1 Introduction

For an r-tuple \(\varvec{\alpha } = (\alpha _{1}, \alpha _{2}, \ldots , \alpha _{r})\) of positive integers with \(\alpha _{r} \ge 2\), a multiple zeta value \(\zeta (\varvec{\alpha })\) and a multiple zeta-star value \(\zeta ^\star (\varvec{\alpha })\) are defined to be [9, 10]

$$\begin{aligned} \zeta (\varvec{\alpha }) = \sum _{1 \le k_{1}< k_{2}< \cdots < k_{r}} k_{1}^{-\alpha _{1}} k_{2}^{-\alpha _{2}} \cdots k_{r}^{-\alpha _{r}}, \end{aligned}$$

and

$$\begin{aligned} \zeta ^{\star }(\varvec{\alpha }) = \sum _{1 \le k_{1} \le k_{2} \le \cdots \le k_{r}} k_{1}^{-\alpha _{1}} k_{2}^{-\alpha _{2}} \cdots k_{r}^{-\alpha _{r}}. \end{aligned}$$

We denote the parameters \(w(\varvec{\alpha })=\mid \varvec{\alpha }\mid = \alpha _{1} + \alpha _{2} + \cdots + \alpha _{r}\), \(d(\varvec{\alpha })=r\), and \(h(\varvec{\alpha })=\#\{i\mid \alpha _i>1, 1\le i\le r\}\), called, respectively, the weight, the depth, and the height of \(\varvec{\alpha }\) (or of \(\zeta (\varvec{\alpha })\), or of \(\zeta ^\star (\varvec{\alpha })\)). In fact, these sums were originally studied by Euler. Beginning in the early 1990s, the works of Hoffman [14] and Zagier [25] began to stimulate research again.

For any multiple zeta value \(\zeta (\varvec{\alpha })\), we put a bar on top of \(\alpha _j\) (\(j=1,2,\ldots ,r\)) if there is a sign \((-1)^{k_j}\) appearing in the numerator of the summand [22, 23]. We call it an alternating multiple zeta value. For example, an alternating multiple zeta value

$$\begin{aligned} \zeta (\overline{1},\{1\}^{m},\overline{2})=\sum _{1\le k_1<k_2<\cdots <k_{m+2}} \frac{(-1)^{k_1+k_{m+2}}}{k_1\cdots k_{m+1}k_{m+2}^2}, \end{aligned}$$

where \(\{a\}^{k}\) is k repetitions of a. Its weight is \(m+3\), the height is one, and the depth is \(m+2\).

Due to Kontsevich, multiple zeta values can be represented by iterated integrals over a simplex of dimension weight:

$$\begin{aligned} \zeta (\alpha _{1}, \alpha _{2}, \ldots , \alpha _{r}) = \int _{E_{\mid \varvec{\alpha }\mid }} \Omega _{1} \Omega _{2} \cdots \Omega _{\mid \varvec{\alpha }\mid } \quad \text {or} \quad \int _{0}^{1} \Omega _{1} \Omega _{2} \cdots \Omega _{\mid \varvec{\alpha }\mid } \end{aligned}$$

with \(E_{\mid \varvec{\alpha }\mid }: 0< t_{1}< t_{2}< \cdots< t_{\mid \varvec{\alpha }\mid } < 1\) and

$$\begin{aligned} \Omega _{j} = {\left\{ \begin{array}{ll} \dfrac{dt_{j}}{1-t_{j}} &{}\text {if } j = 1, \alpha _{1}+1, \alpha _{1}+\alpha _{2}+1, \ldots , \alpha _{1}+\alpha _{2}+\cdots +\alpha _{r-1}+1, \\ \dfrac{dt_{j}}{t_{j}} &{}\text {otherwise}. \end{array}\right. } \end{aligned}$$

