Evaluations of Euler-Type Sums of Weight $\le 5$

Abstract

Let $p,p_1,\ldots ,p_m$ be positive integers with $p_1\le p_2\le \cdots \le p_m$ and $x\in [-1,1)$, define the so-called Euler-type sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right)$, which are the infinite sums whose general term is a product of harmonic numbers of index n, a power of $n^{-1}$ and variable $x^n$, by

$$\begin{aligned} {S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right) : = \sum \limits _{n = 1}^{\infty } {\frac{{H_n^{\left( {{p_1}} \right) }H_n^{\left( {{p_2}} \right) } \ldots H_n^{\left( {{p_m}} \right) }}}{{{n^p}}}{x^n}}\quad (m\in \mathbb {N}:=\{1,2,3,\ldots \}), \end{aligned}$$

where $H_n^{(p)}$ is defined by the generalized harmonic number. Extending earlier work about classical Euler sums, we prove that whenever $p+p_1+\cdots +p_m \le 5$, then all sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( 1/2\right)$ can be expressed as a rational linear combination of products of zeta values, polylogarithms and $\log (2)$. The proof involves finding and solving linear equations which relate the different types of sums to each other.

1 Introduction

This paper is concerned with the discussion of sums of the type

$$\begin{aligned} {S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right) : = \sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( {{p_1}} \right) }H_n^{\left( {{p_2}} \right) } \ldots H_n^{\left( {{p_m}} \right) }}}{{{n^p}}}{x^n}}, \end{aligned}$$

(1.1)

where the notation $H_n^{(p)}$ denotes the generalized harmonic number defined by [2, 5]

$$\begin{aligned} H_n^{(p)}:=\sum \limits _{j=1}^n\frac{1}{j^p} \quad \mathrm{and} \quad H_n=H_n^{(1)}. \end{aligned}$$

For which values of the integer parameters $p,p_j\ (j=1,2,\ldots ,m)$ and $x=\frac{1}{2}$ can these sums be expressed in terms of the simpler values of polylogarithm function $\mathrm{Li}_p(x)$ and Riemann zeta function $\zeta (s)$? The polylogarithm function and Riemann zeta function are defined by [1]

$$\begin{aligned} \begin{array}{ll} \mathrm{Li}{_p}\left( x \right) := \sum \limits _{n = 1}^\infty {\frac{{{x^n}}}{{{n^p}}}},&{}\quad \mathfrak {R}(p)>1,\\ \zeta (s):=\sum \limits _{n = 1}^\infty {\frac{1}{n^{s}}},&{}\quad \mathfrak {R}(s)>1, \end{array} \end{aligned}$$

with $\mathrm{Li_1}=-\log (1-x),\ x\in [-1,1).$ Here, the quantity $w:={p _1} + \cdots + {p _m} + p$ is called the weight, and the quantity m is called the depth of ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right)$. As usual, repeated summands in partitions are indicated by powers so that, for instance

$$\begin{aligned} {S_{{1^2}{2^3}4,p}}\left( x\right) = {S_{112224,p}}\left( x\right) = \sum \limits _{n = 1}^\infty {\frac{{H_n^2[H^{(2)} _n]^3{H^{(4)} _n}}}{{{n^p}}}}x^n. \end{aligned}$$

When $x\rightarrow 1$, then the sum ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right)$ reduces to the classical Euler sum, which is defined by [15, 23, 25]

$$\begin{aligned} {S_{{p_1}{p_2} \ldots {p_m},p}} := \sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( {{p_1}} \right) }H_n^{\left( {{p_2}} \right) } \ldots H_n^{\left( {{p_m}} \right) }}}{{{n^p}}}}\quad (p>1), \end{aligned}$$

(1.2)

which is also called the generalized (nonlinear) Euler sums.

Let $s_1,\ldots ,s_k$ be positive integers. The multiple harmonic sums (MHS) are defined by [25]

$$\begin{aligned} {\zeta _n}\left( {{s_1},{s_2}, \ldots ,{s_k}} \right) : = \sum \limits _{n \ge {n_1}> {n_2}> \cdots > {n_k} \ge 1} {\frac{1}{{n_1^{{s_1}}n_2^{{s_2}} \ldots n_k^{{s_k}}}}}, \end{aligned}$$

(1.3)

when $n<k$, then ${\zeta _n}\left( {{s_1},{s_2}, \ldots ,{s_k}} \right) =0$, and ${\zeta _n}\left( \emptyset \right) =1$. The integers k and $w:=s_1+\cdots +s_k$ are called the depth and the weight of a multiple harmonic sum. For convenience, by ${\left\{ {{s_1}, \ldots ,{s_j}} \right\} _d}$ we denote the sequence of depth dj with d repetitions of ${\left\{ {{s_1}, \ldots ,{s_j}} \right\} }$. For example,

$$\begin{aligned} \zeta _n\left( {5,3,{{\left\{ 1 \right\} }_2}} \right) = \zeta _n\left( {5,3,1,1} \right) ,\;{\zeta _n }\left( {4,2,{{\left\{ 1 \right\} }_3}} \right) = {\zeta _n }\left( {4,2,1,1,1} \right) . \end{aligned}$$

When taking the limit $n\rightarrow \infty$ we get the so-called the multiple zeta value (MZV for short) [7, 9, 13, 30, 31]:

$$\begin{aligned} \zeta \left( {{s_1},{s_2}, \ldots ,{s_k}} \right) = \mathop {\lim }\limits _{n \rightarrow \infty } \zeta _n \left( {{s_1},{s_2}, \ldots ,{s_k}} \right) , \end{aligned}$$

(1.4)

defined for $s_2,\ldots ,s_k\ge 1$ and $s_1\ge 2$ to ensure convergence of the series. It is obvious that $H_n^{(p)}=\zeta _n(p)$. Similarly, we define the multiple polylogarithm function $\mathrm{{L}}{\mathrm{{i}}_{{s_1},{s_2}, \ldots ,{s_m}}}\left( x \right)$ by

$$\begin{aligned} \mathrm{{L}}{\mathrm{{i}}_{{s_1},{s_2}, \ldots ,{s_m}}}\left( x \right) : = \sum \limits _{1 \le {k_m}< \cdots < {k_1}} {\frac{{{x^{{k_1}}}}}{{k_1^{{s_1}}k_2^{{s_2}} \ldots k_m^{{s_m}}}}} ,\quad x \in \left[ { - 1,1} \right) . \end{aligned}$$

(1.5)

Here, $\mathbf{S}:=(s_1,s_2,\ldots ,s_m)\in (\mathbb {N})^m$ in above definition (1.5). Of course, if $s_1>1$, then we can allow $x=1$. In below, we let

$$\begin{aligned} \zeta \left( {{s_1},{s_2}, \ldots ,{s_m};x} \right) : =\mathrm{{L}}{\mathrm{{i}}_{{s_1},{s_2}, \ldots ,{s_m}}}\left( x \right) . \end{aligned}$$

(1.6)

Moreover, we put a bar on top of $s_j\ (j=1,2,\ldots , k)$ if there is a sign $(-1)^{k_j}$ appearing in the denominator on the right of (1.3), the sums are also called the alternating MHS. For example

$$\begin{aligned}&{\zeta _n}\left( {{{\bar{s}}_1},{s_2}, \ldots ,{{\bar{s}}_k}} \right) = \sum \limits _{1 \le {k_m}< \cdots < {k_1} \le n} {\frac{{{{\left( { - 1} \right) }^{{k_1} + {k_m}}}}}{{k_1^{{s_1}} \ldots k_m^{{s_m}}}}},\\&{\zeta _n}\left( {\bar{2},3,\bar{1},4} \right) = \sum \limits _{n \ge {n_1}> {n_2}> {n_3} > {n_4} \ge 1}^{} {\frac{{{{\left( { - 1} \right) }^{{n_1} + {n_3}}}}}{{n_1^2n_2^3{n_3}n_4^4}}}. \end{aligned}$$

The limit cases of alternating MHNs give rise to alternating multiple zeta values, for example

$$\begin{aligned}&{\zeta }\left( {\bar{2},3,5} \right) = \mathop {\lim }\limits _{n \rightarrow \infty } \zeta _n\left( {\bar{2},3,5} \right) = \sum \limits _{{n_1}> {n_2}>1}^{} {\frac{{{{\left( { - 1} \right) }^{{n_1}}}}}{{n_1^2n_2^3n_3^5}}} \\&\zeta \left( {\bar{2},3,\bar{1},4} \right) = \mathop {\lim }\limits _{n \rightarrow \infty } {\zeta _n}\left( {\bar{2},3,\bar{1},4} \right) = \sum \limits _{{n_1}> {n_2}> {n_3} > {n_4} \ge 1}^{} {\frac{{{{\left( { - 1} \right) }^{{n_1} + {n_3}}}}}{{n_1^2n_2^3{n_3}n_4^4}}}. \end{aligned}$$

According to the rules of the “harmonic algebra” or “stuffle product,” it is obvious that the products of any number of harmonic numbers can be expressed in terms of multiple harmonic sums. For example,

$$\begin{aligned} {H_n}H_n^{\left( 3 \right) } = {\zeta _n}\left( \mathrm{{1}} \right) {\zeta _n}\left( \mathrm{{3}} \right) = {\zeta _n}\left( {\mathrm{{1,3}}} \right) + {\zeta _n}\left( {\mathrm{{3,1}}} \right) + {\zeta _n}\left( \mathrm{{4}} \right) . \end{aligned}$$

Hence, all the Euler sums ${S_{{p_1}{p_2} \ldots {p_m},p}}$ and alternating Euler sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( -1 \right)$ are reducible to rational linear combinations of multiple zeta values or alternating multiple zeta values, for which extensive tables (see [7]) are already available. For instance,

$$\begin{aligned}&{S_{1,4}}\left( { - 1} \right) = \sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^4}}}{{\left( { - 1} \right) }^n}} = \zeta \left( {\bar{5}} \right) + \zeta \left( {\bar{4},1} \right) ,\\&{S_{12,3}} = \sum \limits _{n = 1}^\infty {\frac{{{H_n}H_n^{\left( 2 \right) }}}{{{n^3}}}} = \zeta \left( 6 \right) + \zeta \left( {4,2} \right) + \zeta \left( {5,1} \right) + \zeta \left( {3,3} \right) + \zeta \left( {3,2,1} \right) \\&~~~~~~~~~~~~~+ \zeta \left( {3,1,2} \right) . \end{aligned}$$

Similarly, sums of the form ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( \frac{1}{2} \right)$ are reducible to sums of the multiple polylogarithms $\zeta \left( {{s_1},{s_2}, \ldots ,{s_m};\frac{1}{2}} \right)$.

The relations between Euler sums and multiple zeta values have attracted a lot of research in the area in the last two decades. For details and historical introductions, please see [3, 7,8,9,10, 12, 13, 15, 16, 18,19,20,21,22,23, 25, 30, 31] and references therein. The origin of these numbers goes back to the correspondence of Euler with Goldbach in 1742–1743 (see [3, 15]) that appeared in 1776. Euler’s original contribution was a method to reduce double zeta values $\zeta (q,p)$ (or linear sums $S_{p,q}$) to certain rational linear combinations of products of zeta values, and established some important relation formula for them. For example, Euler proved that the linear sums $S_{p,q}$ are reducible to zeta values whenever $p+q$ is less than 7 or when $p+q$ is odd and less than 13, and he proved that

$$\begin{aligned} S_{1,k}=\sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^k}}}} = \frac{1}{2}\left\{ {\left( {k + 2} \right) \zeta \left( {k + 1} \right) - \sum \limits _{i = 1}^{k - 2} {\zeta \left( {k - i} \right) \zeta \left( {i + 1} \right) } } \right\} ,\ k\ge 2, \end{aligned}$$

which, in particular, implies the simplest but nontrivial relation

$$\begin{aligned} S_{1,2}=2\zeta (3)\quad \mathrm{or\ equivalently,} \quad \zeta (2,1)=\zeta (3). \end{aligned}$$

Moreover, he conjectured that the linear sums $S_{p,q}$ would be reducible whenever weight is odd, and even gave what he hoped to be the general formula. The conjecture was first proved by Borwein et al. [8]. So, the linear sums $S_{p,q}$ can be evaluated in terms of zeta values in the following cases: $p=1,p=q,p+q$ odd and $p+q=6$ with $q\ge 2$ (for more details, see [3, 8, 15]). Some examples on linear Euler sums are as follows:

$$\begin{aligned}&{S_{3,2}} = \frac{11}{2}\zeta \left( 5 \right) -2\zeta \left( 2 \right) \zeta \left( 3 \right) ,\\&{S_{2,4}} = -\frac{{1}}{{3}}\zeta \left( 6 \right) + {\zeta ^2}\left( 3 \right) ,\\&{S_{5,2}} = 11\zeta \left( 7 \right) -4\zeta \left( 2 \right) \zeta \left( 5 \right) - 2\zeta \left( 3 \right) \zeta \left( 4 \right) ,\\&{S_{4,3}} = -17\zeta \left( 7 \right) +\zeta \left( 3 \right) \zeta \left( 4 \right) + 10\zeta \left( 2 \right) \zeta \left( 5 \right) ,\\&{S_{4,5}} = - \frac{{125}}{2}\zeta \left( 9 \right) + 35\zeta \left( 2 \right) \zeta \left( 7 \right) + 5\zeta \left( 4 \right) \zeta \left( 5 \right) , \\&{S_{2,7}} = - \frac{{35}}{2}\zeta \left( 9 \right) + 7\zeta \left( 2 \right) \zeta \left( 7 \right) + 2\zeta \left( 3 \right) \zeta \left( 6 \right) + 4\zeta \left( 4 \right) \zeta \left( 5 \right) , \\&{S_{4,7}} = - \frac{{229}}{2}\zeta \left( {11} \right) + 84\zeta \left( 2 \right) \zeta \left( 9 \right) + 21\zeta \left( 4 \right) \zeta \left( 7 \right) + 4\zeta \left( 5 \right) \zeta \left( 6 \right) . \end{aligned}$$

