I want to show the following equality $$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\frac{\vec{p}}{3\epsilon_0}$$
where $\vec{p}$ is the dipole moment of a charge distribution $\rho\left(\vec{r}^{\prime}\right)$ with respect to the origin. The charge distribution is located inside a sphere with radius $R$.
I started with Gauss's Theorem: $$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\int_{\left|\vec{r}\right|<R}d^3r\nabla\Phi\left(\vec{r}\right)=-\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{R}R^2 \Phi\left(\vec{r}\right)$$
Now I can use the Poisson integral: $$\Phi\left(\vec{r}\right)=\frac{1}{4\pi\epsilon_0}\int d^3r^{\prime}\frac{\rho\left(\vec{r}^{\prime}\right)}{\left|\vec{r}-\vec{r}^{\prime}\right|}$$
And thus: $$\int_{\left|\vec{r}\right|<R}d^3r\vec{E}\left(\vec{r}\right)=-\frac{R}{4\pi\epsilon_0}\int d^3r^{\prime}\rho\left(\vec{r}^{\prime}\right)\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}$$
I know that the dipole moment is: $$\vec{p}=\int d^3r^{\prime}\vec{r}^{\prime}\rho\left(\vec{r}^{\prime}\right)$$
Therefore, to obtain the top equation, the solid angle integration has to be: $$\oint_{\left|\vec{r}\right|=R}d\Omega\frac{\vec{r}}{\left|\vec{r}-\vec{r}^{\prime}\right|}\overset{!?}{=}\frac{4\pi}{3R}\vec{r}^{\prime}$$
Does anyone know how to solve this integral?