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

Statement

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: The completed function on $Re s > 1$ .

[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]

The meromorphic continuation theorem yields an entire function $H$ with $Λ (s) = \frac{1}{s (s - 1)} + H (s)$ on $C$ (The Riemann zeta function extends meromorphically to the complex plane with its only pole at $1$ ).

[A1]

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

Proof

technique · direct

1.1

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 $Re s > 1$ .

givenL2L3algebra

2.1

Define $F (s) : = \frac{1}{s (s - 1)} + H (s),$ where $H$ is the entire function from [L4]. By step 1.1, on $Re s > 1$ this equals the explicit right-hand side there. That explicit formula is unchanged when $s$ is replaced by $1 - s$ , because $1 / (s (s - 1)) = 1 / ((1 - s) (- s))$ and the two powers of $t$ are exchanged. Hence $F (s) = F (1 - s) (Re s > 1) .$

step 1.1L4algebra

3.1

On $Re s > 1$ , [L1] names the completed function as $Λ (s)$ , and step 1.1 identifies that same function with the explicit split formula. Combined with [L4], this shows that $F (s) = Λ (s)$ there. Since both $F$ and $Λ$ are meromorphic on $C$ , [A1] gives $F = Λ$ on all of $C$ . Applying [A1] again to the meromorphic functions $F (s)$ and $F (1 - s)$ , which agree on $Re s > 1$ by step 2.1, yields $F (s) = F (1 - s)$ on $C$ . Therefore $Λ (s) = F (s) = F (1 - s) = Λ (1 - s)$ for every $s \in C$ . The explicit pole term in step 1.1 shows that the poles at $0$ and $1$ are simple.

step 1.1step 2.1L1L4A1algebra∎

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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)