TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-24

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The real Gamma functional equation $Γ (s + 1) = s Γ (s)$

Statement

For every $s > 0$ , $Γ (s + 1) = s Γ (s)$ , and $Γ (1) = 1$ .

Facts & Assumptions

Given: A real parameter $s > 0$ and truncation parameters $0 < ε < 1 < R$ .

[F1]

If differentiable $u, v$ on a compact interval have integrable derivatives, then $\int u v^{'} = u v ∣_{a}^{b} - \int u^{'} v$ (If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ ).

[F2]

The Euler integral converges if and only if its real parameter is positive (Euler's Gamma integral converges exactly for positive real parameters).

Proof

technique · direct

1.1givenF1

Apply [F1] on $[ε, R]$ with $u (t) = t^{s}$ and $v^{'} (t) = e^{- t}$ : $\int_{ε}^{R} t^{s} e^{- t} d t = [- t^{s} e^{- t}]_{ε}^{R} + s \int_{ε}^{R} t^{s - 1} e^{- t} d t$ .

2.1step 1.1algebra

Since $s > 0$ , $ε^{s} e^{- ε} \to 0$ as $ε ↓ 0$ . At infinity choose a natural $m > s$ ; then $R^{s} e^{- R} \leq R^{m} e^{- R} \to 0$ by exponential domination.

3.1step 1.1step 2.1F2

Letting the two truncations approach their improper ends in step 1.1 and using [F2] gives $Γ (s + 1) = s Γ (s)$ .

4.1F2algebra∎

At $s = 1$ , $Γ (1) = \int_{0}^{\infty} e^{- t} d t = [- e^{- t}]_{0}^{\infty} = 1$ .

The real Gamma functional equation $Γ (s + 1) = s Γ (s)$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The real Gamma functional equation Γ(s+1)=sΓ(s)Γ(s+1)=sΓ(s)

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

The real Gamma functional equation $Γ (s + 1) = s Γ (s)$