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It is well-known that the volume of the unit ball in n-space is $\pi^{n/2}\\!/\Gamma(n/2+1)$. Do you know of a proof which explains this formula? Any proof which does not treat the cases n even and n odd separately (like using an explicit expression for $\Gamma(n/2+1)$ in terms of factorials) should be fine.

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Let's reserve soft-question tag for stuff without formulas :) –  Ilya Nikokoshev Nov 19 '09 at 20:26

2 Answers 2

It is easier to take the derivative, and consider the volume of the n-1-sphere (i.e., the "surface area" of the boundary of the ball).

Start with the integral $\int\_{\mathbb{R}^n} e^{-x\_1^2 - ... - x\_n^2} dx\_1 \dots dx\_n$. Fubini's theorem lets you decompose this into a product of 1-dimensional integrals, and you get $\pi^{n/2}$. Since the integrand is spherically symmetric, you can change to the integral $\int\_0^\infty vol(S^{n-1}(r)) e^{-r^2} dr$, where $S^{n-1}(r)$ is the unit n-1-sphere of radius r. The volume of this sphere is $r^{n-1}$ times the volume of the unit sphere, so solving for that, you get $\frac{\pi^{n/2}}{\int_0^\infty r^{n-1} e^{-r^2} dr}$. A change of coordinates (u = r2) in the denominator yields the integral defining $\Gamma(n/2)$.

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I like to write it as $\omega_n = \frac{\pi^\frac{n}{2}}{\frac{n}{2}!}$ (I've seen $\omega_n$ used both for the measure of the unit ball and for that of its boundary, but eh, I had to attach some name to it for below). I guess using the factorial notation for non-integers isn't too popular, though.

Alternatively, induction. It's true for $n=1$ (since $\Gamma(\frac{3}{2})=\frac{\sqrt{\pi}}{2}$) and $n=2$.

So then:

$\omega_{n+2} = \int_{x_1^2 + \dots + x_{n+2}^2 \leq 1}dx = \int_{x_{n+1}^2+x_{n+2}^2 \leq 1}\int_{x_1^2 + \dots + x_n^2 \leq 1 - (x_{n+1}^2+x_{n+2}^2)}d(x_1,\dots,x_n)d(x_1,x_2).$

Polar coordinates in the plane give us

$\omega_{n+2} = \int_0^{2\pi}\int_0^1\sqrt{1-r^2}^n\omega_n r dr d\phi = 2\pi\omega_n \int_0^1(1-r^2)^{\frac{n}{2}}rdr = \pi\omega_n \int_0^1(1-r^2)^{\frac{n}{2}}2rdr.$

Substitute $s=1-r^2$ and get

$\omega_{n+2} = \pi\omega_n\int_0^1s^\frac{n}{2}ds = \omega_n \frac{2\pi}{n+2} = \frac{\pi^\frac{n}{2}}{\frac{n}{2}!}\frac{2\pi}{n+2} = \frac{\pi^\frac{n+2}{2}}{\frac{n+2}{2}!}.$

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That's meant to be $d(x_1,\dots,x_n)d(x_{n+1},x_{n+2})$, of course. –  Jakob Katz Nov 19 '09 at 21:07

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