Let $\Omega = \mathbb{R}^3\setminus\{0\}$. Consider the function $$ f_n \colon \Omega \to \mathbb{R},\quad \vec{x} \mapsto \frac{1}{\|\vec{x}\|^n} $$ with $n \in \mathbb{Z}^{+}$.
I want to calculate the laplacian of the function $f$ in a distributional sense.
Here is what I have so far: Consider the associated distribution $$ T_{f_n} \colon \mathcal{D}(\Omega) \to \mathbb{R},\quad \phi \mapsto \int_{\mathbb{R}^3}f_n(\vec{x})\phi(\vec{x})\ d^3x = \int_{\mathbb{R}^3}\frac{\phi(\vec{x})}{\|\vec{x}\|^n}\ d^3x $$ The distributional derivative of $T_{f_n}$ (the laplacian in this case) is given by \begin{align} <\Delta T_{f_n}, \phi> &= (-1)^2<T_{f_n}, \Delta\phi> \\ &= \int_{\mathbb{R}^3}\frac{\Delta\phi(\vec{x})}{\|\vec{x}\|^n}\ d^3x \\ &= \int_{\mathbb{R}^3-B_{\epsilon}(0)}\frac{\Delta\phi(\vec{x})}{\|\vec{x}\|^n}\ d^3x + \int_{B_{\epsilon}(0)}\frac{\Delta\phi(\vec{x})}{\|\vec{x}\|^n}\ d^3x \\ &= \lim_{\epsilon \to 0} \int_{U_\epsilon}\frac{\Delta\phi(\vec{x})}{\|\vec{x}\|^n}\ d^3x \end{align} where $B_{\epsilon}(0) := \{\vec{x} \in \mathbb{R}^3 \mid \|\vec{x}\| < \epsilon\}$ and I have defined $U_\epsilon := \mathbb{R}^3-B_{\epsilon}(0)$. To proceed, I will make use of the following Green's identity: \begin{equation} \int_{U}\psi\Delta\varphi\ dV = \int_{U}\varphi\Delta\psi\ dV + \oint_{\partial U}\varphi\nabla\psi\cdot d\vec{S} + \oint_{\partial U}\psi\nabla\varphi\cdot d\vec{S} \end{equation} We recognize $\psi = 1/\|\vec{x}\|^n$, $\varphi = \phi$ and $U = U_\epsilon$. Thus, \begin{equation} <\Delta T_{f_n}, \phi> = \lim_{\epsilon \to 0} \left[\int_{U_\epsilon}\phi(\vec{x})\Delta\left(\frac{1}{\|\vec{x}\|^n}\right)\ d^3x + \oint_{\partial U_\epsilon}\frac{1}{\|\vec{x}\|^n}\nabla\phi \cdot d\vec{S} - \oint_{\partial U_\epsilon}\phi(\vec{x})\nabla\left(\frac{1}{\|\vec{x}\|^n}\right) \cdot d\vec{S}\right] \end{equation} We can take advantage of the spherical geometry of the region $U_\epsilon$ and use spherical coordinates to compute some elements above. In what follows, $r = \|\vec{x}\|$ and $d\vec{S} = \epsilon^2(-d\Omega\hat{r}) = -\epsilon^2d\Omega\hat{r}$ (with $d\vec{\Omega} = \sin\theta\ d\theta\ d\phi\ \hat{r}$); note that the minus sign in $d\vec{S}$ comes from the fact that a normal vector on the boundary of the region $U_\epsilon$ is given by $\hat{n} = -\hat{r}$. Then \begin{align} <\Delta T_{f_n}, \phi> &= \lim_{\epsilon \to 0} \left[\int_{U_\epsilon}\phi(\vec{x})\left(\frac{n(n-1)}{r^n}\right)\ d^3x + \oint_{S^2}\frac{1}{\epsilon^n}\frac{\partial\phi}{\partial r}\Bigr|_{r=\epsilon}(-\epsilon^2\ d\Omega) - \oint_{\partial U_\epsilon}\phi(\vec{x})|_{r=\epsilon}\left(-\frac{n}{\epsilon^{n+1}}\right) (-\epsilon^2d\Omega)\right] \\ &= n(n-1)\lim_{\epsilon \to 0}\int_{U_\epsilon}\frac{\phi(\vec{x})}{r^n}\ d^3x - \lim_{\epsilon \to 0}\frac{1}{\epsilon^{n-2}}\oint_{S^2}\frac{\partial\phi}{\partial r}\Bigr|_{r=\epsilon}\ d\Omega - n\lim_{\epsilon \to 0}\frac{1}{\epsilon^{n-1}}\oint_{\partial U_\epsilon}\phi(\vec{x})|_{r=\epsilon}\ d\Omega \end{align} Let us analize the behaviour of each term:
- First term: \begin{equation} \lim_{\epsilon \to 0}\int_{U_\epsilon}\frac{\phi(\vec{x})}{r^n}\ d^3x = \int_{\mathbb{R}^3}\frac{\phi(\vec{x})}{r^n}\ d^3x =\; <T_{f_n}, \phi> \end{equation}
- Second term: \begin{equation} \lim_{\epsilon \to 0}\frac{1}{\epsilon^{n-2}}\oint_{S^2}\frac{\partial\phi}{\partial r}\Bigr|_{r=\epsilon}\ d\Omega = \begin{cases} 0 & n = 1 \\ 4\pi\partial_r\phi(0) & n = 2 \\ \color{red}{+\infty} & n \geq 3 \end{cases} \end{equation}
- Third term: \begin{equation} \lim_{\epsilon \to 0}\frac{1}{\epsilon^{n-1}}\oint_{\partial U_\epsilon}\phi(\vec{x})|_{r=\epsilon}\ d\Omega = \begin{cases} 4\pi\phi(0) & n = 1 \\ \color{red}{+\infty} & n \geq 2 \end{cases} \end{equation}
I would like to know if the analysis is ok so far and what is happening in the case $n \geq 2$. I feel like I'm making a mistake somewhere because of the divergences I'm apparently getting.