TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-30

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Gauss's multiplication formula

Statement

For every integer $m \geq 1$ and every $z \in C$ away from the poles of the factors,

$\prod_{k = 0}^{m - 1} Γ ⁣ (z + \frac{k}{m}) = (2 π)^{(m - 1) / 2} m^{1 / 2 - m z} Γ (m z) .$

Facts & Assumptions

Given: An integer $m \geq 1$ and a complex number $z$ off the poles.

[L1]

Euler's limit formula holds for Gamma (Euler's limit formula for Gamma).

[L2]

Real Stirling gives $n! \sim \sqrt{2 π n} (n / e)^{n}$ as $n \to \infty$ (Stirling's formula for factorials).

Proof

technique · direct

1.1L1givenalgebra

Apply [L1] to each factor $Γ (z + k / m)$ and multiply. The denominator collapses by $\prod_{k = 0}^{m - 1} \prod_{j = 0}^{n} (z + \frac{k}{m} + j) = m^{- m (n + 1)} \prod_{r = 0}^{m (n + 1) - 1} (m z + r) .$ Therefore $\prod_{k = 0}^{m - 1} Γ ⁣ (z + \frac{k}{m}) = Γ (m z) \lim_{n \to \infty} \frac{(n!)^{m} m^{m (n + 1)} n^{m z + (m - 1) / 2}}{(m (n + 1) - 1)! (m (n + 1) - 1)^{m z}} .$

2.1step 1.1L2algebra∎

Apply [L2] to the factorial ratio in step 1.1. After the standard cancellation of the exponential and power terms, the limit becomes $(2 π)^{(m - 1) / 2} m^{1 / 2 - m z}$ . Substituting this into step 1.1 yields the displayed multiplication formula.

Depends on

Used by

Legendre's duplication formula Theorem

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 7 §5 (standard reference, not scraped)