Laplacian of the potential function Given a potential function $$\Gamma(x)=\frac{1}{(n-2)|S^{n-1}||x|^{n-2}}$$
where $n\ge 3$ is an integer, $|S^{n-1}|$ is the volume of the sphere, $x$ is in $n$ dimension. Is it true that 
$$-\Delta \Gamma(x)=\delta_{0}(x)$$
where $\delta_0(x)$ is the Dirac delta function.
I am not sure if the constants are right, and I have the following try: Let operator $T_{\Gamma}:\phi \to \int_{\mathbb{R}^n}\Delta\Gamma(x)\phi(x)dx$. Where $\phi\in D(\mathbb{R})$ is a distribution. We obtain
$$\begin{align*}
T_{\Gamma}(\phi)&=\int_{\mathbb{R}^n}\Delta\Gamma(x)\phi(x)dx=\int_{\mathbb{R}^n}\Gamma(x)\Delta\phi(x)dx\\&=\int_{B_{R}(0)}\Gamma(x)\Delta\phi(x)dx=\lim_{\epsilon\to 0}\int_{B_{R}(0)\backslash B_{\epsilon}(0)}\Gamma(x)\Delta\phi(x)dx
\end{align*}$$
where $R$ is a large constant, where $\phi(x)\equiv 0$ when $|x|>R$. This can be satisfied because $\phi \in D(\mathbb{R})$. But I got stuck here. I needs to prove that this equals $-\phi(0)$, but I do not know what to do next. Thanks for your attention.
 A: Use
$$ \int_{B_R} \triangle \Gamma \phi \mathrm{d}x = \int_{B_R\setminus B_\epsilon} \triangle \Gamma \phi \mathrm{d}x + \int_{B_\epsilon} \triangle \Gamma \phi \mathrm{d}x $$
The first term in the domain of integration $\Gamma$ is smooth and $\triangle \Gamma = 0$ by computation, so you have that it goes to 0. For the second term we integrate by parts twice
$$ = - \int_{B_\epsilon} \nabla \Gamma \cdot \nabla \phi \mathrm{d}x + \int_{\partial B_\epsilon} \nabla \Gamma \cdot \vec{n} \phi \mathrm{d}\sigma = \int_{B_\epsilon} \Gamma \triangle \phi \mathrm{d}x + \int_{\partial B_\epsilon} \nabla \Gamma \cdot \vec{n} \phi - \Gamma \nabla \phi\cdot \vec{n}\mathrm{d}\sigma $$
Using that $\Gamma$ is locally integrable you have that as $\epsilon \to 0$ the only non-vanishing term is 
$$ \lim_{\epsilon \to 0} \int_{\partial B_\epsilon} \nabla \Gamma \cdot \vec{n} \phi \mathrm{d}x \tag{1}$$
evaluating 
$$\nabla \Gamma \cdot \vec{n} = \partial_r \Gamma = -\frac{1}{|S^{n-1}| |x|^{n-1}}$$
by continuity of $\phi$ we have that the expression (1) converges to $-\phi(0)$. 
