While trying to complete the calculations in the question : Value of $u(0)$ of the Dirichlet problem for the Poisson equation I came across a point where I need clarification.
$$F(r) := \int_{B(0:r)}{f(x) \over |x|^{n-2}} dx $$
I want to know when is : $$\int_{R^n\backslash\{0\}} {\left|f(x)\right| \over |x|^{n-2}} dx < \infty$$
That is, what conditions should we impose on $f$ so that ${f \over |x|^{n-2}}$ is continuous and summable ?
I want it to be summable because it will then ensure that
${\partial \over \partial r } F(r) = \int_{\partial B(0:r)}{f(x) \over |x|^{n-2}} dx $ , which is required in my calculations.
As such the only conditions on $f$ are that it is continuous and summable. I'm doubtful about summability because $1 \over {|x|^{n-2}}$ shoots up as we near zero.
Outside the origin, $\frac{f(x)}{\lvert x\rvert^{n-2}}$ is continuous as the quotient of two continuous functions. For continuity in the origin, a necessary condition is that $f$ has a zero of "order" at least $n-2$ in $0$, but that is not sufficient. A zero of order strictly greater than $n-2$ would be sufficient, but not necessary.
However, for the result
$$\frac{\partial}{\partial r} \int_{B(0;r)} \frac{f(x)}{\lvert x\rvert^{n-2}}\,dx = \int_{\partial B(0;r)} \frac{f(x)}{\lvert x\rvert^{n-2}}\, dS(x),\; r > 0,\tag{1}$$
it is not necessary that the integrand be continuous in $0$. That $f$ is continuous and integrable is sufficient. The continuity of $f$ means that $f$ is bounded on $\overline{B(0;1)}$, so
$$\int_{\mathbb{R}^n}\left\lvert \frac{f(x)}{\lvert x\rvert^{n-2}}\right\rvert\,dx \leqslant \max \{\lvert f(x)\rvert : \lvert x\rvert \leqslant 1\}\int_{B(0;1)}\frac{dx}{\lvert x\rvert^{n-2}} + \int_{\mathbb{R}^n\setminus B(0;1)} \lvert f(x)\rvert\,dx < \infty$$
since $\lvert x\rvert^{2-n}$ is locally integrable, and $f$ integrable.
Thus $(1)$ holds already under the assumption that $f$ is continuous and integrable.