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Stirling's formula for Gamma

Statement

Fix $δ$ with $0 < δ < π$ . On the closed sector $∣ \arg z ∣ \leq π - δ$ , using the principal logarithm in $z^{z - 1 / 2} : = \exp ((z - 1 / 2) Log z)$ , one has

$Γ (z) = \sqrt{2 π} z^{z - 1 / 2} e^{- z} (1 + O_{δ} (∣ z ∣^{- 1}))$

as $∣ z ∣ \to \infty$ .

Facts & Assumptions

Given: A fixed closed sector $∣ \arg z ∣ \leq π - δ$ .

[L1]

Reciprocal Gamma has the Weierstrass product (The Weierstrass product for reciprocal Gamma).

[L2]

Harmonic numbers satisfy $H_{n} = \log n + γ + o (1)$ (The Euler–Mascheroni constant and the harmonic asymptotic).

[L3]

The real Stirling formula gives $\log (N - 1)! = (N - \frac{1}{2}) \log N - N + \frac{1}{2} \log (2 π) + o (1)$ as $N \to \infty$ through the positive integers (Stirling's formula for factorials).

Proof

technique · direct

1.1

Taking logarithms in [L1] on the chosen sector gives [L1, given, algebra] $\log Γ (z) = - \log z - γ z + \sum_{n \geq 1} (\frac{z}{n} - Log (1 + \frac{z}{n})),$ where $Log$ denotes the principal logarithm.

L1givenalgebra

2.1

For an integer $N \geq 1$ , define [step 1.1, algebra] $I_{N} (z) : = \int_{0}^{N} \frac{⌊ u ⌋ - u + 1 / 2}{u + z} d u .$ On each interval $[n, n + 1]$ with $0 \leq n \leq N - 1$ , one has $\int_{n}^{n + 1} \frac{n - u + 1 / 2}{u + z} d u = (n + \frac{1}{2} + z) Log \frac{n + 1 + z}{n + z} - 1.$ Summing these equalities and telescoping the logarithms yields $I_{N} (z) = (N + z - \frac{1}{2}) Log (N + z) - (z + \frac{1}{2}) Log z - N - \log (N - 1)! - z H_{N - 1} + \sum_{n = 1}^{N - 1} (\frac{z}{n} - Log (1 + \frac{z}{n})) .$

step 1.1algebra

3.1

For fixed $z$ in the sector, [L2] and [L3] give [L2, L3, step 1.1, step 2.1, algebra] $H_{N - 1} = \log N + γ + o (1), \log (N - 1)! = (N - \frac{1}{2}) \log N - N + \frac{1}{2} \log (2 π) + o (1),$ and also $Log (N + z) = \log N + o (1)$ as $N \to \infty$ . Comparing the $N \to \infty$ limit of step 2.1 with the partial sums in step 1.1 gives the Binet-type formula $\log Γ (z) = (z - \frac{1}{2}) Log z - z + \frac{1}{2} \log (2 π) + \int_{0}^{\infty} \frac{⌊ u ⌋ - u + 1 / 2}{u + z} d u .$

L2L3step 1.1step 2.1algebra

4.1

Let [step 3.1, algebra] $Φ (u) : = \int_{0}^{u} (⌊ v ⌋ - v + 1 / 2) d v .$ Then $Φ$ is $1$ -periodic and therefore bounded. Integrating by parts in step 3.1 gives $\int_{0}^{\infty} \frac{⌊ u ⌋ - u + 1 / 2}{u + z} d u = \int_{0}^{\infty} \frac{Φ (u)}{(u + z)^{2}} d u .$ On the closed sector $∣ \arg z ∣ \leq π - δ$ , one has $∣ u + z ∣ \geq c_{δ} (u + ∣ z ∣)$ for a constant $c_{δ} > 0$ , so the integral above is $O_{δ} (∣ z ∣^{- 1})$ uniformly as $∣ z ∣ \to \infty$ . Hence $\log Γ (z) = (z - \frac{1}{2}) Log z - z + \frac{1}{2} \log (2 π) + O_{δ} (∣ z ∣^{- 1}),$ and exponentiating yields $Γ (z) = \sqrt{2 π} z^{z - 1 / 2} e^{- z} (1 + O_{δ} (∣ z ∣^{- 1})) .$

step 3.1algebra∎

Depends on

Used by

Stirling's approximation for 10! Example

Dependency tree · two levels

13 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 §6 (standard reference, not scraped)