$ \int^{\infty}_0 |\frac{1}{(1+x)\sqrt x}|^p ~ \mathrm dx < \infty \implies p=?$

Question

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) $.

Answer 1

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 $$ \frac{2}{3}<p<2. $$

Answer 2

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 $$\int_1^\infty f(x)^p dx > \int_1^\infty \frac{1}{(1+x)^{3p/2}} dx .$$

Answer 3

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.