Newtonian potential of a rotationally-invariant function Lately I read up in the wikipedia article about the Newtonian potential, that for any compactly supported continuous function $f: \mathbb{R}^d \rightarrow \mathbb{R}$ that is rotationally invariant (for simplicity I assume $d>2$) outside of the support of $f$ whe have the equality
$$f*\Gamma(x) =\lambda \Gamma(x),\quad \lambda=\int_{\mathbb{R}^d} f(y)\,dy.$$
where
$$\Gamma(x) = \frac{1}{d(2-d)\omega_d} | x | ^{2-d} $$
and $\omega_d$ is the volume of the unit $d$-ball.
I like the statement and tried to show it but had no success, therefore I looked up the mentioned references in the wikipedia article, but in both books I could not find the theorem. Can anyone provide a proof or better a reference containing a basic proof?
 A: There is probably a slicker proof but here we go.
We write $G$ for $\Gamma$ so as not to confuse it with the gamma function.
We have
$$\begin{eqnarray*}
(f*G)(x) &=& \int dy\, f(y) G(x-y)  \\
&=& \frac{1}{d(2-d)\omega_d}
\int_0^\infty r^{d-1} dr\, f(r) \int d \Omega_d \frac{1}{|x-y|^{d-2}},
\end{eqnarray*}$$
where $r = |y|$.
(Here we have used the rotational invariance of $f$.)
The crux is the integral
$$I = \int d \Omega_d \frac{1}{|x-y|^{d-2}}.$$
Note that
$$\int d \Omega_d = \int_0^\pi d\phi_{d-1} \sin^{d-2}\phi_{d-1}  \int d\Omega_{d-1}.$$
Let $\phi = \phi_{d-1}$.
Align $x$ along the axis $\phi = 0$ so that $x\cdot y = |x y|\cos\phi$.
Then we find
$$I = \frac{\Omega_{d-1}}{|x|^{d-2}} \int_0^\pi d\phi\, \sin^{d-2}\phi
\frac{1}{(1-2t\cos\phi + t^2)^{(d-2)/2}},$$
where $t = |y|/|x| < 1$ and $\Omega_n = 2\pi^{n/2}/\Gamma(n/2)$.
Since $f$ is spherically symmetric, the condition $t<1$ is the condition that the point $x$ is outside of the support of $f$.
Let $u = \cos\phi$.
The integral becomes
$$\frac{\Omega_{d-1}}{|x|^{d-2}} \int_{-1}^1 du\, (1-u^2)^{\alpha-1/2}
\frac{1}{(1-2u t + t^2)^{\alpha}},$$
where $\alpha = (d-2)/2$.
One way to attack this integral is with the Gegenbauer polynomials (also called the ultraspherical harmonics), which are orthogonal polynomials on $[-1,1]$.
They are a natural generalization of the Legendre polynomials.
In fact, they are the Legendre polynomials for $\alpha = 1/2$.
The factor $1/(1-2u t + t^2)^{\alpha}$ is the generating function for the Gegenbauer polynomials.
We find
$$I = \frac{\Omega_{d-1}}{|x|^{d-2}} \int_{-1}^1 du\, (1-u^2)^{\alpha-1/2}
\sum_{n=0}^\infty C_n^{(\alpha)}(u)t^n.$$
But the polynomials are orthogonal with this measure and $C_0^{(\alpha)}(u) = 1$. Therefore we find
$$I = \frac{\Omega_{d-1}}{|x|^{d-2}} \frac{\pi 2^{1-2\alpha} \Gamma(2\alpha)}{\alpha \Gamma(\alpha)^2}.$$
This result arises from the normalization of $C_0^{(\alpha)}$.
Therefore,
$$\begin{eqnarray*}
(f*G)(x) &=& \frac{I}{d(2-d)\omega_d \Omega_d}
\times \Omega_d \int_0^\infty r^{d-1} dr\, f(r).
\end{eqnarray*}$$
But $I/(d(2-d)\omega_d \Omega_d) = G(x)$ and
$\Omega_d \int r^{d-1} dr\, f(r) = \int dy f(y)$, so
$$\begin{eqnarray*}
(f*G)(x) &=& \lambda \Gamma(x).
\end{eqnarray*}$$
Thanks for the interesting question!
A: This is Theorem 9.7 in Analysis by Lieb and Loss. They prove it for d>2 using basic facts about harmonic functions.
