Existence of L2-functions

Question

On the unit interval (0,1), the functions $f(x)=1/x^p$ are in $L^2$ for $p<1/2$. However, the function $f(x)=x^{-1/2}$ is not in $L^2$ since $\int_0^{1} \frac{dx}{x}$ = $ln(1) - ln(0)$. And $ln(0)$ is undefined.
But, can someone explain why $f(x)=1/x^p$ is in $L^2$ when p = 1/3.
Here we get $\int_0^{1} \frac{dx}{x^{2/3}} = [\frac{-2}{3}x^{-5/3}]_0^1$ and here we get $\frac {1}{0}$ when we put $0$ and then it goes to infinity. Then why is this function is $L^2$?

Answer 1

$\int_0^{1}\frac 1 {x^{2/3}}dx= \frac 1 {1-\frac 2 3}x^{1-\frac 2 3}|_0^{1}=3$.