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Example 2.9.7E243D52E8F6A

Example 2.9.7.E243D52E8F6A The area of the unit sphere $S^{d-1}\subset \mathbb R^ d$ is

AC1522EBF063 2.9.8
\begin{equation} s_{d-1}=\frac{2\pi ^{d/2}}{\Gamma (d/2)}. \end{equation}

For $0<s<d$, the probability measure $\mathrm d\mu =\mathrm e^{-\pi |x|^2}\, \mathrm d x$ on $\mathbb R^ d$ satisfies

\[ I_ s(\mu )=\left(\frac\pi 2\right)^{s/2}\frac{\Gamma ((d-s)/2)}{\Gamma (d/2)}. \]

Proof. By lemma 2.9.2 with $\xi =0$ and $t=1$, the Gaussian $\mathrm e^{-\pi |x|^2}$ has integral one. Integrating in polar coordinates and substituting $v=\pi r^2$, we obtain

\[ 1=\int \mathrm e^{-\pi |x|^2}\, \mathrm d x=s_{d-1}\int _0^\infty \mathrm e^{-\pi r^2}r^{d-1}\, \mathrm d r=\frac{s_{d-1}\Gamma (d/2)}{2\pi ^{d/2}}, \]

which proves (2.9.8). By lemma 2.9.2, $\widehat\mu (\xi )=\mathrm e^{-\pi |\xi |^2}$. Hence theorem 2.9.6 and polar coordinates give

\[ I_ s(\mu )=b_{d,s}\int \mathrm e^{-2\pi |\xi |^2}|\xi |^{s-d}\, \mathrm d\xi =b_{d,s}s_{d-1}\int _0^\infty \mathrm e^{-2\pi r^2}r^{s-1}\, \mathrm d r. \]

The substitution $v=2\pi r^2$ evaluates the radial integral as $\Gamma (s/2)/(2(2\pi )^{s/2})$. Inserting the values of $b_{d,s}$ and $s_{d-1}$ proves the formula. $\square$

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