# Laplacian in $\mathbb R^3$ of the function $\frac{1}{|x|}$

In an exercise I find the request of evaluating the Laplacian of $f(x)=\frac{1}{|x|}$ in $\mathbb{R}^3$. But it exists in a classical sense only if $x\neq 0$ otherwise I have to see it in distributional sense, isn't it?

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Simply put, yes, at $x=0$, Laplacian of $f$ can be defined as a constant times Dirac delta in the sense of distribution. –  Shuhao Cao Mar 1 '13 at 21:15
Yes, but if I want it only in the case $x\neq 0$ I can calculate it in the classical sense, isn't it? –  Mario Mar 1 '13 at 21:25
Yes, component-wise. –  Shuhao Cao Mar 1 '13 at 21:32
What you mean for component-wise? –  Mario Mar 1 '13 at 21:36
I mean $x = (x_1,x_2,x_3)$ having three component in $\mathbb{R}^3$, and you wanna compute each direction's partial derivative. –  Shuhao Cao Mar 1 '13 at 21:44
This is Laplacian of $1/r$. Sum of the second derivatives of $1/r$ wrt the components. The Laplacian of Columb potential, so defined in $3d$, is $0$.
Let $r=\sqrt{x_1^2+x_2^2+x_3^2}$, then ${\partial r \over {\partial x_1}}=x_1/r$. Now ${\partial (1/r) \over \partial x_1 }= -1/r^2 {\partial r \over {\partial x_1}}=-x_1/r^3$. Next ${\partial \over \partial x_1}{\partial (1/r) \over \partial x_1 }={{-r^2+3x_1^2}\over{r^5}}$. The sum of three such terms is ${-3r^2+3(x_1^2+3x_2^2+3x_3^2)} \over {r^5}$ which is $0$.
At the origin it is $-4\pi \delta(r)$, where $\delta$ is the Dirac function.
This computation does not explain where $-4\pi \delta$ came from. See here. –  user53153 Mar 2 '13 at 2:51