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Prove the formula $$A_d=\frac{2\pi^{d/2}}{\Gamma (d/2)}$$ where $A_d$ denote the area of the unit sphere in $\mathbb{R}^d$. Use polar coordinates and the fact that $\int_{\mathbb{R}^d}e^{-\pi |x|^2}dx=1.$

So I need to find out $\int_{S^{d-1}} d\sigma(\gamma),$ and I think this corresponds to, using spherical coordinates, $\int_0^\pi \int_0^\pi \cdots \int_0^{2\pi} \sin^{d-2}\theta_1 \sin^{d-3}\theta_2 \cdots \sin \theta_{d-2}d\theta_{d-1}\cdots d\theta_1.$

I'm trying to prove this exercise following the hint, but I don't know how to derive a the integral of a Gaussian in the first place to compute this. I would greatly appreciate any help.

nomadicmathematician
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1 Answers1

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Let $A_d(R)$ be the area of the $d$-dimensional unit sphere with radius $R$.
Obviously $A_d(R) = R^{d-1} A_d$, and by Cavalieri's principle $$ 1 = \int_{\mathbb{R}^d}\exp(-\pi|x|^2)\,d\mu = \int_{0}^{+\infty}A_d(R) e^{-\pi R^2}\,dR=A_d\int_{0}^{+\infty}R^{d-1}e^{-\pi R^2}\,dR\tag{1}$$ where $$ \int_{0}^{+\infty}R^{d-1}e^{-\pi R^2}\,dR = \frac{1}{2}\int_{0}^{+\infty}z^{d/2-1}e^{-\pi z}\,dz = \frac{\Gamma(d/2)}{2\pi^{d/2}}\tag{2} $$ follows from the integral definition of the $\Gamma$ function. By $(1)$ and $(2)$, $$ A_d = \frac{2\pi^{d/2}}{\Gamma(d/2)} \tag{3}$$ as wanted.

Jack D'Aurizio
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