LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Gamma functional equation away from the poles

Statement

For every $z \in C ∖ {0, - 1, - 2, \dots}$ , $Γ (z + 1) = z Γ (z),$ where $Γ$ is the meromorphic continuation of Euler's Gamma function.

Facts & Assumptions

Given: Euler's Gamma integral on $Re z > 0$ and $z \in C ∖ {0, - 1, - 2, \dots}$ .

[F1]

Euler's Gamma function on the right half-plane and Euler's Gamma integral converges locally uniformly on the right half-plane: Euler's integral defines $Γ (w)$ on $Re w > 0$ .

[F2]

Euler's Gamma function is holomorphic on the right half-plane: this integral defines a holomorphic function on the right half-plane.

[F3]

The Gamma functional equation: if $Re w > 0$ , then $Γ (w + 1) = w Γ (w)$ .

[F4]

Dedekind completeness implies the Archimedean property: for every real $r$ , some integer $m \geq 1$ satisfies $r > - m$ .

Proof

1.1givenF1F2algebraconstruct

For each integer $m \geq 0$ , let $V_{m} = {w \in C : Re w > - m - 1}$ and define $G_{m} (w) = Γ (w + m + 1) / (w (w + 1) \dots (w + m))$ as a meromorphic function on $V_{m}$ . Its numerator is holomorphic on this half-plane by [F1] and [F2], and its denominator is a polynomial.

2.1givenF3F4step 1.1algebra

For every $w \in V_{m}$ , $Re (w + m + 1) > 0$ . Applying [F3] at $w + m + 1$ and cancelling the common meromorphic factor $w + m + 1$ gives $G_{m + 1} (w) = G_{m} (w)$ on the overlap $V_{m} \subset V_{m + 1}$ . By [F4], the union of the $V_{m}$ is $C$ , so these compatible meromorphic formulas define a meromorphic continuation of Euler's Gamma function on the whole plane.

3.1givenF4step 2.1algebrachoose∎

Choose $m \geq 1$ large enough that $z \in V_{m}$ ; then $z + 1 \in V_{m - 1}$ . Because $z$ is not a nonpositive integer, the denominator factors below are nonzero. The glued formulas give $Γ (z) = Γ (z + m + 1) / (z (z + 1) \dots (z + m))$ and $Γ (z + 1) = Γ (z + m + 1) / ((z + 1) \dots (z + m))$ . Multiplying the first identity by $z$ yields the second.

Depends on

Used by

Shifted log-Gamma branch is independent of the shift Lemma

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

DLMF, §5.5, Functional Relations (standard reference, not scraped)
N. M. Temme, “Numerical aspects of special functions,” Acta Numerica 17 (2008), §§2 and 3 (standard reference, not scraped)