On Some General Tornheim-Type Series

Abstract

In this paper, we solve the open problem posed by Kuba by expressing ${\sum }_{j,k\ge 1}\frac{H_k^{\left(u\right)}{H}_j^{\left(v\right)}{H}_{j+k}^{\left(w\right)}}{j^r{k}^s{(j+k)}^t}$ as a linear combination of multiple zeta values. These sums include Tornheim’s double series as a special case. Our approach is based on employing two distinct methods to evaluate the specific integral proposed by Yamamoto, which is associated with the two-poset Hasse diagram. We also provide a new evaluation formula for the general Mordell–Tornheim series and some similar types of double and triple series.

1. Introduction

Given an r-tuple $\mathit{\alpha }=({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_r)$ of positive integers with ${\alpha }_r\ge 2$, a multiple zeta value $\zeta \left(\mathit{\alpha }\right)$ (MZV) is defined to be [1,2,3]

$$\zeta \left(\mathit{\alpha }\right)=\sum _{1\le k_1<k_2<\cdots <k_r}\frac{1}{k_1^{{\alpha }_1}{k}_2^{{\alpha }_2}\cdots k_r^{{\alpha }_r}}.$$

The numbers ${\left|\mathit{\alpha }\right|}_r={\alpha }_1+{\alpha }_2+\cdots +{\alpha }_r$ and r are called the weight and the depth of $\zeta \left(\mathit{\alpha }\right)$, respectively. We let ${\left\{a\right\}}^k$ be k repetitions of a such that $\zeta ({\left\{1\right\}}^3,3)=\zeta (1,1,1,3)$ and $\zeta (2,{\left\{3\right\}}^2,5)=\zeta (2,3,3,5)$.

The generalized harmonic numbers are defined by

$$H_0^{\left(s\right)}=0\mathrm{and}{H}_n^{\left(s\right)}=\sum _{j=1}^n\frac{1}{j^s},$$

where s and n are positive integers. In particular, $H_n^{\left(1\right)}=H_n$ is the classical harmonic number. The famous formula:

$$\sum _{n=1}^{\infty }\frac{H_n}{n^2}=2\zeta \left(3\right)$$

was discovered by Euler. Series incorporating harmonic numbers find application in various mathematical disciplines and related fields (Refs. [4,5,6,7]).

For an r-tuple $\mathit{\alpha }=({\alpha }_1,{\alpha }_2,\dots ,{\alpha }_r)$ of positive integers, the Mordell–Tornheim multiple zeta values are defined by

$${\zeta }_{MT}({\alpha }_1,\dots ,{\alpha }_{r-1};{\alpha }_r)=\sum _{m_1,\dots ,m_r\ge 1}\frac{1}{m_1^{{\alpha }_1}\cdots m_{r-1}^{{\alpha }_{r-1}}{(m_1+\cdots +m_{r-1})}^{{\alpha }_r}}.$$

The investigation into the special values of this function, with $r=3$ at positive integer points, initiated with Tornheim [8] and independently by Mordell [9], particularly when ${\alpha }_1={\alpha }_2={\alpha }_3$. These values were also rediscovered by Witten [10] within the context of his volume formula for specific moduli spaces relevant to theoretical physics. Bradley and Zhou [11] subsequently demonstrated the expressibility of ${\zeta }_{MT}({\alpha }_1,\dots ,{\alpha }_{r-1};{\alpha }_r)$ as a linear combination of multiple zeta values. Kuba [12] delved into the exploration of two general infinite series:

$$S:=\sum _{j,k\ge 1}\frac{H_k^{\left(u\right)}{H}_j^{\left(v\right)}}{j^r{k}^s{(j+k)}^t}\mathrm{and}V:=\sum _{j,k\ge 1}\frac{H_{j+k}^{\left(u\right)}}{j^r{k}^s{(j+k)}^t}.$$

To express them, he utilized Euler sums such as:

$$\mathcal{T}(q;a_1,a_2)=\sum _{n\ge 1}\frac{H_{n-1}^{\left(a_1\right)}{H}_{n-1}^{\left(a_2\right)}}{n^q},\mathcal{M}(a,b,c;d)=\sum _{k\ge 1}\frac{H_{k-1}^{\left(d\right)}}{k^a}\sum _{j=1}^{k-1}\frac{1}{j^b}\sum _{\ell =1}^{j-1}\frac{1}{{\ell }^c},$$

in addition to multiple zeta values. In this paper, we provide a novel expression for the S-series, which differs from the one provided by Kuba in [12] (Theorem 7). Our expression directly represents the S-series as a linear combination of MZVs, making it more straightforward and concise.

Kuba indicated that it is interesting to study the sum defined by

$$\Theta (a,b,c,d,t,s):=\sum _{j,k\ge 1}\frac{H_k^{\left(a\right)}{H}_j^{\left(c\right)}{H}_{j+k}^{\left(s\right)}}{j^b{k}^d{(j+k)}^t},$$

which encompasses both S-series and V-series as specific examples. The objective of this paper is to express this generalized Tornheim-type series as a linear combination of multiple zeta values. In recent years, research on Tornheim-type series has continued to attract significant attention. For further information, see [13,14,15,16,17].

We introduce the shuffle algebraic structure for multiple zeta values in Section 2, as originally proposed by Hoffman [18]. Furthermore, we present a combinatorial generalization of the iterated integral associated with a two-poset, represented by a Hasse diagram. These integrals are referred to as Yamamoto’s integrals [19].

Despite the availability of the formula for assessing Tornheim’s double series, the method for computing general Mordell–Tornheim multiple zeta values remains obscure. While Bradley and Zhou [11] proved that this value can be delineated as a linear combination of multiple zeta values, they did not present a specific expression. Therefore, our paper contributes an explicit formula for it, as outlined in Theorem 1 within Section 3.

$$\begin{array}{c}{\displaystyle {\zeta }_{MT}(a_1+1,\dots ,a_n+1;s+1)}\hfill \\ =\sum _{\genfrac{}{}{0pt}{}{{\left|\mathbf{d}\right|}_n={\left|\mathbf{a}\right|}_n}{d_i\ge 0}}\zeta (d_1+1,d_2+1,\dots ,d_{n-1}+1,d_n+s+2)\sum _{\sigma \in S_n}{\sigma }_a\left\{\prod _{j=2}^n\left(\begin{array}{c}\sum _{k=j}^n{d}_k-\sum _{k=j+1}^n{a}_k\\ a_j\end{array}\right)\right\},\hfill \end{array}$$

where $a_1,a_2,\dots ,a_n$ are given nonnegative integers, $S_n$ is the symmetric group of n objects, and ${\sigma }_{\mathbf{a}}$ is the induced permutations of $\sigma \in S_n$ on the set $\{a_1,a_2,\dots ,a_n\}$.

Our method involves using two different approaches to compute the specific integral proposed by Yamamoto [19], which is connected to the two-poset Hasse diagram.

