im trying to find the values of $\alpha \in \mathbb{R}$ for which the function $f: B_2(0,\frac{1}{2}) \rightarrow \mathbb{R} $, $ x \mapsto |\log(\|x\|_2)|^\alpha$ is in $L^2(B_2(0,\frac{1}{2}))$ and in $L^\infty (B_2(0,\frac{1}{2}))$ respectively.
To be in $L^2 (B_2(0,\frac{1}{2}))$ means that :
$||f||^2 _2 =\int_{B_2(0,\frac{1}{2})} |f(x)|^2 dx=\int_{B_2(0,\frac{1}{2})} |\log(\|x\|_2)|^{2\alpha} dx < \infty $
In order to simplify i wrote the integral in polar coordinates so i got:
$2\pi \int_{0}^{\frac{1}{2}} \log(r)^{2\alpha} r dr < \infty$
At this point i thought that a new substitution like $h=log(r)$ may help, because it makes appear a $e^{2h}$ with a $-\infty$ in the place where zero was and then with integration by parts come to something solid (which i couldn't reach) . Another idea was to integrate in some domain $B:=B_2(0,\frac{1}{2})/B_2(0,\varepsilon)$ for some $\varepsilon > 0$. I also looked at this question: https://math.stackexchange.com/questions/1065535/calculate-weak-derivative/1068662#1068662 where the weak derivative of the same function was calculated and the condition $1-\alpha>1$ appeared as necessary for $|\nabla f(x)|\in L^1(B_2(0,\frac{1}{2}))$ which should also mean $f \in W^1 (B_2(0,\frac{1}{2}))$. I will really appreciate any help.
The integral is finite for all $\alpha\in\mathbb{R}$. Observe that although $\log r$ is unbounded near $r=0$, we always have $\lim_{r\to0^+}|\log r|^{2\alpha}r=0$.