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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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    $\begingroup$ Let's reserve soft-question tag for stuff without formulas :) $\endgroup$ Nov 19, 2009 at 20:26
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    $\begingroup$ More memorably, the formula is $\pi^{n/2} \! \left/ \Pi(n/2) \right.$, using Euler's original notation $\Pi(s)$ for what we now call $\Gamma(s+1)$ (so $\Pi(n) = n!$). I learned this from P.X.Gallagher though for all I know it might go back to Euler Himself. $\endgroup$ Oct 30, 2015 at 1:00

5 Answers 5

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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 $\int_{\mathbb{R}^n} e^{-x_1^2 - \cdots - x_n^2} dx_1 \cdots dx_n$. 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 $\int_0^\infty \mathrm{vol}(S^{n-1}(r)) \, e^{-r^2} dr$, 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 ($u = r^2$) in the denominator yields the integral defining $\Gamma(n/2)$.

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  • $\begingroup$ Isn't it true that given any $G(r)$ on $[0,\infty)$ such that $g(x_1,\ldots, x_n) := G(\sqrt{x_1^2+\ldots + x_n^2})$ is a rotationally symmetric probability distribution on $\mathbb R^n$, then the same computation holds, i.e. $1 = \int_{\mathbb R^n} g d x = \int_0^\infty \text{vol}(S^{n-1}(r)) G(r) d r$? In that case, why is the Gaussian the most natural choice? Perhaps it is the fact that the Gaussian is the only distribution that is rotational symmetric and has independent coordinates, so it's the only that gives a sort of "uniform" formula for all $n$? $\endgroup$
    – D.R.
    Feb 24, 2022 at 23:15
  • $\begingroup$ @D.R. The Gaussian lets you reduce to a product of 1-dimensional integrals. This is incredibly powerful. $\endgroup$
    – S. Carnahan
    Feb 28, 2022 at 8:30
  • $\begingroup$ It's certainly a nice property, but not entirely relevant/critical here, right? In fact I don't really know where exactly such a property is "powerful"/useful. $\endgroup$
    – D.R.
    Feb 28, 2022 at 9:05
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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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  • $\begingroup$ That's meant to be $d(x_1,\dots,x_n)d(x_{n+1},x_{n+2})$, of course. $\endgroup$
    – Jakob Katz
    Nov 19, 2009 at 21:07
  • $\begingroup$ I really like this one. I've never seen it proved this way, I've only ever seen the $e^{-x^2}$ way and some even more incredible way my linear algebra teacher did using inner product spaces. $\endgroup$
    – user78249
    Nov 2, 2016 at 20:36
  • $\begingroup$ Years ago P.X.Gallagher noted the even more memorable form $$ \pi^{n/2} \bigl/ \Pi(n/2) \bigr. $$ where $\Pi(x)$ is Euler's original notation for what we now call $\Gamma(x+1)$. $\endgroup$ Nov 2, 2016 at 21:01
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Here is an argument similar to that of S. Carnahan, but more direct (without sphere surface area.)

Denote the volume of a unit ball in $\mathbb{R}^n$ by $c_n$, then the volume of a ball of radius $r$ equals $c_nr^n$.

We use the (Lebesgue-Stieltjes?) formula $\int fd\mu=\int_0^\infty \mu\{x:f(x)>t\}dt$ for a positive function $f$ on a measure space $(X,\mu)$.

Let $(X,\mu)$ be $\mathbb{R}^n$ with Lebesgue measure and $f(x)=e^{-\sum x_i^2}$. Then $$\pi^{n/2}=\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^n=\int fd\mu=\int_0^1 c_n(-\log t)^{n/2}dt=\\=c_n\int_0^\infty s^{n/2}e^{-s}ds=c_n\Gamma(n/2+1),$$ where we substitute $t=e^{-s}$.

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The direct way to do this is to integrate $1$ using the spherical volume element $$ d^nV = r^{n-1} \sin^{n-2}(\phi_1) \, \sin^{n-3}(\phi_2) \, \cdots \sin(\phi_{n-2}) \, dr \, d\phi_1 \, \cdots \, d\phi_{n-1}. $$ Here, for $\phi_k$ goes from $0$ to $\pi$ for $k = 1, \dots, n-2$, and $\phi_{n-1}$ goes from $0$ to $2 \, \pi$.