In Section 4, we present the $\Theta$ series as linear combinations using the following Euler sums:

$$\sum _{n,m\ge 1}\frac{H_{n-1}^{\left(a\right)}{H}_{m-1}^{\left(c\right)}}{n^b{m}^d{(n+m)}^t}\mathrm{and}\sum _{n,m,k\ge 1}\frac{H_{n-1}^{\left(a\right)}{H}_{m-1}^{\left(c\right)}}{n^b{m}^d{k}^e{(n+m)}^t{(k+n+m)}^s}.$$

We delve into the evaluation of these Euler sums in the subsequent five sections, employing various parameters. In the concluding section, we summarize our findings and illustrate their practical implications with examples like:

$$\begin{array}{ccc}\hfill \sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm(n+m)}& =& \frac{767{\pi }^6}{22680}+6\zeta {\left(3\right)}^2,\hfill \\ \hfill \sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm{(n+m)}^2}& =& \frac{11}{8}\zeta \left(7\right)+\frac{7{\pi }^2}{6}\zeta \left(5\right)-\frac{17{\pi }^4}{180}\zeta \left(3\right).\hfill \end{array}$$

2. Algebraic Settings and Integrals Associated with Two-Posets

We revisit the algebraic framework of multiple zeta values as outlined by Hoffman [18]. Consider $\mathbb{Q}\langle x,y\rangle$, the $\mathbb{Q}$-algebra of polynomials in two non-commutative variables, graded by degree (where both x and y are assigned a degree of 1). We identify the algebra $\mathbb{Q}\langle x,y\rangle$ with the graded $\mathbb{Q}$-vector space $\mathfrak{H}$ spanned by the monomials in the variables x and y.

We also introduce the graded $\mathbb{Q}$-vector spaces ${\mathfrak{H}}^0\subset {\mathfrak{H}}^1\subset \mathfrak{H}$ by ${\mathfrak{H}}^1=\mathbb{Q}\mathbf{1}⨁x\mathfrak{H}$, ${\mathfrak{H}}^0=\mathbb{Q}\mathbf{1}⨁x\mathfrak{H}y$, where $\mathbf{1}$ denotes the unit (the empty word of weight 0 and length 0) of the algebra $\mathfrak{H}$. A word starts with x and ends with y, and we refer to such words as “admissible”. In other words, the subalgebra ${\mathfrak{H}}^0$ is generated by admissible words. Let $Z:{\mathfrak{H}}^0\to \mathbb{R}$ be the $\mathbb{Q}$-linear map that assigns to each word $u_1{u}_2\cdots u_k$ in ${\mathfrak{H}}^0$, where $u_i\in \{x,y\}$, the multiple integral

$${\int }_{0<t_1<\cdots <t_k<1}{w}_{u_1}\left(t_1\right)w_{u_2}\left(t_2\right)\cdots w_{u_k}\left(t_k\right).$$

(1)

Here, $w_x\left(t\right)=dt/(1-t)$, $w_y\left(t\right)=dt/t$. As the word $u_1{u}_2\cdots u_k$ is in ${\mathfrak{H}}^0$, we always have $w_{u_1}\left(t\right)=dt/(1-t)$ and $w_{u_k}\left(t\right)=dt/t$, so the integral converges. The space ${\mathfrak{H}}^1$ can be regarded as the subalgebra of $\mathbb{Q}\langle x,y\rangle$ generated by the words $z_s=x{y}^{s-1}$ ($s=1,2,3,\dots$).

Let us define the bilinear product ⧢ (the shuffle product) on $\mathfrak{H}$ by the rules

$$1⧢w=w⧢1=w,$$

(2)

for any word w, and

$$w_1{x}_1⧢w_2{x}_2=(w_1⧢w_2{x}_2)x_1+(w_1{x}_1⧢w_2)x_2,$$

(3)

for any words $w_1$, $w_2$ and any letters $x_i=x$ or y ($i=1,2$), and then, extend the above rules to the whole algebra $\mathfrak{H}$ and the whole subalgebra ${\mathfrak{H}}^1$ by linearity. It is known that each of the above products is commutative and associative. We denote the algebras $({\mathfrak{H}}^1,+,⧢)$ by ${\mathfrak{H}}_{⧢}^1$. By the standard shuffle product identity of iterated integrals, the evaluation map Z is again an algebra homomorphism for the multiplication ⧢ (Ref. [20]):

$$Z(w_1⧢w_2)=Z\left(w_1\right)Z\left(w_2\right).$$

(4)

Yamamoto [19] introduced a combinatorial generalization of the iterated integral, the integral associated with a two-poset. We review the definitions and basic properties of 2-labeled posets (we will call them 2-posets for short in this paper) and the associated integrals first introduced by Yamamoto [19].

Definition 1

([21], Definition 3.1). A two-poset is a pair $(X,{\mu }_X)$, where $X=(X,\le )$ is a finite partially ordered set (poset for short) and ${\mu }_X$ is a map from X to $\{0,1\}$. We often omit ${\mu }_X$ and simply say “a two-poset X”. The ${\mu }_X$ is called the label map of X.

A two-poset $(X,{\mu }_X)$ is called admissible if ${\mu }_X\left(x\right)=0$ for all maximal elements $x\in X$ and ${\mu }_X\left(x\right)=1$ for all minimal elements $x\in X$.

A two-poset X is depicted as a Hasse diagram in which an element x with ${\mu }_X\left(x\right)=0$ (respectively, ${\mu }_X\left(x\right)=1$) is represented by ∘ (respectively, •). For example, the diagram

represents the two-poset $X=\{x_1,x_2,x_3,x_4,x_5\}$ with order $x_1<x_2<x_3>x_4<x_5$ and label $({\mu }_X\left(x_1\right),\dots ,{\mu }_X\left(x_5\right))=(1,0,0,1,0)$.

Definition 2

([21], Definition 3.2). For an admissible two-poset X, we define the associated integral:

$$I\left(X\right)={\int }_{{\Delta }_X}\prod _{x\in X}{\omega }_{{\mu }_X\left(x\right)}\left(t_x\right),$$

(5)

where

$${\Delta }_X=\left\{{\left(t_x\right)}_x\in {[0,1]}^X|t_x<t_{y}ifx<y\right\}and{\omega }_0\left(t\right)=\frac{dt}{t},{\omega }_1\left(t\right)=\frac{dt}{1-t}.$$

Note that the admissibility of a two-poset corresponds to the convergence of the associated integral. We also recall an algebraic setup for two-posets (cf. Remark at the end of §2 of [19]). Let $\mathfrak{P}$ be the $\mathbb{Q}$-algebra generated by the isomorphism classes of 2-posets, whose multiplication is given by the disjoint union of 2-posets. Then, the integral (5) defines a $\mathbb{Q}$-algebra homomorphism $I:{\mathfrak{P}}^0\to \mathbb{R}$ from the subalgebra ${\mathfrak{P}}^0$ of $\mathfrak{P}$ generated by the classes of admissible two-posets. We refer to this type of integral as Yamamoto’s integral.

It is known that [22] there is a $\mathbb{Q}$-linear map:

$$W:\mathfrak{P}\to \mathfrak{H},$$

(6)

which transforms a two-poset into a finite sum of words in x and y. This transformation is characterized by the following two conditions: the first condition states that, for a totally ordered $X=x_1<x_2<\cdots <x_k$, $W\left(X\right)=z_{\mu \left(x_1\right)}{z}_{\mu \left(x_2\right)}\cdots z_{\mu \left(x_k\right)}$, and the second condition asserts that, if a and b are non-comparable in X, then $W\left(X\right)$ can be expressed as $W\left(X_a^b\right)+W\left(X_b^a\right)$, where $X_a^b$ represents the two-poset obtained from X by adjoining a new relation $a<b$. This W sends ${\mathfrak{P}}^0$ onto ${\mathfrak{H}}^0$ and satisfies $I=Z\circ W:{\mathfrak{P}}^0\to \mathbb{R}$.