Once multiple zeta values are expressed as iterated integrals, the shuffle product of two multiple zeta values can be defined as:

$$\begin{aligned} \int _{0}^{1} \Omega _{1} \Omega _{2} \cdots \Omega _{m} \int _{0}^{1} \Omega _{m+1} \Omega _{m+2} \cdots \Omega _{m+n} = \sum _{{\phi }} \int _{0}^{1} \Omega _{{\phi }(1)} \Omega _{{\phi }(2)} \cdots \Omega _{{\phi }(m+n)}, \end{aligned}$$

where \(\phi \) ranges over all permutations on the set \(\{ 1,2,\ldots ,m+n \}\) that preserve the orderings of \(\Omega _{1} \Omega _{2} \cdots \Omega _{m}\) and \(\Omega _{m+1} \Omega _{m+2} \cdots \Omega _{m+n}\). This is equivalent to saying that for all \(1 \le i < j \le m\) or \(m+1 \le i < j \le m+n\), we have \(\phi ^{-1}(i) < \phi ^{-1}(j)\). It is clear that the shuffle product of two multiple zeta values of weight m and n, respectively, will produce \(\left( {\begin{array}{c}m+n\\ n\end{array}}\right) \) multiple zeta values of weight \(m+n\). Transforming the Kontsevich iterated integrals into log-type iterated integrals can make great progress in the research of multiple zeta values, alternating multiple zeta values, and so on, refer to [4, 6, 11, 12, 23].

In this paper, we will investigate the following integral:

$$\begin{aligned} \frac{1}{p!} \int _{\begin{array}{c} 0<t_1<t_2<1\\ 0<u_1<u_2<1 \end{array}} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

We derive our main theorem by finding its different representations.

Main Theorem For integer \(p \ge 0\), we have

$$\begin{aligned}&\frac{(p+1)(p+2)}{2} \zeta (p+4) =\sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \\&\times {\sum _{a+b+c=m}}^{*} \Bigl ( W_{\varvec{\alpha }}(a,b,c) + W_{\varvec{\alpha }}(a,b,0) + W_{\varvec{\alpha }}(0,b,c) + W_{\varvec{\alpha }}(0,m,0) \Bigr ), \end{aligned}$$

where \(W_{\varvec{\alpha }}(a,b,c) = 2^{\sigma (a+b+1) -\sigma (a)-(b+1)} (1-2^{1-\alpha _{a+b+1}})\), with \(\sigma (r) = \sum _{j=0}^{r} \alpha _{j}\), and

$$\begin{aligned} {\sum _{a+b+c=m}}^{*} W_{\varvec{\alpha }}(a,b,0)&=\sum _{a+b=m}W_{\varvec{\alpha }}(a,b,0),\\ {\sum _{a+b+c=m}}^{*} W_{\varvec{\alpha }}(0,b,c)&=\sum _{b+c=m}W_{\varvec{\alpha }}(0,b,c),\\ {\sum _{a+b+c=m}}^{*} W_{\varvec{\alpha }}(0,m,0)&=W_{\varvec{\alpha }}(0,m,0). \end{aligned}$$

For brevity, we hereafter use the lowercase English letters and lowercase Greek letters in the summation, with or without subscripts, to denote the nonnegative and positive integers, unless otherwise specified.

We present our formulas for \(p=0\) and \(p=1\) as follows:

$$\begin{aligned} p&=0,&\zeta (4)&=4\zeta (1,3).\\ p&=1,&3\zeta (5)&=12\zeta (1,4)+4\zeta (2,3)+2\zeta (1,2,2)+6\zeta (1,1,3). \end{aligned}$$

In fact, our main theorem is from the integral

$$\begin{aligned} J_{p} = \frac{1}{p!} \int _{\begin{array}{c} 0<t_1<t_2<1\\ 0<u_1<u_2<1 \end{array}} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

