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The Riemann zeta function has the standard Bernoulli special values at the positive even and nonpositive integers

Statement

Assume countable choice.

For every integer $m \geq 1$ ,

$ζ (2 m) = (- 1)^{m + 1} \frac{B_{2 m} (2 π)^{2 m}}{2 (2 m)!}, ζ (1 - 2 m) = - \frac{B_{2 m}}{2 m}, ζ (- 2 m) = 0,$

and also

$ζ (0) = - \frac{1}{2} .$

Facts & Assumptions

Given: Countable choice, the Bernoulli generating function and the functional equation.

[L1]

Bernoulli numbers satisfy $\frac{t}{e^{t} - 1} = \sum_{n \geq 0} \frac{B_{n}}{n!} t^{n},$ with $B_{1} = - 1 / 2$ (The Bernoulli numbers are defined by the generating series $t / (e^{t} - 1)$ ).

[L2]

The cotangent expansion is $π \cot (π z) = \frac{1}{z} + \sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}}$ (The Mittag-Leffler expansion of pi cotangent).

[L3]

Zeta satisfies the classical functional equation (The Riemann zeta function satisfies the classical sine-gamma functional equation).

[L4]

The meromorphic continuation has a simple residue-one pole at $1$ (The Riemann zeta function extends meromorphically to the complex plane with its only pole at $1$ ).

[L5]

The negative even integers are exactly the trivial zeros of zeta (The only zeros of zeta on the nonpositive real axis are the negative even integers, and every other zero lies in the open critical strip).

[L6]

The complex exponential obeys $e^{u + v} = e^{u} e^{v}$ and $e^{0} = 1$ , so $e^{u} e^{- u} = 1$ and $e^{- u} = 1 / e^{u}$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, The Bernoulli numbers are defined by the generating series $t / (e^{t} - 1)$ ).

[L7]

Complex sine and cosine are $(e^{i u} - e^{- i u}) / (2 i)$ and $(e^{i u} + e^{- i u}) / 2$ , and $\cot u = \cos u / \sin u$ where $\sin u \neq 0$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential, Tangent, cotangent, secant, and cosecant on their exact natural domains).

Proof

technique · direct

1.1L2algebra

For $∣ z ∣ < 1$ , expand the summand in [L2] as $\frac{2 z}{z^{2} - n^{2}} = - \frac{2 z}{n^{2}} \frac{1}{1 - z^{2} / n^{2}} = - 2 \sum_{m \geq 1} \frac{z^{2 m - 1}}{n^{2 m}} .$ Summing over $n$ gives $π z \cot (π z) = 1 - 2 \sum_{m \geq 1} ζ (2 m) z^{2 m} .$

1.2L3L4L5algebra

Let $s \to 0$ in [L3]. One has $\sin (π s / 2) \sim π s / 2$ , $Γ (1 - s) \to 1$ , and [L4] gives $ζ (1 - s) \sim - 1 / s$ . Therefore $ζ (0) = \lim_{s \to 0} 2^{s} π^{s - 1} \sin (π s / 2) Γ (1 - s) ζ (1 - s) = - \frac{1}{2} .$ Also [L5] already gives $ζ (- 2 m) = 0$ for $m \geq 1$ .

2.1step 1.1L1L6L7algebra

Put $F (t) = t / (e^{t} - 1) + t / 2$ using the removable value at $t = 0$ from [L1]. For nonzero $t$ near zero, [L6] gives $F (- t) = \frac{- t}{e^{- t} - 1} - \frac{t}{2} = \frac{t e^{t}}{e^{t} - 1} - \frac{t}{2} = F (t) .$ Since $F$ is analytic at zero, it is even there too; comparing its power series from [L1] shows $B_{2 m + 1} = 0$ for every $m \geq 1$ , using $B_{1} = - 1 / 2$ . For $t = 2 π i z$ and nonzero $z$ near zero, [L6] and [L7] give $π z \cot (π z) = \frac{t}{2} \frac{e^{t} + 1}{e^{t} - 1} = \frac{t}{e^{t} - 1} + \frac{t}{2} = F (t) .$ Both sides extend analytically to $z = 0$ , so their Taylor coefficients agree. Thus $π z \cot (π z) = \sum_{m \geq 0} \frac{B_{2 m} (2 π i z)^{2 m}}{(2 m)!} = 1 + \sum_{m \geq 1} (- 1)^{m} \frac{B_{2 m} (2 π)^{2 m}}{(2 m)!} z^{2 m} .$ Comparing coefficients with step 1.1 gives $ζ (2 m) = (- 1)^{m + 1} \frac{B_{2 m} (2 π)^{2 m}}{2 (2 m)!} (m \geq 1) .$

3.1step 2.1step 1.2L3algebra∎

For $m \geq 1$ , substitute $s = 1 - 2 m$ into [L3]. Since $\sin (π (1 - 2 m) / 2) = (- 1)^{m}$ and $Γ (2 m) = (2 m - 1)!$ , step 2.1 yields $ζ (1 - 2 m) = 2^{1 - 2 m} π^{- 2 m} (- 1)^{m} (2 m - 1)! ζ (2 m) = - \frac{B_{2 m}}{2 m} .$ Together with step 1.2, this gives all the displayed special values.

Depends on

Used by

The analytic continuation of zeta is not the same object as the defining Dirichlet series outside Res>1 Remark

Dependency tree · two levels

37 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 11 §3 (standard reference, not scraped)
Elias M. Stein and Rami Shakarchi, Complex Analysis, Ch. 6 §2.1 (standard reference, not scraped)