Trace Formulas Applied to the Riemann ζ-Function

Abstract

We use a spectral theory perspective to reconsider properties of the Riemann zeta function. In particular, new integral representations are derived and used to present its value at odd positive integers.

Appendix: Basic Formulas for the Riemann ζ-Function

We present a number of formulas for ζ(z) and special values of ζ(⋅). It goes without saying that no such collection can ever attempt at any degree of completeness, and certainly our compilation of formulas is no exception in this context.

Definition

$$\displaystyle \begin{aligned} \zeta(z) &= \sum_{k \in {\mathbb{N}}} k^{-z}, \quad z \in {\mathbb{C}}, \, \operatorname{Re}(z) > 1 {} \end{aligned} $$

(A.1)

(A.2)

(A.3)

Functional Equation

$$\displaystyle \begin{aligned} \zeta(z) = 2^z \pi^{z-1} \sin{}(\pi z/2) \varGamma(1 - z) \zeta(1 - z), \quad z \in {\mathbb{C}}, \, \operatorname{Re}(z) < 0. \end{aligned} $$

(A.4)

Alternative Formulas

$$\displaystyle \begin{aligned} \zeta(z) &= \varGamma(z)^{-1} \int_0^{\infty} dt \, \frac{t^{z - 1}}{e^t - 1}, \quad z \in {\mathbb{C}}, \, \operatorname{Re}(z) > 1 \end{aligned} $$

(A.5)

(A.6)

$$\displaystyle \begin{aligned} &= \varGamma(z)^{-1} [1 - 2^{1 - z}]^{-1} \int_0^{\infty} dt \, \frac{t^{z - 1}}{e^{t} + 1}, \quad z \in {\mathbb{C}}, \, \operatorname{Re}(z) > 0 \end{aligned} $$

(A.7)

(A.8)

where

$$\displaystyle \begin{aligned} \varGamma(z) = \int_0^{\infty} dt \, t^{z - 1} e^{- t}, \quad z \in {\mathbb{C}}, \, \operatorname{Re}(z) > 0. \end{aligned} $$

(A.9)

In addition,

(A.10)

(A.11)

$$\displaystyle \begin{aligned} &= \pi^{z/2} \varGamma(z/2)^{-1} \int_0^{\infty} dt \, t^{(z/2) - 1} \sum_{k \in {\mathbb{N}}} e^{- k^2 \pi t}, \end{aligned} $$

(A.12)

and

(A.13)

$$\displaystyle \begin{aligned} &= \frac{2^{z - 1}}{z - 1} - 2^z \int_0^{\infty} dt \, \frac{\sin{}(z \arctan(t))}{(1 + t^2)^{z/2} (e^{\pi t} + 1)}, \quad z \in {\mathbb{C}}\backslash\{1\}, \end{aligned} $$

(A.14)

(A.15)

(A.16)

(A.17)

(A.18)

$$\displaystyle \begin{aligned} &= \varGamma(z+1)^{-1} 4^{-1} (2a)^{z + 1} \big[1 - 2^{1 - z}\big]^{-1} \int_0^{\infty} dt \, \frac{t^{z}}{[\cosh(at)]^2}, \end{aligned} $$

(A.19)

(A.20)

(A.21)

(A.22)

(A.23)

(A.24)

(A.25)

(A.26)

(A.27)

(A.28)

(A.29)

Specific Values

$$\displaystyle \begin{aligned} \zeta (2n) = \frac{(-1)^{n + 1}(2 \pi)^{2n}B_{2n} }{2 (2n)!}, \quad n \in {\mathbb{N}}_0, {} \end{aligned} $$

(A.30)

where B _m are the Bernoulli numbers generated, for instance, by

$$\displaystyle \begin{aligned} \frac{w}{e^{w} - 1} = \sum_{m \in {\mathbb{N}}_0} B_m \frac{w^m}{m!}, \quad w \in {\mathbb{C}}, \, |w| < 2 \pi, \end{aligned} $$

