# On the solution of a Riesz equilibrium problem
and integral identities for special functions

Djalil Chafaï  CEREMADE, Université Paris-Dauphine, Université PSL, CNRS,
75016 Paris, France  mailto:djalil(at)chafai.net http://djalil.chafai.net/ ,
Edward B. Saff  Center for Constructive Approximation, Vanderbilt University,
Nashville, TN 37240, USA<EMAIL_ADDRESS>https://my.vanderbilt.edu/edsaff/ and  Robert S. Womersley  School of
Mathematics and Statistics, University of New South Wales, Sydney NSW 2052,
Australia<EMAIL_ADDRESS>https://web.maths.unsw.edu.au/~rsw/

(Date: Summer 2021, compiled )

###### Abstract.

The aim of this note is to provide a quadratic external field extension of a
classical result of Marcel Riesz for the equilibrium measure on a ball with
respect to Riesz $s$-kernels, including the logarithmic kernel, in arbitrary
dimensions. The equilibrium measure is a radial arcsine distribution. As a
corollary, we obtain new integral identities involving special functions such
as elliptic integrals and more generally hypergeometric functions. These
identities are not found in the existing tables for series and integrals, and
are not recognized by advanced mathematical software. Among other ingredients,
our proofs involve the Euler – Lagrange variational characterization, the Funk
– Hecke formula, and the Weyl lemma for the regularity of elliptic equations.

###### Key words and phrases:

Potential theory; Equilibrium measure; Log-gas; Riesz kernel; Special
function; Hypergeometric function; Elliptic integral; Funk–Hecke formula;
Landen transform; Projective geometry; Möbius transform

###### 2000 Mathematics Subject Classification:

31A10; 31B10; 44A20; 33C20; 33C75

##  1\. Introduction and main results

The goal of this note is to provide a quadratic external field extension
(Theorem 1.5 below) of a classical result of Marcel Riesz (Theorem 1.1 below)
for the equilibrium measure on a ball with respect to Riesz $s$-kernels,
including the logarithmic kernel, in arbitrary dimensions. The equilibrium
measure is a radial arcsine distribution. As a corollary, we obtain new
integral identities involving special functions such as elliptic integrals and
more generally hypergeometric functions, see, for example, Corollaries 1.3,
1.6, and 1.7 below. These identities are not found in the existing tables for
series and integrals, and are not recognized by advanced mathematical
software.

Before we present our results and identities, we recall some basic notions
from potential theory. Throughout this note, we denote by $d$ the Euclidean
dimension, which is always a positive integer, and by $s\in(-2,+\infty)$ the
Riesz parameter. For all $x\in\mathbb{R}^{d}$, $x\neq 0$, the $s$-Riesz kernel
is defined by

$K_{s}(x):=\begin{cases}\mathrm{sign}(s)\left|x\right|^{-s}&\text{if $-2<s<0$
or $s>0$}\\\ -\log\left|x\right|&\text{if $s=0$}\end{cases},$ (1.1)

where $\left|x\right|^{2}:=x_{1}^{2}+\cdots+x_{d}^{2}$. It is the Coulomb or
Newton kernel if $s=d-2$. For $R>0$, let $\mathcal{M}_{1}(B_{R})$ be the set
of probability measures supported in the closed ball

$B_{R}:=\\{x\in\mathbb{R}^{d}:|x|\leq R\\}.$

Let $V:B_{R}\mapsto(-\infty,+\infty]$ be a lower semicontinuous function,
which will play the role of an external field, taken for simplicity in this
note either of the form $V=0$ or of the form
$V(\cdot)=\gamma\left|\cdot\right|^{2}$, $\gamma>0$. (In a forthcoming work
[6] we deal with more general external fields $V$.) The energy of
$\mu\in\mathcal{M}_{1}(B_{R})$ with external field $V$ is defined by

$\mathrm{I}(\mu):=\iint_{\mathbb{R}^{d}\times\mathbb{R}^{d}}(K_{s}(x-y)+V(x)+V(y))\mu(\mathrm{d}x)\mu(\mathrm{d}y)\in(-\infty,+\infty].$
(1.2)

For $s\in(-2,d)$, with our choices of $V$, the integrand in the double
integral in (1.2) is bounded below, $\mathrm{I}$ is strictly convex111In other
words $K_{s}$ is conditionally strictly positive in the sense of Bochner, see
for instance [2, Section 4.4]. on $\mathcal{M}_{1}(B_{R})$, lower
semicontinuous with compact level sets222We follow the probability theory
standard and equip the convex set $\mathcal{M}_{1}(B_{R})$ with the topology
of weak convergence with respect to continuous and bounded test functions; in
other words the weak-$*$ convergence.. It has a unique global minimizer called
the “equilibrium measure” $\mu_{\mathrm{eq}}\in\mathcal{M}_{1}(B_{R})$; in
other words,

$\mathrm{I}(\mu_{\mathrm{eq}})=\min_{\mu\in\mathcal{M}_{1}(B_{R})}\mathrm{I}(\mu)\quad\text{and}\quad\mathrm{I}(\mu)>\mathrm{I}(\mu_{\mathrm{eq}})\text{
for all $\mu\neq\mu_{\mathrm{eq}}$.}$ (1.3)

We refer to [13] and [2] for more details. If $s<0$, then $K_{s}$ is not
singular and we could have $\mathrm{I}(\mu)<\infty$ for a
$\mu\in\mathcal{M}_{1}(B_{R})$ having Dirac masses; in particular
$\mu_{\mathrm{eq}}$ could conceivably have Dirac masses. In contrast, if
$s\geq 0$ then $K_{s}$ is singular and $\mathrm{I}(\mu)=+\infty$ whenever
$\mu$ has Dirac masses; consequently $\mu_{\mathrm{eq}}$ cannot have such
masses.

We first recall a classical result of M. Riesz for the equilibrium measure
with constant external field. For its statement, we denote by
$\mathbb{S}^{d-1}:=\\{x\in\mathbb{R}^{d}:|x|=1\\}$ the unit sphere and by
$\sigma_{\mathbb{S}^{d-1}}$ the uniform probability measure on it. We recall
that the surface area of $\mathbb{S}^{d-1}$ is
$|\mathbb{S}^{d-1}|=2\pi^{d/2}/\Gamma(d/2)$.

######  Theorem 1.1 (Riesz theorem [21]).

Suppose that $d\geq 2$, $R=1$, and $V=0$.

  * •

If $-2<s\leq d-2$, then $\mu_{\mathrm{eq}}=\sigma_{\mathbb{S}^{d-1}}$,

  * •

If $d-2<s<d$, then $\mu_{\mathrm{eq}}$ is the probability measure

$\mu_{\mathrm{eq}}(\mathrm{d}x)=\frac{\Gamma(1+\frac{s}{2})}{\pi^{\frac{d}{2}}\Gamma(1+\frac{s-d}{2})}\frac{\mathbf{1}_{|x|\leq
1}}{(1-|x|^{2})^{\frac{d-s}{2}}}\mathrm{d}x=2\frac{\Gamma(1+\frac{s}{2})}{\Gamma(1+\frac{s-d}{2})\Gamma(\frac{d}{2})}\frac{r^{d-1}\mathbf{1}_{r\leq
1}}{(1-r^{2})^{\frac{d-s}{2}}}\,\mathrm{d}r\mathrm{d}\sigma_{\mathbb{S}^{d-1}},$
(1.4)

where $\mathrm{d}x$ and $\mathrm{d}r$ denote the Lebesgue measures on
$\mathbb{R}^{d}$ and on $[0,+\infty)$ respectively.
Moreover, the equilibrium potential $U^{\mu_{\mathrm{eq}}}$ satisfies, for
$x\in B_{1}$,

$U^{\mu_{\mathrm{eq}}}(x):=(K_{s}*\mu_{\mathrm{eq}})(x)=\int_{\mathbb{R}^{d}}K_{s}(x-y)\mu_{\mathrm{eq}}(\mathrm{d}y)=\mathrm{I}(\mu_{\mathrm{eq}})=\frac{\Gamma(1+\frac{s}{2})\Gamma(\frac{d-s}{2})}{\Gamma(\frac{d}{2})}.$
(1.5)

Theorem 1.1, stated for the unit ball, has a natural extension to balls of
arbitrary radius $R>0$.

The case $0\leq d-2<s<d$ in Theorem 1.1 is a direct consequence of the
following formula.

######  Lemma 1.2 (Riesz formula [21]).

If $0\leq d-2<s<d$, then for $x\in B_{1}$,

$\int_{\mathbb{R}^{d}}\frac{|x-y|^{-s}}{(1-|y|^{2})^{\frac{d-s}{2}}}\mathbf{1}_{|y|\leq
1}\mathrm{d}y=\frac{\pi^{\frac{d}{2}+1}}{\Gamma(\frac{d}{2})\sin(\frac{\pi}{2}(d-s))}.$
(1.6)

The proof of Theorem 1.1 and Lemma 1.2 can be found, together with some
geometric aspects, in the works of M. Riesz [20, p. 438–439] and [21, §16, Eq.
(1)], where it is mentioned that the cases $d=1,2,3$ were already considered
by Pólya and Szegő in [19]. It can also be found in the book [13, §II.3.13 p.
163–164 and Appendix p. 399 – 400], and is stated in [2, Eq. (4.6.13)]. The
proof sketched by Riesz, with a bit more detail by Landkof, involves first a
geometric inversion transforming the integral on the ball into an integral on
its complement, and second a trigonometric substitution which has a geometric
interpretation, both steps being inspired by the analytic-geometric techniques
used classically for elliptic integrals since the eighteenth century. For the
reader’s convenience, a detailed proof of Lemma 1.2 is given in Appendix B.

Our first result, Corollary 1.3, is a simple consequence of Theorem 1.1. It
relates an equilibrium measure of potential theory with an integral identity
for special functions (here a ${}_{2}F_{1}$ hypergeometric function). Before
stating it, let us recall the _Newton binomial series_ that states that

$\frac{1}{(1-z)^{\alpha}}=\sum_{n=0}^{\infty}(\alpha)_{n}\frac{z^{n}}{n!},\quad\alpha,z\in\mathbb{C},\
|z|<1,$ (1.7)

where $(\alpha)_{n}:=\alpha(\alpha+1)\cdots(\alpha+n-1)$ is the _Pochhammer
symbol_ for the rising factorial, with the convention $(\alpha)_{0}:=1$ if
$\alpha\neq 0$. If $\Re(\alpha)>0$, then
$(\alpha)_{n}=\Gamma(\alpha+n)/\Gamma(\alpha)$. More generally, the
_hypergeometric function_ with parameters
$(a_{1},\ldots,a_{p})\in\mathbb{C}^{p}$ and
$(b_{1},\ldots,b_{q})\in\mathbb{C}^{q}$, at $z\in\mathbb{C}$, $|z|<1$, is
given (when it makes sense) by the series

${}_{p}F_{q}(a_{1},\ldots,a_{p};b_{1},\ldots,b_{q};z):=\sum_{n=0}^{\infty}\frac{(a_{1})_{n}\cdots(a_{p})_{n}}{(b_{1})_{n}\cdots(b_{q})_{n}}\frac{z^{n}}{n!}.$
(1.8)

Special choices of parameters allow us to recover many special functions, for
instance we have ${}_{2}F_{1}(\alpha,\beta;\beta;z)=(1-z)^{-\alpha}$,
${}_{2}F_{1}(1,1;2;-z)=\frac{\log(1+z)}{z}$, and
${}_{2}F_{1}(\frac{1}{2},\frac{1}{2};\frac{3}{2};z^{2})=\frac{\arcsin(z)}{z}$.
Actually one of the main historical motivations for the introduction and study
of hypergeometric functions is the unification of as many as possible special
functions via series expansions. For instance the complete _elliptic integral_
of first and second kind $K$ and $E$ satisfy, for $z\in[0,1]$,

$\displaystyle K(z)$
$\displaystyle:=\int_{0}^{\frac{\pi}{2}}\frac{\mathrm{d}\theta}{\sqrt{1-z\sin^{2}(\theta)}}=\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-zt^{2}}\sqrt{1-t^{2}}}=\frac{\pi}{2}\,{}_{2}F_{1}\Bigr{(}\frac{1}{2},\frac{1}{2};1;z\Bigr{)}$
(1.9) and $\displaystyle E(z)$
$\displaystyle:=\int_{0}^{\frac{\pi}{2}}\sqrt{1-z\sin^{2}(\theta)}\mathrm{d}\theta=\int_{0}^{1}\frac{\sqrt{1-zt^{2}}}{\sqrt{1-t^{2}}}\,\mathrm{d}t=\frac{\pi}{2}\,{}_{2}F_{1}\Bigr{(}-\frac{1}{2},\frac{1}{2};1;z\Bigr{)}.$
(1.10)

They can be extended to the complex plane, with a branch cut discontinuity
running from $1$ to $\infty$. For more basic facts about these functions, we
refer for instance to the classical books [8, 4]. Here we only remark that
$K,E\geq 0$ on the interval $[0,1]$, $K(0)=E(0)=\frac{\pi}{2}$, $K(1)=\infty$,
and $E(1)=1$.

