$ \int^{\infty}_0 |\frac{1}{(1+x)\sqrt x}|^p ~ \mathrm dx < \infty \implies p=?$ 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) $.
 A: 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
$$
\frac{2}{3}<p<2.
$$
A: Hint: Since
$$\int_0^{\infty} f(x)^p dx =\int_0^1 f(x)^p dx+\int_1^{\infty} f(x)^p dx$$
So if $0<p<1$, we can only consider the integral $\int_1^{\infty} f(x)^p dx$, and
$$\int_1^{\infty} f(x)^p dx < \int_1^\infty \left(\frac{1}{x\sqrt{x}}\right)^p dx$$ 
And
$$\int_1^\infty f(x)^p dx > \int_1^\infty \frac{1}{(1+x)^{3p/2}} dx .$$
A: Using the change of variables $ z=\frac{1}{1+x} $ and the beta function
$$ \beta(x,y) = \int_{0}^{1}t^{x-1}(1-t)^{y-1}dt,\quad Re(x),\,Re(y)>0, $$
we have
$$ \int_{0}^{\infty} \frac {1}{(1+x)^p x^{\frac{p}{2}}} dx = \int _{0}^{1}\! \left( 1-z \right) ^{-\frac{p}{2}}{z}^{\frac{3p}{2}-2}{dz}.$$
Comparing with the existence conditions for the beta function implies that 
$$ 1-\frac{p}{2}>0,\quad\frac{3p}{2}-1 > 0. $$
Solving the two inequalities gives the desired result.
