# On some constants related to the harmonic zeta function

Marc-Antoine Coppo, Bernard Candelpergher

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Marc-Antoine Coppo, Bernard Candelpergher. On some constants related to the harmonic zeta function. 2024. hal-03602568v3

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# On some remarkable identities related to the harmonic zeta function

Marc-Antoine Coppo, Bernard Candelpergher

► To cite this version:

Marc-Antoine Coppo, Bernard Candelpergher. On some remarkable identities related to the harmonic zeta function. Research in Number Theory, 2025, 11 (3), pp.10. 10.1007/s40993-025-00642-w. hal-03602568v9

HAL Id: hal-03602568

https://univ-cotedazur.hal.science/hal-03602568v9

Submitted on 4 Jun 2025

HAL is a multi-disciplinary open access
archive for the deposit and dissemination of sci-
entific research documents, whether they are pub-
lished or not. The documents may come from
teaching and research institutions in France or
abroad, or from public or private research centers.

L'archive ouverte pluridisciplinaire HAL, est
destinée au dépôt et à la diffusion de documents
scientifiques de niveau recherche, publiés ou non,
émanant des établissements d'enseignement et de
recherche français ou étrangers, des laboratoires
publics ou privés.

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Distributed under a Creative Commons Attribution 4.0 International License
# On some constants related to the harmonic zeta function

Marc-Antoine Coppo* and Bernard Candelpergher

Université Côte d'Azur, CNRS, LJAD (UMR 7351), Nice, France

**Abstract** We study a range of constants that are closely linked to the harmonic zeta function $\zeta_H$. In addition, by means of a new representation of $\zeta_H$, we derive possibly new integral formulas for $\zeta(n)$ and $\zeta'(n)$ and new evaluations for a variety of integrals of the same type as those previously considered by Glasser and Manna.

**Keywords** Harmonic zeta function; Stieltjes constants; zeta values; digamma function; Glasser-Manna integrals, Ramanujan summation of series.

## 1 Introduction

Let $\zeta_H$ be the harmonic zeta function defined for $\text{Re}(s) > 1$ by

$$ \zeta_H(s) := \sum_{n=1}^{\infty} \frac{H_n}{n^s}, $$

where, for all $n \ge 1$,

$$ H_n = 1 + \frac{1}{2} + \dots + \frac{1}{n} $$

are the classical harmonic numbers. Four decades ago, Apostol and Vu [2] showed that this function could be continued as a meromorphic function with a double pole at $s = 1$, and an infinite number of simple poles at $s = 0$ and $s = 1 - 2k$ for each integer $k \ge 1$. The Laurent expansion of the harmonic zeta function $\zeta_H$ around its double pole $s = 1$ can be written as

$$ \zeta_H(s) = \frac{1}{(s-1)^2} + \frac{\gamma}{s-1} + \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \gamma_H^{(k)}(s-1)^k \quad (0 < |s-1| < 1), $$

<footnote>*Corresponding author. Email address: coppo@unice.fr</footnote>
# On some remarkable identities related to the harmonic zeta function

Marc-Antoine Coppo*

Bernard Candelpergher

Université Côte d'Azur, CNRS, LJAD, Nice, France

Research in Number Theory, 11 (2025)

**Abstract** Using Ramanujan's summation method applied to a certain type of divergent series involving harmonic numbers and powers of logarithms, we consider a variety of interesting constants and study their properties. In particular, we clarify their close connection with the classical Stieltjes constants and their analogues for the harmonic zeta function which allows us to deduce several remarkable identities.

**Keywords** Harmonic numbers; Bernoulli numbers; Stieltjes constants; Gamma function; Harmonic zeta function; Ramanujan summation of divergent series.

## 1 Introduction

The harmonic zeta function $\zeta_H$ is defined for $\mathrm{Re}(s) > 1$ by

$$ \zeta_H(s) := \sum_{n=1}^{\infty} \frac{H_n}{n^s}, $$

where, for all integers $n \ge 1$,

$$ H_n = 1 + \frac{1}{2} + \cdots + \frac{1}{n} $$

are the classical harmonic numbers. Forty years ago, Apostol and Vu [1] and Matsuoka [12], building upon the foundational work of Euler [11], showed that this function extends as a meromorphic function, featuring a double pole at $s = 1$,

<footnote>*Corresponding author. Email address: coppo@unice.fr</footnote>
where $\gamma = -\Gamma'(1)$ is Euler's constant, and the coefficients $\gamma_H^{(k)}$ are called harmonic Stieltjes constants by analogy with the classical Stieltjes constants $\gamma_k$. A common definition of these constants is

$$ \gamma_k = \lim_{s \to 1} \left\{ (-1)^k \zeta^{(k)}(s) - \frac{k!}{(s-1)^{k+1}} \right\} \quad (k \ge 0), $$

where $\zeta^{(k)}(s)$ is the $k$th derivative of the Riemann zeta function. From now on and throughout the text, we will use the lighter notation $\gamma^{(k)}$ in place of $\gamma_H^{(k)}$. An explicit expression of $\gamma^{(0)}$ is given by the following nice formula [10, Equation (6)]:

$$ \gamma^{(0)} = \frac{1}{2}\gamma^2 + \frac{1}{2}\zeta(2) = \frac{1}{2}\Gamma''(1). \quad (1) $$

Incidentally, this formula can also be generalized to the case of height 1 multiple zeta functions $\zeta(s, 1, \dots, 1)$ [17, Corollary 4.1]. Regarding the classical Stieltjes constants, we recall the well-known asymptotic formula

$$ \gamma_k = \lim_{N \to \infty} \left\{ \sum_{n=1}^{N} \frac{(\log n)^k}{n} - \frac{(\log N)^{k+1}}{k+1} \right\} \quad (k \ge 0). $$

From our point of view, it should be noted that $\gamma_k$ is nothing else than the $\mathcal{R}$-sum (i.e. the sum of the series in the sense of Ramanujan's summation method) of the divergent series $\sum_{n \ge 1} \frac{(\log n)^k}{n}$ [9, p. 67]. The existence of a similar asymptotic formula for $\gamma^{(k)}$:

$$ \gamma^{(k)} = \lim_{N \to \infty} \left\{ \sum_{n=1}^{N} \frac{H_n (\log n)^k}{n} - \frac{(\log N)^{k+2}}{k+2} - \gamma \frac{(\log N)^{k+1}}{k+1} \right\} \quad (k \ge 0), $$

which results from an extension of a formula of Briggs and Buschman [4, 8], strongly suggests that the constant $\gamma^{(k)}$ is closely linked to the $\mathcal{R}$-sum of the divergent series $\sum_{n \ge 1} \frac{H_n (\log n)^k}{n}$. This is indeed the case: such a relation exists although it is more intricate than in the classical case. We have already encountered in earlier studies [9, Equation (3.23)], [10, Equation (4)] the intriguing relation:

$$ r_0 = \gamma_1 + \tau_1 + \gamma^{(0)} - \zeta(2) = \gamma_1 + \tau_1 + \frac{1}{2}\gamma^2 - \frac{1}{2}\zeta(2), \quad (2) $$

where

$$ r_0 := \sum_{n \ge 1} \frac{H_n}{n} \quad \text{and} \quad \tau_1 = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\zeta(n+1)}{n} = - \sum_{n=2}^{\infty} \zeta'(n). $$

In the next section, we will see to what extent our formula (2) can be generalized (see Proposition 1). We also mention that another natural generalization of this
and simple poles at $s = 0$ and the odd negative integers. The special values of $\zeta_H$ at even negative integers are given by Matsuoka's formula:

$$2\zeta_H(-2p) = (1-2p)\zeta(1-2p) = (2p-1)\frac{B_{2p}}{2p} \quad (p \ge 1),$$

where the $B_{2p}$ are the classical Bernoulli numbers. The Laurent expansion of the harmonic zeta function $\zeta_H$ around its double pole can be written as

$$\zeta_H(s) = \frac{1}{(s-1)^2} + \frac{\gamma}{s-1} + \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \gamma_{H,k}(s-1)^k \quad (0 < |s-1| < 1),$$

where $\gamma = -\Gamma'(1) = 0.577215...$ is Euler's constant, and the coefficients $\gamma_{H,k}$ are referred to as harmonic Stieltjes constants, drawing an analogy to the classical Stieltjes constants. A common definition of the Stieltjes constants $\gamma_k$ for arbitrary $k$ is

$$\gamma_k = \lim_{s \to 1} \left\{ (-1)^k \zeta^{(k)}(s) - \frac{k!}{(s-1)^{k+1}} \right\} \quad (k \ge 0),$$

where $\zeta^{(k)}(s)$ is the $k$th derivative of the Riemann zeta function [7]. In particular, $\gamma_0$ is Euler's constant $\gamma$. An explicit expression of $\gamma_{H,0}$ is given by the following formula [6, Equation (6)]:

$$\gamma_{H,0} = \frac{1}{2}\gamma^2 + \frac{1}{2}\zeta(2) = \frac{1}{2}\Gamma^{(2)}(1) = 0.989055... \quad (1)$$

As noted by Young, this nice expression derives from a special case of a general formula which applies to height one multiple zeta functions of arbitrary depth [13, Equation (28)].