The following identity derived in Section 2.1 is an unnoticed easy consequence
of Theorem 1.1.

######  Corollary 1.3 (Special function identity).

For all $d\geq 2$, $d-2<s<d$, $\lambda\in[0,1]$,

$\int_{0}^{1}{}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};\frac{4r^{2}\lambda^{2}}{(\lambda^{2}+r^{2})^{2}}\right)\frac{r^{d-1}}{(\lambda^{2}+r^{2})^{\frac{s}{2}}(1-r^{2})^{\frac{d-s}{2}}}\,\mathrm{d}r=\frac{\pi}{2\sin(\frac{\pi}{2}(d-s))}.$
(1.11)

######  Remark 1.4 (Landen transform).

Since $\frac{4r^{2}\lambda^{2}}{(\lambda^{2}+r^{2})^{2}}\leq 1$ with equality
if and only if $\lambda=r$, so that the radius of convergence $1$ of the
${}_{2}F_{1}$ in (1.11) is reached in the interior of the interval of
integration. Note also
$\frac{4r^{2}\lambda^{2}}{(\lambda^{2}+r^{2})^{2}}=\frac{4x}{(1+x)^{2}}$,
where $x:=(\frac{\lambda}{r})^{2}$ and that the map
$x\mapsto\frac{4x}{(1+x)^{2}}$ is the _Landen transform_ which often appears
in the context of elliptic integrals.

Here are some special cases for the hypergeometric function in (1.11) worth
noting:

  * •

for $d=2$ and $s=1$ we get
${}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};z\right)={}_{2}F_{1}\left(\frac{1}{4},\frac{3}{4};1;z\right)=\frac{2K\left(\frac{2\sqrt{z}}{\sqrt{z}+1}\right)}{\pi\sqrt{\sqrt{z}+1}}$.

  * •

for $d=3$ and $s=2$ we get
${}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};z\right)={}_{2}F_{1}\left(\frac{1}{2},1;\frac{3}{2};z\right)=\frac{\tanh^{-1}(\sqrt{z})}{\sqrt{z}}$.

  * •

for $d=5$ and $s=4$ we get
${}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};z\right)={}_{2}F_{1}\left(1,\frac{3}{2};\frac{5}{2};z\right)=\frac{3\left(\sqrt{z}\tanh^{-1}\left(\sqrt{z}\right)-z\right)}{z^{2}}$.

Our main potential theoretic result is the following external field version of
Theorem 1.1.

######  Theorem 1.5 (Main result).

Suppose that $d\geq 2$ and $s=d-3$, namely

$(d,s)\in\\{(2,-1),(3,0),(4,1),\ldots\\}.$

If $V=\gamma\left|\cdot\right|^{2}$, $\gamma>0$, and

$R=\left(\frac{\max(|d-3|,1)\sqrt{\pi}}{4\gamma}\frac{\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{\Gamma\bigr{(}\frac{d+2}{2}\bigr{)}}\right)^{\frac{1}{d-1}},$
(1.12)

then the equilibrium measure $\mu_{\mathrm{eq}}$ for the minimum energy
problem (1.3) on $B_{R}$ with kernel $K_{s}$ and external field $V$ is the
“radial arcsine distribution”,

$\mu_{\mathrm{eq}}(\mathrm{d}x)=\frac{\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{\pi^{\frac{d+1}{2}}R^{d-1}}\frac{\mathbf{1}_{r\leq
R}}{\sqrt{R^{2}-|x|^{2}}}\mathrm{d}x=2\frac{\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{\sqrt{\pi}\Gamma(\frac{d}{2})R^{d-1}}\frac{r^{d-1}\mathbf{1}_{r\leq
R}}{\sqrt{R^{2}-r^{2}}}\,\mathrm{d}r\,\mathrm{d}\sigma_{\mathbb{S}^{d-1}},$
(1.13)

where $\mathrm{d}x$ and $\mathrm{d}r$ are the Lebesgue measures on
$\mathbb{R}^{d}$ and on $[0,\infty)$ respectively.

Moreover, $\mu_{\mathrm{eq}}$ is also the equilibrium measure when $s=d-1$,
$V=0$, and $R$ is given by (1.12), cf. Theorem 1.1.

Theorem 1.5 is proved in Section 2.2.

The non-intuitive choice for $R$ in Theorem 1.5 arose from considering the
equilibrium problem in the full space $\mathbb{R}^{d}$ with same kernel
$K_{s}$, $s=d-3$, and external field $V=\gamma\left|\cdot\right|^{2}$,
$\gamma>0$. We conjecture that the radial arcsine distribution that appears in
Theorem 1.5 also solves this full space problem for all dimensions $d$. The
results of the present note imply that this conjecture is true for
$d\in\\{2,3,4,5\\}$. Several extensions of Theorem 1.5 for more general $V$
and $(d,s)$, and their extension to external field problems on
$\mathbb{R}^{d}$, are considered in a forthcoming article [6].

$d=2,s=-1$ | $d=3,s=0$ | $d=4,s=1$
---|---|---
$\displaystyle\frac{\pi}{8}\approx 0.392699$ | $\displaystyle\frac{1}{\sqrt{3}}\approx 0.57735$ | $\displaystyle\frac{1}{2}\sqrt[3]{\frac{3\pi}{4}}\approx 0.665335$
Table 1. Values of the critical radius $R$ for low dimensions in Theorem 1.5,
with $\gamma=1$.

Table 1 gives values of the ball radius $R$ in (1.12) for $\gamma=1$ and
$d=2,3,4$. For $\gamma=1$, $d\geq 2$, the function $d\mapsto R$ achieves its
minimum $\approx 0.392699$ at $d=2$ ($s=-1$) and its maximum $\approx 1.04747$
at $d=16$ ($s=13$), and these values are the unique critical points. Regarding
high dimensional behavior, $\lim_{d=s+3\to\infty}R=1$.

As we shall verify, Theorem 1.5 yields the following integral formulas.

######  Corollary 1.6 (Integral formula).

Let $d\in\\{2,3,4,\ldots\\}$ and $\lambda\in[0,1]$. Then:

$\int_{0}^{1}S_{d-3}\Bigr{(}\frac{4\lambda
r}{(\lambda+r)^{2}}\Bigr{)}\frac{(\lambda+r)^{3-d}r^{d-1}\,\mathrm{d}r}{\sqrt{1-r^{2}}}=\frac{\pi^{\frac{3}{2}}\Gamma(\frac{d-1}{2})}{2^{d+1}\Gamma(\frac{d}{2})}\Bigr{(}\Bigr{(}\frac{3}{d}-1\Bigr{)}\lambda^{2}+1\Bigr{)}$
(1.14)

where

$S_{s}(z):=\int_{0}^{\frac{\pi}{2}}\\!\frac{\sin^{s+1}(2\alpha)\mathrm{d}\alpha}{2^{s+1}(1-z\sin^{2}(\alpha))^{\frac{s}{2}}}=\int_{0}^{1}\\!\frac{t^{s+1}(1-t^{2})^{\frac{s}{2}}\mathrm{d}t}{(1-zt^{2})^{\frac{s}{2}}}=\frac{\Gamma(\frac{s+2}{2})^{2}}{2\Gamma(s+2)}\,{}_{2}F_{1}\Bigr{(}\frac{s+2}{2},\frac{s}{2};s+2;z\Bigr{)}.$
(1.15)

Equivalently, for $d\in\\{2,3,4,\ldots\\}$ and $\lambda\in[0,1]$,

$\int_{0}^{1}{}_{2}F_{1}\Bigr{(}\frac{d-1}{2},\frac{d-3}{2};d-1;\frac{4\lambda
r}{(\lambda+r)^{2}}\Bigr{)}\frac{(\lambda+r)^{3-d}r^{d-1}}{\sqrt{1-r^{2}}}\,\mathrm{d}r=\frac{\pi}{4}\Bigr{(}\Bigr{(}\frac{3}{d}-1\Bigr{)}\lambda^{2}+1\Bigr{)}.$
(1.16)

Corollary 1.6 is proved in Section 2.3. The formula 1.16 comes from the Euler
– Lagrange equations (Frostman conditions) related to Theorem 1.5. Numerical
experiments suggest that (1.14) and (1.16) remain valid whenever the parameter
$d$ is real.

It is tempting to regard $S_{s}$ as a _special function_ in its own right.
When $(d,s)=(2,-1)$, it becomes the complete elliptic integral of the second
kind, namely $S_{-1}=E$, and (1.14) becomes (1.19) below.

######  Corollary 1.7 (More integral formulas).

For all $\lambda\in[0,1]$,

$\displaystyle\int_{0}^{1}\bigr{(}(\lambda+r)^{2}\log(\lambda+r)-(\lambda-r)^{2}\log|\lambda-r|\bigr{)}\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}$
$\displaystyle=\pi\Bigr{(}\frac{\lambda^{3}}{3}+(1-\log 2)\lambda\Bigr{)},$
(1.17)
$\displaystyle\int_{0}^{1}((\lambda+r)\log(\lambda+r)-(\lambda-r)\log|\lambda-r|)\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}$
$\displaystyle=\frac{\pi}{2}\Bigr{(}\lambda^{2}+\frac{1}{2}-\log 2\Bigr{)},$
(1.18) $\displaystyle\int_{0}^{1}E\left(\frac{4\lambda
r}{(\lambda+r)^{2}}\right)\frac{(\lambda+r)r\,\mathrm{d}r}{\sqrt{1-r^{2}}}$
$\displaystyle=\frac{\pi^{2}}{8}\Bigr{(}\frac{\lambda^{2}}{2}+1\Bigr{)},$
(1.19) $\displaystyle\int_{0}^{1}K\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)\frac{(\lambda-r)r\,\mathrm{d}r}{\sqrt{1-r^{2}}}$
$\displaystyle=\frac{\pi^{2}}{8}\Bigr{(}\frac{3\lambda^{2}}{2}-1\Bigr{)},$
(1.20)
$\displaystyle\Re\int_{0}^{1}E\Bigr{(}\frac{r^{2}}{\lambda^{2}}\Bigr{)}\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}$
$\displaystyle=\frac{\pi^{2}}{8}\lambda,$ (1.21)
$\displaystyle\Re\int_{0}^{1}K\Bigr{(}\frac{r^{2}}{\lambda^{2}}\Bigr{)}\frac{(\lambda^{2}-r^{2})r\,\mathrm{d}r}{\sqrt{1-r^{2}}}$
$\displaystyle=\frac{\pi^{2}}{8}\Bigr{(}\frac{3\lambda^{3}}{2}-\lambda\Bigr{)},$
(1.22)

where $E$ and $K$ are the special functions defined in (1.10) and (1.9).

Corollary 1.7 is proved in Section 2.4 by applying further transformations to
the Euler – Lagrange equations of Theorem 1.5 in the special cases
$(d,s)\in\\{(2,-1),(3,0)\\}$.