As a replacement of shuffle product, we decompose the domain \(\{(t_1,t_2,u_1,u_2)\in \mathbb R^4\mid 0<t_1<t_2<1,0<u_1<u_2<1\}\) into 6 disjoint simplices of dimension 4:

$$\begin{aligned}&D_{1}: 0< t_{1}< t_{2}< u_{1}< u_{2}< 1, \quad D_{2}: 0< u_{1}< u_{2}< t_{1}< t_{2}< 1, \\&D_{3}: 0< t_{1}< u_{1}< t_{2}< u_{2}< 1, \quad D_{4}: 0< t_{1}< u_{1}< u_{2}< t_{2}< 1, \\&D_{5}: 0< u_{1}< t_{1}< u_{2}< t_{2}< 1, \quad D_{6}: 0< u_{1}< t_{1}< t_{2}< u_{2} < 1. \end{aligned}$$

So

$$\begin{aligned} J_{p} = \sum _{j=1}^{6} \mathbb J(j) \end{aligned}$$

with

$$\begin{aligned} \mathbb J(j) = \frac{1}{p!} \int _{D_{j}} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

The main task now is to evaluate \(\mathbb J(j)\) one by one in terms of multiple zeta values. The weighted sum in our main theorem is \(\sum _{j=3}^6\mathbb J(j)\). The crucial step is to prove

$$\begin{aligned} \mathbb J(1) + \mathbb J(2) = J_{p} - \frac{(p+1)(p+2)}{2} \zeta (p+4) \end{aligned}$$
(1.1)

and the resulted shuffle relation is

$$\begin{aligned} \sum _{j=3}^{6} \mathbb J(j) = \frac{(p+1)(p+2)}{2} \zeta (p+4). \end{aligned}$$
(1.2)

We organize this paper as follows: We give some preliminaries and auxiliary tools in Sect. 2. In Sect. 3, we break \(J_p\) into six parts and evaluate \(\mathbb J(1)\) and \(\mathbb J(2)\) in terms of multiple zeta values. In the next section, we express \(\mathbb J(j)\) for \(3\le j\le 6\) as the part of the weighted sum of our main theorem. We prove the crucial step Eq. (1.1) in Sect. 5. In Sect. 6, we give another application of the identity Eq. (1.2) to another sum formula which is also proved by a duality theorem due to Ohno [19]. In Sect. 7, we relate \(\mathbb J(3)\) with the weighted sum

$$\begin{aligned} \sum _{\mid \varvec{\alpha }\mid =n+3}2^{\alpha _2}\zeta ^\star (\alpha _1,\{1\}^m,\alpha _2+1). \end{aligned}$$

In the final section, we will give some formulas related to weighted alternating Euler sums, e.g.,

$$\begin{aligned} \sum _{m+n=p}n\zeta (m+1,\overline{n+1}),\quad \text {and}\quad \sum _{m+n=p}n(n-1)\zeta (m+1,\overline{n+1}). \end{aligned}$$

2 Some Preliminaries and Auxiliary Tools

Here, we list two general integral representations that we will use frequently in the similar tricks.

Proposition 2.1

[11] For nonnegative integers \({r}, b_{1}, b_{2}, \ldots , b_{r}, b_{r+1}\), we have

$$\begin{aligned}&\zeta (b_{1}+1, b_{2}+1, \ldots , b_{r}+1, b_{r+1}+2)\\&= \frac{1}{b_{1}! b_{2}! \cdots b_{r+1}!} \int _{E_{r+2}} \prod _{j=1}^{r+1} \frac{\textrm{d}t_{j}}{1-t_{j}} \left( \log \frac{t_{j+1}}{t_{j}} \right) ^{b_{j}} \frac{\textrm{d}t_{r+2}}{t_{r+2}}, \end{aligned}$$

where \(E_{r+2}\) is a simplex defined as: \(0< t_{1}< t_{2}< \cdots< t_{r+2}{<1}\).