Regarding the classical Stieltjes constants, we recall the well-known asymptotic representation of $\gamma_k$ (see e.g. [7]):

$$\gamma_k = \lim_{N \to \infty} \left\{ \sum_{n=1}^{N} \frac{(\ln n)^k}{n} - \frac{(\ln N)^{k+1}}{k+1} \right\} \quad (k \ge 0).$$

From the point of view adopted in this study, the constant $\gamma_k$ can be interpreted as the $\mathcal{R}$-sum of the divergent series $\sum_{n \ge 1} \frac{(\ln n)^k}{n}$, i.e. the sum of the series in the sense of Ramanujan's summation method, following the exposition in [4] (see also [5, Chapter IX]). Thus, using these notations, we can write [4, p. 67], [5, p. 296]:

$$\sum_{n \ge 1} \frac{\mathcal{R}(\ln n)^k}{n} = \gamma_k.$$
formula with Roman's harmonic numbers $H_{n,k}$ instead of $H_n$ was given by Young [17, Theorem 6.1 (d)]. In the subsequent section, we will use the new representation of $\zeta_H(s)$ provided by our Theorem 1 (which is a refinement of a formula due to Boyadzhiev, Gadiyar and Padma [7]) to obtain new expressions of the harmonic Stieltjes constants $\gamma^{(k)}$ (see Corollary 1). Moreover, we derive from this representation new integral formulas for $\zeta(n)$ and $\zeta'(n)$ (see Corollary 2) and new evaluations for a variety of integrals of the same type as those introduced earlier by Glasser and Manna in relation with the Laplace transform of the digamma function [1, 15] (see Corollaries 4–6).

## 2 How the constants $\sum_{n \ge 1} \frac{H_n (\log n)^k}{n}$ and $\gamma^{(k)}$ are linked together

The forthcoming Proposition 1 successfully generalizes the previous relation (2).

**Proposition 1.** Let $k$ be a natural number. If $r_k$ and $\tau_k$ are respectively defined by

$$r_k := \sum_{n \ge 1} \frac{H_n (\log n)^k}{n} \quad (k \ge 0),$$

and

$$\tau_k := \sum_{n=1}^{\infty} (-1)^{n+k} \frac{\zeta(n+1)}{n^k} \quad (k \ge 1),$$

then

$$r_0 = \gamma_1 + \tau_1 + \gamma^{(0)} - \zeta(2),$$

and the following relations hold true for all positive values of $k$:

$$r_{2k-1} = \frac{\gamma_{2k}}{2k} + (2k-1)! \tau_{2k} + \gamma^{(2k-1)} + \sum_{n=1}^{k} \frac{(2k-1)!}{(2k-2n)!} \times \frac{(2^{2n-1}-1)}{2^{2n-2}} \gamma_{2k-2n} \zeta(2n), \quad (3)$$

and

$$r_{2k} = \frac{\gamma_{2k+1}}{2k+1} + (2k)! \tau_{2k+1} + \gamma^{(2k)} + \sum_{n=1}^{k} \frac{(2k)!}{(2k-2n+1)!} \times \frac{(2^{2n-1}-1)}{2^{2n-2}} \gamma_{2k-2n+1} \zeta(2n) \\ - (2k)! \frac{(2^{2k+1}-1)}{2^{2k}} \zeta(2k+2). \quad (4)$$
The existence of a similar asymptotic representation for $\gamma_{H,k}$:

$$ \gamma_{H,k} = \lim_{N \to \infty} \left\{ \sum_{n=1}^{N} \frac{H_n(\ln n)^k}{n} - \frac{(\ln N)^{k+2}}{k+2} - \gamma \frac{(\ln N)^{k+1}}{k+1} \right\} \quad (k \ge 0), $$

which results from a generalization of a formula of Briggs and Buschman [2, 3], strongly suggests that the harmonic Stieltjes constant $\gamma_{H,k}$ and the $\mathcal{R}$-sum $\sum_{n \ge 1} \frac{H_n(\ln n)^k}{n}$ are closely linked. Although much less simple than in the previous case, such a connection exists: as we will see, each harmonic Stieltjes constant $\gamma_{H,k}$ differs from the corresponding $\mathcal{R}$-sum of $H_n(\ln n)^k/n$ by a quantity specified entirely in terms of the Riemann zeta function (see Proposition 1 below). In earlier studies [6, Equation (4)], [8, Equation (5.1)], we have already encountered and focus on the intriguing identity:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n}{n} = \frac{1}{2}\gamma^2 - \frac{1}{2}\zeta(2) + \gamma_1 + \tau_1 = \gamma_{H,0} + \tau_1 + \gamma_1 - \zeta(2) = 0.529052\dots, \quad (2) $$

where

$$ \tau_1 := \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \zeta(n+1) = 1.257746\dots. $$

it should be noted that this constant $\tau_1$ arises in several contexts in number theory (see e.g. [14, pp. 12–13] for a summary of these occurrences). The main purpose of this article is to determine to what extent this formula can be generalized and the consequences to be drawn. We start with the following definition:

**Definition 1.** For any positive integer $k$, the constant $\tau_k$ is defined by

$$ \tau_k := \sum_{n=1}^{\infty} \frac{(-1)^{n+k}}{n^k} \zeta(n+1). $$

Let us mention that the definition of the constant $\tau_k$ given above is, for $k > 1$, different than the definition used in [6] and [8]. We also give another expression of the constant $\tau_k$ which can be easily deduced from the previous definition:

$$ \tau_k = (-1)^k \sum_{n=1}^{\infty} \frac{1}{n} \mathrm{Li}_k \left(-\frac{1}{n}\right) \quad (k \ge 1), $$

where $\mathrm{Li}_k(z)$ is the polylogarithm function. In particular, for $k = 1$, the known formula

$$ \tau_1 = \sum_{n=1}^{\infty} \frac{1}{n} \ln \left(1 + \frac{1}{n}\right) $$

is regained.
In particular,

$$r_1 = \frac{1}{2}\gamma_2 + \tau_2 + \gamma^{(1)} + \gamma\zeta(2),$$ $$r_2 = \frac{1}{3}\gamma_3 + 2\tau_3 + \gamma^{(2)} + 2\gamma_1\zeta(2) - \frac{7}{2}\zeta(4).$$

**Example 1.** Numerical evaluations of $r_k$, $\tau_k$ and $\gamma^{(k)}$ for small values of $k$ are given below. Note that the constant $\tau_0$ is undefined.

$$r_0 = 0.529052\dots \qquad \tau_1 = 1.257746\dots \qquad \gamma^{(0)} = 0.989055\dots$$ $$r_1 = -0.078850\dots \qquad \tau_2 = -1.424248\dots \qquad \gamma^{(1)} = 0.400761\dots$$ $$r_2 = -0.008095\dots \qquad \tau_3 = 1.523800\dots \qquad \gamma^{(2)} = 0.971304\dots$$

**Remark 1.** Although not used afterwards, we also mention another expression of the constant $\tau_k$ which can be easily deduced from its definition:

$$\tau_k = (-1)^k \sum_{n=1}^{\infty} \frac{1}{n} \mathrm{Li}_k \left(-\frac{1}{n}\right) \quad (k \ge 1),$$

where $\mathrm{Li}_k$ denotes the $k$th polylogarithm. In particular, for $k = 1$, the formula

$$\tau_1 = \sum_{n=1}^{\infty} \frac{1}{n} \log \left(1 + \frac{1}{n}\right)$$

is regained. Note that the definition of $\tau_k$ given here differs from that given in [10] when $k > 1$.