To the best of our knowledge, the formulas provided by Corollaries 1.3, 1.6,
and 1.7 are not found in the existing catalogs of identities and tables for
series and integrals such as [7, 18, 3], and are not recognized by advanced
software such as Maplesoft Maple and Wolfram Mathematica. However it is worth
noting that these softwares do recognize the first parts of (1.17) and (1.18)
in terms of ${}_{3}F_{2}$ functions:

$\int_{0}^{1}(\lambda+r)^{2}\log(\lambda+r)\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}\\\
=\frac{\pi\,_{3}F_{2}\left(\frac{1}{2},1,1;2,3;\frac{1}{\lambda^{2}}\right)}{16\lambda}-\frac{2\,_{3}F_{2}\left(1,1,\frac{3}{2};\frac{5}{2},\frac{7}{2};\frac{1}{\lambda^{2}}\right)}{45\lambda^{2}}+\frac{\pi\lambda}{4}+\frac{3\lambda(2\lambda+\pi)+4}{6}\log(\lambda)+1$
(1.23)

and

$\int_{0}^{1}(\lambda+r)\log(\lambda+r)\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}\\\
=\frac{32\lambda\,_{3}F_{2}\left(\frac{1}{2},1,1;\frac{3}{2},\frac{5}{2};\frac{1}{\lambda^{2}}\right)-3\pi\,_{3}F_{2}\left(1,1,\frac{3}{2};2,3;\frac{1}{\lambda^{2}}\right)+24\lambda^{2}((4\lambda+\pi)\log(\lambda)+\pi)}{96\lambda^{2}}.$
(1.24)

##  2\. Proofs

###  2.1. Proof of Corollary 1.3

Let us consider the settings of Theorem 1.1 in the case $d-2<s<d$. Then, using
the Funk – Hecke formula (see (A.3) in the Appendix), we get, for all
$x\in\mathbb{R}^{d}$, $x\neq 0$, denoting $\lambda:=|x|$,

$\displaystyle U^{\mu_{\mathrm{eq}}}(x)$ $\displaystyle=C\int_{|y|\leq
1}\frac{|x-y|^{-s}}{(1-|y|^{2})^{\frac{d-s}{2}}}\mathrm{d}y$
$\displaystyle=C\int_{0}^{1}\Bigr{(}\int_{\mathbb{S}^{d-1}}\frac{(\lambda^{2}+r^{2}-2\lambda
r\frac{x}{|x|}\cdot
u)^{-\frac{s}{2}}}{(1-r^{2})^{\frac{d-s}{2}}}r^{d-1}\mathrm{d}u\Bigr{)}\,\mathrm{d}r$
$\displaystyle=C\tau_{d-1}|\mathbb{S}^{d-1}|\int_{0}^{1}\Bigr{(}\int_{-1}^{1}\frac{(1-t^{2})^{\frac{d-3}{2}}}{(\lambda^{2}+r^{2}-2\lambda
rt)^{\frac{s}{2}}}\mathrm{d}t\Bigr{)}\frac{r^{d-1}}{(1-r^{2})^{\frac{d-s}{2}}}\,\mathrm{d}r.$

Now, since $\frac{2\lambda r}{(\lambda^{2}+r^{2})}\in[0,1/2]$, using the
Newton binomial series (1.7),

$\displaystyle\int_{-1}^{1}\frac{(1-t^{2})^{\frac{d-3}{2}}}{(\lambda^{2}+r^{2}-2\lambda
rt)^{\frac{s}{2}}}\mathrm{d}t$
$\displaystyle=\frac{1}{(\lambda^{2}+r^{2})^{\frac{s}{2}}}\int_{-1}^{1}\frac{(1-t^{2})^{\frac{d-3}{2}}\theta}{(1-\frac{2\lambda
r}{\lambda^{2}+r^{2}}t)^{\frac{s}{2}}}\mathrm{d}t$
$\displaystyle=\frac{1}{(\lambda^{2}+r^{2})^{\frac{s}{2}}}\sum_{n=0}^{\infty}\frac{(\frac{s}{2})_{2n}}{(2n)!}\Bigr{(}\frac{4r^{2}\lambda^{2}}{(\lambda^{2}+r^{2})^{2}}\Bigr{)}^{n}\int_{-1}^{1}(1-t^{2})^{\frac{d-3}{2}}t^{2n}\mathrm{d}t$
$\displaystyle=\frac{1}{(\lambda^{2}+r^{2})^{\frac{s}{2}}}\sum_{n=0}^{\infty}\frac{(\frac{s}{2})_{2n}}{(2n)!}\Bigr{(}\frac{4r^{2}\lambda^{2}}{(\lambda^{2}+r^{2})^{2}}\Bigr{)}^{n}\frac{\Gamma(\frac{d-1}{2})\Gamma(n+\frac{1}{2})}{\Gamma(\frac{d}{2}+n)}$
$\displaystyle=\frac{\sqrt{\pi}\Gamma(\frac{d-1}{2})}{\Gamma(\frac{d}{2})}\frac{\,{}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};\frac{4r^{2}\lambda^{2}}{\left(r^{2}+\lambda^{2}\right)^{2}}\right)}{(\lambda^{2}+r^{2})^{\frac{s}{2}}},$

where we have use the identities, related to the Legendre duplication formula
(A.2),

$\Bigr{(}\frac{s}{2}\Bigr{)}_{2n}=2^{2n}\Bigr{(}\frac{s}{4}\Bigr{)}_{n}\Bigr{(}\frac{s}{4}+\frac{1}{2}\Bigr{)}_{n}\quad\text{and}\quad
2^{2n}\Bigr{(}\frac{1}{2}\Bigr{)}_{n}=\frac{(2n)!}{n!}.$

Hence, we get, for all $x\in\mathbb{R}^{d}$, $x\neq 0$,

$\displaystyle U^{\mu_{\mathrm{eq}}}(x)$
$\displaystyle=\frac{2\pi^{\frac{d}{2}}}{\Gamma(\frac{d}{2})}\int_{0}^{1}\,{}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};\frac{4r^{2}\lambda^{2}}{(\lambda^{2}+r^{2})^{2}}\right)\frac{r^{d-1}}{(\lambda^{2}+r^{2})^{\frac{s}{2}}(1-r^{2})^{\frac{d-s}{2}}}\,\mathrm{d}r.$

Now, the Euler – Lagrange equations (A.5) and the continuity of
$U^{\mu_{\mathrm{eq}}}$ give that this quantity is constant on $B_{1}$. To
compute the value of the constant, we let $\lambda=|x|\to 0$ and note that

$\int_{0}^{1}\,{}_{2}F_{1}\left(\frac{s}{4},\frac{s+2}{4};\frac{d}{2};0\right)\frac{r^{d-1}}{(0+r^{2})^{\frac{s}{2}}(1-r^{2})^{\frac{d-s}{2}}}\,\mathrm{d}r=\int_{0}^{1}r^{d-1-s}(1-r^{2})^{\frac{s-d}{2}}\,\mathrm{d}r=\frac{\pi}{2\sin(\frac{2}{\pi}(d-s))},$

where the final step comes from the Euler reflection formula (A.1). We could
also have directly used (1.6). ∎

###  2.2. Proof of Theorem 1.5

We split the proof into several subsections.

####  2.2.1. Computation of critical radius and candidate equilibrium measure

From the uniqueness property we know that the equilibrium measure
$\mu_{\mathrm{eq}}$ is radially symmetric. Let us make a succession of
assumptions to extract a candidate for $\mu_{\mathrm{eq}}$, and we will then
check that it indeed satisfies the Euler – Lagrange (Frostman) conditions
(A.5). We start by observing from (A.5) that, for all $x\in
S_{*}:=\mathrm{supp}(\mu_{\mathrm{eq}})$,

$\int K_{s}(x-y)\mu_{\mathrm{eq}}(\mathrm{d}y)+\gamma|x|^{2}=c.$ (2.1)

Applying the Laplacian operator to (2.1) and assuming it can be taken inside
the integral, we get from (2.1) that for all $x$ in the interior of $S_{*}$,

$-c_{d,s}\int K_{s+2}(x-y)\mu_{\mathrm{eq}}(\mathrm{d}y)+2\gamma d=0,$ (2.2)

where we used that $\Delta K_{s}(x)=-c_{d,s}K_{s+2}(x)$, $x\neq 0$, with

$c_{d,s}=\begin{cases}|s|(d-2-s)&\text{if $s\neq 0$}\\\ d-2&\text{if
$s=0$}\end{cases}.$

Note that in our case $s=d-3$, so

$c_{d,s}=c_{d,d-3}=\begin{cases}|d-3|&\text{if $d\neq 3$}\\\ 1&\text{if
$d=3$}\end{cases}.$

Next suppose that $S_{*}=B_{R}$ for some $R>0$. Let $\nu_{R}$ be the
equilibrium measure for the minimum energy problem on $B_{R}$ with kernel
$K_{s+2}=\left|\cdot\right|^{-(s+2)}$ and $V=0$. Observing that $\nu_{R}$ is
the dilation by a factor of $R$ of $\nu_{1}$, we see from Theorem 1.1 that
$\nu_{R}$ is the “radial arcsine distribution”, in other words the measure

$\nu_{R}(\mathrm{d}x)=\frac{C_{d,R}}{\sqrt{R^{2}-|x|^{2}}}\mathbf{1}_{|x|\leq
R}\;\mathrm{d}x,\quad\text{where}\quad
C_{d,R}=\frac{2\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{|\mathbb{S}^{d-1}|\sqrt{\pi}\Gamma\bigr{(}\frac{d}{2}\bigr{)}R^{d-1}}=\frac{\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{R^{d-1}\pi^{\frac{d+1}{2}}}.$
(2.3)

In particular, the support of $\nu_{R}$ is all of $B_{R}$. Next, by definition
of $\nu_{R}$, the associated Euler – Lagrange equations state that, for some
constant $W_{R}$,

$\int K_{s+2}(x-y)\nu_{R}(\mathrm{d}x)=W_{R},\quad\mbox{for all }y\in B_{R}.$
(2.4)

As $d=s+3$ we obtain (using $y=0\in B_{R}$)

$W_{R}=\frac{W_{1}}{R^{s+2}}=\frac{W_{1}}{R^{d-1}}\quad\text{and}\quad
W_{1}=C_{d,1}|\mathbb{S}^{d-1}|\int_{0}^{1}\frac{r^{-(s+2)}r^{d-1}}{\sqrt{1-r^{2}}}\,\mathrm{d}r=\sqrt{\pi}\frac{\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{\Gamma\bigr{(}\frac{d}{2}\bigr{)}}.$
(2.5)

To derive the value of $R$ we first integrate (2.2) with respect to
$\nu_{R}(\mathrm{d}x)$ and swap the integrals, assuming that this is legal,
giving

$-c_{d,s}\int\Bigr{(}\int
K_{s+2}(x-y)\nu_{R}(\mathrm{d}x)\Bigr{)}\mu_{\mathrm{eq}}(\mathrm{d}y)+2\gamma
d=0.$ (2.6)

Then, using (2.4) and (2.5) in (2.6), we get

$\frac{c_{d,s}\sqrt{\pi}\Gamma\bigr{(}\frac{d+1}{2}\bigr{)}}{\Gamma\bigr{(}\frac{d}{2}\bigr{)}}R^{-s-2}=2\gamma
d.$

Finally, from the formula $z\Gamma(z)=\Gamma(z+1)$ with $z=d/2$, we derive the
desired formula for $R$, namely

$R=\left(\frac{(|d-3|\vee
1)\sqrt{\pi}\Gamma(\frac{d+1}{2})}{4\gamma\Gamma(\frac{d+2}{2})}\right)^{\frac{1}{d-1}}=\left(\frac{(|s|\vee
1)\sqrt{\pi}\Gamma(\frac{s+4}{2})}{4\gamma\Gamma(\frac{s+5}{2})}\right)^{\frac{1}{s+2}}.$
(2.7)

####  2.2.2. Euler – Lagrange conditions

We next show that $\nu_{R}$ in (2.3) with $R$ as in (2.7) satisfies the Euler
– Lagrange conditions (A.5) with kernel $K_{s}$ and
$V=\gamma\left|\cdot\right|^{2}$. For this purpose set

$\Phi:=K_{s}*\nu_{R}+\gamma\left|\cdot\right|^{2},$ (2.8)

and notice that $\Phi$ is radially symmetric. The measure $\nu_{R}$ and the
radius $R$ have been chosen in the preceding subsection precisely in such a
way that on $\mathrm{int}(B_{R})$,

$\Delta\Phi=-c_{s,s+3}K_{s+2}*\nu_{R}+2d\gamma=0;$ (2.9)

see Lemma A.1 for $K_{s}*\nu_{R}\in
L^{1}_{\mathrm{loc}}(\mathbb{R}^{d},\mathrm{d}x)$ and the swap of Laplacian
and Riesz potential integral.

