How to prove following for $n\geq0$ ?  

$$\int\_{4\pi}d\vec{\Omega}(\vec{\Omega}\cdot\vec{\nabla})^{2n}f(\vec{r})=\frac{4\pi}{2n+1}\nabla^{2n}f(\vec{r})$$
Where, at any point $\vec{r}$, the $\vec{\Omega}$ can be described by the polar angle $\theta$ measured with respect to the *z* axis and an azimuthal angle $\phi$ measured with respect to the *x* axis, *i.e.*  

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$,  

$\cos\theta=\vec{\Omega}\cdot\hat{k}$,  

$\cos\phi=sin^{-1}\theta(\vec{\Omega}\cdot\hat{i})$  

Here, $\vec{\Omega}$ is a unit vector situated at $\vec{r}$ and pointing in the direction ($\theta,\phi$) which is different than the direction of $\vec{r}$.



Identity involving dot product of solid angle and gradient
Because of isotropy, the differential operator 
$$J\_n=\int\_{|\vec{\Omega}|=1} d\vec{\Omega}\,(\vec{\Omega}\cdot\vec{\nabla})^{2n}=C\_n\Delta^n$$
with $\Delta$ the Laplacian. To find the coefficient $C\_n$, let $J\_n$ act on $z^{2n}$,
$$J\_n z^{2n}=(2n)!\int\_{|\vec{\Omega}|=1} d\vec{\Omega}\,\Omega\_z^{2n}=2\pi(2n)!\int\_0^{\pi}\sin\theta d\theta\,\cos^{2n}\theta=\frac{4\pi}{2n+1}(2n)!=C\_n\Delta^n z^{2n},$$
hence $C\_n=\frac{4\pi}{2n+1}$.


