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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

1.1

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

givenF1

2.1

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.

step 1.1algebra

3.1

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

step 1.1step 2.1F2

4.1

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

F2algebra∎

Depends on

Used by

Gautschi's inequality for the real Gamma function Corollary
The closed form for the volume of the unit n-ball Corollary
The elementary recurrences for the real Beta function Corollary
The unit-ball volume is maximal in dimension five Corollary
The volume of the unit n-ball tends to zero with dimension Corollary
Γ(n+1)=n! for every natural number n Corollary
A positive non-log-convex solution of the Gamma functional equation Example
Special values of the real Gamma and Beta functions Example
The unit-ball volumes through dimension eight from the Gamma formula Example
Bohr--Mollerup characterisation of the real Gamma function Theorem
The real Gamma function has one minimum and diverges at both ends of its domain Theorem
Γ(1/2)=√π by Wallis's product Theorem

Dependency tree · two levels

37 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

University of Toronto MAT237Y1, The Gamma Function and the Beta Function, §1.2 (standard reference, not scraped)
K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 6 §1(iv) (standard reference, not scraped)

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