# 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?

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Hint: what does the Jacobian look like in $d$-dimensions for spherical coordinates. As well, $f$ only depends on radius. – Alex R. Apr 29 '12 at 21:09
The Jacobian of which function do you mean? – Listing Apr 29 '12 at 22:36

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!

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Thank you for the great answer. I will still wait a bit before I accept this, because I hope there is a solution without generating functions. – Listing Apr 30 '12 at 8:29
@Listing: Glad to help. There are other approaches for the $u$, $t$ integral. I'll look into it later. – user26872 Apr 30 '12 at 12:56
@Listing: I looked into this a bit. As far as I can see, getting the integral without invoking the Gegenbauer polynomials is more painful than just using them. They are meant to deal with spherical problems in higher dimensions. Even showing the integral is not a function of $t$ is nontrivial. If we know this, the integral is easy. Let $t=0$, use the fact that the integral is even, and let $z=u^2$ to find the integral is just the beta function. I am interested to see if anyone here can find a simple way to evaluate the $u$ integral. Have you taken a stab at it? – user26872 May 6 '12 at 5:01
Yes but I could not solve it either, I guess there is some trick one needs to know. – Listing May 6 '12 at 9:11

This is Theorem 9.7 in Analysis by Lieb and Loss. They prove it for d>2 using basic facts about harmonic functions.

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Thank you, what a coincidence that I lately looked up something in that book relating a different question :-). I should probably start to completely read it. – Listing Jun 7 '12 at 21:35