Generalizations of Poisson's Equation I'm currently reading a book in Multivariable Calculus, and there is a section on Applications of Calculus to Physics - Poisson's Equation. It states the following:
We have the 3D version: Let $$u(x) = \iiint_{\Bbb{R}^3}\frac{p(x+y)}{|y|}d^3y$$, where $p$ is a $C^2$ function on $\mathbb{R}^3$, that vanished outside a bounded set. Then $u$ is of class $C^2$ and $\nabla^2u = -4\pi p$.
The 2D analog: Let $u(x) = \int p(x+y)\log{|y|}d^2y$, where $p$ is a $C^2$ function on $\mathbb{R}^2$, that vanished outside a bounded set. Then $u$ is of class $C^2$ and $\nabla^2u = 2\pi p$.
I was wondering how we could generalize this to 4 dimensions - or even n dimensions (if its possible).
My conjecture for the 4D version is: Let $$u(x) = \iiiint_{\Bbb{R}^4}\frac{p(x+y)}{|y|^2}d^4y$$, where $p$ is a $C^2$ function on $\mathbb{R}^4$, that vanished outside a bounded set. Then $u$ is of class $C^2$ and $\nabla^2u = -4\pi^2 p$.
Is this conjecture correct? If not, what would the correct analog be, and if so, how could we prove it?

The $n$-dimensional version (for $n \ge 2$) is that
$$
u(x)=\int_{\mathbb{R}^n} p(x+y) \, R(|y|) \, dy
$$
satisfies $\nabla^2 u = p$ if the function $R(r)$ is such that $1/R'(r)$ equals the $(n-1)$-dimensional surface area of the sphere with radius $r$ in $\mathbb{R}^n$. That is,
$$
R'(r) = \frac{1}{A_n r^{n-1}}
,
$$
where $A_n = 2 \pi^{n/2} \, / \, \Gamma(n/2)$ is the area of the unit sphere in $\mathbb{R}^n$ (see this question, for example).
(I won't prove this here, but the phrase to search for if you want to learn more is the fundamental solution for the Laplace operator.)
For the exceptional case $n=2$, this gives $R'(r)=\frac{1}{2\pi r}$ so that $R(r)=\frac{1}{2\pi} \ln r$ (plus a constant, which is conventionally taken to be zero).
For $n \ge 3$, we get instead
$$
R(r) = \frac{-1}{(n-2) A_n r^{n-2}}
$$
(plus a constant, which is conventionally taken to be zero).
So your conjecture for $n=4$ is correct, since we indeed have $(n-2)A_n = 2 A_4 = 2 \cdot 2 \pi^2 = 4 \pi^2$.