Here, we present three known examples in [19,22]: For admissible $\mathit{\alpha }=({\alpha }_1,\dots ,{\alpha }_r)$ and $\mathit{\beta }=({\beta }_1,{\beta }_2,\dots ,{\beta }_m)$, we have ([22], Propositions 2.4 and 2.7)

(7)

The last example is the Mordell–Tornheim multiple zeta value. They have the following integral form ([22], Proposition 2.8):

(8)

3. Explicit Formula for Mordell–Tornheim Series

Bradley and Zhou [11] used the mathematical induction to prove that ${\zeta }_{MT}(s_1,\dots ,s_{r-1};s_r)$ can be written in the form of a linear combination of the multiple zeta values. We could not find any explicit formula for this kind of representation except for the case $r=3$. Therefore, we give an explicit formula for the Mordell–Tornheim multiple zeta values in the following theorem. To save space, we usually write ${\sum }_{{\left|\mathit{\alpha }\right|}_r=w}$, which means ${\sum }_{\genfrac{}{}{0pt}{}{{\alpha }_1+{\alpha }_2+\cdots +{\alpha }_r=w}{{\alpha }_1,{\alpha }_2,\dots ,{\alpha }_r\ge 0}}$. Starting here, we will use this setting as the basis for describing the parameters of summations.

Theorem 1.

Given $n+1$ non-negative integers $a_1,a_2,\dots ,a_n$ and s with ${\sum }_{i=1}^n{a}_i=w$, we have

$$\begin{array}{c}{\displaystyle {\zeta }_{MT}(a_1+1,\dots ,a_n+1;s+1)}\hfill \\ =\sum _{{\left|\mathbf{d}\right|}_n=w}\zeta (d_1+1,d_2+1,\dots ,d_{n-1}+1,d_n+s+2)\sum _{\sigma \in S_n}{\sigma }_a\left\{\prod _{j=2}^n\left(\begin{array}{c}\sum _{k=j}^n{d}_k-\sum _{k=j+1}^n{a}_k\\ a_j\end{array}\right)\right\},\hfill \end{array}$$

(9)

where $S_n$ is the symmetric group of n objects and ${\sigma }_{\mathbf{a}}$ is the induced permutations of $\sigma \in S_n$ on the set $\{a_1,a_2,\dots ,a_n\}$.

Proof. 

Followed by Equation (8), we consider the corresponding integral form

We use the map W, which is defined in Equation (6), to transform this two-poset as $(x{y}^{a_1}⧢x{y}^{a_2}⧢x{y}^{a_n})y^{s+1}$. Since (see [23], Equation (4.6))

$$x{y}^{a_1}⧢x{y}^{a_2}⧢\cdots ⧢x{y}^{a_n}=\sum _{{\left|\mathbf{d}\right|}_n=w}\prod _{\ell =1}^{n}x{y}^{d_{\ell }}\sum _{\sigma \in S_n}{\sigma }_{\mathbf{a}}\left\{\prod _{j=2}^n\left(\genfrac{}{}{0pt}{}{{\sum }_{k=j}^n{d}_k-{\sum }_{k=j+1}^n{a}_k}{a_j}\right)\right\},$$

where $S_n$ is the symmetric group of n objects and ${\sigma }_{\mathbf{a}}$ is the induced permutations of $\sigma \in S_n$ on the set $\{a_1,a_2,\dots ,a_n\}$, then we shuffle them together into a totally ordered diagram. Using Equation (7), we can represent the integral in the desired form. This completes our proof. ☐

Here, we present the representation of a linear combination of multiple zeta values for the Tornheim double series: for non-negative integers $a_1,a_2,s$ with $a_1+a_2=w$ (Ref. [12], Theorem 5),

$${\zeta }_{MT}(a_1+1,a_2+1;s+1)=\sum _{d_1+d_2=w}\zeta (d_1+1,d_2+s+2)\left[\left(\genfrac{}{}{0pt}{}{d_2}{a_1}\right)+\left(\genfrac{}{}{0pt}{}{d_2}{a_2}\right)\right].$$

(10)

4. Decompositions with Relations to $\mathit{M}$-Series and $\mathit{K}$-Series

For non-negative integers $a,b,c,d,e$, and $s,t$, we define

$$\begin{array}{ccc}\hfill M(a,b,c,d,t)& :=& \sum _{n,m\ge 1}\frac{H_{n-1}^{\left(a\right)}{H}_{m-1}^{\left(c\right)}}{n^b{m}^d{(n+m)}^t},\mathrm{and}\hfill \end{array}$$

(11)

$$\begin{array}{ccc}\hfill K(a,b,c,d,e,t,s)& :=& \sum _{n,m,k\ge 1}\frac{H_{n-1}^{\left(a\right)}{H}_{m-1}^{\left(c\right)}}{n^b{m}^d{k}^e{(n+m)}^t{(k+n+m)}^s}.\hfill \end{array}$$

(12)

It is clear that M-series and K-series have the symmetric properties:

$$M(a,b,c,d,t)=M(c,d,a,b,t),\mathrm{and}K(a,b,c,d,e,t,s)=K(c,d,a,b,e,t,s).$$

We will divide the calculation of the series:

$$\sum _{n,m\ge 1}\frac{H_n^{\left(a\right)}{H}_m^{\left(c\right)}{H}_{n+m}^{\left(s\right)}}{n^b{m}^d{(n+m)}^t}$$

into two cases based on the value of s in one of the factors in the numerator, $H_{n+m}^{\left(s\right)}$. The two cases are $s=1$ and $s>1$. We will first discuss the case of $s=1$.

We rewrite the harmonic number $H_{n+m}$ as

$$H_{n+m}=\sum _{k=1}^{\infty }\frac{n+m}{k(k+n+m)}.$$

For non-negative integers $a,b,c,d$, and t, we have

$$\sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}{H}_{n+m}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}=\sum _{n,m,k\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}k{(n+m)}^t(k+n+m)}.$$

Given that $H_n^{\left(a\right)}=H_{n-1}^{\left(a\right)}+\frac{1}{n^a}$, we can reformulate the preceding series as follows:

$$\begin{array}{cc}\hfill & \sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}{H}_{n+m}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}\hfill \\ \hfill & =\sum _{n,m,k\ge 1}\frac{H_{n-1}^{(a+1)}{H}_{m-1}^{(c+1)}}{n^{b+1}{m}^{d+1}k{(n+m)}^t(k+n+m)}+\sum _{n,m,k\ge 1}\frac{H_{n-1}^{(a+1)}}{n^{b+1}{m}^{c+d+2}k{(n+m)}^t(k+n+m)}\hfill \\ \hfill & +\sum _{n,m,k\ge 1}\frac{H_{m-1}^{(c+1)}}{n^{a+b+2}{m}^{d+1}k{(n+m)}^t(k+n+m)}+\sum _{n,m,k\ge 1}\frac{1}{n^{a+b+2}{m}^{c+d+2}k{(n+m)}^t(k+n+m)}.\hfill \end{array}$$

(13)

It is evident that these four series stem from substituting specific values into the K-series. They correspond to $K(a+1,b+1,c+1,d+1,1,t,1)$, $K(a+1,b+1,0,c+d+2,1,t,1)$, $K(0,a+b+2,c+1,d+1,1,t,1)$, and $K(0,a+b+2,0,c+d+2,1,t,1)$. Given the symmetry property in the K-series, we will focus on computing only three general types of K-series in the following three sections.