Proposition 2.2

[11, Proposition 2.1] For integers \(p,q,r,\ell \ge 0\),

$$\begin{aligned}&\quad \sum _{\mid \varvec{\alpha }\mid =q+r+1} \zeta (\{1\}^{p}, \alpha _{0}, \alpha _{1}, \ldots , \alpha _{q}+\ell +1) = \frac{1}{p!q!r!\ell !}\\&\times \int _{0< t_{1}< t_{2} < 1} \left( \log \frac{1}{1-t_{1}} \right) ^{p} \left( \log \frac{1-t_{1}}{1-t_{2}} \right) ^{q} \left( \log \frac{t_{2}}{t_{1}} \right) ^{r} \left( \log \frac{1}{t_{2}} \right) ^{\ell } \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}}. \end{aligned}$$

The well-known sum formula due to Granville [13] asserted that

$$\begin{aligned} \sum _{\mid \varvec{\alpha }\mid =q+r+1} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{q}+1) = \zeta (q+r+2). \end{aligned}$$

If we let \(p = \ell = 0\) in the formula of Proposition 2.2, then we have

$$\begin{aligned}&\sum _{\mid \varvec{\alpha }\mid =q+r+1} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{q}+1) \nonumber \\&= \frac{1}{q!r!} \int _{0< t_{1}< t_{2} < 1} \left( \log \frac{1-t_{1}}{1-t_{2}} \right) ^{q} \left( \log \frac{t_{2}}{t_{1}} \right) ^{r} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}}. \end{aligned}$$
(2.1)

Therefore, the integral

$$\begin{aligned} \frac{1}{q!r!} \int _{0< t_{1}< t_{2} < 1} \left( \log \frac{1-t_{1}}{1-t_{2}} \right) ^{q} \left( \log \frac{t_{2}}{t_{1}} \right) ^{r} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}}=\zeta (q+r+2). \end{aligned}$$

Our main results are obtained by the convolution of the following sums:

$$\begin{aligned} Z_-(m) = \sum _{a+b=m} (-1)^{b} \zeta (\{1\}^{a},b+2) \quad \text {and} \quad Z^\star _+(n) = \sum _{c+d=n} \zeta ^{\star }(\{1\}^{c},d+2). \end{aligned}$$
(2.2)

A special case of results proved by Le and Murakami [18] is stated as follows:

$$\begin{aligned} \sum _{a+b=2w-2}(-1)^{a+1}\zeta (\{1\}^a,b+2) =2\zeta (\overline{2w}). \end{aligned}$$
(2.3)

Both \(Z_{-}(n)\), and \(Z^\star _{+}(n)\) are sums of multiple zeta(-star) values of height one and can be expressed as double integrals (see [6])

$$\begin{aligned} Z_-(n)&= \frac{1}{n!} \int _{E_{2}} \left( \log \frac{1}{1-t_{1}} - \log \frac{t_{2}}{t_{1}} \right) ^{n} \frac{\textrm{d}t_{1} \textrm{d}t_{2}}{(1-t_{1})t_{2}}, \\ Z^\star _+(n)&= \frac{1}{n!} \int _{E_{2}} \left( \log \frac{1}{1-t_{2}} + \log \frac{t_{2}}{t_{1}} \right) ^{n} \frac{\textrm{d}t_{1} \textrm{d}t_{2}}{(1-t_{1})t_{2}}. \end{aligned}$$

Through the duality theorem \(\zeta (\{1\}^{a},b+2) = \zeta (\{1\}^{b},a+2)\), we see that \(Z_-(2m+1) = 0\). Therefore, we combine these with Eq. (2.3), and then, we have

Proposition 2.3

[6, Proposition 3.1] For any nonnegative integer m, we have \(Z_-(2m+1)=0\) and

$$\begin{aligned} Z_-(2m) = \zeta ^{\star }(\{2\}^{m+1}) = 2 \left( 1 - \frac{1}{2^{2m+1}} \right) \zeta (2m+2)=-2\zeta (\overline{2m+2}). \end{aligned}$$

