When are $\frac{1}{|x|^s}$ and $\log|x|$ integrable near the origin? When are $\frac{1}{|x|^s}$ for $s>0$ and $\log|x|$ integrable near the origin? I'm reading Evans PDE and in the construction of the fundamental solution of Poisson's equation, he defines
$$
\Phi(x) = C \log|x|
$$
for $\mathbb{R}^2$ and a suitable $C$, and
$$
\Phi(x) = C \frac{1}{|x|^{n-2}}
$$
for $\mathbb{R}^n$ with $n\geq 3$ and $C=C(n)$ a suitable constant.
The construction then goes about defining $\Phi * f$ for an $f \in C^2_c$. Clearly if $\Phi \in L^1_{\mathrm{loc}}$, then the convolution will be finite a.e., but I worry about  whether this holds because of the blow up near $0$. In general in what dimensions are
$$
\log |x|,~~~~~\frac{1}{|x|^s}
$$
integrable in a ball around $0$?
 A: For the logarithm, the answer is already implicitly given in the comments above. On $\mathbb{R}^{n}$ with $n \geq 2,$ we have, using polar co-ordinates, $$\int_{B_{1}(0)} \log |x| dx = \lim_{\varepsilon \rightarrow 0} \int_{B_{1}(0)\setminus B_{\varepsilon}(0)} \log |x| dx = \lim_{\varepsilon \rightarrow 0} \int_{B_{1}(0)\setminus B_{\varepsilon}(0)}  r^{n-1} \log r \ dr d\Omega_{S^{n-1}}, $$
where $d\Omega_{S^{n-1}}$ is the volume form on $S^{n-1}.$ Now, integrate by parts to explicitly integrate this and check this will always be finite as long as $n \geq 2.$
For $\frac{1}{|x|^{s}}$ would be integrable near the origin of $\mathbb{R}^{n}$ when $s < n. $ This can be easily checked using polar co-ordinates again. 
$$ \int_{B_{1}(0)} \frac{1}{|x|^{s}} dx = \lim_{\varepsilon \rightarrow 0} \int_{B_{1}(0)\setminus B_{\varepsilon}(0)} \frac{1}{|x|^{s}} dx = \lim_{\varepsilon \rightarrow 0} \int_{B_{1}(0)\setminus B_{\varepsilon}(0)} \frac{1}{r^{s}} r^{n-1} dr d\Omega_{S^{n-1}}.$$
 If $s <n,$ integrating, we obtain, 
$$ \int_{B_{1}(0)} \frac{1}{|x|^{s}} dx = \lim_{\varepsilon \rightarrow 0} \left[  \frac{r^{n-s}}{n-s}\right]_{\varepsilon}^{1} .\operatorname{Vol}(S^{n-1}).$$ This limit would be finite since $n-s >0$ and thus $\varepsilon^{n-s} \rightarrow 0$ as $\varepsilon \rightarrow 0.$ The integral will blow up for $s=n,$ since in that case after integrating we would obtain, 
$$ \int_{B_{1}(0)} \frac{1}{|x|^{n}} dx = \lim_{\varepsilon \rightarrow 0} \left[  \log r \right]_{\varepsilon}^{1} .\operatorname{Vol}(S^{n-1}).$$ Now since $-\log \varepsilon \rightarrow + \infty$ as $\varepsilon \rightarrow 0,$ the inetgral would not be finite for $s=n$ ( and obviously also for $s > n$ as well ).
