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The Riemann zeta function has its Euler product on the half-plane $Re s > 1$

Statement

For every $s \in C$ with $Re s > 1$ ,

$ζ (s) = \prod_{p} \frac{1}{1 - p^{- s}},$

where the product ranges over the primes and converges absolutely and locally uniformly on $Re s > 1$ .

Facts & Assumptions

Given: A complex number $s$ with $Re s > 1$ .

[L1]

On $Re s > 1$ , $ζ (s) = \sum_{n \geq 1} n^{- s}$ and this series converges absolutely and locally uniformly (The Riemann zeta function on the half-plane $Re s > 1$ , The Dirichlet series for zeta converges absolutely and locally uniformly on the half-plane $Re s > 1$ ).

[L2]

If $\sum ∣ a_{n} ∣$ converges, then $\prod (1 + a_{n})$ converges; its value is nonzero when every factor is nonzero (Absolute convergence criterion for complex infinite products).

[L3]

Every integer $n \geq 1$ has a unique prime factorization up to order (The fundamental theorem of arithmetic: every integer $n \geq 1$ is a product of primes, and the factorisation is unique up to order — if $\prod_{i < r} p_{i} = \prod_{j < s} q_{j}$ with every $p_{i}$ and $q_{j}$ prime, then $r = s$ and $q_{i} = p_{π (i)}$ for some $π \in Sym (r)$ ).

[L4]

Canonical prime factorization rewrites each integer by the exponents of its prime divisors (For $n \geq 1$ and any injective list $p : r \to Z$ of primes containing every prime divisor of $n$ , one has $n = \prod_{i < r} p_{i}^{v_{p_{i}} (n)}$ ; the exponents are determined by $n$ , and $v_{q} (n) = 0$ for every prime $q$ outside the list).

Proof

technique · direct

1.1givenL1L2algebra

Write $σ : = Re s$ . Since the primes are among the integers at least $2$ , $\sum_{p} ∣ p^{- s} ∣ = \sum_{p} p^{- σ} \leq \sum_{n = 2}^{\infty} n^{- σ} < \infty$ by [L1]. For every prime $p$ , $∣ p^{- s} ∣ = p^{- σ} < 1$ , so $1 - p^{- s} \neq 0$ . Therefore [L2] applies to $a_{p} : = - p^{- s}$ and makes the product $\prod_{p} (1 - p^{- s})$ convergent and nonzero.

1.2L3L4algebra

For a finite set $P$ of primes, $\prod_{p \in P} \frac{1}{1 - p^{- s}} = \prod_{p \in P} \sum_{k \geq 0} p^{- k s} .$ Multiplying out this finite product lists exactly the terms $n^{- s}$ for those integers $n \geq 1$ whose prime divisors all lie in $P$ , and [L3] with [L4] shows that each such integer appears exactly once. Hence $\prod_{p \in P} \frac{1}{1 - p^{- s}} = \sum_{\begin{array}{c} n \geq 1 \\ p ∣ n \Rightarrow p \in P \end{array}} n^{- s} .$

2.1step 1.1step 1.2L1algebra

Let $P_{N}$ be the set of primes at most $N$ . By step 1.2 the corresponding partial products are the partial sums over integers all of whose prime divisors lie in $P_{N}$ . Every fixed integer eventually has this property, and the omitted terms are bounded in absolute value by the tail of the absolutely convergent series in [L1]. Therefore these partial products converge pointwise to $\sum_{n \geq 1} n^{- s} = ζ (s)$ .

3.1step 2.1L1choosealgebra∎

Let $K \subseteq {s : Re s > 1}$ be compact, and choose $σ > 1$ with $Re s \geq σ$ on $K$ . If an integer $n$ is omitted from the partial product over $P_{N}$ , then $n$ has some prime divisor greater than $N$ , hence $n > N$ . So for $s \in K$ , $∣ ζ (s) - \prod_{p \leq N} \frac{1}{1 - p^{- s}} ∣ \leq \sum_{n > N} n^{- σ} .$ The tail on the right tends to $0$ independently of $s$ , so the partial products converge uniformly on $K$ . Thus the Euler product converges locally uniformly on $Re s > 1$ . Combining this with steps 1.1 and 2.1 proves the stated formula together with its absolute and locally uniform convergence.

Depends on

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46 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 (standard reference, not scraped)
K. Chandrasekharan, Lectures on the Riemann Zeta-Function, Lecture 11 §3 (standard reference, not scraped)

The Riemann zeta function has its Euler product on the half-plane Re⁡s>1Res>1