Investigation of Euler sums has a long history, but usually the authors were not aware of Euler’s results so that special instances of Euler’s identities have been independently rediscovered time and again. It was mainly the publication of Berndt’s edition of Ramanujan’s notebooks [5] that served to fit all the scattered individual results into the framework of Euler’s work. Besides the works referred to above, there are many other researches devoted to the Euler sums. For example, in 1994, Bailey et al. [3] proved that all Euler sums of the form $S_{1^p,q}$ for weights $p+q\in \{3,4,5,6,7,9\}$ are reducible to Q-linear combinations of zeta values by using the experimental method. In 1995, Borwein et al. [8] showed that the quadratic sums $S_{ 1^2,q}$ can reduce to linear sums $S_{2,q}$ and polynomials in zeta values. In 1998, Flajolet and Salvy [15] used the contour integral representations and residue computation to show that the quadratic sums $S_{p_1p_2,q}$ are reducible to linear sums and zeta values when the weight $p_1 + p_2 + q$ is even and $p_1,p_2>1$. The best results to date are due to Xu and Wang et al., see the most recent papers [20, 23, 25]. In [23, 25], we proved that all Euler sums of weight $\le 8$ are reducible to $\mathbb {Q}$-linear combinations of single zeta monomials with the addition of $\{S_{2,6}\}$ for weight 8. For weight 9, all Euler sums of the form ${S_{{s_1} \ldots {s_k},q}}$ with $q\in \{4,5,6,7\}$ are expressible polynomially in terms of zeta values. For weight $p_1+p_2+q=10$, all quadratic sums $S_{p_1p_2,q}$ are reducible to $S_{2,6}$ and $S_{2,8}$. Wang et al. [20] showed that all Euler sums of weight $\le 9$ are reducible to zeta values and linear sums. Examples for such evaluations, all due to Xu and Wang, are

$$\begin{aligned}&S_{12^2,3} = - \frac{{6313}}{{288}}\zeta \left( 8 \right) + \frac{{43}}{2}\zeta \left( 3 \right) \zeta \left( 5 \right) + \frac{1}{2}\zeta \left( 2 \right) {\zeta ^2}\left( 3 \right) - \frac{{17}}{4}{S_{2,6}},\\&{S_{{1^2}23,2}} = \frac{{505}}{{36}}\zeta (9) + \frac{7}{4}\zeta (2)\zeta (7) + 3\zeta (3)\zeta (6) - \frac{{37}}{4}\zeta (4)\zeta (5) - \frac{5}{3}{\zeta ^3}(3),\\&{S_{{{12}^2},4}} = - \frac{{775}}{{36}}\zeta \left( 9 \right) + \frac{{85}}{8}\zeta \left( 2 \right) \zeta \left( 7 \right) - \frac{{221}}{{24}}\zeta \left( 3 \right) \zeta \left( 6 \right) + 10\zeta \left( 4 \right) \zeta \left( 5 \right) + 3{\zeta ^3}\left( 3 \right) . \end{aligned}$$

In this paper, we are interested in Euler-type sums with harmonic numbers ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right)$. Such series could be of interest in analytic number theory. We will show that these sums are related to the values of the Riemann zeta function and polylogarithm function when $x=\frac{1}{2}$ and weight $\le 5$ by using the method of based on simple integral representations of logarithms.

2 Some Lemmas

In this section, we give some lemmas which will be useful in the development of the main results.

Lemma 2.1

[22] Let s, t be positive integers with $x\in [-1,1)$. Then, the product of two polylogarithm functions is reducible to Euler-type sums

$$\begin{aligned} \begin{aligned} \mathrm{Li}{_s}\left( x \right) \mathrm{Li}{_t}\left( x \right)&= \sum \limits _{j = 1}^s {A_j^{\left( {s,t} \right) }} \sum \limits _{n = 1}^\infty {\frac{{{H^{(j)} _n}}}{{{n^{s + t - j}}}}} {x^n} + \sum \limits _{j = 1}^t {B_j^{\left( {s,t} \right) }} \sum \limits _{n = 1}^\infty {\frac{{{H^{(j)} _n}}}{{{n^{s + t - j}}}}} {x^n}\\&\quad -\,\left( {\sum \limits _{j = 1}^s {A_j^{\left( {s,t} \right) }} + \sum \limits _{j = 1}^t {B_j^{\left( {s,t} \right) }} } \right) \mathrm{Li}{_{s + t}}\left( x \right) , \end{aligned} \end{aligned}$$

(2.1)

where $A_j^{\left( {s,t} \right) } := \left( {\begin{array}{*{20}{c}} {s + t - j - 1} \\ {s - j} \\ \end{array}} \right) ,\ B_j^{\left( {s,t} \right) } := \left( {\begin{array}{*{20}{c}} {s + t - j - 1} \\ {t - j} \\ \end{array}} \right) .$

Lemma 2.2

[28] For integers $m\ge 1$ and $k\ge 0$, the following identity holds:

$$\begin{aligned}&\begin{aligned}&m!\sum \limits _{l = 0}^k {l!\left( {\begin{array}{*{20}{c}} k \\ l \\ \end{array}} \right) {{\left( {\log \left( 2\right) } \right) }^{k - l}}\zeta \left( {l + 2,{{\left\{ 1 \right\} }_{m - 1}};\frac{1}{2}} \right) }\\&\quad +\,k!\sum \limits _{l = 0}^m {l!\left( {\begin{array}{*{20}{c}} m \\ l \\ \end{array}} \right) {{\left( {\log \left( 2\right) } \right) }^{m - l}}\zeta \left( {l + 1,{{\left\{ 1 \right\} }_{k }};\frac{1}{2}} \right) }\\&\qquad = m!k!\zeta \left( {m + 1,{{\left\{ 1 \right\} }_k}} \right) , \end{aligned} \end{aligned}$$

(2.2)

where $\zeta \left( {{s_1},{s_2}, \ldots ,{s_m};x} \right)$ is defined by (1.6).

Lemma 2.3

[10] For integer $m\in \mathbb {N}_0:=\mathbb {N}\cup \{0\}$, the following identity holds:

$$\begin{aligned} \zeta \left( 2,\{1\}_m;\frac{1}{2}\right) =\mathrm{{L}}{\mathrm{{i}}_{2,{{\left\{ 1 \right\} }_m}}}\left( {\frac{1}{2}} \right) = \zeta \left( {m + 2} \right) - \sum \limits _{l = 0}^{m + 1} {\frac{{{{\left( {\log ( 2)} \right) }^{m + 1 - l}}}}{{\left( {m + 1 - l} \right) !}}\mathrm{{L}}{\mathrm{{i}}_{l + 1}}\left( {\frac{1}{2}} \right) }. \end{aligned}$$

(2.3)

The first published proof of (2.3) is due to Borwein et al [10] in the Transactions of the American Mathematical Society in 2000 (in different notation). After Borwein et al  original work, several other proofs have appeared in the literature, see, for example, [29, 32]. It should be emphasized that References [10, 32] also contain many other types of results.

Lemma 2.4

[25] For integer $k>0$ and $x\in [-1,1)$, we have

$$\begin{aligned}&\begin{aligned} {\log ^k}\left( {1 - x} \right) = {\left( { - 1} \right) ^k}k!\sum \limits _{n = 1}^\infty {\frac{{{x^n}}}{n}{\zeta _{n - 1}}\left( {{{\left\{ 1 \right\} }_{k - 1}}} \right) },\end{aligned}\end{aligned}$$

(2.4)

$$\begin{aligned}&\begin{aligned} s\left( {n,k} \right) = \left( {n - 1} \right) !{\zeta _{n - 1}}\left( {{{\left\{ 1 \right\} }_{k - 1}}} \right) , \end{aligned} \end{aligned}$$

(2.5)

where ${s\left( {n,k} \right) }$ denotes the (unsigned) Stirling number of the first kind (see [14]), and we have

$$\begin{aligned}&s\left( {n,1} \right) = \left( {n - 1} \right) !,\\&s\left( {n,2} \right) = \left( {n - 1} \right) !{H_{n - 1}},\\&s\left( {n,3} \right) = \frac{{\left( {n - 1} \right) !}}{2}\left[ {H_{n - 1}^2 - {H^{(2)} _{n - 1}}} \right] ,\\&s\left( {n,4} \right) = \frac{{\left( {n - 1} \right) !}}{6}\left[ {H_{n - 1}^3 - 3{H_{n - 1}}{H^{(2)} _{n - 1}} + 2{H^{(3)}_{n - 1}}} \right] , \\&s\left( {n,5} \right) = \frac{{\left( {n - 1} \right) !}}{{24}}\left[ {H_{n - 1}^4 - 6{H^{(4)}_{n - 1}} - 6H_{n - 1}^2{H^{(2)}_{n - 1}} + 3(H^{(2)}_{n-1})^2+ 8H_{n - 1}^{}{H^{(3)}_{n - 1}}} \right] . \end{aligned}$$

The Stirling numbers ${s\left( {n,k} \right) }$ of the first kind satisfy a recurrence relation in the form

$$\begin{aligned} s\left( {n,k} \right) = s\left( {n - 1,k - 1} \right) + \left( {n - 1} \right) s\left( {n - 1,k} \right) ,\;\;n,k \in \mathbb {N}, \end{aligned}$$

with $s\left( {n,k} \right) = 0,n < k,s\left( {n,0} \right) = s\left( {0,k} \right) = 0,s\left( {0,0} \right) = 1$.

Lemma 2.5

[22] For integers $n\ge 1$ and $k\ge 0$,

$$\begin{aligned} \int \limits _0^1 {{t^{n - 1}}{{\log }^k}\left( {1 - t} \right) } \mathrm{d}t = {\left( { - 1} \right) ^k}\frac{{{Y_k}\left( n \right) }}{n},\ k,n\in \mathbb {N}, \end{aligned}$$

(2.6)

where ${Y_k}\left( n \right) := {Y_k}\left( {{H _n},1!{H^{(2)} _n},2!{H^{(3)} _n}, \ldots , \left( {k - 1} \right) !{H^{(k)} _n}} \right)$, ${Y_k}\left( {{x_1},{x_2}, \ldots ,x_k } \right)$ stands for the complete exponential Bell polynomial is defined by (see [14])

$$\begin{aligned} \exp \left( {\sum \limits _{m \ge 1}^{} {{x_m}\frac{{{t^m}}}{{m!}}} } \right) = 1 + \sum \limits _{k \ge 1}^{} {{Y_k}\left( {{x_1},{x_2}, \ldots ,x_k} \right) \frac{{{t^k}}}{{k!}}}. \end{aligned}$$

From the definition of the complete exponential Bell polynomial, we know that the Bell number ${Y_k}\left( n \right)$ is a rational linear combination of products of harmonic numbers. We deduce

$$\begin{aligned}&{Y_1}\left( n \right) = {H_n},{Y_2}\left( n \right) = H_n^2 + {H^{(2)} _n},{Y_3}\left( n \right) = H_n^3+ 3{H_n}{H^{(2)} _n}+ 2{H^{(3)} _n},\\&{Y_4}\left( n \right) = H_n^4 + 8{H_n}{H^{(3)} _n} + 6H_n^2{H^{(2)} _n} + 3(H^{(2)} _n)^2 + 6{H^{(4)} _n},\\&{Y_5}\left( n \right) = H_n^5 + 10H_n^3{H^{(2)} _n} + 20H_n^2{H^{(3)}_n} + 15{H_n}({H^{(2)}_n})^2 + 30{H_n}{H^{(4)} _n}\\&\quad +\,20{H^{(2)} _n}{H^{(3)} _n} + 24{H^{(5)}_n}. \end{aligned}$$

Remark 2.1

It should be emphasized that the (unsigned) Stirling number of the first kind s(n, k) and Bell number ${Y_k}\left( n \right)$ can be expressed by

$$\begin{aligned}&s\left( {n,k} \right) = (n-1)!{P_{k - 1}}\left( {{H_{n - 1}}, \ldots ,H_{n - 1}^{\left( {k - 1} \right) }} \right) , \end{aligned}$$

(2.7)

$$\begin{aligned}&{Y_k}\left( n \right) = k!{Q_k}\left( {{H_n}, \ldots ,H_n^{\left( k \right) }} \right) , \end{aligned}$$

(2.8)

where $P_k$ is the polynomial that expresses the kth elementary symmetric function $e_k$ in terms of the power sums $p_i$, i.e., $e_k = P_k (p_1,\ldots ,p_k)$. $Q_k$ is the polynomial that expresses the kth complete symmetric function $h_k$ in terms of the power sums $p_i$, i.e., $h_k = Q_k (p_1,\ldots ,p_k)$. The $P_k$ are simply related to the $Q_k$; e.g.,

$$\begin{aligned} {P_3}\left( {{x_1},{x_2},{x_3}} \right) = \frac{x_1^3 - 3{x_1}{x_2} + 2{x_3}}{6}\quad \mathrm{and}\quad {Q_3}\left( {{x_1},{x_2},{x_3}} \right) =\frac{ x_1^3 + 3{x_1}{x_2} + 2{x_3}}{6}. \end{aligned}$$

These polynomials are discussed in Section 2 of [17, 27].

3 Some Theorems and Proofs

In this section, we will establish some relations between Euler-type sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( 1/2\right)$ and integrals of logarithms by using above lemmas.

