How to show that a certain function is in $L^p$?

Question

How can I show that the function $ F(x)= \dfrac{|x| ^{-n+1}} { \log \frac{1}{|x|} } $, for $ 0 < |x| \leqslant \large\frac{1}{2} $ and $ F(x)=0 $, if $ |x|>\large\frac{1}{2} $, is in $L^p(\mathbb{R}^n)$ for $p \leqslant \large\frac{n}{n-1} $?

I managed to bound $ (F(x))^p $ by $ |x|^{-a} $ for some $a<n$ in order to use a corollary from G.Folland's book that states:

If $ |f(x)| \leqslant C|x|^{-a} $ on B for some $a<n$, then $ f \in L^1(B) $, where $ B= \left\{{x \in \mathbb{R}^n: |x|<c }\right\} $ and $\,c,\, C\,$ are positive constants.

Answer 1

For radial function on $\Bbb R^n$, that is, which can be written by $f(x)=g(|x|)$ where $|\cdot|$ is the euclidian norm, we have the following: for $0<a<b$,
$$\int_{\{x,a<|x|<b\}}f(x)dx=C\int_a^bg(r)r^{n-1}dr,$$ where $C$ is an universal constant (which doesn't depend on $f$). So we have to see whether $$\int_0^{1/2}\frac{(r^{1-n})^p}{(-\log r)^p}r^{n-1}dr$$ is convergent. Work with the integral $\int_\varepsilon^{1/2}\frac{1}{(-\log r)^p}r^{(n-1)(1-p)}dr$, and do the substitution $s=-\log r$ to get an easier integral.