*Proof of Proposition 1.* The key formula to derive the general relations (3) and (4) is the decomposition of $\sum_{n \ge 1} \frac{H_n}{n^s}$ given by [10, Theorem 1] which is reproduced below:

$$\sum_{n \ge 1} \frac{H_n}{n^s} = \frac{\pi}{\sin(\pi s)} \zeta(s) + \int_0^1 \frac{\psi(x+1) + \gamma}{x^s} dx + \zeta_H(s), \quad (5)$$

where $\psi(x) = \frac{d}{dx} \log \Gamma(x)$ is the digamma function. This formula applies to all complex numbers $s$ such that $\mathrm{Re}(s) < 2$ and $s \ne 1, 0, 1 - 2k$ for each integer $k > 0$. Fortunately, the Laurent expansion of each component in (5) can be written explicitly.

a) The expansion of $\frac{\pi}{\sin(\pi s)} \zeta(s)$ at $s = 1$ can be obtained as follows: first, we write
## 2 How the constants $\tau_k$, $\gamma_k$ and $\gamma_{H,k}$ are linked together

In this section, we clarify the relations between the constants defined above. The forthcoming Proposition 1 successfully generalizes formula (2).

**Proposition 1.** For any positive integer $k$, we have the following relations:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n (\ln n)^{2k-1}}{n} = \gamma_{H,2k-1} + (2k-1)! \tau_{2k} \\ + \frac{\gamma_{2k}}{2k} + \sum_{j=1}^{k} \frac{(2k-1)!}{(2k-2j)!} \times \frac{(2^{2j-1}-1)}{2^{2j-2}} \gamma_{2k-2j} \zeta(2j), \quad (3) $$

and

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n (\ln n)^{2k}}{n} = \gamma_{H,2k} + (2k)! \tau_{2k+1} \\ + \frac{\gamma_{2k+1}}{2k+1} + \sum_{j=1}^{k} \frac{(2k)!}{(2k+1-2j)!} \times \frac{(2^{2j-1}-1)}{2^{2j-2}} \gamma_{2k+1-2j} \zeta(2j) \\ - (2k)! \frac{(2^{2k+1}-1)}{2^{2k}} \zeta(2k+2). \quad (4) $$

In particular, we have

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n \ln n}{n} = \gamma_{H,1} + \tau_2 + \frac{1}{2}\gamma_2 + \gamma\zeta(2), $$

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n (\ln n)^2}{n} = \gamma_{H,2} + 2\tau_3 + \frac{1}{3}\gamma_3 + 2\gamma_1\zeta(2) - \frac{7}{2}\zeta(4), $$

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n (\ln n)^3}{n} = \gamma_{H,3} + 6\tau_4 + \frac{1}{4}\gamma_4 + 3\gamma_2\zeta(2) + \frac{21}{2}\gamma\zeta(4). $$

*Proof of Proposition 1.* The key formula to derive the general relations (3) and (4) is the splitting of $\sum_{n \ge 1}^{\mathcal{R}} \frac{H_n}{n^s}$ given by [6, Theorem 1] which is recalled below:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n}{n^s} = \frac{\pi}{\sin(\pi s)} \zeta(s) + \int_0^1 \frac{\psi(x+1) + \gamma}{x^s} dx + \zeta_H(s), \quad (5) $$

where $\psi = \Gamma'/\Gamma$ is the digamma function. This formula applies to all complex numbers $s$ such that $\text{Re}(s) < 2$ and $s \ne 1, 0$, and $1 - 2k$ for each positive integer
the successive equations:

$$
\begin{align*}
\frac{-\pi}{\sin(\pi s)} &= \frac{\pi}{\sin(\pi(s-1))} = \frac{\exp(i\pi(s-1))}{s-1} \frac{2i\pi(s-1)}{\exp(2i\pi(s-1))-1} \\
&= \frac{1}{s-1} \left( \sum_{k \ge 0} i^k \pi^k \frac{1}{k!} (s-1)^k \right) \left( \sum_{k \ge 0} i^k (2\pi)^k \frac{B_k}{k!} (s-1)^k \right) \\
&= \frac{1}{s-1} + \sum_{k \ge 1} (-1)^k \left( \sum_{j=0}^{2k} \binom{2k}{j} 2^j B_j \right) \frac{\pi^{2k}}{(2k)!} (s-1)^{2k-1},
\end{align*}
$$

where $B_j$ are the Bernoulli numbers. Euler's identity:

$$
\zeta(2k) = (-1)^{k+1} 2^{2k-1} B_{2k} \frac{\pi^{2k}}{(2k)!} \quad (k \ge 1),
$$

then allows us to rewrite this expansion as follows:

$$
\frac{-\pi}{\sin(\pi s)} = \frac{1}{s-1} - \sum_{k \ge 1} \frac{2^{1-2k}}{B_{2k}} \sum_{j=0}^{2k} \binom{2k}{j} 2^j B_j \zeta(2k) (s-1)^{2k-1}.
$$

Morover, the latter expression can be simplified thanks to the identity

$$
\sum_{j=0}^{k} \binom{k}{j} 2^j B_j = 2^k B_k \left(\frac{1}{2}\right) = 2(1 - 2^{k-1}) B_k \quad (k \ge 2).
$$

Hence, the expansion of $\frac{\pi}{\sin(\pi s)}$ at $s=1$ is given by

$$
\frac{\pi}{\sin(\pi s)} = -\frac{1}{s-1} - \sum_{k=1}^{\infty} \frac{2^{2k-1}-1}{2^{2k-2}} \zeta(2k) (s-1)^{2k-1}.
$$

On the other hand, the expansion of $\zeta(s)$ at $s=1$ is

$$
\zeta(s) = \frac{1}{s-1} + \gamma + \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \gamma_k (s-1)^k,
$$

where $\gamma_k$ are the Stieltjes constants. The expansion of $\frac{\pi}{\sin(\pi s)} \zeta(s)$ is then obtained
by Cauchy product. We have

$$
\begin{align*}
\frac{\pi}{\sin(\pi s)} \zeta(s) &= -\frac{1}{(s-1)^2} - \frac{\gamma}{(s-1)} + \gamma_1 - \zeta(2) \\
&\quad - \left(\frac{1}{2}\gamma_2 + \gamma\zeta(2)\right) (s-1) + \left(\frac{1}{6}\gamma_3 + \gamma_1\zeta(2) - \frac{7}{4}\zeta(4)\right) (s-1)^2 \\
&\quad - \left(\frac{1}{24}\gamma_4 + \frac{1}{2}\gamma_2\zeta(2) + \frac{7}{4}\gamma\zeta(4)\right) (s-1)^3 \\
&\quad + \left(\frac{1}{120}\gamma_5 + \frac{1}{6}\gamma_3\zeta(2) + \frac{7}{4}\gamma_1\zeta(4) - \frac{31}{16}\zeta(6)\right) (s-1)^4 - \dots \quad (6)
\end{align*}
$$
k. Fortunately, the Laurent expansion of each component in (5) can be written explicitly.

a) The expansion of $\frac{\pi}{\sin(\pi s)} \zeta(s)$ at $s = 1$ can be obtained as follows: first, we write the successive equations:

$$ 
\begin{aligned}
\frac{-\pi}{\sin(\pi s)} &= \frac{\pi}{\sin(\pi(s-1))} = \frac{\exp(i\pi(s-1))}{s-1} \frac{2i\pi(s-1)}{\exp(2i\pi(s-1))-1} \\
&= \frac{1}{s-1} \left( \sum_{k \ge 0} i^k \pi^k \frac{1}{k!} (s-1)^k \right) \left( \sum_{k \ge 0} i^k (2\pi)^k \frac{B_k}{k!} (s-1)^k \right) \\
&= \frac{1}{s-1} + \sum_{k \ge 1} (-1)^k \left( \sum_{j=0}^{2k} \binom{2k}{j} 2^j B_j \right) \frac{\pi^{2k}}{(2k)!} (s-1)^{2k-1},
\end{aligned}
$$

where the $B_j$ are the Bernoulli numbers. Euler's identity:

$$ \zeta(2k) = (-1)^{k+1} 2^{2k-1} B_{2k} \frac{\pi^{2k}}{(2k)!} \quad (k \ge 1), $$

then allows us to rewrite this expansion as follows:

$$ \frac{-\pi}{\sin(\pi s)} = \frac{1}{s-1} - \sum_{k \ge 1} \frac{2^{1-2k}}{B_{2k}} \sum_{j=0}^{2k} \binom{2k}{j} 2^j B_j \zeta(2k) (s-1)^{2k-1}. $$

Morover, the latter expression can be simplified thanks to the identity

$$ \sum_{j=0}^{k} \binom{k}{j} 2^j B_j = 2^k B_k \left(\frac{1}{2}\right) = 2(1 - 2^{k-1}) B_k \quad (k \ge 2). $$

Hence, the expansion of $\frac{\pi}{\sin(\pi s)}$ at $s = 1$ is given by

$$ \frac{\pi}{\sin(\pi s)} = -\frac{1}{s-1} - \sum_{k=1}^{\infty} \frac{2^{2k-1}-1}{2^{2k-2}} \zeta(2k) (s-1)^{2k-1}. $$