Theorem 3.1

For any $m\in \mathbb {N}_0$, the following identity holds:

$$\begin{aligned}&\begin{aligned} S_{1,m+1}(-1)&= \zeta \left( \overline{m + 2} \right) - \frac{{{{\left( { - 1} \right) }^{m + 1}}}}{{m!\left( {m + 2} \right) }}{\left( {\log (2)} \right) ^{m + 2}}\\&\quad -\,\frac{{{{1}}}}{{m!}}\sum \limits _{k = 1}^m \left( {\begin{array}{c}m\\ k\end{array}}\right) {\left( { - 1} \right) ^{ k}}\int \limits _{1/2}^1 {\frac{{{{\log }^{m - k + 1}}(x){{\log }^k}\left( {1 - x} \right) }}{x}{\mathrm{d}x}}. \end{aligned} \end{aligned}$$

(3.1)

Proof

By a direct calculation, we deduce that

$$\begin{aligned} {S_{1,m + 1}}\left( { - 1} \right)&= \sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^{m + 1}}}}{{\left( { - 1} \right) }^n}} = \frac{{{{\left( { - 1} \right) }^m}}}{{m!}}\sum \limits _{n = 1}^\infty {{{\left( { - 1} \right) }^n}{H_n}\int \limits _0^1 {{x^{n - 1}}{{\log }^m}\left( x \right) {\mathrm{d}x}} } \nonumber \\&= \frac{{{{\left( { - 1} \right) }^{m + 1}}}}{{m!}}\int \limits _0^1 {\frac{{{{\log }^m}\left( x \right) \log \left( {1 + x} \right) }}{{\left( {1 + x} \right) x}}} {\mathrm{d}x}\nonumber \\&= \frac{{{{\left( { - 1} \right) }^{m + 1}}}}{{m!}}\left\{ {\int \limits _0^1 {\frac{{{{\log }^m}\left( x \right) \log \left( {1 + x} \right) }}{x}} {\mathrm{d}x} - \int \limits _0^1 {\frac{{{{\log }^m}\left( x \right) \log \left( {1 + x} \right) }}{{1 + x}}} {\mathrm{d}x}} \right\} \nonumber \\&=\frac{{{{\left( { - 1} \right) }^m}}}{{m!}}\sum \limits _{n = 1}^\infty {\frac{{{{\left( { - 1} \right) }^n}}}{n}\int \limits _0^1 {{x^{n - 1}}{{\log }^m}\left( x \right) {\mathrm{d}x}} }\nonumber \\&\quad +\,\frac{{{{\left( { - 1} \right) }^m}}}{{m!}}\int \limits _1^2 {\frac{{{{\log }^m}\left( {t - 1} \right) \log \left( t \right) }}{t}\mathrm{d}t} \nonumber \\&= \zeta \left( \overline{m + 2} \right) - \frac{{{{\left( { - 1} \right) }^m}}}{{m!}}\int \limits _{1/2}^1 {\frac{{{{\log }^m}\left( {\frac{{1 - x}}{x}} \right) \log \left( x \right) }}{x}{\mathrm{d}x}}. \end{aligned}$$

(3.2)

We note that the integral on the right-hand side of (3.2) can be rewritten as

$$\begin{aligned}&\begin{aligned}&\int \limits _{1/2}^1 {\frac{{{{\log }^m}\left( {\frac{{1 - x}}{x}} \right) \log \left( x \right) }}{x}{\mathrm{d}x}} = \sum \limits _{k = 0}^m {\left( {\begin{array}{c}m\\ k\end{array}}\right) {{\left( { - 1} \right) }^{m - k}}\int \limits _{1/2}^1 {\frac{{{{\log }^k}\left( {1 - x} \right) {{\log }^{m - k + 1}}\left( x \right) }}{x}{\mathrm{d}x}} } \\&\quad = \sum \limits _{k = 1}^m {\left( {\begin{array}{c}m\\ k\end{array}}\right) {{\left( { - 1} \right) }^{m - k}}\int \limits _{1/2}^1 {\frac{{{{\log }^k}\left( {1 - x} \right) {{\log }^{m - k + 1}}\left( x \right) }}{x}{\mathrm{d}x}} } - \frac{1}{{m + 2}}{\log ^{m + 2}}\left( 2 \right) . \end{aligned} \end{aligned}$$

(3.3)

Then, substituting identity (3.3) into (3.2) yields the desired result. This completes the proof of Theorem 3.1. $\square$

Theorem 3.2

For integer $m\in \mathbb {N}_0$ and real $x\in [-1,1)$, we have

$$\begin{aligned}&\begin{aligned} \int \limits _0^x {\frac{{{{\log }^m}\left( {1 - t} \right) }}{{1 + t}}\mathrm{d}t}&= {\left( { - 1} \right) ^m}m!\mathrm{{L}}{\mathrm{{i}}_{m + 1}}\left( {\frac{1}{2}} \right) + {\log ^m}\left( {1 - x} \right) \log \left( {\frac{{1 + x}}{2}} \right) \\&\quad +\,\sum \limits _{l = 1}^m {{{\left( { - 1} \right) }^{l + 1}}{{\log }^{m - l}}\left( {1 - x} \right) {{\left( m \right) }_l}\mathrm{Li}{_{l + 1}}\left( {\frac{{1 - x}}{2}} \right) }, \end{aligned} \end{aligned}$$

(3.4)

where $(m)_l:=m(m-1)\ldots (m-l+1)$.

Proof

Changing the variable $t\mapsto 1-u$, then the integral on the left-hand side of (3.4) can be rewritten as

$$\begin{aligned} \int \limits _0^x {\frac{{{{\log }^m}\left( {1 - t} \right) }}{{1 + t}}{\mathrm{d}t}} = \int _{1 - x}^1 {\frac{{{{\log }^m}\left( u \right) }}{{2 - u}}} {\mathrm{d}u} = \sum \limits _{n = 1}^\infty {\frac{1}{{{2^n}}}\int _{1 - x}^1 {{u^{n - 1}}{{\log }^m}\left( u \right) } } {\mathrm{d}t}. \end{aligned}$$

(3.5)

On the other hand, by using integration by parts, we deduce that, for $n,m\in \mathbb {N}$,

$$\begin{aligned} \int \limits _0^x {{t^{n - 1}}{{\left( {\log (t)} \right) }^m}} {\mathrm{d}t} = \sum \limits _{l = 0}^m {l!\left( {\begin{array}{*{20}{c}} m \\ l \\ \end{array}} \right) \frac{{{{\left( { - 1} \right) }^l}}}{{{n^{l + 1}}}}{{\left( {\log (x)} \right) }^{m - l}}{x^n}},\quad x\in (0,1). \end{aligned}$$

(3.6)

Hence, substituting (3.6) into (3.5), by a simple calculation we obtain the formula (3.4). $\square$

Taking $m=1$ and 2 in (3.4), we obtain the following cases

$$\begin{aligned} \int \limits _0^x {\frac{{\log \left( {1 - t} \right) }}{{1 + t}}{\mathrm{d}t}}&= \log \left( {1 - x} \right) \log \left( {\frac{{1 + x}}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{{1 - x}}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) , \end{aligned}$$

(3.7)

$$\begin{aligned} \int \limits _0^x {\frac{{{{\log }^2}\left( {1 - t} \right) }}{{1 + t}}{\mathrm{d}t}}&= 2\mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) + {\log ^2}\left( {1 - x} \right) \log \left( {\frac{{1 + x}}{2}} \right) \nonumber \\&\quad +\,2\log \left( {1 - x} \right) \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{{1 - x}}{2}} \right) - 2\mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{{1 - x}}{2}} \right) . \end{aligned}$$

(3.8)

Noting that from [11], we have

$$\begin{aligned} \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) = \frac{{\zeta \left( 2 \right) - {{\log }^2}\left( 2 \right) }}{2}\quad \mathrm{and}\quad \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) = \frac{7}{8}\zeta \left( 3 \right) - \frac{1}{2}\zeta \left( 2 \right) \log \left( 2 \right) + \frac{1}{6}{\log ^3}\left( 2 \right) . \end{aligned}$$

Corollary 3.3

For any $x\in [-1,1)$, the following identity holds:

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( 2 \right) } - L_n^2\left( 1 \right) }}{{n + 1}}} {x^{n + 1}} = 4\left[ {\mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{{1 - x}}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) } \right] \nonumber \\&\quad -\,2\log \left( {1 - x} \right) \left[ {\mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{{1 - x}}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) } \right] , \end{aligned}$$

(3.9)

where $L_n(p)$ denotes the alternating harmonic number, which is defined by

$$\begin{aligned} L_{n}(p):=\sum \limits _{j=1}^n\frac{(-1)^{j-1}}{j^p},\ p \in \mathbb {N}. \end{aligned}$$

Many papers use the notation ${\bar{H}}^{(p)}_n$ which stands for the alternating harmonic number, for example, see [15].

Proof

In [27], we proved the result

$$\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( 2 \right) } - L_n^2\left( 1 \right) }}{{n + 1}}} {x^{n + 1}} = 2\log \left( {1 - x} \right) \int \limits _0^x {\frac{{\log \left( {1 - t} \right) }}{{1 + t}}{\mathrm{d}t}} - 2\int \limits _0^x {\frac{{{{\log }^2}\left( {1 - t} \right) }}{{1 + t}}{\mathrm{d}t}}. \end{aligned}$$

Applying formulas (3.7) and (3.8) to the above equation, we deduce the desired result. $\square$

Differentiate both sides of (3.9), then

$$\begin{aligned} \sum \limits _{n = 1}^\infty {\left\{ {H_n^{\left( 2 \right) } - L_n^2\left( 1 \right) } \right\} {x^n}} = 2\frac{{\mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{{1 - x}}{2}} \right) - \log \left( {1 - x} \right) \log \left( {\frac{{1 + x}}{2}} \right) }}{{1 - x}},\ x\in (-1,1). \end{aligned}$$

Theorem 3.4

For integers $m\in \mathbb {N}_0$ and $k\in \mathbb {N}$, the following relations hold:

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,m + 1} \right) {Y_k}\left( n \right) }}{{n!{2^n}}}}= \frac{{{{\left( { - 1} \right) }^k}}}{{m!}}\sum \limits _{j = 0}^m {{\left( { - 1} \right) }^j}\left( {\begin{array}{*{20}{c}} m \\ j \\ \end{array}} \right) {{\log }^{m - j}}\left( 2 \right) \nonumber \\&\quad ~~~~~~~\int \limits _0^1 {\frac{{{{\log }^j}\left( {1 + t} \right) {{\log }^k}\left( t \right) }}{{1 + t}}{\mathrm{d}t}}, \end{aligned}$$

(3.10)

$$\begin{aligned}&\int \limits _0^1 {\frac{{{{\log }^j}\left( {1 + t} \right) {{\log }^k}\left( t \right) }}{{1 + t}}{\mathrm{d}t}} = {\left( { - 1} \right) ^{k + j}}k!j!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,j + 1} \right) }}{{n!{n^k}}}{{\left( { - 1} \right) }^{n + 1}}}. \end{aligned}$$

(3.11)

Proof

The identity (3.11) is easily derived. Next, we prove the formula (3.10). By using Lemmas 2.4, 2.5 and considering the following integral

$$\begin{aligned} I(m,k):=\int \limits _0^1 {\frac{{{{\log }^m}\left( {1 - \frac{x}{2}} \right) {{\log }^k}\left( {1 - x} \right) }}{{1 - \frac{x}{2}}}{\mathrm{d}x}}, \end{aligned}$$

we have

$$\begin{aligned}&\begin{aligned} I\left( {m,k} \right)&= {\left( { - 1} \right) ^m}m!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n + 1,m + 1} \right) }}{{n!{2^n}}}\int \limits _0^1 {{x^n}{{\log }^k}\left( {1 - x} \right) {\mathrm{d}x}} }\\&= {\left( { - 1} \right) ^{m + k}}m!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n + 1,m + 1} \right) {Y_k}\left( {n + 1} \right) }}{{n!{2^n}\left( {n + 1} \right) }}}\\&= 2{\left( { - 1} \right) ^{m + k}}m!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,m + 1} \right) {Y_k}\left( n \right) }}{{n!{2^n}}}}. \end{aligned} \end{aligned}$$

(3.12)

On the other hand, applying the change of variable $x\rightarrow 1-t$ to the above integral on the left-hand side of (3.12), which can be rewritten as

$$\begin{aligned}&\begin{aligned} I\left( {m,k} \right)&= 2\int \limits _0^1 {\frac{{{{\log }^m}\left( \frac{1 + t}{2} \right) {{\log }^k}\left( t \right) }}{{1 + t}}{\mathrm{d}t}} \\&= 2\sum \limits _{j = 0}^m {{{\left( { - 1} \right) }^{m - j}}\left( {\begin{array}{*{20}{c}} m \\ j \\ \end{array}} \right) {{\log }^{m - j}}\left( 2 \right) \int \limits _0^1 {\frac{{{{\log }^j}\left( {1 + t} \right) {{\log }^k}\left( t \right) }}{{1 + t}}{\mathrm{d}t}} }\end{aligned} \end{aligned}$$

(3.13)

Thus, combining identities (3.12) and (3.13), we obtain the result. $\square$

Theorem 3.5

For positive integers m and k, the following equation holds:

$$\begin{aligned}&{\left( { - 1} \right) ^{m + k}}m!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,m} \right) {Y_k}\left( n \right) }}{{n!n{2^n}}}} \nonumber \\&\quad = \sum \limits _{j = 0}^m {{{\left( { - 1} \right) }^j}\left( {\begin{array}{*{20}{c}} m \\ j \\ \end{array}} \right) {{\log }^j}\left( 2 \right) \int \limits _0^1 {\frac{{{{\log }^{m - j}}\left( {1 + t} \right) {{\log }^k}\left( t \right) }}{{1 - t}}{\mathrm{d}t}} } \end{aligned}$$

(3.14)

Proof

Similarly as in the proof of Theorem 3.4, considering the integral

$$\begin{aligned} J(m,k):=\int \limits _0^1 {\frac{{{{\log }^m}\left( {1 - \frac{x}{2}} \right) {{\log }^k}\left( {1 - x} \right) }}{x}{\mathrm{d}x}}. \end{aligned}$$

Then, using Lemmas 2.4, 2.5 and applying the change of variable $x\rightarrow 1-t$ to the above integral, we have

$$\begin{aligned}&\begin{aligned} J\left( {m,k} \right)&= {\left( { - 1} \right) ^m}m!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,m} \right) }}{{n!{2^n}}}\int \limits _0^1 {{x^{n - 1}}{{\log }^k}\left( {1 - x} \right) {\mathrm{d}x}} } \\&= {\left( { - 1} \right) ^{m + k}}m!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,m} \right) {Y_k}\left( n \right) }}{{n!n{2^n}}}} \\&= \int \limits _0^1 {\frac{{{{\log }^m}\left( {\frac{{1 + t}}{2}} \right) {{\log }^k}\left( t \right) }}{{1 - t}}{\mathrm{d}t}} \\&= \sum \limits _{j = 0}^m {{{\left( { - 1} \right) }^j}\left( {\begin{array}{*{20}{c}} m \\ j \\ \end{array}} \right) {{\log }^j}\left( 2 \right) \int \limits _0^1 {\frac{{{{\log }^{m - j}}\left( {1 + t} \right) {{\log }^k}\left( t \right) }}{{1 - t}}{\mathrm{d}t}} }.\end{aligned} \end{aligned}$$

