Determine the values of $p>0$ for which the improper integral $ \int_0^{\infty}\frac{dx}{x^p \sqrt{1+x^p}} $ converges. Determine the values of $p>0$ for which the improper integral 
$$
\int_0^{\infty}\frac{dx}{x^p \sqrt{1+x^p}}
$$
converges. I know that
$$
\frac{1}{x^p \sqrt{1+x^p}} \leq \frac{1}{x^p},
$$
so 
$$
\int_1^{\infty}\frac{dx}{x^p \sqrt{1+x^p}}
$$
converges for $p>1$ and
$$
\int_0^1\frac{dx}{x^p \sqrt{1+x^p}}
$$
converges for $p<1$.  
After that I am stuck.
 A: The integral from $1$ to $\infty$ converges when $\frac{3p}{2}\gt 1$, and diverges otherwise.  This can be done by limit comparison with $\frac{1}{x^{3p/2}}$.
So we want $p\gt \frac{2}{3}$. As you pointed out, the integral  from $0$ to $1$ converges if $p\lt 1$. It diverges for $p\ge 1$. This is because on $(0,1]$ we have $\frac{1}{x^p\sqrt{1+x^p}}\ge \frac{1}{\sqrt{2}x^p}$. 
So we have convergence of both integrals precisely if $\frac{2}{3}\lt p\lt 1$.
A: At $x = 0$ there is a vertical asymptote, so you need to split the integral in two:
$$\int_0^{\infty}\frac{\mathrm dx}{x^p \sqrt{1+x^p}} = \int_0^1\frac{\mathrm dx}{x^p \sqrt{1+x^p}} + \int_1^{\infty}\frac{\mathrm dx}{x^p \sqrt{1+x^p}}$$
Let's denote by $f(x)$ the function you want to integrate. We'll use twice the asymptotic comparison test for improper integrals. It states that a function has the same behaviour as its principal part.
For $x \to 0$ you have that
$$f(x)\sim\frac1{x^p\left(1 + \frac12x^p\right)}\sim\frac1{x^p}$$
and thus the first integral converges only if $p < 1$.
For $x \to +\infty$ we have instead
$$f(x) = \frac1{x^px^{p/2}\left(1 + \frac1{x^p}\right)}\sim\frac1{x^{3p/2}}$$
Hence the second integral converges only if $\frac{3p}2 > 1 \implies p > \frac23$.
Therefore we have convergence of both integrals for $\frac23 < p < 1$.