The sum of multiple zeta-star values \(Z^\star _+(n)\) appeared as the principal term of the evaluation of \(\zeta ^{\star }(3,\{2\}^{n})\) (see [5]):

$$\begin{aligned} \zeta ^{\star }(3,\{2\}^{n}) = Z_+^\star (2n+1) - 2 \sum _{a+b=n} \zeta (\{2\}^{a}) \zeta (2b+3). \end{aligned}$$

Arakawa and Kaneko [1] defined the function

$$\begin{aligned} \xi _k(s)=\frac{1}{\Gamma (s)}\int ^\infty _0\frac{t^{s-1}}{e^t-1} {\text {Li}}_k(1-e^{-t})\,dt, \end{aligned}$$

where \({\text {Li}}_k(s)\) denotes the k-th polylogarithm \({\text {Li}}_k(s)=\sum ^\infty _{n=1}\frac{s^n}{n^k}\). It is exactly the multiple zeta-star values of height one

$$\begin{aligned} \xi _k(s)=\zeta ^\star (\{1\}^{s-1},k+1). \end{aligned}$$

Many properties of the generalized Arakawa–Kaneko zeta functions have been discovered recently (ref. [3, 8, 15,16,17, 24]).

Indeed, \(Z^\star _+(n)\) has the generating function

$$\begin{aligned} \sum _{n=0}^{\infty } Z^\star _+(n) (-x)^{n} = \int _{0< t_{1}< t_{2} < 1} (1-t_{2})^{x} \left( \frac{t_{1}}{t_{2}} \right) ^{x} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}}. \end{aligned}$$

The dual of the above is

$$\begin{aligned} \int _{0< u_{1}< u_{2} < 1} \left( \frac{1-u_{2}}{1-u_{1}} \right) ^{x} u_{1}^{x} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

Using the binomial theorem to expand \((1-u_1)^{-x-1}\), we can evaluate the integral as:

$$\begin{aligned} \sum _{k=1}^{\infty } \frac{1}{(k+x) (k+2x)} \frac{\Gamma (k+x)^{2}}{\Gamma (k) \Gamma (k+2x)}. \end{aligned}$$

Fortunately, we have the identity

$$\begin{aligned} \sum _{k=1}^{\infty } \frac{1}{(k+x) (k+2x)} \frac{\Gamma (k+x)^{2}}{\Gamma (k) \Gamma (k+2x)} = \sum _{\ell =1}^{\infty } \frac{2(-1)^{\ell -1}}{(\ell +x)^{2}} \end{aligned}$$

by investigating possible poles of both sides of the meromorphic functions. This leads to the evaluation of \(Z^\star _+(n)\):

$$\begin{aligned} \sum _{a+b=w-2}\zeta ^\star (\{1\}^a,b+2) =2(w-1)(1-2^{1-w})\zeta (w)=-2(w-1)\zeta (\overline{w}), \end{aligned}$$
(2.4)

where \(w>1\) is an integer. This formula was first proved by Ohno [20, Theorem 8] in 2005.

3 The Main Integral \(J_p\) and its \(\mathbb J(1), \mathbb J(2)\) Parts

We begin with the integral

$$\begin{aligned} J_p = \frac{1}{p!} \int _{E_{2} \times E_{2}} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}, \end{aligned}$$

where \(E_{2} \times E_{2} = \{ (t_{1}, t_{2}, u_{1}, u_{2}) \in \mathbb {R}^{4} \mid 0< t_{1}< t_{2}< 1, 0< u_{1}< u_{2} < 1 \}\). As

