MHB Proving $\int_{\mathbb{R}^3}\Delta G(x,y)dx=1$ with Gauss's Theorem

[evinda]

Hello! (Wave)We have $G(x,y)=-\frac{1}{4 \pi} \frac{1}{||\overline{x}-\overline{y}||}$ for $x, y \in \mathbb{R}^3$.I want to show that $\int_{\mathbb{R}^3} \Delta{G(x,y)} dx= 1$.

It suffices to show that $\int_{\mathbb{R}^3} \Delta{G(x,0)} dx= 1$, since setting $\overline{x}=x-y$ we have $G(\overline{x},0)=G(x,y)$, right?

It holds that $\Delta G(x,y)=0$ for $x \neq y$, so $\int_{\mathbb{R}^3} \Delta{G(x,0)} dx= \int_{B(0,r)} \Delta{G(x,0)}$.

How can we apply Gauss's theorem in order to compute the above integral?

 

[I like Serena]

Hello! (Wave)We have $G(x,y)=-\frac{1}{4 \pi} \frac{1}{||\overline{x}-\overline{y}||}$ for $x, y \in \mathbb{R}^3$.I want to show that $\int_{\mathbb{R}^3} \Delta{G(x,y)} dx= 1$.

It suffices to show that $\int_{\mathbb{R}^3} \Delta{G(x,0)} dx= 1$, since setting $\overline{x}=x-y$ we have $G(\overline{x},0)=G(x,y)$, right?

It holds that $\Delta G(x,y)=0$ for $x \neq y$, so $\int_{\mathbb{R}^3} \Delta{G(x,0)} dx= \int_{B(0,r)} \Delta{G(x,0)}$.

How can we apply Gauss's theorem in order to compute the above integral?

Hi evinda! (Smile)

What do $\overline x$ and $\overline y$ represent?
And what is $\Delta{G(x,y)}$, since he Laplacian operator $\Delta$ usually applies only to some $x \in \mathbb R^3$? (Wondering)

Anyway, I think we would have:
$$\int_{\mathbb{R}^3} \Delta{G(x,0)} \,dx
=\lim_{r\to\infty} \int_{B(0,r)} \Delta{G(x,0)} \,dx
=\lim_{r\to\infty} \int_{B(0,r)} \nabla \cdot \nabla {G(x,0)} \,dx
\overset{\text{Gauss}}=\lim_{r\to\infty} \oint_{\partial B(0,r)} \nabla {G(x,0)} \,dS \\
=\lim_{r\to\infty} \oint_{\partial B(0,r)} \nabla {-\frac {1}{4\pi r}} \,dS
=\lim_{r\to\infty} \oint_{\partial B(0,r)} {\frac {1}{4\pi r^2}}\,dS
=\lim_{r\to\infty} {\frac {1}{4\pi r^2}}\cdot 4\pi r^2 = \lim_{r\to\infty} 1 = 1
$$
(Thinking)