Green's Function for Operator I'm trying to show that the Green's function for the Laplace operator $-\nabla^2$ is badly behaved at infinity. I.e. 
$$\int d^dx|G(x,y)|^2=\infty$$ for $d=1,2,3$. What happens when $d>4$?
I know the 1D Green's function is given by
$$G(x,y)=-\frac{|x-y|}{2}$$
but I'm not sure how to generalize this. Could someone push me in the right direction?
Update: for d=1, I have $\int |G(x,y)|^2 dx=\frac{1}{4}\int dx |x-y|^2=\infty$. This is trivial. How do I show this same thing for d=2, 3? And then what happens when $d=4$? I know $G(x, y)\propto|x-y|^{-(d-2)}$ for d>2.
 A: It may be easier to start with vector Green's functions for $\nabla$ and then work up to Green's functions for the Laplacian.
$G_1$, the Green's function for $\nabla$, can be easy found through the Generalized Stokes theorem.
$$\oint G_1(y-x) \wedge d^{N-1}y = \int \delta(y-x) \, d^Ny = i_N$$
For $N=3$, we can pick a ball for the volume integral.  This relates the values of $G_1$ on the boundary of the ball to a volume integral over that ball.  We can conclude that $|G_1|$ should be constant over the surface of the ball and that its direction should be radially outward.  Let the radius of the ball be $R$, and we conclude that
$$\oint G_1(y-x) \wedge d^{N-1}y = i_N S_N |G_1(R)|$$
where $S_N$ is the surface area, and $|G_1(R)|$ signifies the magnitude of $G_1$ for any argument with magnitude $R$ (a slight abuse of notation).  The result is
$$i_N S_N |G_1(R)| = i_N$$
so $|G_1(R)| = 1/S_N$.  In 3d, this would tell us that the magnitude of the Green's function is $1/4\pi R^2$, which is absolutely true.  Only a couple steps remain to build the vector Green's function.  We said the direction had to be radial, so that the result is
$$G_1(x-y) = \frac{1}{S_N(|x-y|)} \frac{x-y}{|x-y|}$$
where $S_N(R)$ is the "surface area" of a ball with radius $R$ in $N$ dimensions.
Now, to find $G_2$, the Green's function for the Laplacian, invoke radial symmetry to find that
$$G_2(x) = \int_\infty^{|x|} \frac{1}{S_N(r)} \, dr$$
(Referencing to infinity here is a choice, but an incredibly convenient one for making the math work out.)  Quick check:  $S_3(r) = 4\pi r^2$.  The result in 3d is then
$$\int_\infty^{|x|} \frac{1}{4\pi r^2} \, dr = \left. -\frac{1}{4\pi r} \right|_\infty^{|x|} = -\frac{1}{4\pi |x|}$$
This is indeed the Green's function for the Laplacian in 3 dimensions.
A: If $\Omega = \{ \vec{X} : \|\vec{X}\| < a \}$ (ball of radius $a$),
$$
G(\vec{X},\vec{Y}) = 
\begin{cases}
\frac{1}{(2-n)\omega_n} \left(\|\vec{X}-\vec{Y}\|^{2-n} - \frac{1}{a^{2-n}}\|\vec{Y}\|^{2-n}\|\vec{X} - \vec{Z}\|^{2-n}\right) & n > 2\\
\frac{1}{2\pi}\log\left(\frac{\|\vec{X} - \vec{Y}\|}{\|\vec{X} - \vec{Z}\|} \frac{a}{\|\vec{Y}\|}\right) & n = 2
\end{cases}
$$
where $\vec{Z} = \frac{a^2\vec{Y}}{\|\vec{Y}\|^2}$ and $\omega_n$ is the area of the unit sphere.