(A.31)

in particular,

$$\displaystyle \begin{aligned} & B_0 = 1, B_1 = - 1/2, \; B_2 = 1/6, \; B_3 = 0, B_4 = - 1/30, B_5 = 0, B_6 = 1/42, \text{etc.,} \end{aligned} $$

(A.32)

$$\displaystyle \begin{aligned} & B_{2k+1} = 0, \; k \in {\mathbb{N}}. {} \end{aligned} $$

(A.33)

Moreover, one has the generating functions for ζ(2n),

$$\displaystyle \begin{aligned} - (\pi z/2) \cot{}(\pi z) &= \sum_{n \in {\mathbb{N}}_0} \zeta(2n) z^{2n}, \quad |z| < 1, \; \zeta(0) = -1/2, {} \end{aligned} $$

(A.34)

$$\displaystyle \begin{aligned} - (\pi z/2) \coth(\pi z) &= \sum_{n \in {\mathbb{N}}_0} (-1)^n \zeta(2n) z^{2n}, \quad |z| < 1, \; \zeta(0) = -1/2, {} \end{aligned} $$

(A.35)

and [32]

$$\displaystyle \begin{aligned} (n!/6)[\zeta(n - 2) - 3 \zeta(n - 1) + 2 \zeta(n)] = \int_0^{\infty} dt \, \frac{t^n e^t}{(e^t - 1)^4}, \quad n \in {\mathbb{N}}, \, n \geq 4. \end{aligned} $$

(A.36)

Choosing k = 2n, $n \in {\mathbb {N}}$, even, employing (A.30) for ζ(2n), ζ(2n − 2), yields a formula for ζ(2n − 1). Moreover,

$$\displaystyle \begin{aligned} \zeta(2n+1) &= \frac{1}{(2n)!} \int_0^{\infty} dt \, \frac{t^{2n}}{e^t - 1}, \quad n \in {\mathbb{N}} \end{aligned} $$

(A.37)

(A.38)

where B _m(⋅) are the Bernoulli polynomials,

$$\displaystyle \begin{aligned} B_m(z) = \sum_{j = 0}^m \begin{pmatrix} m \\ j \end{pmatrix} B_j z^{m - j}, \quad t \in {\mathbb{C}}, \end{aligned} $$

(A.39)

generated, for instance, by

$$\displaystyle \begin{aligned} \frac{w e^{z w}}{e^{w} - 1} = \sum_{m \in {\mathbb{N}}_0} B_m(z) \frac{w^m}{m!}, \quad w \in {\mathbb{C}}, \, |w| < 2 \pi. \end{aligned} $$

(A.40)

Explicitly,

$$\displaystyle \begin{aligned}& \begin{aligned} & B_0(x) = 1, \; B_1(x) = x - (1/2), \; B_2(x) = x^2 - x + (1/6), \\ & B_3(x) = x^3 - (3/2)x^2 + (1/2)x, \; \text{etc.,} {} \end{aligned} \end{aligned} $$

(A.41)

$$\displaystyle \begin{aligned} & B_n(0) = B_n, \; n \in {\mathbb{N}}, \quad B_1(1) = - B_1 = 1/2, \; B_n(1) = B_n, \; n \in {\mathbb{N}}_0 \backslash \{1\}, {} \end{aligned} $$

(A.42)

$$\displaystyle \begin{aligned} & B_n^{\prime}(x) = n B_{n-1}(x), \quad n \in {\mathbb{N}}, \; x \in {\mathbb{R}}. {} \end{aligned} $$

(A.43)

In addition, for $n \in {\mathbb {N}}$,

(A.44)

(A.45)

(A.46)

(A.47)

Just for curiosity,

$$\displaystyle \begin{aligned} \zeta(3) = 1.2020569032 \ldots.. \end{aligned} $$

(A.48)

Apery [1] proved in 1978 that ζ(3) is irrational (see also Beukers [2], van der Poorten [29], Zudilin [35], and [31], [33]).