(3.15)

Thus, the formula (3.14) holds. $\square$

Theorem 3.6

For positive integer m, the following identity holds:

$$\begin{aligned}&\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{{H_n}H_n^{\left( {m + 1} \right) }}}{{n{2^n}}}}&= \sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( {m + 1} \right) }}}{{{n^2}{2^n}}}} + \frac{{{{\left( { - 1} \right) }^{m + 1}}}}{{2\left( {m!} \right) }}\int \limits _0^1 {\frac{{{{\log }^2}\left( {1 + t} \right) {{\log }^m}\left( {1 - t} \right) }}{t}{\mathrm{d}t}}\\&\quad -\,\frac{{{{\left( { - 1} \right) }^{m + 1}}}}{{m!}}\log \left( 2 \right) \int \limits _0^1 {\frac{{\log \left( {1 + t} \right) {{\log }^m}\left( {1 - t} \right) }}{t}{\mathrm{d}t}}.\end{aligned} \end{aligned}$$

(3.16)

Proof

In the same way as in proofs of Theorems 3.4 and 3.5, considering the integral

$$\begin{aligned} \int \limits _0^1 {\frac{{{{\log }^2}\left( {1 - \frac{x}{2}} \right) {{\log }^m}\left( x \right) }}{{1 - x}}{\mathrm{d}x}}. \end{aligned}$$

Then, applying the identity (2.4) with the help of the following elementary integral

$$\begin{aligned} \int \limits _0^1 {\frac{{{x^n}{{\log }^m}x}}{{1 - x}}} {\mathrm{d}x} = {\left( { - 1} \right) ^m}m!\left( {\zeta \left( {m + 1} \right) - {H^{(m+1)} _n}} \right) \;\ (m \in \mathbb {N}), \end{aligned}$$

(3.17)

By a simple calculation, we deduce the desired result. $\square$

4 The Relations Between Mixed Euler-Type Sums and Alternating mzvs

In this section, we will establish some explicit relationships between mixed Euler-type sums and alternating multiple zeta values by using the method of iterated integral representations of series. The main results of this section are the following theorems.

Theorem 4.1

For integers $p,k,m\in \mathbb {N}_0$, we have

$$\begin{aligned}&\begin{aligned}&{\left( { - 1} \right) ^{k + 1}}\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,k + 1} \right) {Y_p}\left( n \right) }}{{n!{n^m}{2^n}}}}\\&\quad =p! \sum \limits _{0 \le {\mathbf{I}_{m + k}} \le p} \zeta \Bigg ( \overline{p + 1 - {\mathbf{I}_{m + k}}},{i_{m + k}} + 1, \ldots ,{i_{m + 1}} \\&\qquad +\,1,{\overline{i_m + 1}},{i_{m - 1}} + 1, \ldots ,{i_1} + 1 \Bigg ), \end{aligned} \end{aligned}$$

(4.1)

where the quantity $\mathbf{I}_{m+k}$ defined by

$$\begin{aligned} {\mathbf{{I}}_{{{m}} + {{k}}}}: = {i_1} + {i_2} + \cdots + {i_{m + k}},\;\left( {{i_j} \ge 0,j = 1,2, \ldots ,m + k} \right) . \end{aligned}$$

Proof

To prove the identity (4.1), we consider the multiple integral

$$\begin{aligned} {I_{p,m,k}}: = \frac{1}{{{2^{k + 1}}}}\int \limits _{0< {t_{m + k + 1}}< \cdots< {t_1} < 1} {\frac{{{{\log }^p}\left( {1 - {t_1}} \right) }}{{{t_1}{t_2} \ldots {t_m}\left( {1 - \frac{{{t_{m + 1}}}}{2}} \right) \ldots \left( {1 - \frac{{{t_{m + k + 1}}}}{2}} \right) }}} \mathrm{d}{t_1} \ldots \mathrm{d}{t_{m + k+1}}. \end{aligned}$$

(4.2)

By a simple calculation, we can find that

$$\begin{aligned}&\begin{aligned}&\int \limits _{0< {t_{m + k + 1}}< \cdots< {t_{m + 1}} < {t_m}} {\frac{1}{{\left( {1 - \frac{{{t_{m + 1}}}}{2}} \right) \ldots \left( {1 - \frac{{{t_{m + k + 1}}}}{2}} \right) }}} \mathrm{d}{t_{m + 1}} \ldots \mathrm{d}{t_{m + k + 1}}\\&\qquad \qquad \qquad = \int \limits _0^{{t_m}} {\frac{1}{{1 - \frac{{{t_{m + 1}}}}{2}}}\mathrm{d}{t_{m + 1}}} \ldots \int \limits _0^{{t_{m + k}}} {\frac{1}{{1 - \frac{{{t_{m + k + 1}}}}{2}}}\mathrm{d}{t_{m + k + 1}}} \\&\qquad \qquad \qquad = {2^{k + 1}}\frac{{{{\left( { - 1} \right) }^{k + 1}}}}{{\left( {k + 1} \right) !}}{\log ^{k + 1}}\left( {1 - \frac{{{t_m}}}{2}} \right) . \end{aligned} \end{aligned}$$

(4.3)

Hence, substituting identity (5.3) into (5.2), then using Lemmas 2.4 and 2.5, we arrive at the conclusion that

$$\begin{aligned}&\begin{aligned} {I_{p,m,k}}&= \sum \limits _{n = 1}^\infty {\frac{{s\left( {n,k + 1} \right) }}{{n!{n^{m - 1}}{2^n}}}\int \limits _0^1 {t_1^{n - 1}{{\log }^p}\left( {1 - {t_1}} \right) \mathrm{d}{t_1}} } \\&= {\left( { - 1} \right) ^p}\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,k + 1} \right) {Y_p}\left( n \right) }}{{n!{n^m}{2^n}}}}.\end{aligned} \end{aligned}$$

(4.4)

On the other hand, applying the change of variables $t_i\mapsto 1-t_{m+k+2-i}$ to the above multiple integral, and using the integral formula (3.6), we can get the following result

$$\begin{aligned}&{I_{p,m,k}}: = \frac{1}{{{2^{k + 1}}}}\int \limits _{0< {t_{m + k + 1}}< \cdots< {t_1} < 1} {\frac{{{{\log }^p}\left( {{t_{m + k + 1}}} \right) }\mathrm{d}{t_1} \ldots \mathrm{d}{t_{m + k + 1}}}{{\left( {1 - {t_{m + k + 1}}} \right) \ldots \left( {1 - {t_{k + 2}}} \right) \left( {\frac{{1 + {t_{k + 1}}}}{2}} \right) \ldots \left( {\frac{{1 + {t_1}}}{2}} \right) }}} \nonumber \\&\quad = \int \limits _0^1 {\frac{1}{{1 + {t_1}}}\mathrm{d}{t_1}} \ldots \int \limits _0^{{t_k}} {\frac{1}{{1 + {t_{k + 1}}}}\mathrm{d}{t_{k + 1}}} \int \limits _0^{{t_{k + 1}}} {\frac{1}{{1 - {t_{k + 2}}}}\mathrm{d}{t_{k + 2}}} \ldots \int \limits _0^{{t_{m + k - 1}}} {\frac{1}{{1 - {t_{m + k}}}}\mathrm{d}{t_{m + k}}}\nonumber \\&\quad \quad \times \int \limits _0^{{t_{m + k}}} {\frac{{{{\log }^p}\left( {{t_{m + k + 1}}} \right) }}{{1 - {t_{m + k + 1}}}}\mathrm{d}{t_{m + k + 1}}}\nonumber \\&\quad = \sum \limits _{{n_{m + k + 1}} = 1}^\infty { \ldots \sum \limits _{{n_1} = 1}^\infty {{{\left( { - 1} \right) }^{{n_{m + 1}} + \cdots + {n_{m + k + 1}} - \left( {k + 1} \right) }}\int \limits _0^1 {t_1^{{n_{m + k + 1}} - 1}\mathrm{d}{t_1}} } } \ldots \int \limits _0^{{t_{m + k - 1}}} {t_{m + k}^{{n_2} - 1}\mathrm{d}{t_{m + k}}}\nonumber \\&\quad \quad \times \int \limits _0^{{t_{m + k}}} {t_{m + k + 1}^{{n_1} - 1}{{\log }^p}\left( {{t_{m + k + 1}}} \right) \mathrm{d}{t_{m + k + 1}}}\nonumber \\&\quad = {\left( { - 1} \right) ^{p + k + 1}}\sum \limits _{0 \le {i_1} + \cdots + {i_{m + k}} \le p} {{\left( p \right) }_{{i_1}}}{{\left( {p - {i_1}} \right) }_{{i_2}}} \ldots \left( p - {i_1} \cdots \right. \nonumber \\&\qquad \left. -\,{i_{m + k - 1}} \right) _{{i_{m + k}}}\left( {p - {i_1} - \cdots - {i_{m + k}}} \right) !\nonumber \\&\quad \quad \times \sum \limits _{{n_1},{n_2}, \ldots ,{n_{m + k + 1}} = 1}^\infty {\frac{{{{\left( { - 1} \right) }^{{n_{m + 1}} + \cdots + {n_{m + k + 1}}}}}}{{n_1^{{i_1} + 1} \ldots {{\left( {{n_1} + \cdots + {n_{m + k}}} \right) }^{{i_{m + k}} + 1}}{{\left( {{n_1} + \cdots + {n_{m + k + 1}}} \right) }^{p+1 - \mathbf{I}_{m+k} }}}}} \nonumber \\&\quad = p!{\left( { - 1} \right) ^{p + k + 1}} \sum \limits _{0 \le {\mathbf{I}_{m + k}} \le p} \zeta \left( \overline{p + 1 - {\mathbf{I}_{m + k}}},{i_{m + k}} + 1, \ldots ,{i_{m + 1}} \right. \nonumber \\&\qquad \left. +\,1,{\overline{i_m + 1}},{i_{m - 1}} + 1, \ldots ,{i_1} + 1 \right) , \end{aligned}$$

(4.5)

where $(x)_l:=x(x-1)\ldots (x-l+1)$ with $(x)_0=1$.

Thus, the relations (4.4) and (4.5) yield the desired result. $\square$

Theorem 4.2

Let $m>1,p$ be positive integers and k be an nonnegative integer. Then

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,k + 1} \right) H_n^{\left( {p + 1} \right) }}}{{n!{n^{m - 1}}{2^n}}}} \nonumber \\&\quad = {\left( { - 1} \right) ^k}\zeta \left( {\bar{1},{{\left\{ 1 \right\} }_k},\bar{1},{{\left\{ 1 \right\} }_{m - 2}},2,{{\left\{ 1 \right\} }_{p - 1}}} \right) + \zeta \left( {p + 1} \right) \zeta \left( {m,{{\left\{ 1 \right\} }_k};\frac{1}{2}} \right) , \end{aligned}$$

(4.6)

and

$$\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{s\left( {n,k + 1} \right) H_n^{\left( {p + 1} \right) }}}{{n!{2^n}}}} = {\left( { - 1} \right) ^k}\zeta \left( {\bar{1},{{\left\{ 1 \right\} }_k},\bar{2},{{\left\{ 1 \right\} }_{p - 1}}} \right) + \zeta \left( {p + 1} \right) \zeta \left( {1,{{\left\{ 1 \right\} }_k};\frac{1}{2}} \right) . \end{aligned}$$

(4.7)

Proof

Similarly as in the proof of Theorem 4.1, we consider the multiple integral

$$\begin{aligned} {J_{p,m,k}}: = \frac{1}{{{2^{k + 1}}}}\int \limits _{0< {t_{m + k + 1}}< \cdots< {t_1} < 1} {\frac{{{{\log }^p}\left( {{t_1}} \right) }}{{\left( {1 - {t_1}} \right) {t_2} \ldots {t_m}\left( {1 - \frac{{{t_{m + 1}}}}{2}} \right) \ldots \left( {1 - \frac{{{t_{m + k + 1}}}}{2}} \right) }}} \mathrm{d}{t_1} \ldots \mathrm{d}{t_{m + k + 1}}. \end{aligned}$$

Then, with the help of formula (3.17), we deduce that

$$\begin{aligned} {J_{p,m,k}} = {\left( { - 1} \right) ^p}p!\sum \limits _{n = 1}^\infty {\frac{{s\left( {n,k + 1} \right) \left( {\zeta \left( {p + 1} \right) - H_n^{\left( {p + 1} \right) }} \right) }}{{n!{n^{m - 1}}{2^n}}}}. \end{aligned}$$

(4.8)

Applying the change of variables $t_i\mapsto 1-t_{m+k+2-i}$ to $J_{p,m,k}$ and using (3.6), we obtain

$$\begin{aligned} {J_{p,m,k}} = {\left( { - 1} \right) ^{p + k + 1}}p!\zeta \left( {\bar{1},{{\left\{ 1 \right\} }_k},\bar{1},{{\left\{ 1 \right\} }_{m - 2}},2,{{\left\{ 1 \right\} }_{p - 1}}} \right) . \end{aligned}$$

(4.9)

Thus, combining (4.8) and (4.9), then formula (4.6) holds. By considering

$$\begin{aligned} {J_{p,1,k}}: = \frac{1}{{{2^{k + 1}}}}\int \limits _{0< {t_{k + 2}}< \cdots< {t_1} < 1} {\frac{{{{\log }^p}\left( {{t_1}} \right) }}{{\left( {1 - {t_1}} \right) \left( {1 - \frac{{{t_{2}}}}{2}} \right) \ldots \left( {1 - \frac{{{t_{ k + 2}}}}{2}} \right) }}} \mathrm{d}{t_1} \ldots \mathrm{d}{t_{k + 2}} \end{aligned}$$

we deduce the formula (4.7). $\square$

From Theorems 4.1 and 4.2, we can get the following examples.