On the other hand, the expansion of $\zeta(s)$ at $s = 1$ is

$$ \zeta(s) = \frac{1}{s-1} + \gamma + \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \gamma_k (s-1)^k, $$
In (6), the coefficient of $(s-1)^{2k}$ and $(s-1)^{2k-1}$ are respectively

$$ \frac{\gamma_{2k+1}}{(2k+1)!} + \sum_{j=1}^{k} \frac{\gamma_{2k-2j+1}}{(2k-2j+1)!} \times \frac{(2^{2j-1}-1)}{2^{2j-2}} \zeta(2j) - \frac{(2^{2k+1}-1)}{2^{2k}} \zeta(2k+2), $$

and

$$ -\frac{\gamma_{2k}}{(2k)!} - \sum_{j=1}^{k} \frac{\gamma_{2k-2j}}{(2k-2j)!} \times \frac{(2^{2j-1}-1)}{2^{2j-2}} \zeta(2j). $$

b) The well-known Taylor series expansion of the digamma function:

$$ \psi(x+1) + \gamma = \sum_{n=1}^{\infty} (-1)^{n+1} \zeta(n+1) x^n \quad (|x| < 1), $$

allows us write

$$ \int_0^1 \frac{\psi(x+1) + \gamma}{x^s} dx = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\zeta(n+1)}{n - (s-1)}. $$

This leads to the following expansion:

$$ \int_0^1 \frac{\psi(x+1) + \gamma}{x^s} dx = \sum_{k=0}^{\infty} \left( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\zeta(n+1)}{n^{k+1}} \right) (s-1)^k \\ = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \times k! \tau_{k+1} (s-1)^k \quad (|s-1| < 1). \quad (7) $$

c) To obtain the expansion of $\sum_{n \ge 1} \frac{H_n}{n^s}$ at $s = 1$, we proceed as follows: first we write the identities

$$ \frac{H_n}{n^s} = \frac{H_n}{n} e^{-(s-1) \log n} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \times \frac{H_n(\log n)^k}{n} (s-1)^k, $$

so that

$$ \sum_{n \ge 1} \frac{H_n}{n^s} = \sum_{n \ge 1} \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \times \frac{H_n(\log n)^k}{n} (s-1)^k \right). $$

Moreover, [9, Theorem 9] allows us interchange $\sum_{n \ge 1}^{\mathcal{R}}$ and $\sum_{k=0}^{\infty}$.

$$ \sum_{n \ge 1} \frac{H_n}{n^s} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left( \sum_{n \ge 1} \frac{H_n(\log n)^k}{n} \right) (s-1)^k = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} r_k (s-1)^k. \quad (8) $$

The desired formulas (3) and (4) can then be easily obtained by combining the previous expansions (6), (7), and (8). $\square$
where $\gamma_k$ are the Stieltjes constants. The expansion of $\frac{\pi}{\sin(\pi s)} \zeta(s)$ is then obtained by Cauchy product. We have

$$ \frac{\pi}{\sin(\pi s)} \zeta(s) = -\frac{1}{(s-1)^2} - \frac{\gamma}{(s-1)} + \gamma_1 - \zeta(2) \\ - \left(\frac{1}{2}\gamma_2 + \gamma\zeta(2)\right) (s-1) + \left(\frac{1}{6}\gamma_3 + \gamma_1\zeta(2) - \frac{7}{4}\zeta(4)\right) (s-1)^2 \\ - \left(\frac{1}{24}\gamma_4 + \frac{1}{2}\gamma_2\zeta(2) + \frac{7}{4}\gamma\zeta(4)\right) (s-1)^3 \\ + \left(\frac{1}{120}\gamma_5 + \frac{1}{6}\gamma_3\zeta(2) + \frac{7}{4}\gamma_1\zeta(4) - \frac{31}{16}\zeta(6)\right) (s-1)^4 - \dots \quad (6) $$

In (6), the coefficient of $(s-1)^{2k}$ and $(s-1)^{2k-1}$ are respectively

$$ \frac{\gamma_{2k+1}}{(2k+1)!} + \sum_{j=1}^{k} \frac{\gamma_{2k-2j+1}}{(2k-2j+1)!} \times \frac{(2^{2j-1}-1)}{2^{2j-2}} \zeta(2j) - \frac{(2^{2k+1}-1)}{2^{2k}} \zeta(2k+2), $$

and

$$ -\frac{\gamma_{2k}}{(2k)!} - \sum_{j=1}^{k} \frac{\gamma_{2k-2j}}{(2k-2j)!} \times \frac{(2^{2j-1}-1)}{2^{2j-2}} \zeta(2j). $$

b) The well-known Taylor series expansion of the digamma function:

$$ \psi(x+1) + \gamma = \sum_{n=1}^{\infty} (-1)^{n+1} \zeta(n+1) x^n \quad (|x| < 1), $$

allows us write

$$ \int_0^1 \frac{\psi(x+1) + \gamma}{x^s} dx = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\zeta(n+1)}{n - (s-1)}. $$

This leads to the following expansion:

$$ \int_0^1 \frac{\psi(x+1) + \gamma}{x^s} dx = \sum_{k=0}^{\infty} \left( \sum_{n=1}^{\infty} (-1)^{n+1} \frac{\zeta(n+1)}{n^{k+1}} \right) (s-1)^k \\ = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \times k! \tau_{k+1} (s-1)^k \quad (|s-1| < 1). \quad (7) $$

c) To obtain the expansion of $\sum_{n \ge 1} \frac{H_n}{n^s}$ at $s=1$, we proceed as follows: first we write the identities

$$ \frac{H_n}{n^s} = \frac{H_n}{n} e^{-(s-1) \ln n} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \times \frac{H_n (\ln n)^k}{n} (s-1)^k, $$
### 3 A new expression of $\zeta_H(s)$ and its implications

In this section, we give a new expression of the harmonic zeta function (Theorem 1) from which follows a new evaluation of the harmonic Stieltjes constants (Corollary 1). By means of this new representation of $\zeta_H$, we can also derive new integral formulas for $\zeta(n)$ and $\zeta'(n)$ (Corollary 2), and new evaluations for Glasser-Manna type integrals (Corollaries 4–6).

**Theorem 1.** For all complex numbers $s$ in $\mathbb{C} \setminus \mathbb{Z}$, we have

$$ \zeta_H(s) = \pi \cot(\pi s) \zeta(s) + \zeta(s+1) + \Gamma(1-s)\Phi(s) \quad (9) $$

with

$$ \Phi(s) := -\frac{1}{2i\pi} \int_{-\pi}^{\pi} x (\operatorname{Log}(1+e^{ix}))^{s-1} dx, \quad (10) $$

where $\operatorname{Log}$ denotes the principal value of the complex logarithm.

*Proof.* Let $z = \operatorname{Log}(1+e^{ix}) = \log(2\cos(\frac{x}{2})) + \frac{ix}{2}$, $-\pi < x < \pi$. When $x$ varies from $-\pi$ to $\pi$, the variable $z$ traverses the path $L$ defined by the parametric equations $\operatorname{Re}z = \log(2\cos(x/2))$ and $\operatorname{Im}z = x/2$. This path extends from $-\infty$ below the line $\operatorname{Im}z = 0$, passes through the point $(\log 2, 0)$, then extends back to $-\infty$ above the line $\operatorname{Im}z = 0$. Since $L$ is homotopic to the Hankel contour $C$ defined in [6, 7], then by means of [7, Equation (25)], we have, for $s \in \mathbb{C} \setminus \mathbb{Z}$, the representation

$$ \zeta_H(s) = \pi \cot(\pi s) \zeta(s) + \zeta(s+1) + \Gamma(1-s)\phi(s) - \psi(1-s)\zeta(s) - \zeta'(s), \quad (11) $$

where

$$ \phi(s) := \frac{1}{2i\pi} \int_L \frac{z^{s-1}e^z}{e^z - 1} \operatorname{Log} \left( \frac{e^z - 1}{z} \right) dz. $$

We now show that the equations (9) and (11) are equivalent. To do this, we proceed as follows. Differentiation of the integral representation

$$ \zeta(s) = \frac{\Gamma(1-s)}{2i\pi} \int_L \frac{z^{s-1}e^z}{1-e^z} dz $$

[6, Equation (2.4)], leads to the following identity:

$$ \psi(1-s)\zeta(s) + \zeta'(s) = \frac{\Gamma(1-s)}{2i\pi} \int_L \frac{z^{s-1}e^z \operatorname{Log}(z)}{1-e^z} dz \quad (s \neq 1, 2, 3, \dots) \quad (12) $$

[6, Equation (2.7)]. Splitting the right-hand member of (12), we can write

$$ \frac{1}{2i\pi} \int_L \frac{z^{s-1}e^z \operatorname{Log}(z)}{1-e^z} dz = \phi(s) - \Phi(s), \quad (13) $$
so that

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n}{n^s} = \sum_{n \ge 1}^{\mathcal{R}} \left( \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \times \frac{H_n(\ln n)^k}{n} (s-1)^k \right). $$

Moreover, [4, Theorem 9] allows us interchange $\sum_{n \ge 1}^{\mathcal{R}}$ and $\sum_{k=0}^{\infty}$. This leads to the expansion:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n}{n^s} = \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \left( \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n(\ln n)^k}{n} \right) (s-1)^k. \quad (8) $$

By combining the previous expansions (6), (7), and (8), we recover our formula (2) and we also obtain the desired formulas (3) and (4). □

### 3 Explicit expression of $\gamma_{H,k}$

The previously established Proposition 1 will enable us to derive an explicit expression of the constant $\gamma_{H,k}$ for each positive integer $k$.

