frac{n+1}{2}\right)\Gamma\left(\frac{2m-n-1}{2}\right)}{2\cdot \Gamma\left(m\right)}}.$$

A (not so) rough estimation comes from:
$$\begin{eqnarray*} I(n,m)&=&\int_{0}^{1}\frac{x^n}{(1+x^2)^m}\,dx+\int_{1}^{+\infty}\frac{x^n}{(1+x^2)^m}\,dx\\ &\leq& \int_{0}^{1} x^n\,dx + \int_{1}^{+\infty}x^{n-2m}\,dx = \frac{1}{n+1}+\frac{1}{2m-n-1}.\end{eqnarray*}$$
A: If $n<2m-1$, we can just directly bound the integrand: 
$$\frac{x^n}{(1+x^2)^m}\leq \frac{x^n}{x^{2m}+1}.$$