At this step, let us remark that when $d<6$, Lemma A.1 _(iv)_ gives that
$K_{s}*\nu_{R}$ is continuous on $\mathbb{R}^{d}$ since
$\frac{\mathrm{d}\nu_{R}}{\mathrm{d}x}\in L^{p}(\mathbb{R}^{d},\mathrm{d}x)$,
$2>p>d/(d-s)=d/3$ (recall that $s=d-3$). Actually, by using the definition of
$\nu_{R}$, we will prove now that $K_{s}*\nu_{R}$ is $\mathcal{C}^{\infty}$ on
the interior of $B_{R}$.

We now appeal to Lemma 2.1 below which implies that for some constant $c$,

$\Phi=c\quad\text{on}\quad\mathrm{int}(B_{R}),$ (2.10)

where $\mathrm{int}(B_{R})=\\{x\in\mathbb{R}^{d}:|x|<R\\}$. That is, the
expected Euler – Lagrange condition (A.5) holds, verifying that $\nu_{R}$ is
indeed the equilibrium measure for the extremal problem of Theorem 1.5.

######  Lemma 2.1 (Liouville Lemma on ball).

Let $\Phi:\mathrm{int}(B_{R})\to(-\infty,+\infty]$, $d\geq 1$, $R>0$. If

  * •

(local integrability) $\Phi\in L^{1}_{\mathrm{loc}}(\mathrm{d}x)$;

  * •

(weak harmonicity) $\Delta\Phi=0$ in the sense of Schwartz distributions;

  * •

(radial symmetry) $\Phi$ is equal to a constant on
$\\{x\in\mathbb{R}^{d}:|x|=r\\}$, for all $r<R$;

then $\Phi$ is equal to a constant $c$.

Related statements can be found in [22, Sec. 0.3] for $d=2$, and in [13, Th.
3.3, Ch. III, p. 183].

###### Proof of Lemma 2.1.

First by the Weyl Lemma333See for instance [26] for a short and nice proof,
close to the potential theoretic original proof given by Weyl in [27]. This is
an important basic result in the regularity theory of elliptic partial
differential equations, which can be seen nowadays as a special case of the
Hörmander theory of hypoelliptic partial differential operators. We need here
only the version in which $\Delta\Phi=0$, considered for instance in [10, Th.
11.1.1 (iii)–(v)]. , $\Phi$ is $\mathcal{C}^{\infty}(\mathrm{int}(B_{R}))$.
Next, set $x=r\hat{x}$, $r=|x|$, $\hat{x}\in\mathbb{S}^{d-1}$. Denoting
$\Delta_{\mathbb{S}^{d-1}}$ the Laplacian on $\mathbb{S}^{d-1}$,

$\Delta=\partial^{2}_{r}+\frac{d-1}{r}\partial_{r}+\frac{1}{r^{2}}\Delta_{\mathbb{S}^{d-1}}.$
(2.11)

Since $\Phi$ is radially symmetric, we have $\Delta_{\mathbb{S}^{d-1}}\Phi=0$,
and setting $\psi(r):=\Phi(x)$, we get for all $r\in(0,R)$,

$\Delta\Phi(x)=\psi^{\prime\prime}(r)+\frac{d-1}{r}\psi^{\prime}(r)=\frac{1}{r^{d-1}}(r^{d-1}\psi^{\prime}(r))^{\prime}.$

But $\Delta\Phi=0$ on $\\{x\in\mathbb{R}^{d}:0<|x|<R\\}$, and so
$(r^{d-1}\psi^{\prime})^{\prime}=0$ on $(0,R)$. Hence
$r^{d-1}\psi^{\prime}=c_{0}$ for some constant $c_{0}$ on $(0,R)$, which
implies that for some constant $c_{1}$ and $r\in(0,R)$

$\psi(r)=\begin{cases}c_{0}\log r+c_{1}&\text{if $d=2$}\\\
\frac{c_{0}}{2-d}\frac{1}{r^{d-2}}+c_{1}&\text{if $d\geq 3$}\end{cases}.$

Finally, since $\psi$ is bounded, we get $c_{0}=0$ and $\psi=c_{1}$ on
$(0,R)$. ∎

To complete the proof of Theorem 1.5 note that we can reinterpret (2.2) as an
Euler – Lagrange equation (A.5) in the sense that $\mu_{\mathrm{eq}}=\nu_{R}$
is seen as an equilibrium measure for the kernel $\widetilde{K}=K_{s+2}$ with
external field $\widetilde{V}=0$. ∎

###  2.3. Proof of Corollary 1.6

Let $d$, $s$, $\mu_{\mathrm{eq}}=\nu_{R}$, $R$, and
$\sigma_{\mathbb{S}^{d-1}}$ be as in Theorem 1.5. For $x\in\mathbb{R}^{d}$,
$|x|\leq 1$, the quantity $\Phi(x)$ depends only on $\lambda:=|x|/R$ and we
can define

$\varphi(\lambda):=\Phi(x)=U^{\mu_{\mathrm{eq}}}(x)+\gamma\left|x\right|^{2}=U^{\mu_{\mathrm{eq}}}(x)+\gamma
R^{2}\lambda^{2}.$ (2.12)

The function $\varphi$ is continuous on $[0,1]$ and differentiable on $(0,1)$.
The Euler – Lagrange conditions (2.10) states that $\varphi$ is constant and
equal to $\varphi(0)$ on $\lambda\in[0,1]$, namely
$\varphi^{\prime}(\lambda)=0$ for $\lambda\in(0,1)$.

The formula 1.14 in Corollary 1.6 comes from the equation
$\varphi(\lambda)=\varphi(0)$ and the formulas for $\varphi$ provided by Lemma
2.2 and Lemma 2.3. Note that (1.14) is trivial when $d=3$. The formula (1.16)
is obtained from (1.14) by using (1.15) and the Legendre duplication formula
(A.2).

######  Lemma 2.2 (Integral formula for potential).

Let $\varphi$, $d$, and $R$ be as in (2.12). For all $\lambda\in[0,1]$,

$\varphi(\lambda)=\begin{cases}\displaystyle\frac{1}{R^{d-3}}\left(c_{d}\int_{0}^{1}\Bigr{(}\int_{0}^{\pi}\frac{\sin^{d-2}(\theta)}{(\lambda^{2}-2r\lambda\cos(\theta)+r^{2})^{\frac{d-3}{2}}}\mathrm{d}\theta\Bigr{)}\frac{r^{d-1}\,\mathrm{d}r}{\sqrt{1-r^{2}}}+\gamma
R^{d-1}\lambda^{2}\right)&\text{if $d\neq 3$}\\\\[10.00002pt]
\displaystyle-\int_{0}^{1}\\!\frac{(\lambda+r)^{2}\log(\lambda+r)-(\lambda-r)^{2}\log|\lambda-r|}{\pi\lambda}\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}-\log
R+\frac{1}{2}+\gamma R^{2}\lambda^{2}&\text{if $d=3$}\end{cases},$

where $\displaystyle
c_{d}:=\frac{2\,\mathrm{sign}(d-3)\Gamma(\frac{d+1}{2})}{\pi\Gamma(\frac{d-1}{2})}$.

Note that $\gamma R^{d-1}$ does not depend on $\gamma$.

###### Proof of Lemma 2.2.

By the Funk – Hecke formula (A.3), for $x\in\mathbb{R}^{d}$, $s\neq 0$, with
$C_{d}:=\mathrm{sign}(s)\frac{\Gamma(\frac{d+1}{2})}{\pi^{\frac{d+1}{2}}}$,

$\displaystyle U^{\mu_{\mathrm{eq}}}(x)$ $\displaystyle=C_{d}\int_{|y|\leq
1}\frac{\mathrm{d}y}{|x-Ry|^{s}\sqrt{1-|y|^{2}}}$
$\displaystyle=C_{d}|\mathbb{S}^{d-1}|\int_{0}^{1}\int_{\mathbb{S}^{d-1}}\frac{\sigma_{\mathbb{S}^{d-1}}(\mathrm{d}y)r^{d-1}\,\mathrm{d}r}{(|x|^{2}-2rR\langle
x,y\rangle+r^{2}R^{2})^{\frac{s}{2}}\sqrt{1-r^{2}}}$
$\displaystyle=\frac{C_{d}|\mathbb{S}^{d-1}|\tau_{d-1}}{R^{s}}\int_{0}^{1}\Bigr{(}\int_{-1}^{1}\frac{(1-t^{2})^{\frac{s}{2}}}{(\lambda^{2}-2r\lambda
t+r^{2})^{\frac{s}{2}}}\mathrm{d}t\Bigr{)}\frac{r^{d-1}}{\sqrt{1-r^{2}}}\,\mathrm{d}r$
$\displaystyle=\frac{2\mathrm{sign}(s)\Gamma(\frac{d+1}{2})}{\pi\Gamma(\frac{d-1}{2})R^{d-3}}\int_{0}^{1}\Bigr{(}\int_{0}^{\pi}\frac{\sin^{s+1}(\theta)}{(\lambda^{2}-2r\lambda\cos(\theta)+r^{2})^{\frac{s}{2}}}\mathrm{d}\theta\Bigr{)}\frac{r^{d-1}}{\sqrt{1-r^{2}}}\,\mathrm{d}r,$
(2.13)

while if $s=0$,

$\displaystyle U^{\nu_{R}}(x)$ $\displaystyle=-\frac{1}{\pi^{2}}\int_{|y|\leq
1}\frac{\log\left|x-Ry\right|}{\sqrt{1-|y|^{2}}}\mathrm{d}y$
$\displaystyle=-\frac{2}{\pi}\int_{0}^{1}\Bigr{(}\int_{\mathbb{S}^{2}}\log\bigr{(}\lambda^{2}R^{2}-2rR\langle
x,y\rangle+r^{2}R^{2}\bigr{)}\sigma_{\mathbb{S}^{d-1}}(\mathrm{d}y)\Bigr{)}\frac{r^{2}}{\sqrt{1-r^{2}}}\,\mathrm{d}r$
$\displaystyle=-\frac{1}{\pi}\int_{0}^{1}\Bigr{(}4\log
R+\int_{-1}^{1}\log(\lambda^{2}-2r\lambda
t+r^{2})\mathrm{d}t\Bigr{)}\frac{r^{2}}{\sqrt{1-r^{2}}}\,\mathrm{d}r.$

Finally we observe that

$\int_{-1}^{1}\log(\lambda^{2}-2r\lambda
t+r^{2})\mathrm{d}t=\frac{(\lambda+r)^{2}\log(\lambda+r)-(\lambda-r)^{2}\log|\lambda-r|}{r\lambda}-2.$

∎

######  Lemma 2.3 (Landen transform and a special function).

For all integer $d\geq 2$, $\lambda\geq 0$, and $r\in[0,1]$,

$\int_{0}^{\pi}\frac{\sin^{d-2}(\theta)}{(\lambda^{2}-2r\lambda\cos(\theta)+r^{2})^{\frac{d-3}{2}}}\mathrm{d}\theta=\frac{2^{d-1}}{(\lambda+r)^{d-3}}S_{d-3}\left(\frac{4\lambda
r}{(\lambda+r)^{2}}\right),$

where for $z\in[0,1]$,

$S_{s}(z):=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{s+1}(\alpha)\cos^{s+1}(\alpha)}{(1-z\sin^{2}(\alpha))^{\frac{s}{2}}}\mathrm{d}\alpha=\int_{0}^{1}\frac{t^{s+1}(1-t^{2})^{\frac{s}{2}}}{(1-zt^{2})^{\frac{s}{2}}}\mathrm{d}t=\frac{\Gamma(\frac{s}{2}+1)^{2}}{2\Gamma(s+2)}\,{}_{2}F_{1}\Bigr{(}\frac{s}{2}+1,\frac{s}{2};s+2;z\Bigr{)}.$

###### Proof.

We set $\rho_{1}:=\frac{2\lambda r}{\lambda^{2}+r^{2}}$,
$\rho_{2}:=\frac{2\rho_{1}}{1+\rho_{1}}=\frac{4\lambda r}{(\lambda+r)^{2}}$,
which gives $(\lambda^{2}+r^{2})(1+\rho_{1})=(\lambda+r)^{2}$. Using the
change of variable $\theta=2\alpha$, and $\cos(\theta)=1-2\sin^{2}(\alpha)$,
$\sin(\theta)=2\sin(\alpha)\cos(\alpha)$, we get