If $b_n$ denotes the Bernoulli numbers of the second kind defined by means of their generating function

$$ \frac{x}{\ln(x+1)} = \sum_{n=0}^{\infty} b_n x^n = 1 + \frac{x}{2} - \frac{x^2}{12} + \frac{x^3}{24} - \frac{19x^4}{720} + \dots \quad (|x| < 1), $$

then the following formula is well-known [4, Equation 4.29], [10, Equation 16]:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n}{n} = \sum_{n=1}^{\infty} \frac{|b_n|}{n} \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} H_j = \sum_{n=1}^{\infty} \frac{|b_n|}{n^2} = 0.529052\dots $$

Hence formula (2) can be rewritten as follows:

$$ \gamma_{H,0} = \sum_{n=1}^{\infty} \frac{|b_n|}{n^2} - \gamma_1 - \tau_1 + \zeta(2). $$

More generally, for each natural number $k$, we derive from [10, Equation (13)] the following identity:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_n(\ln n)^k}{n} = \sum_{n=1}^{\infty} \frac{|b_n|}{n} \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} H_j (\ln j)^k \quad (k \ge 0). \quad (9) $$

Using (9) and Proposition 1, we then obtain an expression of $\gamma_{H,k}$ to be compared with a similar (but simpler) formula for $\gamma_k$ given long ago [9], namely:

$$ \gamma_k = \sum_{n=1}^{\infty} \frac{|b_n|}{n} \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} (\ln j)^k \quad (k \ge 0). $$
with

$$ \Phi(s) = \frac{1}{2i\pi} \int_L \frac{z^{s-1}e^z}{e^z - 1} \mathrm{Log}(e^z - 1) dz. $$

From (12) and (13), we deduce the identity

$$ \Gamma(1-s)\phi(s) - \psi(1-s)\zeta(s) - \zeta'(s) = \Gamma(1-s)\Phi(s). $$

Moreover, the substitution

$$ z = \mathrm{Log}(1 + e^{ix}), \quad -\pi < x < \pi, $$

in the contour integral defining $\Phi$ gives this function the equivalent form (10), proving that (9) and (11) are equivalent equations. $\square$

## 3.1 New formulas for $\gamma^{(k)}$

We deduce from (9) an expression of $\gamma^{(k)}$ which seems to us more meaningful than the one given by Kargin et al. [16, Equation (21)].

**Corollary 1.** For any non-negative integer $n$, let $\mathcal{L}_n$ and $\xi_n$ be defined as follows:

$$ \mathcal{L}_n := \Phi^{(n)}(1) = -\frac{1}{2i\pi} \int_{-\pi}^{\pi} x \mathrm{Log}^n(\mathrm{Log}(1+e^{ix})) dx, $$

and

$$ \xi_n := (-1)^n \Gamma^{(n)}(1) = (-1)^n \int_0^{+\infty} e^{-x} \mathrm{Log}^n(x) dx. $$

Then we have

$$ \gamma^{(0)} = -\mathcal{L}_1 - \gamma_1 - \zeta(2), \quad (14) $$

and, for all positive integers $k$, the following general relations hold true:

$$ \gamma^{(2k-1)} = -\zeta^{(2k-1)}(2) - \frac{\gamma_{2k}}{2k} + \frac{1}{2k} \sum_{n=0}^{2k-1} \binom{2k}{n} \xi_n \mathcal{L}_{2k-n} \\ + \sum_{n=1}^{k} \frac{2(2k-1)!}{(2k-2n)!} \gamma_{2k-2n} \zeta(2n), \quad (15) $$

and

$$ \gamma^{(2k)} = \zeta^{(2k)}(2) - \frac{\gamma_{2k+1}}{2k+1} - \frac{1}{2k+1} \sum_{n=0}^{2k} \binom{2k+1}{n} \xi_n \mathcal{L}_{2k+1-n} \\ + \sum_{n=1}^{k} \frac{2(2k)!}{(2k+1-2n)!} \gamma_{2k+1-2n} \zeta(2n) - 2(2k)! \zeta(2k+2). \quad (16) $$
**Corollary 1.** For $k$ odd, we have

$$ \gamma_{H,k} = \sum_{n=1}^{\infty} \frac{|b_n|}{n} \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} H_j (\ln j)^k - k! \tau_{k+1} - \frac{\gamma_{k+1}}{k+1} - \sum_{j=1}^{k+1/2} \frac{(2j-1)!(2^{2j-1}-1)}{2^{2j-2}} \binom{k}{2j-1} \gamma_{k+1-2j} \zeta(2j), \quad (10) $$

and, for $k$ even, we have

$$ \gamma_{H,k} = \sum_{n=1}^{\infty} \frac{|b_n|}{n} \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} H_j (\ln j)^k - k! \tau_{k+1} - \frac{\gamma_{k+1}}{k+1} - \sum_{j=1}^{k/2} \frac{(2j-1)!(2^{2j-1}-1)}{2^{2j-2}} \binom{k}{2j-1} \gamma_{k+1-2j} \zeta(2j) + k! \frac{(2^{k+1}-1)}{2^k} \zeta(k+2). \quad (11) $$

**Example 1.**

$$ \gamma_{H,1} = \sum_{n=1}^{\infty} \frac{|b_n|}{n} \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} H_j \ln j - \frac{1}{2}\gamma_2 - \tau_2 - \gamma\zeta(2) = 0.400761\dots $$

# 4 Another possible generalization

In this additional section, we examine the possibility of extending formula (2) by replacing the classical harmonic numbers $H_n$ with the Roman harmonic numbers $H_{n,k}$ [10, 14]. This path was recently explored by Young [13]. More precisely, for an arbitrary $k \ge 1$, let

$$ H_{n,k} := \sum_{n \ge n_1 \ge \dots \ge n_k \ge 1} \frac{1}{n_1 n_2 \dots n_k} = \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j} j^{-k}, $$

and

$$ \tau_1^{[k]} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} \zeta(n+1, \underbrace{1, \dots, 1}_{k-1}) = \sum_{n=1}^{\infty} \frac{|s(n,k)|}{n!} \ln \left(1 + \frac{1}{n}\right) $$

(where $s(n,k)$ are the Stirling numbers of the first kind), in such a way that $H_{n,1} = H_n$ and $\tau_1^{[1]} = \tau_1$. If $\gamma_1^{[k]}$ denotes the first Stieltjes constant of the height one zeta function of depth $k$, $\zeta(s, \underbrace{1, \dots, 1}_{k-1})$, then the following beautiful relation is

nothing else than a reformulation of [13, Theorem 6.1 d) and Corollary 5.2]:

$$ \sum_{n \ge 1}^{\mathcal{R}} \frac{H_{n,k}}{n} = P_{k+1}(\gamma, -\zeta(2), \dots, (-1)^k \zeta(k+1)) + \gamma_1^{[k]} + \tau_1^{[k]} = \sum_{n=1}^{\infty} \frac{|b_n|}{n^{k+1}}, \quad (12) $$
Moreover, it follows from (14) and (1) the explicit evaluation:

$$ \mathcal{L}_1 = -\frac{1}{2i\pi} \int_{-\pi}^{\pi} x \operatorname{Log}(\operatorname{Log}(1 + e^{ix})) dx = -\frac{3}{2}\zeta(2) - \frac{1}{2}\gamma^2 - \gamma_1. \quad (17) $$