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{{Y_p}\left( n \right) }}{{n{2^n}}}} = - p!\zeta \left( \overline{p + 1} \right) , \\&\sum \limits _{n = 1}^\infty {\frac{{{H_{n - 1}}{Y_p}\left( n \right) }}{{n{2^n}}}} = p!\sum \limits _{i = 0}^p {\zeta \left( {{\overline{p - i + 1}},i + 1} \right) }, \\&{S_{1,m + 1}}\left( {\frac{1}{2}} \right) = \mathrm{{L}}{\mathrm{{i}}_{m + 2}}\left( {\frac{1}{2}} \right) + \zeta \left( {\bar{1},1,\bar{1},{{\left\{ 1 \right\} }_{m - 1}}} \right) , \\&{S_{2,2}}\left( {\frac{1}{2}} \right) = \zeta \left( {\bar{1},\bar{1},2} \right) + \zeta \left( 2 \right) \zeta \left( {2;\frac{1}{2}} \right) , \\&{S_{3,1}}\left( {\frac{1}{2}} \right) = \zeta \left( {\bar{1},\bar{2},1} \right) + \zeta \left( 3 \right) \zeta \left( {1;\frac{1}{2}} \right) ,\\&{S_{12,1}}\left( {\frac{1}{2}} \right) - {S_{2,2}}\left( {\frac{1}{2}} \right) = - \zeta \left( {\bar{1},1,\bar{2}} \right) + \zeta \left( 2 \right) \zeta \left( {1,1;\frac{1}{2}} \right) ,\\&{S_{2,3}}\left( {\frac{1}{2}} \right) = \zeta \left( {\bar{1},\bar{1},1,2} \right) + \zeta \left( 2 \right) \zeta \left( {3;\frac{1}{2}} \right) , \\&{S_{3,2}}\left( {\frac{1}{2}} \right) = \zeta \left( {\bar{1},\bar{1},2,1} \right) + \zeta \left( 3 \right) \zeta \left( {2;\frac{1}{2}} \right) , \\&{S_{4,1}}\left( {\frac{1}{2}} \right) = \zeta \left( {\bar{1},\bar{2},1,1} \right) + \zeta \left( 4 \right) \zeta \left( {1;\frac{1}{2}} \right) , \\&{S_{12,2}}\left( {\frac{1}{2}} \right) - {S_{2,3}}\left( {\frac{1}{2}} \right) = - \zeta \left( {\bar{1},1,\bar{1},2} \right) + \zeta \left( 2 \right) \zeta \left( {2,1;\frac{1}{2}} \right) , \\&{S_{13,1}}\left( {\frac{1}{2}} \right) - {S_{3,2}}\left( {\frac{1}{2}} \right) = - \zeta \left( {\bar{1},1,\bar{2},1} \right) + \zeta \left( 3 \right) \zeta \left( {1,1;\frac{1}{2}} \right) , \\&\sum \limits _{n = 1}^\infty {\frac{{{Y_p}\left( n \right) }}{{{n^2}{2^n}}}} = - p!\sum \limits _{i = 0}^p {\zeta \left( {\overline{p - i + 1},\overline{i + 1}} \right) } \\&\quad \quad \quad \quad \quad \quad = - \frac{{p!}}{2}\left\{ {\sum \limits _{k = 1}^{p + 1} {\zeta \left( {\bar{k}} \right) \zeta \left( \overline{p + 2 - k} \right) } - \left( {p + 1} \right) \zeta \left( {p + 2} \right) } \right\} , \\&\sum \limits _{n = 1}^\infty {\frac{{{Y_p}\left( n \right) }}{{{n^{m + 1}}{2^n}}}} = -p! \sum \limits _{0 \le {\mathbf{{I}}_\mathbf{{m}}} \le p} {\zeta \left( {\overline{p + 1 - {\mathbf{{I}}_m}},{\overline{{i_m} + 1}},{i_{m - 1}} + 1, \ldots ,{i_1} + 1} \right) }, \\&\sum \limits _{n = 1}^\infty {\frac{{{H_{n - 1}}{Y_p}\left( n \right) }}{{{n^{m + 1}}{2^n}}}}\\&\quad \quad \quad =p! \sum \limits _{0 \le {\mathbf{{I}}_{m + 1}} \le p} {\zeta \left( {\overline{ {p + 1 - {\mathbf{{I}}_{m + 1}}}},{i_{m + 1}} + 1,{\overline{{i_m} + 1}},{i_{m - 1}} + 1, \ldots ,{i_1} + 1} \right) },\\&{S_{{1^2},m + 1}}\left( {\frac{1}{2}} \right) = \mathrm{{L}}{\mathrm{{i}}_{m + 3}}\left( {\frac{1}{2}} \right) + \zeta \left( {\bar{1},1,\bar{1},{{\left\{ 1 \right\} }_m}} \right) \\&\qquad +\,\zeta \left( {\bar{2},1,\bar{1},{{\left\{ 1 \right\} }_{m - 1}}} \right) + \zeta \left( {\bar{1},1,\bar{2},{{\left\{ 1 \right\} }_{m - 1}}} \right) \\&\qquad +\,\zeta \left( {\bar{1},2,\bar{1},{{\left\{ 1 \right\} }_{m - 1}}} \right) + \sum \limits _{i = 1}^{m - 1} {\zeta \left( {\bar{1},1,\bar{1},{{\left\{ 1 \right\} }_{m - i - 1}},2,{{\left\{ 1 \right\} }_{i - 1}}} \right) }. \end{aligned}$$

Some above results are already in the literature, e.g., the first and twelfth equations can be found in Batir [4] and Zlobin [32] (in different notation), respectively.

5 Some Results on Integral of Logarithms

In [21, 24, 26, 27], we obtain numerous results of some alternating Euler sums of weight $\le 6$. We can use these results to find some nice evaluations of integral of logarithms. Hence, in this section, we will give many closed-form representations of logarithms’ integrals. By using Lemma 2.4, Lemma 2.5 and formula (3.6) with the help of results of References [21, 24, 26, 27], the following identities are easily derived

$$\begin{aligned}&\int \limits _0^1 \frac{{{{\log }^m}\left( {1 - x} \right) }}{x}{\mathrm{d}x} = {{\left( { - 1} \right) }^m}m!\zeta \left( {m + 1} \right) , \end{aligned}$$

(5.1)

$$\begin{aligned}&\int \limits _0^1 {\frac{{{{\log }^m}\left( {1 - x} \right) }}{{1 - \frac{x}{2}}}} {\mathrm{d}x} = 2{\left( { - 1} \right) ^m}m!\left( {1 - \frac{1}{{{2^m}}}} \right) \zeta \left( {m + 1} \right) , \end{aligned}$$

(5.2)

$$\begin{aligned}&\int \limits _{1/2}^1 {\frac{{\log \left( x \right) {{\log }^2}\left( {1 - x} \right) }}{x}{\mathrm{d}x}} = - 2\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{1}{2}\zeta \left( 4 \right) + \frac{1}{4}\zeta \left( 3 \right) \log \left( 2 \right) \nonumber \\&\quad -\,\frac{1}{3}{\log ^4}\left( 2 \right) + 2\sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}{2^n}}}} , \end{aligned}$$

(5.3)

$$\begin{aligned}&\int \limits _{1/2}^1 {\frac{{{{\log }^2}\left( x \right) \log \left( {1 - x} \right) }}{x}{\mathrm{d}x}} = 2\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - 2\zeta \left( 4 \right) + \frac{7}{4}\zeta \left( 3 \right) \log \left( 2 \right) \nonumber \\&\quad -\,\frac{1}{2}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) - \frac{1}{6}{\log ^4}\left( 2 \right) ,\end{aligned}$$

(5.4)

$$\begin{aligned}&\int \limits _0^1 {\frac{{\log \left( {1 + x} \right) {{\log }^2}\left( {1 - x} \right) }}{x}{\mathrm{d}x}} = 2\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{5}{8}\zeta \left( 4 \right) + \frac{7}{4}\zeta \left( 3 \right) \log \left( 2 \right) \nonumber \\&\quad -\,\frac{1}{2}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) + \frac{1}{{12}}{\log ^4}\left( 2 \right) , \end{aligned}$$

(5.5)

$$\begin{aligned}&\int \limits _0^1 {\frac{{\log \left( {1 - \frac{x}{2}} \right) {{\log }^3}\left( {1 - x} \right) }}{x}{\mathrm{d}x}} = 12\zeta \left( 5 \right) - \frac{{21}}{4}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{9}{4}\zeta \left( 2 \right) \zeta \left( 3 \right) , \end{aligned}$$

(5.6)

$$\begin{aligned}&\int \limits _0^1 {\frac{{\log \left( {1 + t} \right) {{\log }^3}\left( t \right) }}{{1 - t}}} {\mathrm{d}t} = 12\zeta \left( 5 \right) - \frac{{45}}{4}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{9}{4}\zeta \left( 2 \right) \zeta \left( 3 \right) , \end{aligned}$$

(5.7)

$$\begin{aligned}&\int \limits _0^1 {\frac{{{{\log }^2}\left( {1 + x} \right) {{\log }^2}\left( x \right) }}{{1 + x}}{\mathrm{d}x}} = 8\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 8\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) \nonumber \\&\quad -\,\frac{{33}}{8}\zeta \left( 5 \right) - 2\zeta \left( 2 \right) \zeta \left( 3 \right) +\frac{7}{2}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{4}{3}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{4}{{15}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(5.8)

$$\begin{aligned}&\int \limits _0^1 {\frac{{{{\log }^2}\left( {1 + x} \right) {{\log }^2}\left( {1 - x} \right) }}{x}{\mathrm{d}x}} = 4\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 4\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{{25}}{8}\zeta \left( 5 \right) \nonumber \\&\quad +\,\frac{7}{4}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{2}{3}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{2}{{15}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(5.9)

$$\begin{aligned}&\int \limits _{1/2}^1 {\frac{{\log \left( x \right) {{\log }^3}\left( {1 - x} \right) }}{x}{\mathrm{d}x}} = - 6\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + \frac{3}{4}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{3}{8}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) \nonumber \\&\quad +\,\frac{1}{4}{\log ^5}\left( 2 \right) + 6\sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^4}{2^n}}}}, \end{aligned}$$

(5.10)

$$\begin{aligned}&\int \limits _{1/2}^1 {\frac{{{{\log }^2}\left( x \right) {{\log }^2}\left( {1 - x} \right) }}{x}{\mathrm{d}x}} = 4\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 8\zeta \left( 5 \right) - 4\zeta \left( 2 \right) \zeta \left( 3 \right) - \frac{1}{2}\zeta \left( 4 \right) \log \left( 2 \right) \nonumber \\&\quad +\,\frac{1}{4}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{6}{\log ^5}\left( 2 \right) - 4\sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^4}{2^n}}}},\end{aligned}$$

(5.11)

$$\begin{aligned}&\int \limits _{1/2}^1 {\frac{{{{\log }^3}\left( x \right) \log \left( {1 - x} \right) }}{x}{\mathrm{d}x}} = - 6\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 6\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + 6\zeta \left( 5 \right) \nonumber \\&\quad -\,\frac{{21}}{8}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) +\zeta \left( 2 \right) {\log ^3}\left( 2 \right) . \end{aligned}$$

(5.12)

Next, we only prove the formulas (5.5) and (5.9). From Lemmas 2.4 and 2.5, we deduce that

$$\begin{aligned} \int \limits _0^1 {\frac{{\log \left( {1 + x} \right) {{\log }^m}\left( {1 - x} \right) }}{x}{\mathrm{d}x}}&= \sum \limits _{n = 1}^\infty {\frac{{{{\left( { - 1} \right) }^{n - 1}}}}{n}} \int \limits _0^1 {{x^{n - 1}}{{\log }^m}\left( {1 - x} \right) {\mathrm{d}x}} \nonumber \\&= {\left( { - 1} \right) ^m}\sum \limits _{n = 1}^\infty {\frac{{{Y_m}\left( n \right) }}{{{n^2}}}} {\left( { - 1} \right) ^{n - 1}}, \end{aligned}$$

(5.13)

$$\begin{aligned} \int \limits _0^1 {\frac{{{{\log }^2}\left( {1 + x} \right) {{\log }^m}\left( {1 - x} \right) }}{x}{\mathrm{d}x}}&= 2\sum \limits _{n = 1}^\infty {\left\{ {\frac{{{H_n}}}{n} - \frac{1}{{{n^2}}}} \right\} {{\left( { - 1} \right) }^n}} \int \limits _0^1 {{x^{n - 1}}{{\log }^m}\left( {1 - x} \right) {\mathrm{d}x}} \nonumber \\&= 2{\left( { - 1} \right) ^{m + 1}}\sum \limits _{n = 1}^\infty {\left\{ {\frac{{{H_n}{Y_m}\left( n \right) }}{{{n^2}}} - \frac{{{Y_m}\left( n \right) }}{{{n^3}}}} \right\} {{\left( { - 1} \right) }^{n - 1}}}. \end{aligned}$$

(5.14)

Setting $m=2$ in (5.13) and (5.14) yield

$$\begin{aligned} \int \limits _0^1 {\frac{{\log \left( {1 + x} \right) {{\log }^2}\left( {1 - x} \right) }}{x}{\mathrm{d}x}}&= \sum \limits _{n = 1}^\infty {\frac{{H_n^2 + H_n^{\left( 2 \right) }}}{{{n^2}}}} {\left( { - 1} \right) ^{n - 1}}, \end{aligned}$$

(5.15)

$$\begin{aligned} \int \limits _0^1 {\frac{{{{\log }^2}\left( {1 + x} \right) {{\log }^2}\left( {1 - x} \right) }}{x}{\mathrm{d}x}}&= 2\sum \limits _{n = 1}^\infty {\left\{ {\frac{{H_n^2 + H_n^{\left( 2 \right) }}}{{{n^3}}} - \frac{{H_n^3 + {H_n}H_n^{\left( 2 \right) }}}{{{n^2}}}} \right\} {{\left( { - 1} \right) }^{n - 1}}}. \end{aligned}$$

