TITLE: $ \int^{\infty}_0 |\frac{1}{(1+x)\sqrt x}|^p ~ \mathrm dx < \infty \implies p=?$
QUESTION [5 upvotes]: If $ f(x) = \frac{1}{(1+x)\sqrt x} $ how to find all $ p > 0 $ such that
$$ \int^{\infty}_0 |f(x)|^p dx < \infty $$
The integral is with respect to lebesgue measure. Any solution or hints would be helpful. The answer is the integral converges iff $ p\in (\frac{2}{3}, 2) $.

REPLY [6 votes]: At $+\infty$, you have
$$
|f(x)|^p\sim \frac{1}{x^{3p/2}}
$$
which converges if and only if $3p/2>1$.
At $0$, 
$$
|f(x)|^p\sim\frac{1}{x^{p/2}}
$$
which converges if and only if $p/2<1$.
So your integral converges if and only if
$$
\frac{2}{3}<p<2.
$$