Computing the Laplacian of $\frac{\mu\ \cdot\ \mathbf{r}}{\|\mathbf{r}\|^3}$ How does one compute the Laplacian of
$$\dfrac{(\mathbf \mu \cdot \mathbf r)}{r^3} \;\; \text{where} \;\; r = \Vert \mathbf r \Vert?$$
I am aware that the Laplacian is defined to be $\Delta f=\sum_i \partial_i^2f$ but am a little confused about this computationally in this case.
I have got as far as saying $\partial_i^2[1/r^3(\mu_1r_1+\mu_2r_2+...+\mu_nr_n)]$, but don't know how to compute this.
 A: WARNING: GRAPHIC CONTENT! It's not going to be pretty as Joonas' answer (btw thank you for the correction!)
Let's do it in cartesian coordinates. The function is
\begin{equation}
u(x_1,\dots,x_n)=\frac{\sum_i\mu_ix_i}{\left(\sum_j x_j^2\right)^{3/2}}=\frac{\mu\cdot\mathbf{r}}{r^3}
\end{equation}
We have that
\begin{equation}
\frac{\partial\ r^s}{\partial x_k}=sx_k\ r^{s-2}
\end{equation}
and also
\begin{equation}
\frac{\partial\ (\mu\cdot\mathbf{r})}{\partial x_k}=\mu_k
\end{equation}
so,
\begin{equation}
\frac{\partial u}{\partial x_k}=\frac{\mu_kr^3-3x_kr(\mu\cdot\mathbf{r})}{r^6}=\frac{1}{r^5}\left(\mu_kr^2-3(\mu\cdot\mathbf{r})x_k\right)
\end{equation}
and
\begin{equation}
\frac{\partial^2 u}{\partial x_k^2}=
\frac{1}{r^{10}}\left[(2\mu_kx_k-3\mu_kx_k-3(\mu\cdot\mathbf{r}))r^5-5x_kr^3(\mu_kr^2-3(\mu\cdot\mathbf{r})x_k)\right]=\\
=\frac{1}{r^7}\left[-\mu_kx_kr^2-3(\mu\cdot\mathbf{r})r^2-5\mu_kx_kr^2+15(\mu\cdot\mathbf{r})x_k^2\right]=\\
=\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})(5x_k^2-r^2)-2\mu_kx_kr^2\right]
\end{equation}
Summing everything over $k$ we get
\begin{equation}
\Delta u = \sum_k\frac{\partial^2u}{\partial x_k^2}=\frac{3}{r^7}\sum_k\left[(\mu\cdot\mathbf{r})(5x_k^2-r^2)-2\mu_kx_kr^2\right]=\\
=\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})\sum_k(5x_k^2-r^2)-2r^2\sum_k\mu_kx_k\right]=\\
=\frac{3}{r^7}\left[(\mu\cdot\mathbf{r})(5r^2-nr^2)-2r^2(\mu\cdot\mathbf{r})\right]=\\
=3(3-n)\frac{(\mu\cdot\mathbf{r})}{r^5}
\end{equation}
A: If you have two functions $f$ and $g$, then $\Delta(fg)=f\Delta g+g\Delta f+2\nabla f\cdot\nabla g$.
Now we obviously want to choose $f(x)=|x|^{-3}$ and $g(x)=\mu\cdot x$.
Recall that $\nabla |x|^\alpha=\alpha|x|^{\alpha-2}x$; to remember this rule, remember that the gradient is a vector and the formula should hold true in dimension one.
These choices give
\begin{eqnarray}
\nabla g(x)&=&\mu\\
\Delta g(x)&=&0\\
\nabla f(x)&=&-3|x|^{-5}x\\
\Delta f(x)&=&-3(\nabla|x|^{-5})\cdot x-3|x|^{-5}\text{div(x)}
\\&&\quad=15|x|^{-5}-3n|x|^{-5}
\\&&\quad=(15-3n)|x|^{-5},
\end{eqnarray}
whence
\begin{eqnarray}
\Delta(fg)
&=&
f\Delta g+g\Delta f+2\nabla f\cdot\nabla g
\\&=&
|x|^{-3}0+(\mu\cdot x)(15-3n)|x|^{-5}-6|x|^{-5}x\cdot\mu
\\&=&
(9-3n)|x|^{-5}\mu\cdot x.
\end{eqnarray}