**Example 2.** Applying formulas (15), (16) to the case $k = 1$ and using (17) above, we then obtain

$$ \gamma^{(1)} = -\zeta'(2) - \frac{1}{2}\gamma^3 - \frac{1}{2}\gamma_2 + \frac{1}{2}\mathcal{L}_2 + \frac{1}{2}\gamma\zeta(2) - \gamma\gamma_1, $$

and

$$ \gamma^{(2)} = \zeta''(2) + \frac{1}{2}\gamma^4 - \frac{1}{3}\gamma_3 - \frac{1}{3}\mathcal{L}_3 - \gamma\mathcal{L}_2 + (2\gamma^2 + 5\gamma_1)\zeta(2) + \gamma^2\gamma_1 - \frac{1}{10}(\zeta(2))^2. $$

Numerical evaluations of the constants $\mathcal{L}_2$ and $\mathcal{L}_3$ are

$$ \mathcal{L}_2 = -1.924491\dots \quad \text{and} \quad \mathcal{L}_3 = 7.158075\dots $$

*Proof of Corollary 1.* The key formula to derive the relations (14), (15) and (16) is the representation of $\zeta_H(s)$ given by (9). Fortunately, the Laurent series expansion of each component in (9) can be written explicitly.

a) The Laurent expansions of $\pi \cot(\pi s)$ and $\zeta(s)$ at $s = 1$ are respectively

$$ \pi \cot(\pi s) = \frac{1}{s-1} - \sum_{k \ge 1} 2\zeta(2k)(s-1)^{2k-1}, $$

and

$$ \zeta(s) = \frac{1}{s-1} + \gamma + \sum_{k \ge 1} \frac{(-1)^k}{k!} \gamma_k (s-1)^k, $$

where the coefficients $\gamma_k$ are the Stieltjes constants. The expansion of $\pi \cot(\pi s) \zeta(s)$ is then obtained by Cauchy product:

$$ \begin{aligned} \pi \cot(\pi s) \zeta(s) = & \frac{1}{(s-1)^2} + \frac{\gamma}{s-1} - \gamma_1 - 2\zeta(2) - \left(-\frac{1}{2}\gamma_2 + 2\gamma\zeta(2)\right)(s-1) \\ & + \frac{1}{2}\left(-\frac{1}{3}\gamma_3 + 4\zeta(2)\gamma_1 - 4\zeta(4)\right)(s-1)^2 \\ & - \frac{1}{6}\left(-\frac{1}{4}\gamma_4 + 6\zeta(2)\gamma_2 + 12\gamma\zeta(4)\right)(s-1)^3 + \dots \end{aligned} \quad (18) $$

b) It follows from the definition of $\xi_k$ as $(-1)^k \Gamma^{(k)}(1)$ that the Laurent expansion of $\Gamma(1-s)$ at $s=1$ is given by

$$ \Gamma(1-s) = -\frac{1}{s-1} - \gamma - \sum_{k=2}^{\infty} \frac{\xi_k}{k!} (s-1)^{k-1} \quad (|s-1| < 1), $$
where the $P_n$ are the modified Bell polynomials defined by means of their generating function:

$$ \begin{aligned} \sum_{n=0}^{\infty} P_n(x_1, \dots, x_n) t^n &= \exp \left( \sum_{k=1}^{\infty} x_k \frac{t^k}{k} \right) \\ &= 1 + x_1 t + \frac{1}{2}(x_1^2 + x_2) t^2 + \frac{1}{6}(x_1^3 + 3x_1x_2 + 2x_3) t^3 + \dots \end{aligned} $$

In the simplest case $k = 1$, formula (12) is nothing more than the original formula (2); for $k = 2$, since $\gamma_1^{[2]} = \gamma_{H,1} + \zeta'(2)$ and $|s(n, 2)| = (n - 1)! H_{n-1}$, formula (12) translates into the following noteworthy identity:

$$ \begin{aligned} \sum_{n \ge 1}^{\mathcal{R}} \frac{1}{n} \sum_{k=1}^{n} \frac{H_k}{k} &= \gamma_{H,1} + \frac{1}{6}\gamma^3 - \frac{1}{2}\gamma\zeta(2) + \frac{1}{3}\zeta(3) \\ &\quad + \sum_{n=1}^{\infty} \frac{H_{n-1}}{n} \ln \left(1 + \frac{1}{n}\right) + \zeta'(2) = 0.512673\dots \end{aligned} $$

**Acknowledgements** The authors are grateful to the reviewer for his valuable comments.

## References

[1] T. Apostol and T. H. Vu, Dirichlet series related to the Riemann zeta function, *J. Number Theory* **19** (1984), 85–102.

[2] U. Balakrishnan, Laurent Expansion of Dirichlet Series-II, *Monatshefte für Mathematik*, **103** (1987), 1–5.

[3] W. E. Briggs and R. G. Buschman, The power series coefficients of functions defined by Dirichlet series, *Illinois J. Math.* **5** (1961), 43–44.

[4] B. Candelpergher, *Ramanujan Summation of Divergent Series*, Lecture Notes in Math. 2185, Springer, 2017.

[5] B. Candelpergher, *Les séries divergentes, d'Euler à Ramanujan (Divergent series, from Euler to Ramanujan)*, Calvage & Mounet, Paris, 2023.

[6] B. Candelpergher and M-A. Coppo, Laurent expansion of harmonic zeta functions, *Journal of Mathematical Analysis and Applications* **491** (2020), 124309.

[7] B. K. Choudhury, The Riemann zeta-function and its derivatives, *Proceedings Royal Society London* **450** (1995), 477–499.
On the other hand, the function $\Phi$ defined by (10) is an entire function of $s$ with $\Phi(1) = \mathcal{L}_0 = 0$, and the definition of $\mathcal{L}_k$ as $\Phi^{(k)}(1)$ implies that the Taylor expansion of $\Phi(s)$ at $s = 1$ is given by

$$ \Phi(s) = \sum_{k=1}^{\infty} \frac{\mathcal{L}_k}{k!} (s-1)^k . $$

The Laurent expansion of $\Gamma(1-s)\Phi(s)$ then follows by Cauchy product:

$$ \Gamma(1-s)\Phi(s) = -\mathcal{L}_1 - \left(\frac{1}{2}\mathcal{L}_2 + \gamma\mathcal{L}_1\right)(s-1) \\ + \frac{1}{2}\left(-\frac{1}{3}\mathcal{L}_3 - \gamma\mathcal{L}_2 - 2\gamma^{(0)}\mathcal{L}_1\right)(s-1)^2 \\ - \frac{1}{6}\left(\frac{1}{4}\mathcal{L}_4 + \gamma\mathcal{L}_3 + 3\gamma^{(0)}\mathcal{L}_2 + (2\zeta(3) + 3\gamma\zeta(2) + \gamma^3)\mathcal{L}_1\right)(s-1)^3 + \dots \quad (19) $$

c) The Taylor expansion of $\zeta(s+1)$ at $s = 1$ can be written as follows:

$$ \zeta(s+1) = \zeta(2) + \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} (-1)^k \zeta^{(k)}(2) (s-1)^k . \quad (20) $$

By assembling equations (18), (19) and (20) above, we obtain, by identifying the constant term, the relation (14). In the same way, the general formulae (15) and (16) are derived by identifying the coefficients of higher degree. $\square$

**Remark 2.** a) The constants $\xi_n$ involved in formulae (15)–(16) have a closed form expression in terms of Euler’s constant $\gamma$ and $\zeta(2), \dots, \zeta(n)$ [3, 11]. More precisely, we have $\xi_0 = 1$, $\xi_1 = \gamma$, $\xi_2 = \gamma^2 + \zeta(2) = 2\gamma^{(0)}$, and by the recursion formula [11, Equation (2.2)]

$$ \xi_{n+1} = \gamma\xi_n + \sum_{k=1}^{n} \frac{n!}{(n-k)!} \zeta(k+1) \xi_{n-k} \quad (n \ge 1), $$

we also have

$$ \xi_3 = \gamma^3 + 3\gamma\zeta(2) + 2\zeta(3), \quad \xi_4 = \gamma^4 + 6\gamma^2\zeta(2) + 8\gamma\zeta(3) + 3(\zeta(2))^2 + 6\zeta(4), \text{ etc.} $$

A clever reformulation of this result is

$$ \xi_n = P_n(\gamma, \zeta(2), \dots, \zeta(n)) \quad (n \ge 2), $$

where the polynomials $P_n$ are defined by the exponential generating function

$$ \exp\left(\sum_{k=1}^{\infty} x_k \frac{t^k}{k}\right) = \sum_{n=0}^{\infty} P_k(x_1, x_2, \dots, x_n) \frac{t^n}{n!}. $$
[8] M-A. Coppo and B. Candelpergher, On some formulae related to Euler sums, *Lithuanian Mathematical Journal*, 63 (2023), 31–43.