$\displaystyle\int_{0}^{\pi}\frac{\sin^{s+1}(\theta)}{(\lambda^{2}-2r\lambda\cos(\theta)+r^{2})^{\frac{s}{2}}}\mathrm{d}\theta$
$\displaystyle=\frac{1}{(\lambda^{2}+r^{2})^{\frac{s}{2}}}\int_{0}^{\pi}\frac{\sin^{s+1}(\theta)}{(1-\rho_{1}\cos(\theta))^{\frac{s}{2}}}\mathrm{d}\theta$
$\displaystyle=\frac{2^{s+2}}{(\lambda^{2}+r^{2})^{\frac{s}{2}}}\int_{0}^{\frac{\pi}{2}}\frac{\sin^{s+1}(\alpha)\cos^{s+1}(\alpha)}{(1-\rho_{1}(1-2\sin^{2}(\alpha)))^{\frac{s}{2}}}\mathrm{d}\alpha$
$\displaystyle=\frac{2^{s+2}}{(\lambda^{2}+r^{2})^{\frac{s}{2}}(1-\rho_{1})^{\frac{s}{2}}}\int_{0}^{\frac{\pi}{2}}\frac{\sin^{s+1}(\alpha)\cos^{s+1}(\alpha)}{(1+\frac{2\rho_{1}}{1-\rho_{1}}\sin^{2}(\alpha))^{\frac{s}{2}}}\mathrm{d}\alpha$
$\displaystyle=\frac{2^{s+2}}{(\lambda+r)^{s}}\frac{(1+\rho_{1})^{\frac{s}{2}}}{(1-\rho_{1})^{\frac{s}{2}}}S_{s}\Bigr{(}-\frac{2\rho_{1}}{1-\rho_{1}}\Bigr{)}.$

But for $z\in[0,1]$,

$S_{s}(-z)=\int_{0}^{\frac{\pi}{2}}\frac{\sin^{s+1}(\alpha)\cos^{s+1}(\alpha)}{(1+z\sin^{2}(\alpha))^{\frac{s}{2}}}\mathrm{d}\alpha=\frac{1}{(1+z)^{\frac{s}{2}}}\int_{0}^{\frac{\pi}{2}}\frac{\sin^{s+1}(\alpha)\cos^{s+1}(\alpha)}{(1-\frac{z}{1+z}\cos^{2}(\alpha))^{\frac{s}{2}}}\mathrm{d}\alpha=\frac{1}{(1+z)^{\frac{s}{2}}}S_{s}\Bigr{(}\frac{z}{1+z}\Bigr{)}.$

In particular, with $z=\frac{2\rho_{1}}{1-\rho_{1}}$, we get
$1+z=\frac{1+\rho_{1}}{1-\rho_{1}}$ and
$\frac{z}{1+z}=\frac{2\rho_{2}}{1+\rho_{1}}=\rho_{2}$; therefore

$S_{s}\Bigr{(}-\frac{2\rho_{1}}{1-\rho_{1}}\Bigr{)}=\frac{(1-\rho_{1})^{\frac{s}{2}}}{(1+\rho_{1})^{\frac{s}{2}}}S_{s}(\rho_{2})$

which gives the desired integral formula in terms of $S_{d-3}$. It remains to
derive the hypergeometric formula for $S_{s}$. The Newton binomial series
(1.7) gives
$\frac{1}{(1-zt^{2})^{\frac{s}{2}}}=\sum_{n=0}^{\infty}\frac{(\frac{s}{2})_{n}z^{n}t^{2n}}{n!};$
hence

$\displaystyle S_{s}(z)$
$\displaystyle=\sum_{n=0}^{\infty}\frac{(\frac{s}{2})_{n}}{n!}z^{n}\int_{0}^{1}t^{s+1+2n}(1-t^{2})^{\frac{s}{2}}\mathrm{d}t$
$\displaystyle=\frac{\Gamma(\frac{s}{2}+2)}{2}\sum_{n=0}^{\infty}(\frac{s}{2})_{n}\frac{z^{n}}{n!}\frac{\Gamma(n+\frac{s}{2}+1)}{\Gamma(n+s+2)}$
$\displaystyle=\frac{\Gamma(\frac{s}{2}+1)^{2}}{2\Gamma(s+2)}\,{}_{2}F_{1}\Bigr{(}\frac{s}{2},\frac{s}{2}+1;s+2;z\Bigr{)}.$

∎

###  2.4. Proof of Corollary 1.7

Let us keep the notation used in the proof of Corollary 1.6.

First of all, the formulas (1.17) and (1.18) in Corollary 1.7 come from the
Euler – Lagrange equation $\varphi(\lambda)=\varphi(0)$ and its reformulation
$\varphi^{\prime}(\lambda)=0$, and the formula for $\varphi$ provided by Lemma
2.2.

The formula (1.19) in Corollary 1.7 is obtained by further reformulating
$\varphi$ when $(d,s)=(2,-1)$ in terms of special functions using Lemma 2.4
below. The formula for $\varphi^{\prime}$ provided by this lemma gives

$\int_{0}^{1}\left[(\lambda+r)E\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)+(\lambda-r)K\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)\right]\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}=\frac{\pi^{2}}{4}\lambda^{2}.$
(2.14)

Next, following [12, 11, 1], the Landen transform for $E$ and $K$ gives, for
$z\in[-1,1]$,

$K\Bigr{(}\frac{4z}{(1+z)^{2}}\Bigr{)}=(1+z)K(z^{2})\quad\text{and}\quad
E\Bigr{(}\frac{4z}{(1+z)^{2}}\Bigr{)}=\frac{2}{1+z}E(z^{2})-(1-z)K(z^{2}).$
(2.15)

Now, the formula (1.20) of Corollary 1.6 comes by combining (1.19) and (2.14)
with $z=\frac{r}{\lambda}$.

Finally, it can be checked that the Landen transform formulas in (2.15) remain
valid for $z\in[-1,+\infty)$ provided that we take the real part and a good
determination of the square root. Used with
$z=\frac{r}{\lambda}\in[0,+\infty)$ they give (1.21) and (1.22). Note that we
can also rewrite (2.14) as

$\lambda\int_{0}^{1}\Re\Bigr{[}2E\Bigr{(}\frac{r^{2}}{\lambda^{2}}\Bigr{)}-\Bigr{(}1-\frac{r^{2}}{\lambda^{2}}\Bigr{)}K\Bigr{(}\frac{r^{2}}{\lambda^{2}}\Bigr{)}\Bigr{]}\frac{r\,\mathrm{d}r}{\sqrt{1-r^{2}}}=\frac{\pi^{2}}{8}\Bigr{(}\frac{\lambda^{2}}{2}+1\Bigr{)}.$
(2.16)

######  Lemma 2.4 ($d=2,s=-1$).

For all $\lambda\in[0,1]$,

$\displaystyle\varphi(\lambda)$
$\displaystyle=-\frac{1}{4\gamma}\int_{0}^{1}(\lambda+r)E\left(\frac{4\lambda
r}{(\lambda+r)^{2}}\right)\frac{r}{\sqrt{1-r^{2}}}\,\mathrm{d}r+\frac{\pi^{2}}{64\gamma}\lambda^{2},$
$\displaystyle\varphi^{\prime}(\lambda)$
$\displaystyle=-\frac{1}{8\gamma}\int_{0}^{1}\left[\Bigr{(}1+\frac{r}{\lambda}\Bigr{)}E\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)+\Bigr{(}1-\frac{r}{\lambda}\Bigr{)}K\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)\right]\frac{r}{\sqrt{1-r^{2}}}\,\mathrm{d}r+\frac{\pi^{2}}{32\gamma}\lambda.$

###### Proof.

Note that $R=\frac{\pi}{8\gamma}$. Let us define

$f(r):=\frac{r}{\sqrt{1-r^{2}}}.$

By combining Lemma 2.2 and Lemma 2.3, both with $d=2$, we get, for
$\lambda\geq 0$,

$\varphi(\lambda)=-\frac{R}{2\pi}\int_{0}^{1}(\lambda+r)E\left(\frac{4\lambda
r}{(\lambda+r)^{2}}\right)f(r)\,\mathrm{d}r+\gamma R^{2}\lambda^{2}.$

Finally, by using the well-known ordinary differential equations (for $0<z<1$)

$K^{\prime}(z)=\frac{E(z)-(1-z)K(z)}{2(1-z)z}\quad\text{and}\quad
E^{\prime}(z)=\frac{E(z)-K(z)}{2z}$

we get, after some algebra,

$\varphi^{\prime}(\lambda)=-\frac{R}{\pi}\int_{0}^{1}\left[\Bigr{(}1+\frac{r}{\lambda}\Bigr{)}E\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)+\Bigr{(}1-\frac{r}{\lambda}\Bigr{)}K\left(\frac{4\lambda
r}{(r+\lambda)^{2}}\right)\right]f(r)\,\mathrm{d}r+2\gamma R^{2}\lambda.$

∎

######  Lemma 2.5 ($d=3$, $s=0$).

For all $\lambda\in[0,1]$,

$\displaystyle\varphi(\lambda)$
$\displaystyle=\frac{1+\log(3\gamma)}{2}-\frac{1}{2\pi\lambda}\int_{0}^{1}\Bigr{(}(\lambda+r)^{2}\log((\lambda+r)^{2})-(\lambda-r)^{2}\log\left((\lambda-r)^{2}\right)\Bigr{)}\frac{r}{\sqrt{1-r^{2}}}\,\mathrm{d}r+\frac{\lambda^{2}}{3}.$

###### Proof.

From Lemma 2.2 and with the formulas ($b\in(-a,a)$)

$\int_{-1}^{1}\log(a-bt)\mathrm{d}t=\frac{(a+b)\log(a+b)-(a-b)\log(a-b)}{b}-2\quad\text{and}\quad\int_{0}^{1}\frac{r^{2}}{\sqrt{1-r^{2}}}\,\mathrm{d}r=\frac{\pi}{4}$

with $a=\lambda^{2}+r^{2}$ and $b=2r\lambda$, we obtain, for all
$\lambda\in[0,+\infty)$,

$\varphi(\lambda)=\frac{1}{2}-\log(R)-\int_{0}^{1}\frac{(\lambda+r)^{2}\log((\lambda+r)^{2})-(\lambda-r)^{2}\log\left((\lambda-r)^{2}\right)}{2\pi\lambda}\frac{r}{\sqrt{1-r^{2}}}\,\mathrm{d}r+\gamma
R^{2}\lambda^{2}.$

∎

##  Appendix A Useful tools

Let us recall the Euler reflection formula for the Gamma function, valid for
$z\not\in\\{-1,-2,\ldots\\}$,

$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}.$ (A.1)

Let us recall also the Legendre duplication formula, valid for
$2z\not\in\\{-0,-1,-2,-3,\ldots\\}$,

$\sqrt{\pi}\Gamma(2z)=2^{2z-1}\Gamma(z)\Gamma\Bigr{(}z+\frac{1}{2}\Bigr{)}.$
(A.2)

###  A.1. Funk – Hecke formula

Let $d\geq 2$ and $\mu_{1}$ denote the uniform probability measure on the unit
centered sphere $\mathbb{S}^{d-1}=\\{x\in\mathbb{R}^{d}:|x|=1\\}$. Then, for
all $z\in\mathbb{R}^{d}$ with $|z|=1$,

$\int_{\mathbb{S}^{d-1}}f(z\cdot
x)\mu_{1}(\mathrm{d}x)=\tau_{d-1}\int_{0}^{\pi}f(\cos(\theta))\sin^{d-2}(\theta)\mathrm{d}\theta=\tau_{d-1}\int_{-1}^{1}f(t)(1-t^{2})^{\frac{d-3}{2}}\mathrm{d}t$
(A.3)

where

$\tau_{d-1}:=\int_{0}^{\pi}\sin^{d-2}(\theta)\mathrm{d}\theta=\left(\int_{-1}^{1}(1-t^{2})^{\frac{d-3}{2}}\mathrm{d}t\right)^{-1}=\frac{\Gamma(\frac{d}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{d-1}{2})}.$
(A.4)

The Funk – Hecke formula (A.3) is a useful tool to reduce multivariate
integrals into univariate integrals. It gives the projection on any diameter
of the uniform law on the sphere. If $X$ is a random vector in
$\mathbb{R}^{d}$ uniformly distributed on $\mathbb{S}^{d-1}$ then for all
$z\in\mathbb{S}^{d-1}$, the law of $z\cdot X$ has density
$\tau_{d-1}(1-t^{2})^{\frac{d-3}{2}}\mathbf{1}_{t\in[-1,1]}$. This is an
arcsine law when $d=2$, a uniform law when $d=3$, a semicircle law when $d=4$,
and more generally, for an arbitrary $d\geq 2$, the image by the map
$u\mapsto\sqrt{u}$ of the beta law $\mathrm{Beta}(\frac{1}{2},\frac{d-1}{2})$.
We refer to [17, p, 18] or [2, Eq. (5.1.9) p. 197] for a proof.