(5.16)

From [3, 15, 24, 27], we know that

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{H_n^2}}{{{n^2}}}} {\left( { - 1} \right) ^{n - 1}} = \frac{{41}}{{16}}\zeta \left( 4 \right) + \frac{1}{2}\zeta \left( 2 \right) {\log ^2}\left( 2\right) - \frac{1}{{12}}{\log ^4}\left( 2\right) \nonumber \\&\quad -\,\frac{7}{4}\zeta \left( 3 \right) \log \left( 2\right) - 2\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) ,\nonumber \\&\sum \limits _{n = 1}^\infty {\frac{{{H^{(2)} _n}}}{{{n^2}}}{{\left( { - 1} \right) }^{n - 1}}} = - \frac{{51}}{{16}}\zeta (4) + 4\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \frac{7}{2}\zeta (3)\log \left( 2\right) - \zeta (2){\log ^2}\left( 2\right) + \frac{1}{6}{\log ^4}\left( 2\right) ,\nonumber \\&\sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( 2 \right) }}}{{{n^3}}}} {\left( { - 1} \right) ^{n - 1}} = \frac{5}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) - \frac{{11}}{{32}}\zeta \left( 5 \right) ,\\&\sum \limits _{n = 1}^\infty {\frac{{H_n^2}}{{{n^3}}}} {\left( { - 1} \right) ^{n - 1}} = 4\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 4\log \left( 2\right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \frac{2}{{15}}{\log ^5}\left( 2\right) + \frac{7}{4}\zeta \left( 3 \right) {\log ^2}\left( 2\right) \\&\quad -\,\frac{{19}}{{32}}\zeta \left( 5 \right) - \frac{2}{3}\zeta \left( 2 \right) {\log ^3}\left( 2\right) - \frac{{11}}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) ,\\&\sum \limits _{n = 1}^\infty {\frac{{H_n^3}}{{{n^2}}}} {\left( { - 1} \right) ^{n - 1}} = 6\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 6\log \left( 2\right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \frac{1}{5}{\log ^5}\left( 2\right) + \frac{{21}}{8}\zeta \left( 3 \right) {\log ^2}\left( 2\right) \\&\quad -\,\frac{9}{4}\zeta \left( 5 \right) - \zeta \left( 2 \right) {\log ^3}\left( 2\right) - \frac{{27}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) ,\\&\sum \limits _{n = 1}^\infty {\frac{{{H_n}{H^{(2)}_n}}}{{{n^2}}}{{\left( { - 1} \right) }^{n - 1}}} = - 4\mathrm{Li}{_5}\left( {\frac{1}{2}} \right) - 4\log \left( 2\right) \mathrm{Li}{_4}\left( {\frac{1}{2}} \right) - \frac{2}{{15}}{\log ^5}\left( 2\right) \\&\quad -\,\frac{7}{4}\zeta \left( 3 \right) {\log ^2}\left( 2\right) +\frac{{23}}{8}\zeta \left( 5 \right) + \frac{2}{3}\zeta \left( 2 \right) {\log ^3}\left( 2\right) + \frac{{15}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) . \end{aligned}$$

Hence, substituting the above identities into Eqs. (5.15) and (5.16), by a direct calculation, we can obtain the desired results.

6 Some Evaluation of Euler-Type Sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( 1/2\right)$

We have used our equation system to obtain explicit evaluation for all sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( 1/2\right)$ with weight less than or equal to five. In this section, we only prove the results of all sums weight $= 5$. The formulas of weight $\le 4$ are easily obtained.

6.1 Weight $\le 4$

$$\begin{aligned}&S_{1,1}\left( \frac{1}{2}\right) = \frac{1}{2}\zeta \left( 2 \right) , \\&S_{2,1}\left( \frac{1}{2}\right) = \frac{5}{8}\zeta \left( 3 \right) , \\&S_{1^2,1}\left( \frac{1}{2}\right) = \frac{7}{8}\zeta \left( 3 \right) , \\&S_{1,2}\left( \frac{1}{2}\right) = \zeta \left( 3 \right) - \frac{1}{2}\zeta \left( 2 \right) \log \left( 2 \right) , \\&S_{1,3}\left( \frac{1}{2}\right) = \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \mathrm{{ + }}\frac{1}{8}\zeta \left( 4 \right) - \frac{1}{8}\zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\&S_{2,2}\left( \frac{1}{2}\right) = \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \frac{1}{{16}}\zeta (4) + \frac{1}{4}\zeta (3)\log \left( 2 \right) - \frac{1}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\&S_{3,1}\left( \frac{1}{2}\right) = \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{5}{{16}}\zeta \left( 4 \right) + \frac{7}{8}\zeta \left( 3 \right) \log \left( 2 \right) - \frac{1}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\&S_{1^2,2}\left( \frac{1}{2}\right) = - \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \; + \frac{{37}}{{16}}\zeta \left( 4 \right) - \frac{7}{4}\zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) - \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\&S_{12,1}\left( \frac{1}{2}\right) = \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{1}{8}\zeta \left( 4 \right) + \frac{7}{8}\zeta \left( 3 \right) \log \left( 2 \right) - \frac{1}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\&S_{1^3,1}\left( \frac{1}{2}\right) = - 5\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \frac{{25}}{4}\zeta \left( 4 \right) - \frac{{35}}{8}\zeta \left( 3 \right) \log \left( 2 \right) \mathrm{{ + }}\frac{5}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) - \frac{5}{{24}}{\log ^4}\left( 2 \right) . \end{aligned}$$

6.2 Weight $= 5$

$$\begin{aligned} S_{1,4}\left( \frac{1}{2}\right)&= 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{1}{{32}}\zeta \left( 5 \right) - \frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{1}{8}\zeta \left( 4 \right) \log \left( 2 \right) + \frac{1}{2}\zeta \left( 3 \right) {\log ^2}(2) - \frac{1}{6}\zeta \left( 2 \right) {\log ^3}(2) + \frac{1}{{40}}{\log ^5}(2), \end{aligned}$$

(6.1)

$$\begin{aligned} S_{2,3}\left( \frac{1}{2}\right)&= - 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{23}}{{64}}\zeta \left( 5 \right) + \frac{{23}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{1}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{{23}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{7}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{{13}}{{120}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.2)

$$\begin{aligned} S_{3,2}\left( \frac{1}{2}\right)&= 4\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{{81}}{{64}}\zeta \left( 5 \right) - \frac{7}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad +\,\frac{5}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) + \frac{7}{8}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{5}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{{11}}{{120}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.3)

$$\begin{aligned} S_{4,1}\left( \frac{1}{2}\right)&= - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{27}}{{32}}\zeta \left( 5 \right) + \frac{7}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{7}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{{30}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.4)

$$\begin{aligned} S_{1^2,3}\left( \frac{1}{2}\right)&= - 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{279}}{{64}}\zeta \left( 5 \right) - \frac{9}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{{37}}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) + \frac{7}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{{40}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.5)

$$\begin{aligned} S_{1^3,2}\left( \frac{1}{2}\right)&= - 14\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 9\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{279}}{{16}}\zeta \left( 5 \right) - \frac{7}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{{25}}{4}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{7}{4}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{{13}}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{{31}}{{120}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.6)

$$\begin{aligned} S_{12,2}\left( \frac{1}{2}\right)&= 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{{31}}{{32}}\zeta \left( 5 \right) + \frac{1}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad +\,\frac{1}{8}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{1}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{40}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.7)

$$\begin{aligned} S_{13,1}\left( \frac{1}{2}\right)&=3\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{{31}}{{64}}\zeta \left( 5 \right) - \frac{7}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad +\,\frac{{21}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{1}{2}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{10}}{\log ^5}\left( 2 \right) ,\end{aligned}$$

(6.8)

$$\begin{aligned} S_{1^22,1}\left( \frac{1}{2}\right)&= 3\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{{31}}{{32}}\zeta \left( 5 \right) - \frac{7}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad +\,\frac{{21}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{1}{2}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{10}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.9)

$$\begin{aligned} S_{1^4,1}\left( \frac{1}{2}\right)&=- 15\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 15\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{341}}{{16}}\zeta \left( 5 \right) - \frac{{35}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{{105}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{5}{2}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{2}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.10)

$$\begin{aligned} S_{2^2,1}\left( \frac{1}{2}\right)&=- 7\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 7\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{31}}{{16}}\zeta \left( 5 \right) \mathrm{{ + }}\frac{{49}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad -\,\frac{{49}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{7}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{7}{{30}}{\log ^5}\left( 2 \right) , \end{aligned}$$

(6.11)

$$\begin{aligned} S_{1,5}\left( \frac{1}{2}\right)&= 3\mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{1}{2}\zeta \left( {\bar{5},1} \right) - \frac{{51}}{{32}}\zeta \left( 6 \right) - \frac{1}{4}{\zeta ^2}\left( 3 \right) \nonumber \\&\quad -\,\frac{1}{{32}}\zeta \left( 5 \right) \log \left( 2 \right) + \frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{{16}}\zeta \left( 4 \right) {\log ^2}\left( 2 \right) \nonumber \\&\quad -\,\frac{1}{6}\zeta \left( 3 \right) {\log ^3}\left( 2 \right) + \frac{1}{{24}}\zeta \left( 2 \right) {\log ^4}\left( 2 \right) - \frac{1}{{240}}{\log ^6}\left( 2 \right) . \end{aligned}$$

(6.12)

6.3 Proof of All Euler-Type Sums of $w=5$

In [15], Flajolet and Salvy gave an explicit formula for alternating Euler sums ${\bar{S}}_{1,m}:=S_{1,m}\left( -1\right)$ in terms of zeta values, polylogarithms and $\log \left( 2\right)$ when m is a even by using the method of contour integral representations and residue computation. Hence, we deduce the result

$$\begin{aligned} \sum \limits _{n\mathrm{{ = }}1}^\infty {\frac{{{H_n}}}{{{n^4}}}{{\left( { - 1} \right) }^{n - 1}} = } \frac{{59}}{{32}}\zeta \left( 5 \right) - \frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) . \end{aligned}$$

(6.13)

Letting $m=3$ in (3.1) and combining formulas (5.10)–(5.12), we have

$$\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^4}}}} {\left( { - 1} \right) ^n}&= 3{S_{1,4}}\left( {\frac{1}{2}} \right) - 6\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) \nonumber \\&\quad -\,\frac{{31}}{{16}}\zeta \left( 5 \right) + 2\zeta \left( 2 \right) \zeta \left( 3 \right) \nonumber \\&\quad +\,\frac{3}{8}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{3}{2}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{2}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{3}{{40}}{\log ^5}\left( 2 \right) . \end{aligned}$$

(6.14)

Thus, applying (6.13)–(6.14), the result is (6.1).

Next, we prove the identities (6.2)–(6.4). Multiplying (3.9) by $\frac{\log \left( 1-x\right) }{x}$ and integrating over the interval (0, 1), and using (2.6), we obtain

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{{H_n}L_n^2\left( 1 \right) }}{{{n^2}}}} - 2\sum \limits _{n = 1}^\infty {\frac{{{H_n}{L_n}\left( 1 \right) }}{{{n^3}}}{{\left( { - 1} \right) }^{n - 1}}} + 2\sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^4}}}} - \sum \limits _{n = 1}^\infty {\frac{{{H_n}H_n^{\left( 2 \right) }}}{{{n^2}}}} \nonumber \\&\quad = \int \limits _0^1 {\frac{{4\left( {\mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{{1 - x}}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) } \right) - 2\log \left( {1 - x} \right) \left( {\mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{{1 - x}}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) } \right) }}{x}} \log \left( {1 - x} \right) {\mathrm{d}x} \nonumber \\&\quad \mathop = \limits ^{x = 1 - t} \int \limits _0^1 {\frac{{4\left( {\mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{t}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) } \right) - 2\log \left( t \right) \left( {\mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{t}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) } \right) }}{{1 - t}}} \log \left( t \right) {\mathrm{d}t}\nonumber \\&\quad = 4\int \limits _0^1 {\frac{{\log \left( t \right) \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{t}{2}} \right) }}{{1 - t}}} {\mathrm{d}t} - 4\mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) \int \limits _0^1 {\frac{{\log \left( t \right) }}{{1 - t}}} {\mathrm{d}t}\nonumber \\&\qquad -\,2\mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) \int \limits _0^1 {\frac{{{{\log }^2}\left( t \right) }}{{1 - t}}} {\mathrm{d}t} - 2\int \limits _0^1 {\frac{{{{\log }^2}\left( t \right) \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{t}{2}} \right) }}{{1 - t}}} {\mathrm{d}t}. \end{aligned}$$

(6.15)

Furthermore, by using (3.17) and (5.1), then the integrals on the right-hand side of (6.15) can be rewritten as

$$\begin{aligned}&\int \limits _0^1 {\frac{{\log \left( t \right) }}{{1 - t}}} {\mathrm{d}t} = - \zeta \left( 2 \right) ,\end{aligned}$$

(6.16)

$$\begin{aligned}&\int \limits _0^1 {\frac{{{{\log }^2}\left( t \right) }}{{1 - t}}} {\mathrm{d}t} = 2\zeta \left( 3 \right) ,\end{aligned}$$

(6.17)

$$\begin{aligned}&\int \limits _0^1 {\frac{{\log \left( t \right) \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{t}{2}} \right) }}{{1 - t}}} {\mathrm{d}t} = {S_{2,3}}\left( {\frac{1}{2}} \right) - \zeta \left( 2 \right) \mathrm{{L}}{\mathrm{{i}}_3}\left( {\frac{1}{2}} \right) ,\end{aligned}$$

(6.18)

$$\begin{aligned}&\int \limits _0^1 {\frac{{{{\log }^2}\left( t \right) \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{t}{2}} \right) }}{{1 - t}}} = 2\zeta \left( 3 \right) \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) - 2{S_{3,2}}\left( {\frac{1}{2}} \right) . \end{aligned}$$

(6.19)

