I want to show that $u(x)= \log \frac{1}{|x|}$ and $v(x)= \log \log \frac{1}{|x|}$ are in $L^q(\Omega)$ for all $1 \leq q < \infty$ for $\Omega = B_1(0) \subset \mathbb R^n$, $\Omega = B_{1/e}(0) \subset \mathbb R^n$ respectively.
(I posted the question yesterday, and I think I figured how it works.)
$\newcommand{\dd}{\mathrm{d}}$
For $r =1$ and $r =0$ we have $\log r = 0$ and $r^{n-1}= 0$ respectively, so we can integrate by parts after changing to polar coordinates. Then
$$\int_\Omega |u|^q \, \dd x = C \int_0^1 |\log r|^q \, r^{n-1} \, \dd r = - C \int_0^1 |\log r|^{q-1} \, \frac{r^{n-1}}{n} \, \dd r \\ = \cdots = (-1)^{[q]} \, C \int_0^1 |\log r|^{q - [q]} \, \frac{r^{n-1}}{n^{[q]}} \, \dd r $$ where $[\cdot]$ is the floor function
Let $\alpha =: q - [q] \in [0,1]$ hence we can split and estimate
$$ C \int_0^\frac{1}{e} \underbrace{|\log r|^\alpha}_{\leq |\log r|} \, \, r^{n-1} \, \dd r \leq C \int_0^\frac{1}{e} r^{n-1} \, \dd r < \infty $$
$$ C \int_\frac{1}{e}^1 \underbrace{|\log r|^\alpha}_{\leq |\log r|^0} \, \, r^{n-1} \, \dd r \leq C \int_\frac{1}{e}^1 r^{n-1} \, \dd r < \infty $$
Hence $\log \frac{1}{|x|} \in \bigcap_{1 \leq q < \infty} L^q(\Omega)$
$\,$
essentially the same as above, change to polar coordinates, then again we are able to integrate by parts, after a substitution:
$$\int_\Omega |v|^q \, \dd x = C \int_0^\frac{1}{e} \Big(\log \log \frac{1}{r}\Big)^q \, r^{n-1} \, \dd r = C \int_1^\infty ( \log x )^q \, e^{-nx} \, \dd x \\ = - C \int_1^\infty (\log x)^{q-1} \, \frac{e^{-nx}}{-nx} \, \dd x = \cdots = (-1)^{[q]} C \int_1^\infty (\log x)^{q-[q]} \, \frac{e^{-nx}}{(-nx)^{[q]}} \, \dd x$$
and $\alpha = q- [q] \in [0,1]$ hence $(\log x)^\alpha \leq \log x$ for $x \in [1,\infty)$ and the integral converges