Moreover,

(A.49)

(A.50)

(A.51)

(A.52)

(A.53)

(A.54)

(A.55)

(A.56)

(A.57)

(A.58)

(A.59)

(A.60)

(A.61)

(A.62)

$$\displaystyle \begin{aligned} \zeta(3) &= -\dfrac{2}{7} \pi^2 \operatorname{ln}(2) - \dfrac{16}{7} \, \int_0^1 dt \, \dfrac{\mathrm{arctanh}(t) \operatorname{ln} (t)}{t(1-t^2)} {} \end{aligned} $$

(A.63)

$$\displaystyle \begin{aligned} &= - \dfrac{4}{3} \, \int_0^1 dt \, \dfrac{\operatorname{ln}(t) \operatorname{ln}(1+t)}{t} \end{aligned} $$

(A.64)

$$\displaystyle \begin{aligned} & = - 8 \, \int_0^1 dt \, \dfrac{\operatorname{ln}(t) \operatorname{ln}(1+t)}{1+t} \end{aligned} $$

(A.65)

$$\displaystyle \begin{aligned} &= \int_0^1 dt \, \dfrac{\operatorname{ln}(t) \operatorname{ln}(1-t)}{1-t} = \int_0^1 dt \, \dfrac{\operatorname{ln}(t) \operatorname{ln}(1-t)}{t} \end{aligned} $$

(A.66)

$$\displaystyle \begin{aligned} &= \dfrac{1}{4} \pi^2 \operatorname{ln}(2) + \int_0^1 dt \, \dfrac{\operatorname{ln}(t) \operatorname{ln}(1+t)}{1-t} \end{aligned} $$

(A.67)

$$\displaystyle \begin{aligned} &= \dfrac{2}{13} \pi^2 \operatorname{ln}(2) + \dfrac{8}{13} \, \int_0^1 dt \, \dfrac{\operatorname{ln}(t) \operatorname{ln}(1-t)}{1+t} \end{aligned} $$

(A.68)

$$\displaystyle \begin{aligned} &= \dfrac{2}{7} \, \int_0^{\pi/2} dt \, \dfrac{t (\pi - t)}{\sin{}(t)}. {} \end{aligned} $$

(A.69)

Formulas (A.63)–(A.69) were provided by Glasser and Ruehr and can be found in [21, Problem 80-13]. Finally, we also recall,

(A.70)

$$\displaystyle \begin{aligned} &= \dfrac{6}{7} + \dfrac{2}{7} \,\int_0^\infty dt \, \dfrac{\sin{}(3 \arctan(2t))}{[(1/4) + t^2]^{3/2}} \, \dfrac{1}{e^{2\pi t} - 1} {} \end{aligned} $$

(A.71)

$$\displaystyle \begin{aligned} &= \dfrac{6}{7} + \dfrac{8}{7} \,\int_0^\infty dt \, \dfrac{\sin{}(3 \arctan(t))}{(1 + t^2)^{3/2}} \, \dfrac{1}{e^{\pi t} - 1} {} \end{aligned} $$

(A.72)

$$\displaystyle \begin{aligned} &= 2 - 8 \,\int_0^\infty dt \, \dfrac{\sin{}(3 \arctan(t))}{(1 + t^2)^{3/2}} \, \dfrac{1}{e^{\pi t} + 1} {} \end{aligned} $$

(A.73)

$$\displaystyle \begin{aligned} &= 1 + 2 \,\int_0^\infty dt \, \dfrac{\sin{}(3 \arctan(t))}{(1 + t^2)^{3/2}} \, \dfrac{1}{e^{2\pi t} - 1}. {} \end{aligned} $$

(A.74)

Formulas (A.71)–(A.73) are due to Jensen (1895) and are special cases of results to be found in [30, p. 279] (cf. (A.16), (A.29)); finally, (A.74) is a consequence of (A.72) and (A.73).

For more on ζ(3) see also [12, p. 42–45].

For a wealth of additional formulas, going beyond what is recorded in this appendix, we also refer to [24] and [27].