Trying to derive by myself the Poisson integral formula in a unit ball. I should get
$$\Delta u=0 \,\text{ in } B(0,1), \,\,\, u(x)=\varphi(x)\,\,\text{at } \partial B(0,1) \Longrightarrow \\$$$$u(x) = \frac{1-|x|^2}{n\alpha(n)}\int_{\partial B}\frac{\varphi(y)}{|x-y|^n}dy,\tag{1}$$
where $\alpha(n)$ is the volume of the unit ball. I start with $$u=-\int_{\partial B}\varphi(y)\langle \nabla_y G(x,y), \nu\rangle \,dy = -\int_{\partial B}\varphi(y)\langle \nabla_y G(x,y), y\rangle \,dy \tag{2}$$
where $G(x,y)$ is a function such that $\Delta_y G=\delta_x$ and $G|_{\partial B}\equiv 0$, and where $\nu$ is the unit normal to the surface of the sphere. I construct such $G(x,y)$ by taking the fundamental solution $$\Phi(y) = \frac{1}{n(n-2)\alpha(n)}\frac1{|y|^{n-2}}$$
and taking $$G(x,y)=\Phi(y-x)-\Phi(|x|(y-\tilde{x})),$$ where $\tilde{x} = \frac{1}{|x|^2}x$. (Here we use that $|y-x|=|x||y-\tilde{x}|$ for $x\in B(0,1)\setminus \{0\}$, $y\in \partial B(0,1)$.)
Then (2) becomes (if I have taken the gradient correctly) $$\frac{1}{n\alpha(n)}\int_{\partial B} \varphi(y) \left( \frac{1}{|y-x|^n}(1-\langle x,y \rangle) - \frac{1}{|x|^{n-2}}\frac{1}{|y-\tilde{x}|^{n}} \left(1-\langle \tilde{x},y \rangle\right) \right) \,dy.$$
Now how can I simplify this further to get (1)?
OK, I figured it out. Just have to use the relation $|y-x|=|x||y-\tilde{x}|$ one more time.