Secondly, we will calculate the series:

$$\sum _{n,m\ge 1}\frac{H_n^{\left(a\right)}{H}_m^{\left(c\right)}{H}_{n+m}^{\left(s\right)}}{n^b{m}^d{(n+m)}^t}$$

with the parameter $s>1$.

For a positive integer $s\ge 2$, we rewrite the generalized harmonic number $H_{n+m}^{\left(s\right)}$ as

$$H_{n+m}^{\left(s\right)}=\zeta \left(s\right)-\sum _{k=1}^{\infty }\frac{1}{{(k+n+m)}^s}.$$

For non-negative integers $a,b,c,d$, and $s,t$, we have

$$\begin{array}{cc}\hfill & \sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}{H}_{n+m}^{(s+2)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}\hfill \\ \hfill & =\zeta (s+2)\sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}-\sum _{n,m,k\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}.\hfill \end{array}$$

(14)

The series

$$\sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}$$

was studied by Kuba [12], and he denoted this series as $S(b+1,d+1,t+1,a+1,c+1)$. Nonetheless, our approach will be utilized to offer an alternative expression, achieved through a linear combination of multiple zeta values. We write this series as follows.

$$\begin{array}{cc}\hfill & \sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}\hfill \\ \hfill & =\sum _{n,m\ge 1}\frac{H_{n-1}^{(a+1)}{H}_{m-1}^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}+\sum _{n,m\ge 1}\frac{H_{n-1}^{(a+1)}}{n^{b+1}{m}^{c+d+2}{(n+m)}^{t+1}}\hfill \\ \hfill & +\sum _{n,m\ge 1}\frac{H_{m-1}^{(c+1)}}{n^{a+b+2}{m}^{d+1}{(n+m)}^{t+1}}+\sum _{n,m\ge 1}\frac{1}{n^{a+b+2}{m}^{c+d+2}{(n+m)}^{t+1}}.\hfill \end{array}$$

(15)

The four aforementioned series are evidently derived by substituting specific values into the M-series. These values correspond to $M(a+1,b+1,c+1,d+1,t+1)$, $M(a+1,b+1,0,c+d+2,t+1)$, $M(0,a+b+2,c+1,d+1,t+1)$, and $M(0,a+b+2,0,c+d+2,t+1)$. Notably, the last M-series corresponds to Tornheim’s double series ${\zeta }_{MT}(a+b+2,c+d+2;t+1)$. Additionally, due to the symmetric property of the M-series, our analysis will focus solely on the form $M(a+1,b+1,0,c+1,t+1)$ for the following two M-series: $M(a+1,b+1,0,c+d+2,t+1)$, $M(0,a+b+2,c+1,d+1,t+1)$. These will be computed in Section 8.

The second series in the right-hand side of Equation (14) is written as

$$\begin{array}{cc}\hfill & \sum _{n,m,k\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}\hfill \\ \hfill & =\sum _{n,m,k\ge 1}\frac{H_{n-1}^{(a+1)}{H}_{m-1}^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}+\sum _{n,m,k\ge 1}\frac{H_{n-1}^{(a+1)}}{n^{b+1}{m}^{c+d+2}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}\hfill \\ \hfill & +\sum _{n,m,k\ge 1}\frac{H_{m-1}^{(c+1)}}{n^{a+b+2}{m}^{d+1}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}+\sum _{n,m,k\ge 1}\frac{1}{n^{a+b+2}{m}^{c+d+2}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}.\hfill \end{array}$$

(16)

The four series mentioned earlier can also be generated by substituting specific values into the K-series. These values correspond to $K(a+1,b+1,c+1,d+1,0,t+1,s+2)$, $K(a+1,b+1,0,c+d+2,0,t+1,s+2)$, $K(0,a+b+2,c+1,d+1,0,t+1,s+2)$, and $K(0,a+b+2,0,c+d+2,0,t+1,s+2)$. These formulations will be explored in Section 9.

5. Evaluations of $\mathit{K}(\mathbf{0},\mathit{a}+\mathbf{1},\mathbf{0},\mathit{b}+\mathbf{1},\mathit{c}+\mathbf{1},\mathit{t},\mathit{s}+\mathbf{1})$

Theorem 2.

Given five non-negative integers $a,b,c,t,s$, we have

$$\begin{array}{c}{\displaystyle \sum _{m,n,k\ge 1}\frac{1}{m^{a+1}{n}^{b+1}{k}^{c+1}{(n+m)}^t{(n+m+k)}^{s+1}}}\hfill \\ {\displaystyle =\sum _{\genfrac{}{}{0pt}{}{{\alpha }_1+{\alpha }_2=a+b}{{\beta }_1+{\beta }_2+{\beta }_3=a+b+c+t}}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b}\right)\right]\left\{\left(\genfrac{}{}{0pt}{}{{\beta }_3}{c}\right){\delta }_{{\alpha }_1,{\beta }_1}+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2+t}\right)\right.}\\ \hfill \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2+t}\right)\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+s+2).\end{array}$$

Proof. 

First, we evaluate Yamamoto’s integral:

Since W sends ${\mathfrak{P}}^0$ onto ${\mathfrak{H}}^0$ and $I=Z\circ W:{\mathfrak{P}}^0\to \mathbb{R}$, the integral corresponding to the two-poset Hasse diagram shown above can be written as

$$Z\left(((x{y}^a⧢x{y}^b)y^t⧢x{y}^c)y^{s+1}\right).$$

Using the formula (Ref. [23], Equation (4.2)):

$$x{y}^p⧢x{y}^q=\sum _{{\alpha }_1+{\alpha }_2=p+q}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{p}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{q}\right)\right]x{y}^{{\alpha }_1}x{y}^{{\alpha }_2},$$

(17)

and the formula (Ref. [24], Equation (2.9)):

$$x{y}^{a_1}x{y}^{a_2}⧢x{y}^b=\sum _{\genfrac{}{}{0pt}{}{{\alpha }_1+{\alpha }_2+{\alpha }_3}{=a_1+a_2+b}}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right){\delta }_{{\alpha }_1,a_1}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_2}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a_1-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_2}\right)\right\}x{y}^{{\alpha }_1}x{y}^{{\alpha }_2}x{y}^{{\alpha }_3},$$

(18)

we transform $((x{y}^a⧢x{y}^b)y^t⧢x{y}^c)y^{s+1}$ into a linear combination form in terms of Lyndon words in ${\mathfrak{H}}^0$ and apply the Z mapping, yielding the following result:

$$\begin{array}{c}{\displaystyle Z\left(((x{y}^a⧢x{y}^b)y^t⧢x{y}^c)y^{s+1}\right)}\hfill \\ {\displaystyle =\sum _{\genfrac{}{}{0pt}{}{{\alpha }_1+{\alpha }_2=a+b}{{\beta }_1+{\beta }_2+{\beta }_3=a+b+c+t}}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b}\right)\right]}\hfill & \left\{\left(\genfrac{}{}{0pt}{}{{\beta }_3}{c}\right){\delta }_{{\alpha }_1,{\beta }_1}+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2+t}\right)\right.\hfill \\ & \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2+t}\right)\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+s+2).\hfill \end{array}$$

Combining the infinite series expression we obtained at the beginning, we complete the proof of this theorem. ☐

If $t=0$ in Theorem 2, then this K-series is the Mordell–Tornheim series with $r=4$.