###  A.2. Euler – Lagrange (Frostman) characterization of equilibrium measure

The Euler – Lagrange characterization of the equilibrium measure
$\mu_{\mathrm{eq}}$ states that a necessary and sufficient condition for an
element $\mu$ of $\mathcal{M}_{1}(B_{R})$ to be an equilibrium measure is that
for some finite constant $c$ we have (see, for example, [13])

$U^{\mu}+V\begin{cases}\leq c&\mbox{ for all }x\in\mbox{supp}(\mu)\\\ \geq
c&\mbox{ for quasi-every }x\in B_{R}\end{cases};$ (A.5)

by “quasi-every” we mean almost surely for all probability measures of finite
energy. This condition holds everywhere when $V$ is continuous. In particular,
a probability measure with support equal to $B_{R}$ is an equilibrium measure
on $B_{R}$ if and only if $U^{\mu}+V$ is constant quasi-everywhere on $B_{R}$.

###  A.3. Integrability and regularity of Riesz potentials

The following Lemma summarizes key regularity properties of the Riesz kernel,
some of which are classical. We give a proof for the reader’s convenience. On
this topic, we also refer to the works of Yoshihiro Mizuta such as [14, 15,
16].

######  Lemma A.1 (Integrability and regularity of Riesz potentials).

  1. (i)

$K_{s}\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{d},\mathrm{d}x)$ if and only if
$s=0$ or $s\neq 0$ and $s<d$.

  2. (ii)

If $s<d-2$ then, in the sense of distributions,

$\Delta K_{s}=-c_{d,s}K_{s+2}\quad\text{where}\quad
c_{d,s}:=\begin{cases}|s|(d-2-s)&\text{if $s\neq 0$}\\\ d-2&\text{if
$s=0$}\end{cases}.$ (A.6)

  3. (iii)

Suppose that $s=0$ or $s\neq 0$ and $s<d$. Let $\mu$ be a compactly supported
probability measure on $\mathbb{R}^{d}$. Then the following function is well
defined and belongs to $L^{1}_{\mathrm{loc}}(\mathbb{R}^{d},\mathrm{d}x)$:

$x\in\mathbb{R}^{d}\mapsto(K_{s}*\mu)(x):=\int K_{s}(x-y)\mu(\mathrm{d}y).$
(A.7)

Moreover, in the sense of distributions,

$\Delta(K_{s}*\mu)=(\Delta K_{s})*\mu=-c_{d,s}K_{s+2}*\mu.$ (A.8)

  4. (iv)

Suppose that $s=0$ or $s\neq 0$ and $s<d$. If $\mu$ is a compactly supported
probability measure on $\mathbb{R}^{d}$ such that
$\mathrm{d}\mu(x)=f(x)\mathrm{d}x$, $f\in L^{p}(\mathbb{R}^{d},\mathrm{d}x)$,
and $p>d/(d-s)$; then $K_{s}*\mu$ is continuous on $\mathbb{R}^{d}$.

Note that $K_{s+2}\in L^{1}_{\mathrm{loc}}(\mathbb{R}^{d},\mathrm{d}x)$
implies $s+2<d$; in other words $s<d-2$. Furthermore, the condition $s<d-2$ is
sharp for (A.6), indeed; in the sense of distributions, we have $\Delta
K_{d-2}=-c_{d}\delta_{0}$ (Coulomb kernel). This suggests defining
$K_{d}:=\delta_{0}$ to make the formula (A.6) valid for the critical case
$s=d-2$, provided that we also set $c_{d,d-2}:=c_{d}$.

We remark that (A.6) is a special case of (A.8) which corresponds to taking
$\mu=\delta_{0}$ and that (A.8) goes beyond [25, Eq. (7) p. 118] and [9, Eq.
(85) p. 136]. Note also that the distribution $\Delta\mu$ equals the
convolution $(\Delta\delta_{0})*\mu$, see [24, 23, end of Ch. VI, Sec. 3;
notably eq. (VI, 3; 34) and (VI, 3; 35)]. From this point of view, it follows
that (A.8) is a consequence of the associative law for convolution of three
distributions, two of which have compact support, see [24, 23, Ch. VI, Sec. 3,
Th. VII] and [13, Lemma 0.6]. We give however a direct short proof of (A.8)
below.

###### Proof.

_Proof of (i)._ It suffices to check local integrability in the neighborhood
of the origin, namely

$\int_{|x|\leq 1}|K_{s}(x)|\mathrm{d}x=\begin{cases}\displaystyle
2\pi\int_{0}^{1}r^{-s+d-1}\mathrm{d}r<\infty&\text{ if $s\neq 0$ and
$d>s$}\\\\[5.0pt] \displaystyle
2\pi\int_{0}^{1}\log(r)r^{d-1}\mathrm{d}r<\infty&\text{ if $s=0$}\end{cases}.$

_Proof of (ii)._ On $\mathbb{R}^{d}\setminus\\{0\\}$, the function $K_{s}$ is
$\mathcal{C}^{\infty}$ and a computation reveals that

$\Delta K_{s}=-c_{d,s}K_{s+2}.$

It follows that this equality also holds in the sense of distributions for
test functions supported away from the origin. For general test functions, we
proceed by integration by parts outside a centered ball of small radius.
Namely, let $\varphi$ be a compactly supported $\mathcal{C}^{\infty}$ test
function, and let $\varepsilon>0$. By the Green integration by parts formula
for the open set $\\{x\in\mathbb{R}^{d}:|x|>\varepsilon\\}$, denoting
$n(x)=-x|x|^{-1}$ the inner unit normal vector to the sphere
$\\{x\in\mathbb{R}^{d}:|x|=\varepsilon\\}$ at the point $x$,

$\int_{|x|\geq\varepsilon}\Delta\varphi(x)K_{s}(x)\mathrm{d}x-\int_{|x|\geq\varepsilon}\varphi(x)\Delta
K_{s}(x)\mathrm{d}x\\\ =\int_{|x|=\varepsilon}K_{s}(x)\nabla\varphi(x)\cdot
n(x)\mathrm{d}\sigma_{\varepsilon}(x)-\int_{|x|=\varepsilon}\varphi(x)\nabla
K_{s}(x)\cdot n(x)\mathrm{d}\sigma_{\varepsilon}(x).$

If $s\neq 0$ and $d>s+1$ then

$\displaystyle\int_{|x|=\varepsilon}K_{s}(x)\nabla\varphi(x)\cdot
n(x)\mathrm{d}\sigma_{\varepsilon}(x)$
$\displaystyle=\varepsilon^{-s}\int_{|x|=\varepsilon}\nabla\varphi(x)\cdot
n(x)\mathrm{d}\sigma_{\varepsilon}(x)$
$\displaystyle=\varepsilon^{-s}O(\varepsilon^{d-1})=O(\varepsilon^{-s+d-1})=o_{\varepsilon\to
0^{+}}(1),$

while, using $\nabla K_{s}(x)=-sx|x|^{-(s+2)}=-sxK_{s+2}(x)$ and $x\cdot
n_{x}=-|x|$, if $d>s+2$,

$\displaystyle\int_{|x|=\varepsilon}\varphi(x)\nabla K_{s}(x)\cdot
n(x)\mathrm{d}\sigma_{\varepsilon}(x)$
$\displaystyle=s\varepsilon^{1-(s+2)}\int_{|x|=\varepsilon}\varphi(x)\mathrm{d}\sigma_{\varepsilon}(x)$
$\displaystyle=s\varepsilon^{1-(s+2)}O(\varepsilon^{d-1})=O(\varepsilon^{d-(s+2)})=o_{\varepsilon\to
0^{+}}(1).$

Finally a careful analysis reveals that the conditions on $d$ are the same in
the case $s=0$.

_Proof of (iii)._ If $s<0$, then $K_{s}\leq 0$, and hence $K_{s}*\mu$ is well
defined and takes its values in $[-\infty,0]$. Similarly, if $s>0$, then
$K_{s}\geq 0$, and hence $K_{s}*\mu$ is well defined and takes its values in
$[0,+\infty]$. If $s=0$ then $K_{0}\mathbf{1}_{\left|\cdot\right|\leq 1}\geq
0$ while $\sup_{\mathbb{R}^{d}}K_{0}\mathbf{1}_{\left|\cdot\right|\geq
1}/\log(1+\left|\cdot\right|)<\infty$, hence $K_{0}*\mu$ is well defined and
takes its values in $(-\infty,+\infty]$. Next, by the Fubini – Tonelli
theorem, for all $R>0$, using _(i)_ and the compactness of support of $\mu$
(note that this can be weakened into integrability of
$\log(1+\left|\cdot\right|)\mathbf{1}_{s=0}$),

$\iint|K_{s}(x-y)|\mathbf{1}_{|x|\leq
R}\mu(\mathrm{d}y)\mathrm{d}x=\int\Bigr{(}\int|K_{s}(x)|\mathbf{1}_{|x+y|\leq
R}\mathrm{d}x\Bigr{)}\mu(\mathrm{d}y)<\infty.$

It follows that $K_{s}*\mu$ belongs to
$L^{1}_{\mathrm{loc}}(\mathbb{R}^{d},\mathrm{d}x)$.

For the differentiability, let $\varphi:\mathbb{R}^{d}\to\mathbb{R}$ be a
$\mathcal{C}^{\infty}$ and compactly supported test function. By the Fubini –
Tonelli theorem and _(ii)_ ,

$\displaystyle\int(K_{s}*\mu)(x)\Delta\varphi(x)\mathrm{d}x$
$\displaystyle=\int\Bigr{(}\int
K_{s}(x-y)f(y)\mathrm{d}y\Bigr{)}\Delta\varphi(x)\mathrm{d}x$
$\displaystyle=\iint K_{s}(x-y)f(y)\Delta\varphi(x)\mathrm{d}x\mathrm{d}y$
$\displaystyle=\int f(y)\Bigr{(}\int
K_{s}(x-y)\Delta\varphi(x)\mathrm{d}x\Bigr{)}\mathrm{d}y$
$\displaystyle=-c_{d,s}\int f(y)\Bigr{(}\int
K_{s+2}(x-y)\varphi(x)\mathrm{d}x\Bigr{)}\mathrm{d}y$
$\displaystyle=-c_{d,s}\int\varphi(x)\Bigr{(}\int
K_{s+2}(x-y)f(y)\mathrm{d}y\Bigr{)}\mathrm{d}x$
$\displaystyle=-c_{d,s}\int\varphi(x)(K_{s+2}*\mu)(x)\mathrm{d}x.$

_Proof of (iv)._ For the continuity, let us use the cutoff argument used in
[5, Lemma 4.3], see also [14, Th. 1], [15], and [16, Sec. 5.3]. Namely, let us
consider first the case $s>0$. For all $n\geq 1$ and $x\in\mathbb{R}^{d}$, let
us define

$R_{n}(x):=\int f(y)K_{s}(x-y)\mathbf{1}_{|K_{s}(x-y)|\leq n}\mathrm{d}y$

and

$S_{n}(x):=(K_{s}*\mu)(x)-R_{n}(x)=\int
f(y)K_{s}(x-y)\mathbf{1}_{|K_{s}(x-y)|\geq n}\mathrm{d}y.$

By the dominated convergence theorem, $R_{n}$ is continuous on
$\mathbb{R}^{d}$. Let us show now that $\lim_{n\to\infty}S_{n}=0$ uniformly on
compact subsets, which will prove the continuity of $K_{s}*\mu$. Let
$q:=p/(p-1)$ be the Hölder conjugate exponent of $p$. Now, by the Hölder
inequality, using the fact that $K_{s}=\left|\cdot\right|^{s}$, $s>0$,

