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The completed zeta function satisfies $Λ (s) = Λ (1 - s)$

Statement

Assume countable choice.

The completed zeta function extends meromorphically to $C$ , has simple poles at $0$ and $1$ , and satisfies

$Λ (s) = Λ (1 - s) .$

More explicitly,

$Λ (s) = \frac{1}{s (s - 1)} + \frac{1}{2} \int_{1}^{\infty} (θ (t) - 1) (t^{s / 2 - 1} + t^{- s / 2 - 1 / 2}) d t,$

and the right-hand side is symmetric under $s \mapsto 1 - s$ .

Facts & Assumptions

Given: Countable choice and the completed function on $Re s > 1$ .

[A2]

Countable choice is The Axiom of Countable Choice ( ${A C}_{ω}$ ); it supplies the hypotheses of [L2] and [L4].

[L1]

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

[L2]

The theta transformation is $θ (t) = t^{- 1 / 2} θ (1 / t)$ (The Jacobi theta function satisfies $θ (t) = t^{- 1 / 2} θ (1 / t)$ ).

[L3]

On $Re s > 1$ , $Λ (s) = \frac{1}{2} \int_{0}^{\infty} (θ (t) - 1) t^{s / 2 - 1} d t$ (The completed zeta function has its Mellin-theta integral representation on $Re s > 1$ ).

[L4]

Under countable choice, the meromorphic continuation theorem constructs the entire function $H (s) = \frac{1}{2} \int_{1}^{\infty} (θ (t) - 1) (t^{s / 2 - 1} + t^{- s / 2 - 1 / 2}) d t$ and gives $Λ (s) = 1 / (s (s - 1)) + H (s)$ on $C$ (The Riemann zeta function extends meromorphically to the complex plane with its only pole at $1$ , Proof 2.1).

[A1]

Two meromorphic functions on a connected domain that agree on a nonempty open subset agree everywhere on that domain.

Proof

technique · direct

1.1givenA2L2L3algebra

Under [A2], repeating the split-at- $1$ calculation from the Mellin integral in [L3] and using [L2] on $(0, 1)$ gives $Λ (s) = \frac{1}{s (s - 1)} + \frac{1}{2} \int_{1}^{\infty} (θ (t) - 1) (t^{s / 2 - 1} + t^{- s / 2 - 1 / 2}) d t$ for $Re s > 1$ .

1.2L4algebra

Define $F (s) : = 1 / (s (s - 1)) + H (s)$ on $C$ , using the globally convergent integral for $H$ in [L4]. For every $s \in C$ , its two powers of $t$ exchange under $s \mapsto 1 - s$ , and $1 / (s (s - 1)) = 1 / ((1 - s) (- s))$ . Therefore $F (s) = F (1 - s)$ as meromorphic functions on $C$ . This conclusion uses the global integral identity, not merely its initial validity on $Re s > 1$ .

2.1step 1.1step 1.2L1L4A1algebra∎

On $Re s > 1$ , [L1] names the completed function as $Λ (s)$ and step 1.1 identifies it with $F$ . By [L4] this is its meromorphic continuation, so [A1] identifies $F$ with $Λ$ on all of $C$ . Step 1.2 now gives $Λ (s) = F (s) = F (1 - s) = Λ (1 - s)$ . Since $H$ is entire, the nonzero simple principal parts of $1 / (s (s - 1))$ give simple poles at $0$ and $1$ .

Depends on

Used by

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Sources

Elias M. Stein and Rami Shakarchi, Complex Analysis, Theorem 2.3 (standard reference, not scraped)
K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 12 §7 (standard reference, not scraped)

The completed zeta function satisfies Λ(s)=Λ(1−s)Λ(s)=Λ(1−s)