$$\begin{array}{ccc}& {\zeta }_{MT}& (a+1,b+1,c+1;s+1)=K(0,a+1,0,b+1,c+1,0,s+1)\hfill \\ & =& \sum _{\genfrac{}{}{0pt}{}{{\alpha }_1+{\alpha }_2=a+b}{{\beta }_1+{\beta }_2+{\beta }_3=a+b+c}}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+s+2)\hfill \\ & & \times \left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b}\right)\right]\left\{\left(\genfrac{}{}{0pt}{}{{\beta }_3}{c}\right){\delta }_{{\alpha }_1,{\beta }_1}+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\right\}.\hfill \end{array}$$

The formula provided here differs from the result we substituted in Theorem 1 previously. Below, we have also written the corresponding expression obtained from Theorem 1.

$$\begin{array}{c}{\zeta }_{MT}(a+1,b+1,c+1;s+1)=\sum _{{\alpha }_1+{\alpha }_2+{\alpha }_3=a+b+c}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+s+2)\hfill \\ \times \left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_2+{\alpha }_3-c}{b}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2+{\alpha }_3-c}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2+{\alpha }_3-a}{b}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\right.\\ \hfill \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_2+{\alpha }_3-b}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2+{\alpha }_3-a}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2+{\alpha }_3-b}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)\right\}.\end{array}$$

We list some examples:

$$\begin{array}{ccc}\hfill {\zeta }_{MT}(1,1,1;1)& =& 6\zeta (1,1,2)=6\zeta \left(4\right),\hfill \\ \hfill {\zeta }_{MT}(1,1,2;1)& =& 6\zeta (1,1,3)+4\zeta (1,2,2)+2\zeta (2,1,2),\hfill \\ \hfill \sum _{m,n,k\ge 1}\frac{1}{mnk(m+n)(m+n+k)}& =& 6\zeta (1,1,3)+2\zeta (1,2,2).\hfill \end{array}$$

6. Evaluations of $\mathit{K}(\mathit{a}+\mathbf{1},\mathit{b}+\mathbf{1},\mathbf{0},\mathit{c}+\mathbf{1},\mathit{d}+\mathbf{1},\mathit{t},\mathit{s}+\mathbf{1})$

Theorem 3.

Given six non-negative integers $a,b,c,d,t,s$, we have

$$\begin{array}{c}{\displaystyle \sum _{m,n,k\ge 1}\frac{H_{n-1}^{(a+1)}}{n^{b+1}{m}^{c+1}{k}^{d+1}{(n+m)}^t{(n+m+k)}^{s+1}}}\hfill \\ =\sum _{\genfrac{}{}{0pt}{}{{\left|\mathit{\alpha }\right|}_3=a+b+c}{{\left|\mathit{\beta }\right|}_4=a+b+c+d+t}}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)\right]\hfill \\ \times \left\{\left(\genfrac{}{}{0pt}{}{{\beta }_4}{d}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}+\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2-{\beta }_2}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3+t}\right){\delta }_{{\beta }_1,{\alpha }_1}\right.\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3+t}\right)\left[\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\right]\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+1,{\beta }_4+s+2).\hfill \end{array}$$

Proof. 

First, we evaluate Yamamoto’s integral:

Since W sends ${\mathfrak{P}}^0$ onto ${\mathfrak{H}}^0$ and $I=Z\circ W:{\mathfrak{P}}^0\to \mathbb{R}$, the integral corresponding to the two-poset Hasse diagram shown above can be written as

$$Z\left(((x{y}^{a}x{y}^b⧢x{y}^c)y^t⧢x{y}^d)y^{s+1}\right).$$

Using Equation (18) and the formula (Ref. [24], Equation (2.10)):

$$\begin{array}{c}x{y}^{a_1}x{y}^{a_2}x{y}^{a_3}⧢x{y}^b=\sum _{{\left|\mathit{\alpha }\right|}_4=a_1+a_2+a_3+b}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,a_1}{\delta }_{{\alpha }_2,a_2}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_2-{\alpha }_2}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{a_3}\right){\delta }_{{\alpha }_1,a_1}\right.\hfill \\ \hfill \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_2}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{a_3}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a_1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a_1-{\alpha }_1}\right)\right]\right\}x{y}^{{\alpha }_1}x{y}^{{\alpha }_2}x{y}^{{\alpha }_3}x{y}^{{\alpha }_4},\end{array}$$

(19)

we transform $((x{y}^{a}x{y}^b⧢x{y}^c)y^t⧢x{y}^d)y^{s+1}$ into a linear combination form in terms of Lyndon words in ${\mathfrak{H}}^0$ and apply the Z mapping, yielding the following result:

$$\begin{array}{cc}\hfill & Z\left(((x{y}^{a}x{y}^b⧢x{y}^c)y^t⧢x{y}^d)y^{s+1}\right)\hfill \\ \hfill & {\displaystyle =\sum _{\genfrac{}{}{0pt}{}{{\left|\alpha \right|}_3=a+b+c}{{\left|\mathit{\beta }\right|}_4=a+b+c+d+t}}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)\right]}\hfill \\ \hfill & \times \left\{\left(\genfrac{}{}{0pt}{}{{\beta }_4}{d}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}+\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2-{\beta }_2}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3+t}\right){\delta }_{{\beta }_1,{\alpha }_1}\right.\hfill \\ \hfill & {\displaystyle \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3+t}\right)\left[\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\right]\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+1,{\beta }_4+s+2).}\hfill \end{array}$$

Combining the infinite series expression we obtained at the beginning, we complete the proof of this theorem. ☐

We give some evaluations:

$$\begin{array}{ccc}\hfill \sum _{m,n,k\ge 1}\frac{H_{n-1}}{mnk(n+m+k)}& =& 12\zeta (1,1,1,2)=12\zeta \left(5\right),\hfill \\ \hfill \sum _{m,n,k\ge 1}\frac{H_{n-1}}{mnk(n+m)(n+m+k)}& =& 12\zeta (1,1,1,3)+3\zeta (1,1,2,2).\hfill \end{array}$$

7. Evaluations of $\mathit{K}(\mathit{a}+\mathbf{1},\mathit{b}+\mathbf{1},\mathit{c}+\mathbf{1},\mathit{d}+\mathbf{1},\mathit{e}+\mathbf{1},\mathit{t},\mathit{s}+\mathbf{1})$

Theorem 4.

Given seven non-negative integers $a,b,c,d,e,t,s$, we have

$$\begin{array}{cc}\hfill & \sum _{m,n,k\ge 1}\frac{H_{n-1}^{(a+1)}{H}_{m-1}^{(c+1)}}{n^{b+1}{m}^{d+1}{k}^{e+1}{(n+m)}^t{(n+m+k)}^{s+1}}\hfill \\ \hfill & =\sum _{\genfrac{}{}{0pt}{}{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}{{\left|\mathit{\beta }\right|}_5=a+b+c+d+e+t}}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.\hfill \\ \hfill & \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\times \left\{\left(\genfrac{}{}{0pt}{}{{\beta }_5}{e}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}{\delta }_{{\beta }_3,{\alpha }_3}\right.\hfill \\ \hfill & +\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3-{\beta }_3}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}+\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2-{\beta }_2}\right){\delta }_{{\beta }_1,{\alpha }_1}\hfill \\ \hfill & \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\left[\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\right]\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+1,{\beta }_4+1,{\beta }_5+s+2).\hfill \end{array}$$

Proof. 

