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

Statement

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: 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).

Proof

technique · direct

1.1

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 $π z \cot (π z) = 1 - 2 \sum_{m \geq 1} ζ (2 m) z^{2 m} .$

L2algebra

1.2

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 $m \geq 1$ .

L3L4L5algebra

2.1

From [L1], $\frac{t}{e^{t} - 1} + \frac{t}{2} = \sum_{n \geq 0} \frac{B_{n}}{n!} t^{n} + \frac{t}{2}$ is even, so $B_{2 m + 1} = 0$ for every $m \geq 1$ . Setting $t = 2 π i z$ and simplifying yields $π z \cot (π z) = \sum_{m \geq 0} \frac{B_{2 m} (2 π i z)^{2 m}}{(2 m)!} = 1 + \sum_{m \geq 1} (- 1)^{m} \frac{B_{2 m} (2 π)^{2 m}}{(2 m)!} z^{2 m} .$ Comparing coefficients with step 1.1 gives $ζ (2 m) = (- 1)^{m + 1} \frac{B_{2 m} (2 π)^{2 m}}{2 (2 m)!} (m \geq 1) .$

step 1.1L1algebra

3.1

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.

step 2.1step 1.2L3algebra∎

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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)