The problem is to prove the following for $n \geq 3$ $$u(0)=\frac{1}{n\alpha (n) r^{n-1}}\int_{\partial B(0,r)} g dS +\frac{1}{n(n-2)\alpha (n)} \int_{B(0,r)} (\frac{1}{|x|^{n-2}} - \frac{1}{r^{n-2}})f dx $$ provided $-\Delta u=f $ in $B^0 (0,r)$ $u=g$ on $\partial B(0,r)$ My attempt: From mean value formula we have for $n \geq 3$ $$\Phi (r) = \frac{1}{n\alpha (n) r^{n-1}} \int_{\partial B(0,r)} g dS$$
and also $$ \Phi '(r) = \frac{1}{n\alpha (n) r^{n-1}} \int_{B(0,r)} f dy$$
$$\Phi (\epsilon) = \Phi(r)- \int_{\epsilon}^{r} \Phi '(t) dt $$ $$\int_{\epsilon}^{r} \Phi '(t) dt = \frac{-1}{n\alpha (n) }\int_{\epsilon}^{r} \frac{1}{t^{n-1}}\int_{B(0,r)} f dy dt$$ using integration by parts
$$= \frac{-1}{n\alpha (n) }[[ \frac{-1}{(n-2) t^{n-2}}\int_{B(0,t)} f dy]_{\epsilon}^{r} - \int_{\epsilon}^{r}\frac{-1}{(n-2) t^{n-2}} \int_{\partial B(0,t)} f dS dt ]$$ $$= \frac{1}{n(n-2)\alpha (n) }[ \frac{1}{ r^{n-2}}\int_{B(0,r)} f dy - \frac{1}{ \epsilon ^{n-2}}\int_{B(0,\epsilon)} f dy -\int_{\epsilon}^{r}\frac{1}{ t^{n-2}} \int_{\partial B(0,t)} f dS dt ]$$
so now,by taking $\epsilon \rightarrow 0$ $$\frac{1}{ \epsilon ^{n-2}}\int_{B(0,\epsilon)} f dy = 0 $$
I realise that the following limit tends to the expression on right hand side.but I cannot show that rigorously.
$$\int_{\epsilon}^{r}\frac{1}{ t^{n-2}} \int_{\partial B(0,t)} f dS dt = \int_{B(o,r)} \frac{1}{|x|^{n-2}} f(x) dx$$,
I am new to pde's and real analysis,so please give precise and detailed(step by step) steps to show that limit?