Local integrability of two functions Why $\log|x|$ is a locally integrable function and $1/|x|$ is not? Id know how their graphs look like but I don't know what is the exact difference causing local integrability of the first one.

To show that $\log|x|$ is locally integrable in $\mathbf R^d$ for $d \geqslant 2$, we can use a change of variables to spherical coordinates to deal with the singularity at $0$. Let $B$ be the ball of radius $1$ centered at $0$ in $\mathbf R^d$, and $\sigma$ the usual surface measure on the sphere $S^{d-1}\subset\mathbf R^d$. Then we have
\begin{align*}
\int_B\log|x|\,\mathrm dx &= \int_{S^{d-1}}\int_0^1\log (r)\, r^{d-1}\,\mathrm dr\,\mathrm d\sigma \\
&= \sigma\big(S^{d-1}\big)\int_0^1\log(r)\,r^{d-1}\,\mathrm dr.
\end{align*}
Because $d \geqslant 2$, we have $d-1 \geqslant 1$, hence $r^{d-1} \leqslant r$ for all $r\in [0,1]$. Hence for all $r\in (0,1]$, we have the upper bound
$$0\leqslant -\log(r)\,r^{d-1} \leqslant -\log(r)\,r,$$
so we can conclude that the integral $\log |x|$ converges absolutely on $B$, and quickly conclude that $\log |x|$ is locally integrable on $\mathbf R^d$ for $d\geqslant 2$.