Since $$ \int_0^\pi \sin^k \phi \, d\phi = B\left(\frac12, \frac{k+1}{2} \right) =\frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{k+1}{2}\right) } { \Gamma\left(\frac{k+2}{2} \right) }, $$ where $B(p,q)$ is the beta function, we get $$ \begin{align} \int d^nV &= \int_0^1 r^{n-1} \, dr \, \frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{n-1}{2}\right) } { \Gamma\left(\frac{n}{2} \right) } \frac{ \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{n-2}{2}\right) } { \Gamma\left(\frac{n-1}{2} \right) } \cdots \frac{ 2 \, \Gamma\left(\frac{1}{2} \right) \Gamma\left(\frac{1}{2}\right) } { \Gamma\left(\frac{2}{2} \right) } \\ &= \frac{2}{n} \frac{ \left[ \Gamma\left(\frac{1}{2} \right)\right]^n } { \Gamma\left(\frac{n}{2} \right) } = \frac{ {\sqrt\pi}^n } { \Gamma\left(\frac{n}{2} + 1\right) }. \end{align} $$


Another way is to use the formula for the Fourier transform of a spherical function $$ \tilde f(k) = \int f(r) \, e^{i\mathbf k \cdot \mathbf r} \, d\mathbf r = \frac{(2\pi)^{n/2}} { k^{n/2 - 1} } \int_0^\infty f(r) J_{n/2-1}(k\,r)\, r^{n/2} \, dr, $$ where $J_{n/2-1}(k \, r)$ is the Bessel function.

In our case $f(r) = \Theta(1-r)$ is the step function. Then, by using the recurrence relation $$ \frac{d}{dx} \left[ r^{\nu} J_{\nu}(x) \right] = r^{\nu} J_{\nu-1}(x), $$ with $\nu = n/2$, we get $$ \tilde f(k) = \frac{(2\pi)^{n/2}} { k^{n/2} } J_{n/2}(k). $$ For the limit of $k\rightarrow 0$, we get the volume $$ \lim_{k \rightarrow 0} \tilde f(k) = \frac{(2\pi)^{n/2}} { k^{n/2} } \frac{(k/2)^{n/2}}{\Gamma(\frac{n}{2}+1)} = \frac{\pi^{n/2}} {\Gamma(\frac{n}{2}+1)}. $$

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Adding to what Jakob Katz wrote, where the volume of the unit ball $B_{2n}$ in $\mathbb{C}^n$ is $\pi^n/n!$, the $\pi^n$ calls to mind the volume of a polydisc $B_2^n \subset (\mathbb{C}^1)^n = \mathbb{C}^n$, on which the symmetric group $S_n$ acts by permuting coordinates. The permutation action is volume-preserving, and so $\pi^n/n!$ is naturally interpreted as the volume of the orbit space $B_2^n/S_n$. Can this be somehow related to the volume of the $n$-ball?

Andreas Blass and Stephen Schanuel cooked up such an explanation:

Theorem: (Blass-Schanuel) Given $(z_1, \ldots, z_n) \in \mathbb{C}^n$, write coordinates $z_j$ in polar coordinate form $z_j = r_j e^{i \theta_j}$, and define an $S_n$-invariant map $\phi \colon B_2^n \to B_{2 n}$ by first permuting the $z_j$ so that $r_1 \geq r_2 \geq \ldots \geq r_n$ and then mapping $(z_1, \ldots, z_n)$ to $$(\sqrt{r_1^2 - r_2^2}e^{i\theta_1}, \sqrt{r_2^2 - r_3^2}e^{i(\theta_1 + \theta_2)}, \ldots, \sqrt{r_{n-1}^2-r_n^2}e^{i(\theta_1 + \theta_2 + \ldots + \theta_{n-1})}, r_n e^{i(\theta_1 + \theta_2 + \ldots + \theta_n)})$$ Then $\phi$ induces a continuous well-defined map $B_2^n/S_n \to B_{2 n}$. Furthermore, when restricted to the set $P_n$ of $(z_1, \ldots, z_n)$ for which the $r_j$ are all distinct, $\phi$ induces a smooth symplectic isomorphism mapping $P_n/S_n$ onto the set $Q_n$ of $(w_1, \ldots, w_n) \in B_{2 n}$ for which $w_j \neq 0$ for $1 \leq j \leq n-1$.

In other words, writing $z_j = x_j + i y_j$ the symplectic 2-form

$$\sum_{j=1}^n d x_j \wedge d y_j = \sum_{j=1}^n r_j d r_j \wedge d\theta_j$$

is preserved by pulling back along $\phi \colon P_n/S_n \to Q_n$. Since symplectic maps are locally volume-preserving, and since $P_n$ and $Q_n$ are almost all of $B_2^n$ and $B_{2 n}$ respectively, this gives a proof that the volume of $B_{2 n}$ is $\pi^n/n!$ (alternate to standard purely computational proofs).

Reference:

  • Andreas Blass, Stephen Schanuel, On the volumes of balls (ps).
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    $\begingroup$ Thanks for publicizing my paper with Steve Schanuel. The paper also contains a slightly simpler version of the map, preserving volumes but not the symplectic structure. While Steve and I were procrastinating plans for amplifying our paper, that simpler map was independently found by Omar Hijab; see "The volume of the unit ball in $C^n$," Amer. Math. Monthly107 (2000) 259. (When Steve found this paper, he notified me with an email that began "If one postpones a task long enough, eventually it is no longer necessary.") $\endgroup$ Oct 30, 2015 at 10:45

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