LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-30

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Euler's Gamma integral converges locally uniformly on the right half-plane

Statement

The improper integral

$\int_{0}^{\infty} t^{z - 1} e^{- t} d t$

converges locally uniformly on ${z \in C : Re z > 0}$ .

Facts & Assumptions

Given: A compact set $K \subseteq {z : Re z > 0}$ .

[L1]

The real Euler integral $\int_{0}^{\infty} t^{s - 1} e^{- t} d t$ converges exactly for $s > 0$ (Euler's Gamma integral converges exactly for positive real parameters).

Proof

technique · direct

1.1

Since $K$ is compact in the open right half-plane, choose real numbers $0 < a \leq b$ with $a \leq Re z \leq b$ for every $z \in K$ . For $0 < t \leq 1$ one has $∣ t^{z - 1} e^{- t} ∣ = t^{Re z - 1} e^{- t} \leq t^{a - 1}$ , and for $t \geq 1$ one has $∣ t^{z - 1} e^{- t} ∣ \leq t^{b - 1} e^{- t}$ .

givenchoosealgebra

2.1

By [L1], both majorants from step 1.1 have convergent improper integrals on their respective ranges. The Weierstrass M-test on compact subsets therefore gives local-uniform convergence of the truncated integrals to the Gamma integral on $K$ .

step 1.1L1algebra∎

Depends on

Used by

Euler's Gamma function is holomorphic on the right half-plane Theorem
The Gamma functional equation Theorem

Cited to discharge well-definedness by Euler's Gamma function on the right half-plane.

Dependency tree · two levels

11 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 6 §1(ii) (standard reference, not scraped)
M. Weber, Complex Analysis, §3.7 (standard reference, not scraped)