$0\leq S_{n}(x)=\int f(y)\frac{\mathbf{1}_{|x-y|\leq
n^{-1/s}}}{|x-y|^{s}}\mathrm{d}y\leq\left\|f\right\|_{p,B(x,1)}\varepsilon_{n}^{1/q}$

where $B(x,r):=\\{x\in\mathbb{R}^{d}:|x|\leq r\\}$ is the closed centered ball
of radius $r$, where $\left\|\cdot\right\|_{p,C}$ denotes the $L^{p}$ norm
with respect to the trace of the Lebesgue measure on $C$, where

$\varepsilon_{n}:=\sigma_{d}\int_{0}^{n^{-1/s}}\frac{\mathrm{d}r}{r^{qs-d+1}},$

and where $\sigma_{d}$ is the surface area of the unit sphere
$\\{x\in\mathbb{R}^{d}:|x|=1\\}$. The condition $p>d/(d-s)$, which is
equivalent to $qs-d+1<1$, ensures that $\varepsilon_{n}$ is finite for all $n$
and that $\lim_{n\to\infty}\varepsilon_{n}=0$. Hence, if
$C\subset\mathbb{R}^{d}$ is a compact set, then, denoting
$C_{1}:=\\{x\in\mathbb{R}^{d}:\mathrm{dist}(x,C)\leq 1\\}$, we have

$\sup_{x\in
C}|S_{n}(x)|\leq\left\|f\right\|_{p,K_{1}}\varepsilon_{n}^{1/q}\underset{n\to\infty}{\longrightarrow}0,$

which completes the proof of the continuity of $K_{s}*\mu$. The case $s<0$ is
entirely similar up to a sign. It remains to examine the case $s=0$. Let us
write $K_{0}=K_{0}^{+}-K_{0}^{-}$ with $K_{0}^{\pm}\geq 0$, namely
$K_{0}^{+}=-\log\left|\cdot\right|\mathbf{1}_{\left|\cdot\right|\leq 1}$ and
$K_{0}^{-}=\log\left|\cdot\right|\mathbf{1}_{\left|\cdot\right|>1}$. To
establish the continuity of $K_{0}^{+}*\mu$ we write

$0\leq S_{n}^{+}(x)=\int f(y)\log\frac{1}{|x-y|}\mathbf{1}_{|x-y|\leq
1}\mathbf{1}_{|x-y|\leq\mathrm{e}^{-n}}\mathrm{d}y\\\
\leq\left\|f\right\|_{p,B(x,1)}(\varepsilon_{n}^{+})^{1/q}$

where

$\varepsilon_{n}^{+}=-\sigma_{d}\int_{0}^{\mathrm{e}^{-n}}r^{d-1}\log(r)\mathrm{d}r\underset{n\to\infty}{\longrightarrow}0.$

On the other hand, the continuity of $K_{0}^{-}*\mu$ follows from that of
$K_{0}^{-}$. Hence $K_{0}*\mu$ is continuous. ∎

##  Appendix B Proof of Riesz formula

###  B.1. Cross-ratio

Recall that in projective geometry, the _cross-ratio_ (_birapport_ in French)
of four distinct points $z_{1},z_{2},z_{3},z_{4}$ on the Riemann sphere
$\mathbb{C}\cup\\{\infty\\}$ is defined by

$[z_{1},z_{2};z_{3},z_{4}]=\frac{z_{3}-z_{1}}{z_{3}-z_{2}}/\frac{z_{4}-z_{1}}{z_{4}-z_{2}}=\frac{(z_{3}-z_{1})(z_{4}-z_{2})}{(z_{3}-z_{2})(z_{4}-z_{1})},$

where each length is removed from the formula if it involves the point at
infinity. The following lemma is a classical and important result of
projective geometry.

######  Lemma B.1 (Cross-ratio invariance).

The cross-ratio is invariant under the Möbius transform

$z\mapsto\frac{az+b}{cz+d},\quad ad-bc\neq 0,$

and thus its modulus is invariant under the “conjugated Möbius transform”
$\displaystyle z\mapsto\frac{a\bar{z}+b}{c\bar{z}+d}$, $ad-bc\neq 0$.

###  B.2. Inversions

In $\mathbb{R}^{d}$, $d\geq 1$, the inversion with center $x_{0}$ and radius
$R>0$ is the transform that maps $x\neq x_{0}$ to $T(x)$ on the half line
started from $x_{0}$ and passing through $x$, in such a way that

$\left|x-x_{0}\right|\;\left|T(x)-x_{0}\right|=R^{2}.$

The circle centered at $x_{0}$ and of radius $R$ is pointwise invariant under
the transformation in the sense that all its elements are fixed points of the
transformation. The transformation maps the interior of this circle to its
exterior, and vice versa. In projective geometry, this transformation is
extended to the $d$-dimensional sphere by mapping $x_{0}$ to the point at
infinity $\infty$, and vice versa. We have

$T(x)-x_{0}=\frac{R^{2}}{|x-x_{0}|^{2}}(x-x_{0}),$

which exchanges $x_{0}$ and $\infty$. In dimension $d=2$, using complex
numbers, $T(z)-z_{0}=R^{2}/(\overline{z-z_{0}})$, which is a special case of
the conjugated Möbius transform
$z\mapsto\frac{\alpha\bar{z}+\beta}{\gamma\bar{z}+\delta}$. It is worth
mentioning that inversions are geometric transformations at the basis of the
Kelvin transform of functions $\mathbb{R}^{d}\to\mathbb{R}$.

######  Lemma B.2 (Classical properties of inversions).

Let $T$ be the inversion of $\mathbb{R}^{d}$, $d\geq 1$, with center
$x_{0}\in\mathbb{R}^{d}$ and radius $R>0$. Then we have the following
properties.

  1. (1)

For all $x$,
$\displaystyle\left|x-T(x)\right|=\frac{|R^{2}-\left|x-x_{0}\right|^{2}|}{\left|x-x_{0}\right|}$.

  2. (2)

For all $x,y$,
$\displaystyle\left|T(x)-T(y)\right|=R^{2}\frac{\left|x-y\right|}{\left|x-x_{0}\right|\left|y-x_{0}\right|}$.

  3. (3)

As differential forms
$\displaystyle\frac{\mathrm{d}T(x)}{|T(x)-x_{0}|^{d}}=\frac{\mathrm{d}x}{|x-x_{0}|^{d}}$.

  4. (4)

The modulus of the cross-ratio of distinct coplanar points
$x_{1},x_{2},x_{3},x_{4}$ is invariant under $T$.

###### Proof.

We can assume without loss of generality that $x_{0}=0$.

  1. (1)

Since $0,x,T(x)$ are aligned with $0$ at the edge we have

$|x-T(x)|=||x|-|T(x)||=\left||x|-\frac{R^{2}}{|x|}\right|=\frac{||x|^{2}-R^{2}|}{|x|}.$

  2. (2)

We have

$\displaystyle|T(x)-T(y)|^{2}$ $\displaystyle=|T(x)|^{2}+|T(y)|^{2}-2\langle
T(x),T(y)\rangle$
$\displaystyle=\frac{R^{4}}{|x|^{2}}+\frac{R^{4}}{|y|^{2}}-2\frac{R^{4}}{|x|^{2}|y|^{2}}\langle
x,y\rangle=\frac{R^{4}}{|x|^{2}|y|^{2}}|x-y|^{2}.$

  3. (3)

We have $\displaystyle\mathrm{Jac}(T)(x)=\frac{R^{2}}{|x|^{2}}(I_{d}+u\otimes
v)$, $\displaystyle u=\frac{x}{|x|}$, $\displaystyle v=-2\frac{x}{|x|}$, which
gives then

$|\det\mathrm{Jac}(T)(x)|=\left(\frac{R^{2}}{|x|^{2}}\right)^{d}=\left(\frac{|T(x)|}{|x|}\right)^{d},$

via the “matrix determinant lemma” $\det(A+u\otimes v)=(1+u\cdot
A^{-1}v)\det(A)$, the determinant analogue of the Sherman – Morrison formula
$(A+u\otimes v)^{-1}=A^{-1}-\frac{A^{-1}u\otimes vA^{-1}}{1+v\cdot A^{-1}u}$.

  4. (4)

Follows from the fact that $T$ restricted to the plane is a conjugated Möbius
transform.

∎

###  B.3. Intersecting chords

The _intersecting chords theorem_ in Euclidean (planar) geometry states that
if $AA^{*}$ and $BB^{*}$ are two chords of a circle, intersecting at the point
$M$, see Figure 1, then

$\mathrm{AM}\times\mathrm{MA^{*}}=\mathrm{BM}\times\mathrm{MB^{*}}.$

Indeed, the triangles $A^{*}MB$ and $AMB^{*}$ are similar, identical up to
rotation and scaling, more precisely they have two equal angles:
$\widehat{A^{*}MB}=\widehat{AMB^{*}}$ (opposite angles) and
$\widehat{MA^{*}B}=\widehat{MB^{*}A}$ (subtend the same arc).

$A^{*}$$A$$B^{*}$$B$$M$

$A^{*}$$A$$B^{*}$$B$$M$$O$

$x^{*}$$x$$z^{*}$$z$$y$$O$

Figure 1. Intersecting chords of a circle, $AA^{*}$ and $BB^{*}$ in the first
two pictures, $xx^{*}$ and $zz^{*}$ for the third. On the two last pictures,
the chords $BB^{*}$ and $zz^{*}$ are diameters of the circle. On the right,
$x,y\in\mathbb{R}^{d}$, $d\geq 2$, $|x|=r$, $|y|<r$, $x^{*}$ is aligned with
$x$ and $y$, $y$ separates $x$ and $x^{*}$.

Suppose now that the circle has center $O$, radius $r$, that $BB^{*}$ is a
diameter, and that $M$ belongs to the segment $OB$ (instead $OB^{*}$). Then
$\mathrm{BM}=r-\mathrm{OM}$ while $\mathrm{MB^{*}}=\mathrm{OM}+r$ and thus

$\text{BM}\times\mathrm{MB^{*}}=(r-\text{OM})(\mathrm{OM}+r)=r^{2}-\mathrm{OM}^{2}.$

In Euclidean geometry, this quantity is known as the Laguerre power of the
point $M$ with respect to the circle. We deduce immediately the following
lemma.

######  Lemma B.3 (Intersecting chords).

For every chord $AA^{*}$ of a circle with center $O$ and radius $r$,
intersecting an arbitrary diameter at point $M$, see Figure 1, we have,

$\mathrm{AM}\times\mathrm{MA^{*}}=r^{2}-\mathrm{OM}^{2}.$

###  B.4. Riesz geometric argument

The argument is essentially two-dimensional and involves projective geometry.
Fix $r>0$ and $x,y\in\mathbb{R}^{d}$, $d\geq 2$, with $|y|<r$. Let us define
the map $S:x\mapsto S(x)=x^{*}$ where $x^{*}\in\mathbb{R}^{d}$ is the point
aligned with $x,y$ such that $y$ separates $x$ and $x^{*}$ and

$|x-y|\;|y-x^{*}|=r^{2}-|y|^{2}.$

The map $S$ is the composition on an inversion centered at $y$ of radius
$\sqrt{r^{2}-|y|^{2}}$ and the central symmetry centered at $y$ (recall that
$y$ separates $x$ and $x^{*}$). Moreover, by Lemma B.3, see also Figure 1, we
have $|x|=r$ if and only if $|x^{*}|=r$, namely the centered sphere of radius
$r$ is globally invariant under $S$. The points $y$ and $\infty$ are mapped to
each other by $S$.

Let $T$ be the inversion centered at the origin and with radius $r$. By Lemma
B.2, the modulus of the cross-ratio of the coplanar points $x,T(y),y,T(x)$
satisfies

$|[x,T(y);y,T(x)]|=\frac{\left|x-y\right|\left|T(x)-T(y)\right|}{\left|x-T(x)\right|\left|y-T(y)\right|}=\frac{|x-y|^{2}r^{2}}{\left|r^{2}-|x|^{2}\right|\left|r^{2}-|y|^{2}\right|}.$

Note that since $x,y,x^{*}$ are aligned, the points $x,y,x^{*},T(x),T(y)$ are
coplanar.

######  Lemma B.4 (Commutation).

$S$ and $T$ commute.