[9] M-A. Coppo, Nouvelles expressions des constantes de Stieltjes, *Expositiones mathematicae*, **17** (1999), 349-358.

[10] M-A. Coppo, Miscellaneous series identities with Cauchy and harmonic numbers, and their interpretation as Ramanujan summation, *Integers* **23** (2023), #A61.

[11] L. Euler, Meditationes circa singulare serierum genus, *Novi Comm. Acad. Sci. Petropol. 20* (1776), 140–186. Reprinted in *Opera Omnia* ser. I, vol. 15. B. G. Teubner, Berlin, 1927, pp. 217–267.

[12] Y. Matsuoka, On the values of a certain Dirichlet series at rational integers, *Tokyo Journal of Mathematics* **5** (1982), 399–403.

[13] P. T. Young, Global series for height 1 multiple zeta functions, *European Journal of Mathematics* **9** (2023), 9:99.

[14] P. T. Young, Series of height one multiple zeta functions, *Integers* **24** (2024), #A43.
b) By means of a theorem of Ramanujan [12, Theorem 4] whose proof is given in [5, p. 224], we can write the following identity:

$$(-1)^n \zeta^{(n)}(2) = n! + \gamma_n + \sum_{k=1}^{\infty} \frac{(-1)^k}{k!} \gamma_{k+n} \quad (n \ge 1).$$

## 3.2 New formulas for $\zeta(n)$ and $\zeta'(n)$

The relation between $\Phi(s)$ and $\zeta_H(s)$ resulting from formula (9) when written in a neighborhood of $s = n$ (for an integer $n \ge 2$) enables to derive new formulas for $\zeta(n)$ and $\zeta'(n)$.

**Corollary 2.** For all integers $n \ge 2$, we have

$$\zeta(n) = \frac{(-1)^n}{(n-1)!} \times \frac{1}{2i\pi} \int_{-\pi}^{\pi} x \left(\mathrm{Log}(1+e^{ix})\right)^{n-1} dx, \quad (21)$$

and

$$\zeta'(n) = \frac{(-1)^n}{(n-1)!} \times \frac{1}{2i\pi} \int_{-\pi}^{\pi} x \left(\mathrm{Log}(\mathrm{Log}(1+e^{ix}))\right) \left(\mathrm{Log}(1+e^{ix})\right)^{n-1} dx \\ + \zeta_H(n) - \zeta(n) \psi(n) - \zeta(n+1). \quad (22)$$

*Proof of Corollary 2.* For $n \ge 2$, we can write

$$\pi \cot(\pi s) \zeta(s) = \frac{\zeta(n)}{s-n} + \zeta'(n) + O(s-n), \\ \Gamma(1-s) = \frac{(-1)^n}{(n-1)!} \left( \frac{1}{s-n} + (\gamma - H_{n-1}) \right) + O(s-n),$$

and

$$\Gamma(1-s) \Phi(s) = \frac{(-1)^n \Phi(n)}{(n-1)! s - n} + \frac{(-1)^n}{(n-1)!} (\Phi'(n) + \Phi(n)(\gamma - H_{n-1})) + O(s-n).$$

By applying formula (9) around $s = n$, we deduce the equation

$$\zeta_H(n) + O(s-n) = \left( \zeta(n) + \frac{(-1)^n}{(n-1)!} \Phi(n) \right) \frac{1}{s-n} \\ + \frac{(-1)^n}{(n-1)!} \left( \Phi'(n) + \Phi(n)(\gamma - H_{n-1}) \right) + \zeta'(n) + \zeta(n+1) + O(s-n).$$

This leads to the identities

$$\zeta(n) + \frac{(-1)^n}{(n-1)!} \Phi(n) = 0,$$
and

$$ \zeta_H(n) = \frac{(-1)^n}{(n-1)!} (\Phi'(n) + \Phi(n)(\gamma - H_{n-1})) + \zeta'(n) + \zeta(n+1), $$

thus proving formulas (21) and (22).

<figure><img src="image_2.png" /></figure>

**Remark 3.** It should be noted that our formula (21) for $\zeta(n)$ is similar to but different from [6, Equation (2.6)] which, after substitution $z = \text{Log}(1+e^{ix})$, translates into

$$ \zeta(n) = \frac{(-1)^{n-1}}{(n-1)!} \times \frac{1}{2\pi} \int_{-\pi}^{\pi} \text{Log}(\text{Log}(1+e^{ix})) (\text{Log}(1+e^{ix}))^{n-1} dx. $$

**Corollary 3.** For all positive integers $n$, we have

$$ \Phi(2n+1) = (2n)! \zeta(2n+1), \qquad (23) $$

$$ (-1)^n \Phi(2n) = (-1)^{n-1} (2n-1)! \zeta(2n) = \frac{B_{2n}}{4n} (2\pi)^{2n}, \qquad (24) $$

and

$$ \Phi'(2n) = (2n-1)! [n\zeta(2n+1) - \zeta(2n)\psi(2n) - \sum_{k=1}^{n-1} \zeta(2n-k)\zeta(k+1) - \zeta'(2n)]. \quad (25) $$

*Proof of Corollary 3.* By Euler's formula [5, p. 252], we have

$$ \zeta_H(2n) = (n+1)\zeta(2n+1) - \sum_{k=1}^{n-1} \zeta(2n-k)\zeta(k+1), $$

and thus formulas (23)–(25) derive from (21) and (22).

<figure><img src="image_3.png" /></figure>

**Example 3.** Applying (23), (24) and (25) to the case $n = 1$, we obtain

$$ \begin{aligned} \Phi(3) &= 2\zeta(3) = \zeta_H(2) = -\frac{1}{2i\pi} \int_{-\pi}^{\pi} x \text{Log}^2(1+e^{ix}) dx, \\ -\Phi(2) &= \zeta(2) = \frac{1}{2i\pi} \int_{-\pi}^{\pi} x \text{Log}(1+e^{ix}) dx = \frac{\pi^2}{6}, \\ \Phi'(2) &= \zeta(3) - \zeta(2)(1-\gamma) - \zeta'(2). \end{aligned} $$

### 3.3 New formulas for Glasser-Manna type integrals

The relation between $\Phi(s)$ and $\zeta_H(s)$ resulting from formula (9) when written in a neighborhood of $s = -n$ (for a non-negative integer $n$) enables to derive new formulas for Glasser-Manna type integrals.
**Corollary 4.** We have

$$ \Phi(0) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log(1+e^{ix})} dx = \frac{1}{2}\log(2\pi) - \frac{1}{2}\gamma + \frac{1}{2}, \quad (26) $$

and

$$ \begin{aligned} \Phi'(0) &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ix \log(\log(1+e^{ix}))}{\log(1+e^{ix})} dx \\ &= \frac{1}{2}\gamma_1 + \frac{1}{4}\log^2(2\pi) + \frac{1}{4}\gamma^2 - \frac{1}{2}\gamma - \frac{1}{2}\gamma \log(2\pi) - \frac{7}{8}\zeta(2) + \beta, \end{aligned} \quad (27) $$

where $\beta$ is the linear coefficient in the Laurent expansion of $\zeta_H$ at $s = 0$.

*Proof of Corollary 4.* Around 0, we have the decomposition given by (9):

$$ \zeta_H(s) = \zeta(s+1) + \pi \cot(\pi s) \zeta(s) + \Gamma(1-s)\Phi(s), $$

and the expansion [10, Equation (8)]

$$ \zeta_H(s) = \frac{1}{2s} + \frac{1}{2}\gamma + \frac{1}{2} + \beta s + O(s^2). $$

On the other side, we have the expansions

$$ \zeta(s+1) = \frac{1}{s} + \gamma - \gamma_1 s + O(s^2), $$

$$ \pi \cot(\pi s) \zeta(s) = -\frac{1}{2s} - \frac{1}{2} \log(2\pi) + \left( \frac{1}{2}\gamma_1 - \frac{1}{4}\log^2(2\pi) + \frac{1}{4}\gamma^2 + \frac{7}{8}\zeta(2) \right)s + O(s^2), $$

$$ \Gamma(1-s)\Phi(s) = \Phi(0) + (\Phi'(0) + \gamma\Phi(0)) + O(s^2). $$

By identifying the constant coefficient in the expansion of the right-hand member of (9), we deduce the equation

$$ \frac{1}{2}\gamma + \frac{1}{2} = \gamma - \frac{1}{2}\log(2\pi) + \Phi(0), $$

which is equivalent to (26). In the same way, formula (27) is derived by identifying the linear coefficient in the expansion of the right-hand member of (9). $\square$