First, we evaluate Yamamoto’s integral:

Since W sends ${\mathfrak{P}}^0$ onto ${\mathfrak{H}}^0$ and $I=Z\circ W:{\mathfrak{P}}^0\to \mathbb{R}$, the integral corresponding to the two-poset Hasse diagram shown above can be written as

$$Z\left(((x{y}^{a}x{y}^b⧢x{y}^{c}x{y}^d)y^t⧢x{y}^e)y^{s+1}\right).$$

We use the following equation (Ref. [24], Equation (2.11)) in the above formula:

$$\begin{array}{c}{\displaystyle x{y}^{a_1}x{y}^{a_2}⧢x{y}^{b_1}x{y}^{b_2}}\hfill \\ =\sum _{{\left|\mathit{\alpha }\right|}_4=a_1+a_2+b_1+b_2}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{a_2}\right){\delta }_{{\alpha }_1,b_1}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b_2}\right){\delta }_{{\alpha }_1,a_1}\right.\hfill \\ +\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_2+b_2-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{a_2}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b_1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b_1-{\alpha }_1}\right)\right]\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a_2+b_2-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b_2}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a_1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a_1-{\alpha }_1}\right)\right]\right\}x{y}^{{\alpha }_1}x{y}^{{\alpha }_2}x{y}^{{\alpha }_3}x{y}^{{\alpha }_4},\hfill \end{array}$$

(20)

and the formula (Ref. [24], Equation (2.7)):

$$\begin{array}{c}{\displaystyle x{y}^a⧢x{y}^{b_1}x{y}^{b_2}x{y}^{b_3}x{y}^{b_4}}\hfill \\ =\sum _{{\left|\mathit{\alpha }\right|}_5=a+b_1+b_2+b_3+b_4}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_5}{a}\right){\delta }_{{\alpha }_1,b_1}{\delta }_{{\alpha }_2,b_2}{\delta }_{{\alpha }_3,b_3}+\left(\genfrac{}{}{0pt}{}{{\alpha }_5}{b_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b_3-{\alpha }_3}\right){\delta }_{{\alpha }_1,b_1}{\delta }_{{\alpha }_2,b_2}\right.\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b_3}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_5}{b_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b_2-{\alpha }_2}\right){\delta }_{{\alpha }_1,b_1}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b_2}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b_3}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_5}{b_4}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b_1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b_1-{\alpha }_1}\right)\right]\right\}\hfill \\ x{y}^{{\alpha }_1}x{y}^{{\alpha }_2}x{y}^{{\alpha }_3}x{y}^{{\alpha }_4}x{y}^{{\alpha }_5},\hfill \end{array}$$

(21)

we transform $((x{y}^{a}x{y}^b⧢x{y}^{c}x{y}^d)y^t⧢x{y}^e)y^{s+1}$ into a linear combination form in terms of Lyndon words in ${\mathfrak{H}}^0$ and apply the Z mapping, yielding the following result:

$$\begin{array}{cc}\hfill & Z\left(((x{y}^{a}x{y}^b⧢x{y}^{c}x{y}^d)y^t⧢x{y}^e)y^{s+1}\right)\hfill \\ \hfill & =\sum _{\genfrac{}{}{0pt}{}{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}{{\left|\mathit{\beta }\right|}_5=a+b+c+d+e+t}}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.\hfill \\ \hfill & \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\times \left\{\left(\genfrac{}{}{0pt}{}{{\beta }_5}{e}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}{\delta }_{{\beta }_3,{\alpha }_3}\right.\hfill \\ \hfill & +\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3-{\beta }_3}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}+\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2-{\beta }_2}\right){\delta }_{{\beta }_1,{\alpha }_1}\hfill \\ \hfill & \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\left[\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\right]\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+1,{\beta }_4+1,{\beta }_5+s+2).\hfill \end{array}$$

Combining the infinite series expression we obtained at the beginning, we complete the proof of this theorem. ☐

We give some evaluations:

$$\begin{array}{ccc}\hfill \sum _{m,n,k\ge 1}\frac{H_{n-1}{H}_{m-1}}{mnk(n+m+k)}& =& 30\zeta ({\left\{1\right\}}^4,2)=30\zeta \left(6\right),\hfill \end{array}$$

(22)

$$\begin{array}{ccc}\hfill \sum _{m,n,k\ge 1}\frac{H_{n-1}{H}_{m-1}}{mnk(n+m)(n+m+k)}& =& 30\zeta ({\left\{1\right\}}^4,3)+6\zeta ({\left\{1\right\}}^3,2,2).\hfill \end{array}$$

(23)

We can now write Equation (13) as a linear combination of multiple zeta values. According to Theorems 2–4, we obtain the following result:

$$\begin{array}{c}{\displaystyle \sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}{H}_{n+m}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}}\hfill \\ {\displaystyle =\sum _{\genfrac{}{}{0pt}{}{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}{{\left|\mathit{\beta }\right|}_5=a+b+c+d+t}}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\times \left\{{\beta }_5{\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}{\delta }_{{\beta }_3,{\alpha }_3}\right.\hfill \\ +\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3-{\beta }_3}\right){\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}+\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2-{\beta }_2}\right){\delta }_{{\beta }_1,{\alpha }_1}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_5}{{\alpha }_4+t}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\left[\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\right]\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+1,{\beta }_4+1,{\beta }_5+2)\hfill \\ {\displaystyle +\sum _{\genfrac{}{}{0pt}{}{{\left|\mathit{\alpha }\right|}_3=a+b+c+d+1}{{\left|\mathit{\beta }\right|}_4=a+b+c+d+t+1}}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a+b+1}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c+d+1}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right]\times \left\{{\delta }_{{\beta }_1,{\alpha }_1}{\delta }_{{\beta }_2,{\alpha }_2}+\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2-{\beta }_2}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3+t}\right){\delta }_{{\beta }_1,{\alpha }_1}\right.\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_4}{{\alpha }_3+t}\right)\left[\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\right]\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+1,{\beta }_4+2)\hfill \\ {\displaystyle +\sum _{\genfrac{}{}{0pt}{}{{\left|\mathit{\alpha }\right|}_2=a+b+c+d+2}{{\left|\mathit{\beta }\right|}_3=a+b+c+d+t+2}}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a+b+1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c+d+1}\right)\right]}\hfill \\ \times \left\{{\delta }_{{\alpha }_1,{\beta }_1}+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2+t}\right)+\left(\genfrac{}{}{0pt}{}{{\beta }_2}{{\alpha }_1-{\beta }_1}\right)\left(\genfrac{}{}{0pt}{}{{\beta }_3}{{\alpha }_2+t}\right)\right\}\zeta ({\beta }_1+1,{\beta }_2+1,{\beta }_3+2).\hfill \end{array}$$

(24)

8. Another Expression of $\mathit{S}$-Series

In this section, we will give another expression of S-series.

Firstly, the $M(a+1,b+1,c+1,d+1,t+1)$ series is evaluated by Yamamoto’s integral and the corresponding shuffle relation:

The proof process is the same as that demonstrated in Theorems 1–4. Therefore, we only list the corresponding result.

Theorem 5.

For non-negative integers $a,b,c,d$, and t, we have

$$\begin{array}{cc}\hfill & \sum _{n,m\ge 1}\frac{H_{n-1}^{(a+1)}{H}_{m-1}^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}\hfill \\ \hfill & =\sum _{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.\hfill \\ \hfill & \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+1,{\alpha }_4+t+2).\hfill \end{array}$$

The second series $M(a+1,b+1,0,c+1,t+1)$ is evaluated by Yamamoto’s integral and the corresponding shuffle relation:

Therefore, we have the following theorem.

