Pome69D141CB2664Theorem 2.9.6.69D141CB2664 Let $\mu $ be a finite positive Borel measure on $\mathbb R^ d$, and let $0<s<d$. Then
where $\Gamma (a)=\int _0^\infty v^{a-1}\mathrm e^{-v}\, \mathrm d v$, $a>0$, is Euler's Gamma function. Both sides may be infinite.
Proof. For $z\in \mathbb R^ d$, the substitution $v=\pi t|z|^2$ gives
and both sides are $+\infty $ when $z=0$. By lemma 2.9.2 with $t$ replaced by $1/t$, we have $\mathrm e^{-\pi t|z|^2}=t^{-d/2}\int \mathrm e^{-\pi |\xi |^2/t}\mathrm e^{-2\pi \mathrm i z\cdot \xi }\, \mathrm d\xi $. Taking $z=x-y$ and integrating with respect to $\mu \times \mu $, we obtain
Fubini's theorem applies here because the Gaussian in $\xi $ is integrable and $\mu \times \mu $ is finite. Now insert the integral formula for $|x-y|^{-s}$ into $I_ s(\mu )$. All integrands are nonnegative, so Tonelli's theorem allows us to integrate first in $(x,y)$, then use the last identity, and finally integrate in $t$ before $\xi $, even when the integrals are infinite:
For $\xi \neq 0$, the substitution $v=\pi |\xi |^2/t$ evaluates the inner integral as
and the point $\xi =0$ is a Lebesgue null set. Multiplying the constants gives the formula. $\square$
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