**Remark 4.** We derive from [17, Corollary 4.2] the following evaluation of the constant $\beta$ which appears in formula (27):

$$ \beta = -\gamma_1^{[2]}(0) - \gamma_1 = 1 + \frac{1}{2}\gamma - \frac{1}{4}\gamma^2 - \gamma_1 + \frac{1}{4}\zeta(2) - \sum_{n=2}^{\infty} \frac{|b_n|}{(n-1)^2} = 1.589935\dots, $$

where $b_n$ are the Bernoulli numbers of the second kind defined by way of their generating function

$$ \frac{x}{\log(x+1)} = \sum_{n=0}^{\infty} b_n x^n \quad (|x| < 1). $$
**Remark 5.** Formula (26) appears in a slightly different but equivalent form in [15, Proposition 3.4], due to the identities

$$ \Phi(0) = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{ix \, dx}{2 \log(2 \cos(\frac{x}{2})) + ix} = \frac{2}{\pi} \int_{0}^{\pi} \frac{x^2}{x^2 + 4 \log^2(2 \cos(\frac{x}{2}))} \, dx. $$

**Corollary 5.** We have

$$ \Phi(-1) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log^2(1+e^{ix})} dx = \log(\mathcal{A}) - \frac{1}{12}\gamma + \frac{7}{24}, \quad (28) $$

where $\mathcal{A} = \exp(\frac{1}{12} - \zeta'(-1))$ is the Glaisher-Kinkelin constant, and for all integers $n \ge 2$,

$$ (2n-1)! \Phi(1-2n) = \log(\mathcal{A}_n) - \frac{B_{2n}}{2n}\gamma + \sum_{k=0}^{2n-1} \binom{2n-1}{k} \frac{B_k B_{2n-k}}{(2n-k)^2}, \quad (29) $$

$$ \text{with } \log(\mathcal{A}_n) := \frac{H_{2n-1}B_{2n}}{2n} - \zeta'(1-2n). $$

**Example 4.**

$$ \Phi(-3) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log^4(1+e^{ix})} dx = \frac{1}{6}\log(\mathcal{A}_2) + \frac{1}{720}\gamma + \frac{1}{320}. $$

*Proof of Corollary 5.* Around $s = -1$, we have the expansion [10, Equation (9)]

$$ \zeta_H(s) = -\frac{1}{12(s+1)} - \frac{1}{12}\gamma - \frac{1}{8} + O(s+1). $$

On the other side, we have

$$ \begin{aligned} \zeta(s+1) &= -\frac{1}{2} + O(s+1), \\ \pi \cot(\pi s) \zeta(s) &= \zeta'(-1) + O(s+1), \\ \Gamma(1-s)\Phi(s) &= \Phi(-1) + O(s+1). \end{aligned} $$

Formula (28) is then deduced by identifying the constant coefficient in the expansion of the right-hand member of (9). In the same way, from [10, Proposition 2], the Laurent expansion around $s = 1 - 2n$ for $n \ge 2$ is given by

$$ \zeta_H(s) = \frac{\zeta(1-2n)}{s+2n-1} - \frac{B_{2n}}{2n}\gamma + \frac{H_{2n-1}B_{2n}}{2n} + \sum_{k=0}^{2n-1} \binom{2n-1}{k} \frac{B_k B_{2n-k}}{(2n-k)^2} + O(s+2n-1) $$

which, by the same method, allows us to deduce formula (29).

<figure><img src="image_4.png" /></figure>

<footnote n="1">¹The numbers $\mathcal{A}_n$ are called generalized Glaisher-Kinkelin constants (cf. [13]).</footnote>
**Corollary 6.** For all positive integers $n$, we have

$$ (2n)! \Phi(-2n) = -\zeta'(-2n) + (2n+1) \frac{B_{2n}}{4n}, \qquad (30) $$

from which follows the reflection formula:

$$ 2(2n)! \Phi(-2n) = (-1)^{n+1} (2\pi)^{-2n} \Phi(2n+1) + (2n+1) \frac{B_{2n}}{2n}. \qquad (31) $$

**Example 5.**

$$ 2\Phi(-2) = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{ix}{\mathrm{Log}^3(1+e^{ix})} dx = -\zeta'(-2n) + \frac{1}{8} = \frac{1}{4\pi^2} \zeta(3) + \frac{1}{8}. \qquad (32) $$

*Proof of Corollary 6.* Around $s = -2n$, we have

$$ \zeta_H(s) = -\frac{B_{2n}}{4n} + \frac{B_{2n}}{2} + O(s + 2n). $$

On the other side, we have the expansions

$$ \begin{aligned} \zeta(s+1) &= -\frac{B_{2n}}{2n} + O(s+2n), \\ \pi \cot(\pi s) \zeta(s) &= \zeta'(-2n) + O(s+2n), \\ \Gamma(1-s)\Phi(s) &= (2n)!\Phi(-2n) + O(s+2n). \end{aligned} $$

By identifying the constant coefficient in the expansion of the right-hand member of (9), we obtain formula (30). Moreover, the well-known identity

$$ -2\zeta'(-2n) = (-1)^{n+1} (2\pi)^{-2n} (2n)! \zeta(2n+1) $$

and the relation

$$ \Phi(2n+1) = (2n)! \zeta(2n+1) $$

given by (23) enable to deduce formula (31) from (30). $\square$
# Appendix: Evaluation of Glasser-Manna integrals with shifted Mascheroni series

For any natural number $k$, let us consider the series

$$ \sigma_k := \sum_{n=k+1}^{\infty} \frac{|b_n|}{n-k}, $$

where $b_n$ are the Bernoulli numbers of the second kind defined by means of their generating function

$$ \frac{x}{\log(x+1)} = \sum_{n=0}^{\infty} b_n x^n = 1 + \frac{x}{2} - \frac{x^2}{12} + \frac{x^3}{24} - \frac{19x^4}{720} + \dots $$

For positive values of $k$, the series $\sigma_k$ are called *shifted Mascheroni series* [14]. The nice identities $\sigma_0 = \gamma$ and $\sigma_1 = \frac{1}{2}\log(2\pi) - \frac{1}{2}\gamma - \frac{1}{2}$ are well-known [14, Proposition 2]. The first one is a classical result due to Mascheroni, and the second allows us to deduce from formula (26) the following evaluation:

$$ \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log(1+e^{ix})} dx = \frac{4}{\pi} \int_{0}^{\pi/2} \frac{x^2}{x^2 + \log^2(2\cos(x))} dx = \sigma_1 + 1 = 1.13033\dots \quad (A1) $$

This formula was first given by Glasser and Manna [15, Proposition 3.3] in a slightly different but equivalent form. Moreover, thanks to the simplest special cases of the following formula arising from [14, Proposition 3]:

$$ \sum_{k=1}^{n} (-1)^{n+k} k! S(n, k) \sigma_{k+1} = -\zeta'(-n) - \frac{B_{n+1}}{n+1} (\gamma + H_{n+1} - H_n) \quad (n \ge 1), \quad (A2) $$

with $S(n, k)$ denoting the Stirling numbers of the second kind, we can easily derive from formulas (28) and (32) above further evaluations of the same kind, such as

$$ \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log^2(1+e^{ix})} dx = \sigma_2 + \frac{5}{12} = 0.49232\dots \quad (A3) $$

and

$$ \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{ix}{\log^3(1+e^{ix})} dx = -\sigma_2 + 2\sigma_3 + \frac{1}{8} = 0.15544\dots \quad (A4) $$

More generally, it is possible to deduce from (A2) the following identities that are nothing more than reinterpretations of formulas (29) and (30) in terms of shifted
Mascheroni series $\sigma_k$. These are

$$ \frac{(2n-1)!}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log^{2n}(1+e^{ix})} dx = \sum_{k=1}^{2n-1} (-1)^{k+1} k! S(2n-1, k) \sigma_{k+1} \\ + \frac{B_{2n} H_{2n}}{2n} + \sum_{k=0}^{2n-1} \binom{2n-1}{k} \frac{B_k B_{2n-k}}{(2n-k)^2} \quad (n \ge 2), \quad (A5) $$

and

$$ \frac{(2n)!}{2\pi} \int_{-\pi}^{\pi} \frac{ix}{\log^{2n+1}(1+e^{ix})} dx = \sum_{k=1}^{2n} (-1)^k k! S(2n, k) \sigma_{k+1} + (2n+1) \frac{B_{2n}}{4n} \quad (n \ge 1). \quad (A6) $$

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