Theorem 6.

For non-negative integers $a,b,c$, and t, we have

$$\begin{array}{c}\sum _{n,m\ge 1}\frac{H_{n-1}^{(a+1)}}{n^{b+1}{m}^{c+1}{(n+m)}^{t+1}}\hfill \\ \hfill =\sum _{{\left|\mathit{\alpha }\right|}_3=a+b+c}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+t+2).\end{array}$$

Applying Theorems 5 and 6 to Equation (15), we have

$$\begin{array}{c}{\displaystyle \sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}}\hfill \\ {\displaystyle =\sum _{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+1,{\alpha }_4+t+2)\hfill \\ {\displaystyle +\sum _{{\left|\mathit{\alpha }\right|}_3=a+b+c+d+1}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a+b+1}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c+d+1}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+t+2)\hfill \\ {\displaystyle +\sum _{{\left|\mathit{\alpha }\right|}_2=a+b+c+d+2}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a+b+1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c+d+1}\right)\right]\zeta ({\alpha }_1+1,{\alpha }_2+t+2).}\hfill \end{array}$$

(25)

Here, we provide a novel expression for the S-series, which differs from the one provided by Kuba in [12] (Theorem 7). Kuba’s expression not only uses MZVs, but also extensively employs his defined $\mathcal{T}$ and $\mathcal{M}$ functions, each of which is a sum of three MZVs. Our expression directly represents the S-series as a linear combination of MZVs, making it more straightforward and concise.

Here are some examples illustrating the application of Equation (25).

$$\begin{array}{c}{\displaystyle \sum _{n,m\ge 1}\frac{H_n{H}_m}{nm(n+m)}}\hfill \\ =6\zeta ({\left\{1\right\}}^3,2)+6\zeta (1,1,3)+4\zeta (1,2,2)+2\zeta (2,1,2)+4\zeta (1,4)+2\zeta (2,3),\hfill \end{array}$$

(26)

$$\begin{array}{c}{\displaystyle \sum _{n,m\ge 1}\frac{H_n^{\left(2\right)}{H}_m}{nm(n+m)}}\hfill \\ =3\zeta (1,1,2,2)+4\zeta (1,2,1,2)+3\zeta (2,1,1,2)+3\zeta (1,1,4)+6\zeta (1,3,2)\hfill \\ +\zeta (3,1,2)+4\zeta (1,2,3)+3\zeta (2,1,3)+3\zeta (2,2,2)+6\zeta (1,5)+3\zeta (2,4)+\zeta (3,3).\hfill \end{array}$$

(27)

9. The Case $\mathit{s}\ge \mathbf{2}$

The series $K(a+1,b+1,c+1,d+1,0,t,s+2)$ in Equation (16) is evaluated using Yamamoto’s integral and its corresponding shuffle relation:

Theorem 7.

For non-negative integers $a,b,c,d,t,s$, we have

$$\begin{array}{c}{\displaystyle \sum _{n,m,k\ge 1}\frac{H_{n-1}^{(a+1)}{H}_{m-1}^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^t{(k+n+m)}^{s+2}}}\hfill \\ =\sum _{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+1,{\alpha }_4+t+1,s+2).\hfill \end{array}$$

The series $K(a+1,b+1,0,c+1,0,t,s+2)$ is evaluated by Yamamoto’s integral and the corresponding shuffle relation:

Theorem 8.

For non-negative integers $a,b,c,t,s$, we have

$$\begin{array}{c}\sum _{n,m,k\ge 1}\frac{H_{n-1}^{(a+1)}}{n^{b+1}{m}^{c+1}{(n+m)}^t{(k+n+m)}^{s+2}}\hfill \\ \hfill =\sum _{{\left|\mathit{\alpha }\right|}_3=a+b+c}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+t+1,s+2).\end{array}$$

The last series $K(0,a+1,0,b+1,0,t,s+2)$ is evaluated by Yamamoto’s integral and the corresponding shuffle relation:

Theorem 9.

For non-negative integers $a,b,t,s$, we have

$$\begin{array}{c}\sum _{n,m,k\ge 1}\frac{1}{n^{a+1}{m}^{b+1}{(n+m)}^t{(k+n+m)}^{s+2}}\hfill \\ \hfill =\sum _{{\alpha }_1+{\alpha }_2=a+b}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{b}\right)\right]\zeta ({\alpha }_1+1,{\alpha }_2+t+1,s+2).\end{array}$$

Applying Theorems 7–9 to Equation (16), we obtain the following result:

$$\begin{array}{c}{\displaystyle \sum _{n,m,k\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}{(k+n+m)}^{s+2}}}\hfill \\ {\displaystyle =\sum _{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+1,{\alpha }_4+t+2,s+2)\hfill \\ {\displaystyle +\sum _{{\left|\mathit{\alpha }\right|}_3=a+b+c+d+1}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a+b+1}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c+d+1}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right\}\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+t+2,s+2)\hfill \\ {\displaystyle +\sum _{{\alpha }_1+{\alpha }_2=a+b+c+d+2}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a+b+1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c+d+1}\right)\right]\zeta ({\alpha }_1+1,{\alpha }_2+t+2,s+2).}\hfill \end{array}$$

(28)

Finally, we integrate Equations (25) and (28) with Equation (14). The explicit formula is stated as follows: for any non-negative integers $a,b,c,d,s$, and t, we have

$$\begin{array}{c}{\displaystyle \sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}{H}_{n+m}^{(s+2)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}}}\hfill \\ {\displaystyle =\sum _{{\left|\mathit{\alpha }\right|}_4=a+b+c+d}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{b}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\right]\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b+d-{\alpha }_4}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_4}{d}\right)\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\right]\right\}\hfill \\ \times (\zeta (s+2)\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+1,{\alpha }_4+t+2)-\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+1,{\alpha }_4+t+2,s+2))\hfill \\ {\displaystyle +\sum _{{\left|\mathit{\alpha }\right|}_3=a+b+c+d+1}\left\{\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{a+b+1}\right){\delta }_{{\alpha }_1,c}+\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{c+d+1}\right){\delta }_{{\alpha }_1,a}+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right.}\hfill \\ \left.+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{b}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c-{\alpha }_1}\right)\left(\genfrac{}{}{0pt}{}{{\alpha }_3}{d}\right)\right\}\hfill \\ {\displaystyle \times (\zeta (s+2)\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+t+2)-\zeta ({\alpha }_1+1,{\alpha }_2+1,{\alpha }_3+t+2,s+2))}\hfill \\ {\displaystyle +\sum _{{\left|\mathit{\alpha }\right|}_2=a+b+c+d+2}\left[\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{a+b+1}\right)+\left(\genfrac{}{}{0pt}{}{{\alpha }_2}{c+d+1}\right)\right]}\hfill \\ {\displaystyle \times (\zeta (s+2)\zeta ({\alpha }_1+1,{\alpha }_2+t+2)-\zeta ({\alpha }_1+1,{\alpha }_2+t+2,s+2)).}\hfill \end{array}$$

(29)

We then present our main result and summarize it in the following theorem.

Theorem 10.

