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The closed form for the volume of the unit $n$ -ball

Statement

For every $n \geq 1$ , $V_{n} (1) = π^{n / 2} / Γ (n / 2 + 1)$ . Here $n$ is an integer.

Facts & Assumptions

Given: Unit-ball volumes $V_{n} (1)$ in positive integer dimensions.

[F1]

One has $V_{1} (r) = 2 r$ for $r \geq 0$ , and for $n \geq 2$ , $V_{n} (1) = V_{n - 1} (1) \int_{- 1}^{1} (1 - t^{2})^{(n - 1) / 2} d t$ (Closed Euclidean balls are Jordan measurable and their volumes satisfy the slicing recursion).

[F2]

For $p, q > 0$ , $B (p, q) = Γ (p) Γ (q) / Γ (p + q)$ (The real Beta--Gamma identity).

[F3]

$Γ (1 / 2) = \sqrt{π}$ ( $Γ (1 / 2) = \sqrt{π}$ from the Gaussian integral).

[F4]

For $s > 0$ , $Γ (s + 1) = s Γ (s)$ (The real Gamma functional equation $Γ (s + 1) = s Γ (s)$ ).

Proof

technique · induction

1.1F1F3F4basealgebra

In dimension $1$ , [F1] gives $V_{1} (1) = 2$ , while [F3] and [F4] give $π^{1 / 2} / Γ (3 / 2) = \sqrt{π} / ((1 / 2) \sqrt{π}) = 2$ .

1.2ihF1F2F3algebra

Assume the formula in dimension $n - 1$ . In the integral of [F1], symmetry followed by $u = t^{2}$ gives $\int_{- 1}^{1} (1 - t^{2})^{(n - 1) / 2} d t = B (1 / 2, (n + 1) / 2)$ . By [F2] and [F3], this equals $\sqrt{π} Γ ((n + 1) / 2) / Γ (n / 2 + 1)$ .

2.1step 1.2discharge-inductionalgebra∎

Multiply the expression in step 1.2 by the induction value $π^{(n - 1) / 2} / Γ ((n + 1) / 2)$ ; the common Gamma factor cancels, giving $V_{n} (1) = π^{n / 2} / Γ (n / 2 + 1)$ . This discharges the induction.

The closed form for the volume of the unit $n$ -ball

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The closed form for the volume of the unit nn-ball

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The closed form for the volume of the unit $n$ -ball