Therefore, combining formulas (6.15)–(6.19), by a simple calculation, we obtain the following equation

$$\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{{H_n}L_n^2\left( 1 \right) }}{{{n^2}}}} - 2\sum \limits _{n = 1}^\infty {\frac{{{H_n}{L_n}\left( 1 \right) }}{{{n^3}}}{{\left( { - 1} \right) }^{n - 1}}}&= 4{S_{2,3}}\left( {\frac{1}{2}} \right) +\,4{S_{3,2}}\left( {\frac{1}{2}} \right) \nonumber \\&\quad +\, 4\zeta \left( 3 \right) {\log ^2}\left( 2 \right) -5\zeta \left( 5 \right) - \zeta \left( 2 \right) \zeta \left( 3 \right) . \end{aligned}$$

(6.20)

On the other hand, from (3.13) of [26], we deduce that

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{{H_n}L_n^2\left( 1 \right) + {H_n}H_n^{\left( 2 \right) } + H_n^2{L_n}\left( 1 \right) + {L_n}\left( 1 \right) {L_n}\left( 2 \right) }}{{{n^2}}}} \nonumber \\&\quad = 2\zeta \left( 2 \right) \zeta \left( 3 \right) + 2\log \left( 2 \right) \sum \limits _{n = 1}^\infty {\frac{{H_n^2 + {H_n}{L_n}\left( 1 \right) }}{{{n^2}}}} - 2\log \left( 2 \right) \sum \limits _{n = 1}^\infty {\frac{{{H_n}}}{{{n^3}}}\left( {1 + {{\left( { - 1} \right) }^{n - 1}}} \right) } \nonumber \\&\qquad +\,2\sum \limits _{n = 1}^\infty {\frac{{{H_n}{L_n}\left( 1 \right) }}{{{n^3}}}{{\left( { - 1} \right) }^{n - 1}}} - 2\zeta \left( 2 \right) \sum \limits _{n = 1}^\infty {\frac{{{L_n}\left( 1 \right) }}{{{n^2}}}} + 2\sum \limits _{n = 1}^\infty {\frac{{{H_n}{L_n}\left( 1 \right) }}{{{n^3}}}}. \end{aligned}$$

(6.21)

Hence, substituting (6.20) into (6.21) with the help of the results of alternating Euler sums in Reference [27], we can get the following relation

$$\begin{aligned} {S_{2,3}}\left( {\frac{1}{2}} \right) + {S_{3,2}}\left( {\frac{1}{2}} \right)&= 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \frac{{29}}{{32}}\zeta \left( 5 \right) + \frac{9}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) + \frac{1}{4}\zeta \left( 4 \right) \log \left( 2 \right) \nonumber \\&\quad -\,\frac{9}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{{60}}{\log ^5}\left( 2 \right) . \end{aligned}$$

(6.22)

Taking $(s,t)=(2,3)$ and (1, 4) in (2.1), then letting $x=\frac{1}{2}$, we have

$$\begin{aligned}&3{S_{2,3}}\left( {\frac{1}{2}} \right) + {S_{3,2}}\left( {\frac{1}{2}} \right) = - 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 6\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{3}{{16}}\zeta \left( 5 \right) \nonumber \\&\quad +\,\frac{{55}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) + \frac{1}{8}\zeta \left( 4 \right) \log \left( 2 \right) -\frac{{55}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{4}{3}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) \nonumber \\&\quad - \frac{7}{{30}}{\log ^5}\left( 2 \right) ,\end{aligned}$$

(6.23)

$$\begin{aligned}&2{S_{1,4}}\left( {\frac{1}{2}} \right) + {S_{2,3}}\left( {\frac{1}{2}} \right) + {S_{3,2}}\left( {\frac{1}{2}} \right) + {S_{4,1}}\left( {\frac{1}{2}} \right) = 5\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) . \end{aligned}$$

(6.24)

Combining Eqs. (6.22)–(6.24), we can prove the formulas (6.2)–(6.4).

Now, we establish enough equations of Euler-type sums to prove the identities (6.5)–(6.11). In (2.2), letting $m=k=2$, then

$$\begin{aligned}&2{\log ^2}\left( 2 \right) \zeta \left( {2,1;\frac{1}{2}} \right) + 4\log \left( 2 \right) \zeta \left( {3,1;\frac{1}{2}} \right) + 4\zeta \left( {4,1;\frac{1}{2}} \right) \nonumber \\&\quad +\,2{\log ^2}\left( 2 \right) \zeta \left( {{{\left\{ 1 \right\} }_3};\frac{1}{2}} \right) + 4\log \left( 2 \right) \zeta \left( {2,1,1;\frac{1}{2}} \right) + 4\zeta \left( {3,1,1;\frac{1}{2}} \right) \nonumber \\&\qquad = 4\zeta \left( {3,1,1} \right) . \end{aligned}$$

(6.25)

By the definition of Euler-type sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( x \right)$ and multiple polylogarithm function $\zeta \left( {{s_1},{s_2}, \ldots ,{s_m};x} \right)$, it is easily seen that

$$\begin{aligned} {S_{p,q}}\left( {\frac{1}{2}} \right) = \zeta \left( {q,p;\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_{p + q}}\left( {\frac{1}{2}} \right) . \end{aligned}$$

Noting that the multiple zeta value $\zeta (m+1,\{1\}_{k-1} )$ can be represented as a polynomial of zeta values with rational coefficients (see [9, 25]), and using (2.2), we arrive at the conclusion that

$$\begin{aligned}&\begin{aligned} \zeta \left( {3,1,1;\frac{1}{2}} \right)&= - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + \frac{{63}}{{32}}\zeta \left( 5 \right) - \frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) - \zeta \left( 4 \right) \log \left( 2 \right) \\&\quad +\,\frac{7}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{1}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{60}}{\log ^5}\left( 2 \right) . \end{aligned} \end{aligned}$$

(6.26)

Similarly, putting $m=3$ in Lemma 2.3, we conclude that

$$\begin{aligned} \zeta \left( {2,{{\left\{ 1 \right\} }_3};\frac{1}{2}} \right)&= - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \log \left( 2 \right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \zeta \left( 5 \right) - \frac{7}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) \nonumber \\&\quad +\,\frac{1}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{{24}}{\log ^5}\left( 2 \right) . \end{aligned}$$

(6.27)

Furthermore, from the definition of $\zeta \left( {{s_1},{s_2}, \ldots ,{s_m};x} \right)$, we know that

$$\begin{aligned}&\begin{aligned} \zeta \left( {3,1,1;\frac{1}{2}} \right)&= \sum \limits _{n = 1}^\infty {\frac{1}{{{n^3}{2^n}}}{\zeta _{n - 1}}\left( {1,1} \right) }\\&= \sum \limits _{n = 1}^\infty {\frac{1}{{{n^3}{2^n}}}\left\{ {\frac{{H_n^2 - H_n^{\left( 2 \right) }}}{2} - \frac{{{H_n}}}{n} + \frac{1}{{{n^2}}}} \right\} }\\&=\frac{1}{2}{S_{{1^2},3}}\left( {\frac{1}{2}} \right) - \frac{1}{2}{S_{2,3}}\left( {\frac{1}{2}} \right) - {S_{1,4}}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) , \end{aligned}\end{aligned}$$

(6.28)

$$\begin{aligned}&\begin{aligned} \zeta \left( {2,{{\left\{ 1 \right\} }_3};\frac{1}{2}} \right)&= \sum \limits _{n = 1}^\infty {\frac{1}{{{n^2}{2^n}}}{\zeta _{n - 1}}\left( {{{\left\{ 1 \right\} }_3}} \right) } \\&= \frac{1}{6}\sum \limits _{n = 1}^\infty \frac{1}{{{n^2}{2^n}}}\left\{ H_n^3 - 3{H_n}H_n^{\left( 2 \right) } + 2H_n^{\left( 3 \right) }\right. \\&\quad \left. - 3\frac{{H_n^2 - H_n^{\left( 2 \right) }}}{n} + 6\frac{{{H_n}}}{{{n^2}}} - 6\frac{1}{{{n^3}}} \right\} \\&= \frac{1}{6}\left[ {{S_{{1^3},2}}\left( {\frac{1}{2}} \right) - 3{S_{12,2}}\left( {\frac{1}{2}} \right) + 2{S_{3,2}}\left( {\frac{1}{2}} \right) } \right] \\&\quad -\,\frac{1}{2}\left[ {{S_{{1^2},3}}\left( {\frac{1}{2}} \right) - {S_{2,3}}\left( {\frac{1}{2}} \right) } \right] + {S_{1,4}}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) . \end{aligned} \end{aligned}$$

(6.29)

Setting $m=1,k=3$ in (3.14) with the help of (5.7), we obtain

$$\begin{aligned}&\begin{aligned} {S_{{1^3},2}}\left( {\frac{1}{2}} \right) + 3{S_{12,2}}\left( {\frac{1}{2}} \right)&= - 8\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 6\log \left( 2 \right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \\&\quad +\,\frac{{465}}{{32}}\zeta \left( 5 \right) - \frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\quad -\, \frac{{47}}{8}\zeta \left( 4 \right) \log \left( 2 \right) - \frac{7}{4}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{5}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) \\&\quad -\,\frac{{11}}{{60}}{\log ^5}\left( 2 \right) . \end{aligned} \end{aligned}$$

(6.30)

Thus, the relations (6.26)–(6.30) yield the results (6.5)–(6.7). Then, by using formulas (2.4), (2.5) and the recurrence relation of Stirling numbers of the first kind, we can find that

$$\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{s\left( {n + 1,k} \right) }}{{n!{n^m}{2^n}}} = \zeta \left( {m + 1,{{\left\{ 1 \right\} }_{k - 2}};\frac{1}{2}} \right) } + \zeta \left( {m,{{\left\{ 1 \right\} }_{k - 1}};\frac{1}{2}} \right) ,\;k \ge 2,m \ge 1. \end{aligned}$$

(6.31)

Letting $k=5,m=1$ in above equation, we get the relation

$$\begin{aligned}&\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{H_n^4 - 6H_n^2H_n^{\left( 2 \right) } + 8{H_n}H_n^{\left( 3 \right) } + 3{{\left( {H_n^{\left( 2 \right) }} \right) }^2} - 6H_n^{\left( 4 \right) }}}{{n{2^n}}}}\\&\quad = - 24\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 24\log \left( 2 \right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + 24\zeta \left( 5 \right) \\&\qquad - \frac{{21}}{2}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + 4\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{4}{5}{\log ^5}\left( 2 \right) .\end{aligned} \end{aligned}$$

(6.32)

In Theorem 3.4, taking $(m,k)=(0,4)$ and (1, 3) with the help of formulas (5.2) and

$$\begin{aligned} \int \limits _0^1 {\frac{{\log \left( {1 + t} \right) {{\log }^3}\left( t \right) }}{{1 + t}}{\mathrm{d}t}} = \frac{{87}}{{16}}\zeta \left( 5 \right) - 3\zeta \left( 2 \right) \zeta \left( 3 \right) , \end{aligned}$$

we obtain the following two equations

$$\begin{aligned}&\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{H_n^4 + 6H_n^2H_n^{\left( 2 \right) } + 8{H_n}H_n^{\left( 3 \right) } + 3{{\left( {H_n^{\left( 2 \right) }} \right) }^2} + 6H_n^{\left( 4 \right) }}}{{n{2^n}}}} = \frac{{45}}{2}\zeta \left( 5 \right) , \end{aligned} \end{aligned}$$

(6.33)

$$\begin{aligned}&\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{H_n^4 + 3H_n^2H_n^{\left( 2 \right) } + 2{H_n}H_n^{\left( 3 \right) }}}{{n{2^n}}}} = \frac{{279}}{{16}}\zeta \left( 5 \right) - \frac{{21}}{4}\zeta \left( 2 \right) \zeta \left( 3 \right) . \end{aligned} \end{aligned}$$

(6.34)

Furthermore, setting $m=2$ in (3.16) and using the formulas (5.5) and (5.9), by a simple calculation, we deduce the identity (6.8), namely

$$\begin{aligned}&\begin{aligned} \sum \limits _{n = 1}^\infty {\frac{{{H_n}H_n^{\left( 3 \right) }}}{{n{2^n}}}}&= \sum \limits _{n = 1}^\infty {\frac{{H_n^{\left( 3 \right) }}}{{{n^2}{2^n}}}} - \frac{1}{4}\int \limits _0^1 {\frac{{{{\log }^2}\left( {1 + t} \right) {{\log }^2}\left( {1 - t} \right) }}{t}{\mathrm{d}t}}\\&\quad +\,\frac{1}{2}\log \left( 2 \right) \int \limits _0^1 {\frac{{\log \left( {1 + t} \right) {{\log }^2}\left( {1 - t} \right) }}{t}{\mathrm{d}t}} \\&= 3\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 3\log \left( 2 \right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{{31}}{{64}}\zeta \left( 5 \right) - \frac{7}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\quad +\,\frac{{21}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{1}{2}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{10}}{\log ^5}\left( 2 \right) . \end{aligned} \end{aligned}$$

(6.35)

Finally, combining (6.32)–(6.35), we can prove the identities (6.9) and (6.11). The proofs of formulas (6.1)–(6.11) are finished.

Letting $m=4$ in (3.1), by a similar argument as in the proof of (6.1), we have the formula (6.12).

Moreover, by (3.9), we can evaluate the following two alternating Euler sums

$$\begin{aligned}&\sum \limits _{n = 1}^\infty {\frac{{{H_n}L_n^2\left( 1 \right) }}{n}} {\left( { - 1} \right) ^{n - 1}} = \frac{7}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) - \frac{1}{4}{\log ^4}\left( 2 \right) + \frac{3}{8}\zeta (3)\log \left( 2 \right) ,\\&\sum \limits _{n = 1}^\infty {\frac{{{H_n}L_n^2\left( 1 \right) }}{{{n^2}}}} = 12\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \frac{{53}}{4}\zeta \left( 5 \right) +\,\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\ \quad + 9\zeta \left( 4 \right) \log \left( 2 \right) + \zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{{10}}{\log ^5}\left( 2 \right) . \end{aligned}$$

The proofs of above identities are left to the readers.