This is related to the fact that $S$ leaves globally invariant the fixed
points (circle) of $T$.

###### Proof.

Using complex coordinates $T(z)=r^{2}/\overline{z}$ while
$T(z)-z_{0}=-(r^{2}-|z_{0}|^{2})/(\overline{z-z_{0}})$, where $z_{0}$ stands
for $y$. Now we have

$T(S(z))=\frac{r^{2}}{\overline{z_{0}}-\frac{r^{2}-|z_{0}|^{2}}{z-z_{0}}}=\frac{r^{2}(z-z_{0})}{\overline{z_{0}}z-r^{2}}\quad\text{and}\quad
S(T(z))=z_{0}-\frac{r^{2}-|z_{0}|^{2}}{\overline{\frac{r^{2}}{\overline{z}}-z_{0}}}=\frac{r^{2}(z_{0}-z)}{r^{2}-\overline{z_{0}}z}.$

∎

Since $S$ is the composition of an inversion and a central symmetry, it is a
special case of a conjugate Möbius transform, and then, by Lemma B.1,
$|[x,T(y);y,T(x)]|=|[S(x),S(T(y));S(y),S(T(x))]|.$ Since $S$ and $T$ commute
(Lemma B.4), we have, using Lemma B.2 for the final step,

$\displaystyle|[x,T(y);y,T(x)]|$
$\displaystyle=|[S(x),T(S(y));S(y),T(S(x))]|=|[x^{*},T(\infty);\infty,T(x^{*})]|=|[x^{*},0;\infty,T(x^{*})]|$
$\displaystyle=\frac{|T(x^{*})|}{|T(x^{*})-x^{*}|}=\frac{|T(x^{*})||x^{*}|}{|r^{2}-|x^{*}|^{2}|}=\frac{r^{2}}{|r^{2}-|x^{*}|^{2}|}.$

It follows that in the case $|x|<r$ (in other words $|x^{*}|>r$) we get
(recall that $|y|<r$)

$\frac{|x-y|^{2}}{(r^{2}-|x|^{2})(r^{2}-|y|^{2})}=\frac{1}{|x^{*}|^{2}-r^{2}}\quad\text{hence}\quad\frac{1}{(r^{2}-|x|^{2})^{\frac{\alpha}{2}}|x-y|^{-\alpha}}=\frac{(r^{2}-|y|^{2})^{\frac{\alpha}{2}}}{(|x^{*}|^{2}-r^{2})^{\frac{\alpha}{2}}}.$

Finally, using this formula, we get, for all $y\in\mathbb{R}^{d}$, $|y|\leq
r$, and all $\alpha\geq 0$, $d\geq 2$,

$I(y):=\int_{|x|\leq
r}\frac{\mathrm{d}x}{(r^{2}-|x|^{2})^{\frac{\alpha}{2}}|x-y|^{d-\alpha}}=(r^{2}-|y|^{2})^{\frac{\alpha}{2}}\int_{|x^{*}|\geq
r}\frac{\mathrm{d}x^{*}}{(|x^{*}|^{2}-r^{2})^{\frac{\alpha}{2}}|x^{*}-y|^{d}},$

where the differential identity
$\frac{\mathrm{d}x}{|x-y|^{d}}=\frac{\mathrm{d}x^{*}}{|x^{*}-y|^{d}}$ comes
from Lemma B.2 applied to $S$ which is not an inversion but which is the
composition of an inversion with an isometry (central symmetry).

Using spherical coordinates with $\rho=|x_{*}|$ and the Funk – Hecke formula
(A.3) we get

$\displaystyle I(y)$
$\displaystyle=(r^{2}-|y|^{2})^{\frac{\alpha}{2}}\int_{|x^{*}|\geq
r}\frac{\mathrm{d}x^{*}}{(|x^{*}|^{2}-r^{2})^{\frac{\alpha}{2}}(|x^{*}|^{2}-2x^{*}\cdot
y+|y|^{2})^{\frac{d}{2}}}$
$\displaystyle=|\mathbb{S}^{d-1}|\frac{\Gamma(\frac{d}{2})}{\sqrt{\pi}\Gamma(\frac{d-1}{2})}(r^{2}-|y|^{2})^{\frac{\alpha}{2}}\int_{r}^{\infty}\int_{0}^{\pi}\frac{\rho^{d-1}\sin^{d-2}(\theta)\mathrm{d}\rho\mathrm{d}\theta}{(\rho^{2}-r^{2})^{\frac{\alpha}{2}}(\rho^{2}-2\rho|y|\cos(\theta)+|y|^{2})^{\frac{d}{2}}}$
$\displaystyle=|\mathbb{S}^{d-1}|\frac{\Gamma(\frac{d}{2})}{\sqrt{\pi}\Gamma(\frac{d-1}{2})}(r_{1}^{2}-1)^{\frac{\alpha}{2}}\int_{r_{1}}^{\infty}\frac{\rho_{1}^{d-1}}{(\rho_{1}^{2}-r_{1}^{2})^{\frac{\alpha}{2}}}\Bigr{(}\int_{0}^{\pi}\frac{\sin^{d-2}(\theta)\mathrm{d}\theta}{(\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1)^{\frac{d}{2}}}\Bigr{)}\mathrm{d}\rho_{1}$
(B.1)

where $r:=r_{1}|y|$ and $\rho:=\rho_{1}|y|$. Note that $r_{1}\geq 1$ and
$\rho_{1}\geq 1$.

###  B.5. Trigonometric change of variable

Let us show that for $d>1$ and $\rho_{1}>1$,

$i_{d}:=\int_{0}^{\pi}\frac{\sin^{d-2}(\theta)}{\left(\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1\right)^{\frac{d}{2}}}\mathrm{d}\theta=\frac{\rho_{1}^{2-d}}{\rho_{1}^{2}-1}\int_{0}^{\pi}\sin^{d-2}(\alpha)\mathrm{d}\alpha=\frac{\rho_{1}^{2-d}}{\rho_{1}^{2}-1}\sqrt{\pi}\frac{\Gamma\left(\frac{d-1}{2}\right)}{\Gamma\left(\frac{d}{2}\right)}.$
(B.2)

The second equality in (B.2) follows from the fact that the middle integral
becomes an Euler beta integral after the change of variable $u=\sin(\alpha)$.
To prove the first equality in (B.2), we follow [13, p. 400], and we use the
change of variable

$\frac{\sin(\theta)}{\sqrt{\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1}}=\frac{\sin(\alpha)}{\rho_{1}},$

see Figure 2 for a geometric interpretation444It is mentioned in [13, p. 400]
that this change of variable was suggested S.I. Greenberg. Nevertheless such
geometric reasoning goes back at least to the works on elliptic integrals of
the 19-th century, see [8].. Following this figure, we have the identity

$\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1=A(\alpha)^{2}\quad\text{where}\quad
A(\alpha)=\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}+\cos(\alpha),$

hence
$2\rho_{1}\sin(\theta)\mathrm{d}\theta=2A(\alpha)A^{\prime}(\alpha)\mathrm{d}\alpha$
and by using the formula for the change of variable this gives

$\mathrm{d}\theta=\frac{A^{\prime}(\alpha)}{\sin(\alpha)}\mathrm{d}\alpha=\frac{-\sin(\alpha)-\frac{\sin(\alpha)\cos(\alpha)}{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}}{\sin(\alpha)}\mathrm{d}\alpha=-\left(\frac{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}+\cos(\alpha)}{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}\right)\mathrm{d}\alpha.$

Therefore, we obtain, noting that $\theta=0\Longleftrightarrow\alpha=\pi$ and
$\theta=\pi\Longleftrightarrow\alpha=0$ (see Figure 2),

$\displaystyle i_{d}$
$\displaystyle=+\int_{0}^{\pi}\Bigr{(}\frac{\sin(\alpha)}{\rho_{1}}\Bigr{)}^{d-2}\frac{1}{\left(\cos(\alpha)+\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}\right)^{2}}\frac{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}+\cos(\alpha)}{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}\mathrm{d}\alpha$
$\displaystyle=+\int_{0}^{\pi}\Bigr{(}\frac{\sin(\alpha)}{\rho_{1}}\Bigr{)}^{d-2}\frac{1}{\cos(\alpha)+\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}\frac{1}{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}\mathrm{d}\alpha$
$\displaystyle=+\int_{0}^{\pi}\Bigr{(}\frac{\sin(\alpha)}{\rho_{1}}\Bigr{)}^{d-2}\frac{\cos(\alpha)-\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}{\cos^{2}(\alpha)-(\rho_{1}^{2}-\sin^{2}(\alpha))}\frac{1}{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}\mathrm{d}\alpha$
$\displaystyle=+\frac{1}{\rho_{1}^{d-2}(1-\rho_{1}^{2})}\int_{0}^{\pi}(\sin(\alpha))^{d-2}\left(\frac{\cos(\alpha)}{\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}}-1\right)\mathrm{d}\alpha$
$\displaystyle=\frac{1}{\rho_{1}^{d-2}(\rho_{1}^{2}-1)}\int_{0}^{\pi}(\sin(\alpha))^{d-2}\mathrm{d}\alpha,$

where the last equality follows from the antisymmetry of $\cos$ around
$\pi/2$. This proves (B.2).

CBAPQ$\theta$$\alpha$ Figure 2.  Geometric interpretation of the $\theta$ to
$\alpha$ change of variables for $i_{d}$. The angles and distances are
$ACB=\theta$, $CBA=\alpha$, $CB=1$ and $CA=\rho_{1}$. The right-angled
triangle $ABQ$ has hypotenuse $AB$, thus
$AB^{2}=BQ^{2}+AQ^{2}=\sin^{2}(\theta)+(AC-
QC)^{2}=\sin^{2}(\theta)+(\rho_{1}-\cos(\theta))^{2}=\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1.$
The sine rule then gives
$\frac{\sin(\alpha)}{\rho_{1}}=\frac{\sin(\theta)}{\sqrt{\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1}}.$
On the other hand, we also have
$\sqrt{\rho_{1}^{2}-2\rho_{1}\cos(\theta)+1}=AB=AP+PB=\sqrt{\rho_{1}^{2}-\sin^{2}(\alpha)}+\cos(\alpha).$

###  B.6. Conclusion

By combining (B.1) and (B.2), using the successive changes of variables
$t=\rho_{1}^{2}-r_{1}^{2}$, $t_{1}=t/(r_{1}^{2}-1)$, and $u=1/(1+t_{1})$, and
the Euler reflection formula (A.1), we get

$\displaystyle I(y)$
$\displaystyle=|\mathbb{S}^{d-1}|(r_{1}^{2}-1)^{\frac{\alpha}{2}}\int_{r_{1}}^{\infty}\frac{\rho_{1}\mathrm{d}\rho_{1}}{(\rho_{1}^{2}-r_{1}^{2})^{\frac{\alpha}{2}}(\rho_{1}^{2}-1)}$
$\displaystyle=|\mathbb{S}^{d-1}|\frac{(r_{1}^{2}-1)^{\frac{\alpha}{2}}}{2}\int_{0}^{\infty}\frac{\mathrm{d}t}{t^{\frac{\alpha}{2}}(t+r_{1}^{2}-1)}$
$\displaystyle=\frac{|\mathbb{S}^{d-1}|}{2}\int_{0}^{\infty}\frac{\mathrm{d}t_{1}}{t_{1}^{\frac{\alpha}{2}}(t_{1}+1)}$
$\displaystyle=\frac{|\mathbb{S}^{d-1}|}{2}\int_{0}^{1}\frac{u^{\frac{\alpha}{2}-1}\mathrm{d}u}{(1-u)^{\frac{\alpha}{2}}}=\frac{|\mathbb{S}^{d-1}|\Gamma(\frac{\alpha}{2})\Gamma(1-\frac{\alpha}{2})}{2}=\frac{\pi^{\frac{d}{2}+1}}{\Gamma(\frac{d}{2})\sin(\frac{\pi}{2}\alpha)}.$

This completes the proof of (1.6) and thus of Lemma 1.2.

Acknowledgments. The authors are grateful to Bent Fuglede and Wolfgang
Wendland for helpful comments on the convolution of distributions, to Franck
Wielonsky for references to Riesz’s work, and to Ekatrina Karatsuba for the
references to tables and related literature. The authors are also indebted to
Doug Hardin for the suggestion of using the spherical Laplacian.

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