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The Riemann zeta function satisfies the classical sine-gamma functional equation

Statement

Assume countable choice.

For all $s \in C$ ,

$ζ (s) = 2^{s} π^{s - 1} \sin (π s / 2) Γ (1 - s) ζ (1 - s),$

as an identity of meromorphic functions.

Facts & Assumptions

Given: Countable choice and the completed functional equation.

[L1]

The completed zeta function satisfies $π^{- s / 2} Γ (s / 2) ζ (s) = π^{- (1 - s) / 2} Γ ((1 - s) / 2) ζ (1 - s)$ (The completed zeta function satisfies $Λ (s) = Λ (1 - s)$ ).

[L2]

Euler's reflection formula is $Γ (z) Γ (1 - z) = \frac{π}{\sin (π z)}$ (Euler's reflection formula).

[L3]

Legendre's duplication formula is $Γ (z) Γ (z + 1 / 2) = 2^{1 - 2 z} \sqrt{π} Γ (2 z)$ (Legendre's duplication formula).

Proof

technique · direct

1.1L1givenalgebra

Rearranging [L1] gives $ζ (s) = π^{s - 1 / 2} \frac{Γ ((1 - s) / 2)}{Γ (s / 2)} ζ (1 - s) .$

1.2L2L3algebra

Apply [L3] with $z = (1 - s) / 2$ to obtain $Γ ((1 - s) / 2) Γ (1 - s / 2) = 2^{s} \sqrt{π} Γ (1 - s) .$ Apply [L2] with $z = s / 2$ to obtain $Γ (s / 2) Γ (1 - s / 2) = \frac{π}{\sin (π s / 2)} .$ Dividing the first identity by the second yields $π^{s - 1 / 2} \frac{Γ ((1 - s) / 2)}{Γ (s / 2)} = 2^{s} π^{s - 1} \sin (π s / 2) Γ (1 - s) .$

2.1step 1.1step 1.2algebra∎

Substitute the factor identity from step 1.2 into step 1.1. This gives $ζ (s) = 2^{s} π^{s - 1} \sin (π s / 2) Γ (1 - s) ζ (1 - s),$ which is the classical functional equation.

Depends on

Used by

Dependency tree · two levels

17 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

Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 6 §2.1 (standard reference, not scraped)
K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 12 §7 (standard reference, not scraped)