6.4 Some Special Values of Alternating mzvs with $w\le 5$

It is easily seen that applying the relations between mixed Euler-type sums and alternating mzvs in Sect. 4, and using the results in Sects. 6.1 and 6.2, we can obtain many special values of alternating mzv. Some examples on alternating mzvs are as follows:

$$\begin{aligned} \zeta \left( {\bar{1},1,\bar{1},1} \right)&= \frac{1}{8}\zeta \left( 4 \right) - \frac{1}{8}\zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\ \zeta \left( {\bar{1},1,\bar{2}} \right)&= \frac{3}{{16}}\zeta \left( 4 \right) - \frac{5}{8}\zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{2}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) , \\ \zeta \left( {\bar{1},\bar{2},1} \right)&= \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{5}{{16}}\zeta \left( 4 \right) - \frac{1}{8}\zeta \left( 3 \right) \log \left( 2 \right) - \frac{1}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) \\&\quad + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\ \zeta \left( {\bar{1},\bar{1},2} \right)&= \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) - \frac{{19}}{{16}}\zeta \left( 4 \right) + \frac{1}{4}\zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{4}\zeta \left( 2 \right) {\log ^2}\left( 2 \right) \\&\quad + \frac{1}{{24}}{\log ^4}\left( 2 \right) , \\ \zeta \left( {\bar{1},\bar{1},1,2} \right)&= - 2\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{23}}{{64}}\zeta \left( 5 \right) \\&\quad +\,\frac{9}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) + \frac{19}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) \\&\quad -\,\frac{{23}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{5}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{{13}}{{120}}{\log ^5}\left( 2 \right) , \\ \zeta \left( {\bar{1},\bar{1},2,1} \right)&= 4\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) + 3\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{{81}}{{64}}\zeta \left( 5 \right) - \frac{{11}}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\quad +\,\frac{5}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) \\&\quad +\,\frac{{11}}{8}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{5}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{{11}}{{120}}{\log ^5}\left( 2 \right) , \\ \zeta \left( {\bar{1},\bar{2},1,1} \right)&= - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{27}}{{32}}\zeta \left( 5 \right) \\&\quad +\,\frac{7}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) - \zeta \left( 4 \right) \log \left( 2 \right) \\&\quad -\,\frac{7}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{1}{{30}}{\log ^5}\left( 2 \right) , \\ \zeta \left( {\bar{1},1,\bar{2},1} \right)&= \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - \frac{{25}}{{32}}\zeta \left( 5 \right) \\&\quad +\,\frac{5}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) + \frac{1}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) \\&\quad +\,\frac{1}{{12}}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) -\,\frac{1}{{120}}{\log ^5}\left( 2 \right) , \\ \zeta \left( {\bar{1},1,\bar{1},2} \right)&= - 4\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) - 4\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{{85}}{{64}}\zeta \left( 5 \right) + \frac{{23}}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\quad -\,\frac{3}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) \\&\quad -\,\frac{{23}}{{16}}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) + \frac{1}{2}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) - \frac{2}{{15}}{\log ^5}\left( 2 \right) , \\ \zeta \left( {\bar{1},1,1,\bar{2}} \right)&= \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \mathrm{{ + L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{1}{8}\zeta \left( 5 \right) - \frac{7}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\quad -\,\frac{3}{{16}}\zeta \left( 4 \right) \log \left( 2 \right) \\&\quad +\,\frac{3}{4}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{1}{3}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{30}}{\log ^5}\left( 2 \right) , \\ \zeta \left( {\bar{1},1,\bar{1},1,1} \right)&= \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \mathrm{{ + L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{1}{{32}}\zeta \left( 5 \right) - \frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) \\&\quad -\,\frac{1}{8}\zeta \left( 4 \right) \log \left( 2 \right) \\&\quad +\,\frac{1}{2}\zeta \left( 3 \right) {\log ^2}\left( 2 \right) - \frac{1}{6}\zeta \left( 2 \right) {\log ^3}\left( 2 \right) + \frac{1}{{40}}{\log ^5}\left( 2 \right) . \end{aligned}$$

We have used the mathematical tool EZ Face (an abbreviation for Euler Zetas interFace) at the URL http://wayback.cecm.sfu.ca/projects/EZFace/ to check numerically each of the specific identities listed. We confirm that they are correct. Moreover, from [7], we know that all alternating mzvs of weight $\le 5$ can be expressed as a rational linear combination of products of $\log (2)$, polylogarithms and zeta values. So far, we can give all closed forms of alternating mzvs of weight $\le 4$ and part of weight 5. It should be emphasized that the closed forms of alternating mzvs of weight $\le 4$ were also given in [6] by another method.

7 Conclusion

In this paper, we have proved the conclusion: All Euler-type sums ${S_{{p_1}{p_2} \ldots {p_m},p}}\left( 1/2\right)$ of weight $\le 5$ are reducible to Q-linear combinations of single zeta values, polylogarithms and $\log (2)$. Based on the above discussion, we conjectured that all such sums with $w=6$ satisfy a relation involving homogeneous combinations of these constants

$$\begin{aligned}&\mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) ,\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) ,\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) {\log ^2}\left( 2 \right) ,\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \zeta \left( 2 \right) ,\\&\zeta \left( 6 \right) ,{\zeta ^2}\left( 3 \right) ,\zeta \left( 5 \right) \log \left( 2 \right) ,\zeta \left( 2 \right) \zeta \left( 3 \right) \log \left( 2 \right) ,\zeta \left( 4 \right) {\log ^2}\left( 2 \right) ,\\&\zeta \left( 3 \right) {\log ^3}\left( 2 \right) ,\zeta \left( 2 \right) {\log ^4}\left( 2 \right) ,{\log ^6}\left( 2 \right) . \end{aligned}$$

However, we have been unable, so far, to prove the conjecture. By using the method of this paper, we can establish some identities involving two or more Euler-type sums of the weight $=6$. Some of these relations are shown in following

$$\begin{aligned}&\mathrm{{Li}}_3^2\left( {\frac{1}{2}} \right) = 2{S_{3,3}}\left( {\frac{1}{2}} \right) + 6{S_{2,4}}\left( {\frac{1}{2}} \right) + 12{S_{1,5}}\left( {\frac{1}{2}} \right) - 20\mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) , \\&{\mathrm{L}}{{\mathrm{i}}_2}\left( {\frac{1}{2}} \right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) = {S_{4,2}}\left( {\frac{1}{2}} \right) + 2{S_{3,3}}\left( {\frac{1}{2}} \right) + 4{S_{2,4}}\left( {\frac{1}{2}} \right) \\&\quad +\,8{S_{1,5}}\left( {\frac{1}{2}} \right) - 15\mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) , \\&\mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) = {S_{5,1}}\left( {\frac{1}{2}} \right) + {S_{4,2}}\left( {\frac{1}{2}} \right) + {S_{3,3}}\left( {\frac{1}{2}} \right) + {S_{2,4}}\left( {\frac{1}{2}} \right) \\&\quad +\,2{S_{1,5}}\left( {\frac{1}{2}} \right) - 6\mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) ,\\&\zeta \left( {4,1,1;\frac{1}{2}} \right) = \frac{1}{2}{S_{{1^2},4}}\left( {\frac{1}{2}} \right) - \frac{1}{2}{S_{2,4}}\left( {\frac{1}{2}} \right) - {S_{1,5}}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) \\&\quad = \frac{{23}}{{32}}\zeta \left( 6 \right) - \frac{1}{2}{\zeta ^2}\left( 3 \right) \mathrm{{ + L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) + \frac{1}{2}\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) {\log ^2}\left( 2 \right) \\&\quad -\,\frac{{63}}{{32}}\zeta \left( 5 \right) \log \left( 2 \right) +\frac{1}{2}\zeta \left( 2 \right) \zeta \left( 3 \right) \log \left( 2 \right) + \frac{1}{2}\zeta \left( 4 \right) {\log ^2}\left( 2 \right) \\&\quad -\,\frac{1}{{24}}\zeta \left( 2 \right) {\log ^4}\left( 2 \right) + \frac{1}{{90}}{\log ^6}\left( 2 \right) . \\&\zeta \left( {3,{{\left\{ 1 \right\} }_3};\frac{1}{2}} \right) = \frac{{75}}{{32}}\zeta \left( 6 \right) - 2\mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) \\&\quad -\,\frac{1}{4}{\zeta ^2}\left( 3 \right) + \frac{1}{2}\zeta \left( {\bar{5},1} \right) - \zeta \left( 5 \right) \log \left( 2 \right) \\&\quad +\,\frac{7}{{48}}\zeta \left( 3 \right) {\log ^3}\left( 2 \right) - \frac{1}{{24}}\zeta \left( 2 \right) {\log ^4}\left( 2 \right) + \frac{1}{{144}}{\log ^6}\left( 2 \right) \\&\quad =\,\frac{1}{6}\left\{ {{S_{{1^3},3}}\left( {\frac{1}{2}} \right) - 3{S_{12,3}}\left( {\frac{1}{2}} \right) + 2{S_{3,3}}\left( {\frac{1}{2}} \right) } \right\} - \zeta \left( {4,1,1;\frac{1}{2}} \right) ,\\&\zeta \left( {2,{{\left\{ 1 \right\} }_4};\frac{1}{2}} \right) = \zeta \left( 6 \right) - \mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) - \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{1}{2}\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) {\log ^2}\left( 2 \right) \\&\quad -\,\frac{7}{{48}}\zeta \left( 3 \right) {\log ^3}\left( 2 \right) +\frac{1}{{16}}\zeta \left( 2 \right) {\log ^4}\left( 2 \right) - \frac{{11}}{{720}}{\log ^6}\left( 2 \right) \\&\quad = \frac{1}{{24}}\left\{ {S_{{1^4},2}}\left( {\frac{1}{2}} \right) - 6{S_{{1^2}2,2}}\left( {\frac{1}{2}} \right) + 8{S_{13,2}}\left( {\frac{1}{2}} \right) +3{S_{{2^2},2}}\left( {\frac{1}{2}} \right) - 6{S_{4,2}}\left( {\frac{1}{2}} \right) \right\} \\&\quad -\,\zeta \left( {3,{{\left\{ 1 \right\} }_3};\frac{1}{2}} \right) , \\&{S_{{1^4},2}}\left( {\frac{1}{2}} \right) + 6{S_{{1^2}2,2}}\left( {\frac{1}{2}} \right) + 8{S_{13,2}}\left( {\frac{1}{2}} \right) + 3{S_{{2^2},2}}\left( {\frac{1}{2}} \right) + 6{S_{4,2}}\left( {\frac{1}{2}} \right) \\&\quad = \frac{{333}}{8}\zeta \left( 6 \right) - \frac{{45}}{2}\zeta \left( 5 \right) \log \left( 2 \right) - \frac{{27}}{4}{\zeta ^2}\left( 3 \right) ,\\&\frac{1}{2}{S_{{2^2},2}}\left( {\frac{1}{2}} \right) - \frac{1}{2}{S_{4,2}}\left( {\frac{1}{2}} \right) - {S_{2,4}}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) \\&\quad = \frac{5}{{128}}\zeta \left( 6 \right) - \frac{1}{{32}}{\zeta ^2}\left( 3 \right) - \frac{3}{{16}}\zeta \left( 5 \right) \log \left( 2 \right) + \frac{1}{8}\zeta \left( 2 \right) \zeta \left( 3 \right) \log \left( 2 \right) \\&\quad \,-\, \frac{1}{{32}}\zeta \left( 4 \right) {\log ^2}\left( 2 \right) \\&\quad \,-\,\frac{1}{{24}}\zeta \left( 3 \right) {\log ^3}\left( 2 \right) + \frac{1}{{48}}\zeta \left( 2 \right) {\log ^4}\left( 2 \right) - \frac{1}{{720}}{\log ^6}\left( 2 \right) . \end{aligned}$$

From formulas above, we obtain the closed form of Euler-type sum ${S_{5,1}}\left( {\frac{1}{2}} \right)$ as follows:

$$\begin{aligned} {S_{5,1}}\left( {\frac{1}{2}} \right)&= \mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \mathrm{{L}}{\mathrm{{i}}_2}\left( {\frac{1}{2}} \right) \mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) + \frac{1}{2}\mathrm{{Li}}_3^2\left( {\frac{1}{2}} \right) \\&= \mathrm{{L}}{\mathrm{{i}}_6}\left( {\frac{1}{2}} \right) + \mathrm{{L}}{\mathrm{{i}}_5}\left( {\frac{1}{2}} \right) \log \left( 2 \right) - \frac{1}{2}\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) \zeta \left( 2 \right) + \frac{1}{2}\mathrm{{L}}{\mathrm{{i}}_4}\left( {\frac{1}{2}} \right) {\log ^2}\left( 2 \right) \\&\quad +\,\frac{{49}}{{128}}{\zeta ^2}\left( 3 \right) - \frac{7}{{16}}\zeta \left( 2 \right) \zeta \left( 3 \right) \log \left( 2 \right) + \frac{5}{{16}}\zeta \left( 4 \right) {\log ^2}\left( 2 \right) \\&\quad +\,\frac{7}{{48}}\zeta \left( 3 \right) {\log ^3}\left( 2 \right) - \frac{1}{{12}}\zeta \left( 2 \right) {\log ^4}\left( 2 \right) + \frac{1}{{72}}{\log ^6}\left( 2 \right) . \end{aligned}$$

More general, we have (see Corollary 1 in Reference [32])

$$\begin{aligned} {S_{2n - 1,1}}\left( z \right) = \mathrm{{L}}{\mathrm{{i}}_{2n}}\left( z \right) + \frac{1}{2}\sum \limits _{k = 1}^{2n - 1} {{{\left( { - 1} \right) }^{k + 1}}\mathrm{{L}}{\mathrm{{i}}_k}\left( z \right) } \mathrm{{L}}{\mathrm{{i}}_{2n - k}}\left( z \right) , \end{aligned}$$

Where $n\in \mathbb {N}, z\in D:=\mathbb {C}{\setminus }\{z{:}\,\left| {\arg \left( {1 - z} \right) } \right| < \pi \}$.