For non-negative integers $a,b,c,d$, and $s,t$, we have the series:

$$\sum _{n,m\ge 1}\frac{H_n^{(a+1)}{H}_m^{(c+1)}{H}_{n+m}^{(s+1)}}{n^{b+1}{m}^{d+1}{(n+m)}^{t+1}},$$

which can be written in the form of a linear combination of the multiple zeta values with the same weight. Moreover, the formula for $s=0$ is given by Equation (24), while the formula for $s\ge 1$ is provided in Equation (29).

10. Examplesand Conclusions

We use some examples to explain how to evaluate $\Theta$-series. The first example is to calculate $\Theta (1,1,1,1,1,1)$.

$$\begin{array}{ccc}\hfill \sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm(n+m)}& =& \sum _{n,m,k\ge 1}\frac{H_n{H}_m}{nmk(k+n+m)}\hfill \\ & =& \sum _{n,m,k\ge 1}\frac{H_{n-1}{H}_{m-1}}{nmk(n+m+k)}+\sum _{n,m,k\ge 1}\frac{H_{n-1}}{n{m}^{2}k(n+m+k)}\hfill \\ & & +\sum _{n,m,k\ge 1}\frac{H_{m-1}}{n^{2}mk(n+m+k)}+\sum _{n,m,k\ge 1}\frac{1}{n^2{m}^{2}k(n+m+k)}.\hfill \end{array}$$

The first term is evaluated in Equation (22). The second and third terms are $K(1,1,0,2,1,0,1)$. We use Theorem 3 and obtain its value:

$$\sum _{n,m,k\ge 1}\frac{H_{n-1}}{n{m}^{2}k(n+m+k)}=12\zeta (1,1,1,3)+9\zeta (1,1,2,2)+6\zeta (1,2,1,2)+3\zeta (2,1,1,2).$$

The last term is $K(0,2,0,2,1,0,1)$. We use Theorem 2 and obtain its value:

$$\sum _{n,m,k\ge 1}\frac{1}{n^2{m}^{2}k(n+m+k)}=12\zeta (1,1,4)+4\zeta (1,3,2)+8\zeta (1,2,3)+4\zeta (2,1,3)+2\zeta (2,2,2).$$

We put all these together and obtain

$$\begin{array}{c}\sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm(n+m)}=30\zeta \left(6\right)+12\zeta (1,1,4)+4\zeta (1,3,2)+8\zeta (1,2,3)+4\zeta (2,1,3)\hfill \\ \hfill +2\zeta (2,2,2)+24\zeta (1,1,1,3)+18\zeta (1,1,2,2)+12\zeta (1,2,1,2)+6\zeta (2,1,1,2).\end{array}$$

From the known values of multiple zeta values with a weight of six (Ref. [25]), we obtain

$$\sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm(n+m)}=\frac{767{\pi }^6}{22680}+6\zeta {\left(3\right)}^2.$$

(30)

The second example is to evaluate $\Theta (1,1,1,1,2,1)$.

$$\begin{array}{ccc}& & \sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm{(n+m)}^2}=\sum _{n,m,k\ge 1}\frac{H_n{H}_m}{nmk(n+m)(k+n+m)}\hfill \\ & & =\sum _{n,m,k\ge 1}\frac{H_{n-1}{H}_{m-1}}{nmk(n+m)(k+n+m)}\hfill \\ & & +2\sum _{n,m,k\ge 1}\frac{H_{n-1}}{n{m}^{2}k(n+m)(k+n+m)}+\sum _{n,m,k\ge 1}\frac{1}{n^2{m}^{2}k(n+m)(k+n+m)}.\hfill \end{array}$$

We apply Theorem 4 to calculate the first term, then we obtain

$$\sum _{n,m,k\ge 1}\frac{H_{n-1}{H}_{m-1}}{nmk(n+m)(k+n+m)}=30\zeta ({\left\{1\right\}}^4,3)+6\zeta (1,1,1,2,2).$$

(31)

To calculate the second term, we use Theorem 3 and obtain

$$\begin{array}{c}\sum _{n,m,k\ge 1}\frac{H_{n-1}}{n{m}^{2}k(n+m)(k+n+m)}=\zeta (2,1,2,2)+3\zeta (2,1,1,3)\hfill \\ \hfill +6\zeta (1,2,1,3)+2\zeta (1,2,2,2)+9\zeta (1,1,2,3)+3\zeta (1,1,3,2)+12\zeta (1,1,1,4).\end{array}$$

(32)

Applying Theorem 2 to evaluate the last term, we have

$$\begin{array}{c}\sum _{n,m,k\ge 1}\frac{1}{n^2{m}^{2}k(n+m)(k+n+m)}=12\zeta (1,1,5)+8\zeta (1,2,4)+4\zeta (1,3,3)\hfill \\ \hfill +4\zeta (1,4,2)+4\zeta (2,1,4)+2\zeta (2,2,3)+2\zeta (2,3,2).\end{array}$$

(33)

Combining these results, we have

$$\begin{array}{c}\sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm{(n+m)}^2}\hfill \\ =30\zeta ({\left\{1\right\}}^4,3)+6\zeta ({\left\{1\right\}}^3,2,2)+2\zeta (2,1,2,2)+6\zeta (2,1,1,3)+12\zeta (1,2,1,3)\\ +4\zeta (1,2,2,2)+18\zeta (1,1,2,3)+6\zeta (1,1,3,2)+24\zeta (1,1,1,4)+12\zeta (1,1,5)\\ \hfill +8\zeta (1,2,4)+4\zeta (1,3,3)+4\zeta (1,4,2)+4\zeta (2,1,4)+2\zeta (2,2,3)+2\zeta (2,3,2).\end{array}$$

(34)

From the known values of multiple zeta values with a weight of seven (ref. [25]), we deduce that

$$\sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}}{nm{(n+m)}^2}=\frac{11}{8}\zeta \left(7\right)+\frac{7{\pi }^2}{6}\zeta \left(5\right)-\frac{17{\pi }^4}{180}\zeta \left(3\right).$$

(35)

The third example is to evaluate $\Theta (1,1,1,1,1,2)$.

$$\begin{array}{c}{\displaystyle \sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}^{\left(2\right)}}{nm(n+m)}}\hfill \\ =\zeta \left(2\right)\left(6\zeta ({\left\{1\right\}}^3,2)+6\zeta (1,1,3)+4\zeta (1,2,2)+2\zeta (2,1,2)+4\zeta (1,4)+2\zeta (2,3)\right)\\ \hfill -\left(6\zeta ({\left\{1\right\}}^3,2,2)+6\zeta (1,1,3,2)+4\zeta (1,{\left\{2\right\}}^3)+2\zeta (2,1,2,2)+4\zeta (1,4,2)+2\zeta (2,3,2)\right).\end{array}$$

Therefore,

$$\sum _{n,m\ge 1}\frac{H_n{H}_m{H}_{n+m}^{\left(2\right)}}{nm(n+m)}=\frac{51}{4}\zeta \left(7\right)+\frac{11{\pi }^4}{60}\zeta \left(3\right)-\frac{11{\pi }^2}{6}\zeta \left(5\right).$$

(36)

In this paper, we successfully solved the problem proposed by Kuba and provided a new and concise representation of the general Mordell–Tornheim multiple zeta values as a linear combination of multiple zeta values.

Our method primarily involves evaluating Yamamoto’s integral in two different ways: one using shuffle relations to derive a linear combination of multiple zeta values and the other by expanding the integral as a power series. This approach elegantly leads to our results. A similar method was used in [26] by the author and Yang to derive infinite series involving harmonic numbers and the reciprocals of binomial coefficients.