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Euler's reflection formula

Statement

For every $z \in C ∖ Z$ ,

$Γ (z) Γ (1 - z) = \frac{π}{\sin (π z)} .$

The identity extends meromorphically to all $z \in C$ .

Facts & Assumptions

Given: The reciprocal-Gamma product and the sine product.

[L1]

Reciprocal Gamma has the product $1 / Γ (z) = z e^{γ z} \prod_{n \geq 1} (1 + z / n) e^{- z / n}$ (The Weierstrass product for reciprocal Gamma).

[L2]

Sine has the product $\sin (π z) = π z \prod_{n \geq 1} (1 - z^{2} / n^{2})$ (The Weierstrass product for sine).

[L3]

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

Proof

technique · direct

1.1L1L3algebra

Apply [L1] at $1 - z$ . For the $N$ th partial product, $\frac{1}{Γ (1 - z)} = (1 - z) e^{γ (1 - z)} \lim_{N \to \infty} \prod_{n = 1}^{N} (1 + \frac{1 - z}{n}) e^{- (1 - z) / n} .$ The identity $(1 - z) \prod_{n = 1}^{N} (1 + \frac{1 - z}{n}) = (N + 1) \prod_{n = 1}^{N + 1} (1 - \frac{z}{n})$ therefore gives $\frac{1}{Γ (1 - z)} = e^{- γ z} \lim_{N \to \infty} (N + 1) e^{γ - H_{N}} e^{- z / (N + 1)} \prod_{n = 1}^{N + 1} (1 - \frac{z}{n}) e^{z / n} .$ By [L3], the scalar prefactor tends to $1$ , so $\frac{1}{Γ (1 - z)} = e^{- γ z} \prod_{n \geq 1} (1 - \frac{z}{n}) e^{z / n} .$

2.1step 1.1L1L2algebra∎

Multiplying step 1.1 by the product for $1 / Γ (z)$ from [L1], the exponential factors cancel and one gets $\frac{1}{Γ (z) Γ (1 - z)} = z \prod_{n \geq 1} (1 - \frac{z^{2}}{n^{2}}) .$ By [L2], the product on the right is $\sin (π z) / π$ . Therefore $\frac{1}{Γ (z) Γ (1 - z)} = \frac{\sin (π z)}{π}$ on $C ∖ Z$ , which is equivalent to the displayed formula.

Depends on

Used by

The value of Gamma at one half Corollary
Checking the reflection formula at z=1/2 Example
Hankel's representation for reciprocal Gamma Theorem
The Riemann zeta function satisfies the classical sine-gamma functional equation Theorem

Dependency tree · two levels

16 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 §2 (standard reference, not scraped)
M. Weber, Complex Analysis, §3.7 (standard reference, not scraped)