$$\begin{aligned}&\quad \frac{1}{p!} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \\&= \frac{1}{p!} \left\{ \left( -\log \frac{1}{1-t_{1}} + \log \frac{t_{2}}{t_{1}} \right) + \left( \log \frac{1}{1-u_{2}} + \log \frac{u_{2}}{u_{1}} \right) \right\} ^p \\&= \sum _{m+n=p} \frac{1}{m!n!} \left( -\log \frac{1}{1-t_{1}} + \log \frac{t_{2}}{t_{1}} \right) ^{m} \left( \log \frac{1}{1-u_{2}} + \log \frac{u_{2}}{u_{1}} \right) ^{n}, \end{aligned}$$

so we have the evaluation

$$\begin{aligned} J_p = \sum _{m+n=p} (-1)^{m} Z_-(m) Z_+^\star (n){,} \end{aligned}$$

where \(Z_-(m)\) and \(Z_+^\star (n)\) are defined in Eq (2.2). By Proposition 2.3\(Z_-(2m+1)=0\), thus

$$\begin{aligned} J_p=\sum _{m=0}^{[p/2]} Z_-(2m) Z_+^\star (p-2m)=\sum _{m+n=p}Z_-(m)Z_+^\star (n). \end{aligned}$$

We got an expression of the finite convolution of \(Z_-(m)\) and \(Z_+^\star (n)\) in [6, Corollary 5.4]:

$$\begin{aligned} J_{p}=2((-1)^p-1)\zeta (p+2,\overline{2})+(p+2)(p+1+2^{-p-2})\zeta (p+4). \end{aligned}$$

In this paper, we consider another expression of \(J_p\). When we decompose \(E_{2} \times E_{2}\) into 6 simplices of dimension 4:

$$\begin{aligned}&D_{1}: 0< t_{1}< t_{2}< u_{1}< u_{2}< 1, \quad D_{2}: 0< u_{1}< u_{2}< t_{1}< t_{2}< 1, \\&D_{3}: 0< t_{1}< u_{1}< t_{2}< u_{2}< 1, \quad D_{4}: 0< t_{1}< u_{1}< u_{2}< t_{2}< 1, \\&D_{5}: 0< u_{1}< t_{1}< u_{2}< t_{2}< 1, \quad D_{6}: 0< u_{1}< t_{1}< t_{2}< u_{2} < 1 \end{aligned}$$

and let

$$\begin{aligned} \mathbb J(j) = \frac{1}{p!} \int _{D_{j}} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

Then, \(J_p\) is broken into 6 parts:

$$\begin{aligned} J_{p} = \sum _{j=1}^{6} \mathbb J(j). \end{aligned}$$

In the following, we will evaluate \(\mathbb J(j)\), for \(1\le j\le 6\), in terms of multiple zeta values.

Proposition 3.1

Given any nonnegative integer p, we have

$$\begin{aligned} \mathbb J(1) = \sum _{m+n=p} \sum _{a+b+c=m} \sum _{u+v=n} \sum _{\begin{array}{c} \mid \varvec{\alpha }\mid =a+u+1 \\ \mid \varvec{\beta }\mid =c+v+1 \end{array}} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{a}+1, \{1\}^{b}, \beta _{0}, \ldots , \beta _{c}+1). \end{aligned}$$

Proof

We first expand the integrand of \(\mathbb J(1)\) as:

$$\begin{aligned} \sum _{m+n=p}\frac{1}{m!n!}\left( \log \frac{1-t_1}{1-u_2}\right) ^m \left( \log \frac{t_2}{t_1}+\log \frac{u_2}{u_1}\right) ^n. \end{aligned}$$

Since \(\mathbb J(1)\) is an integral on \(D_1: 0<t_1<t_2<u_1<u_2<1\), we replace the factor

$$\begin{aligned} \log \frac{1-t_{1}}{1-u_{2}} \end{aligned}$$

by its equal

$$\begin{aligned} \log \frac{1-t_{1}}{1-t_{2}} + \log \frac{1-t_{2}}{1-u_{1}} + \log \frac{1-u_{1}}{1-u_{2}}. \end{aligned}$$

Then, we expand the integrand of \(\mathbb J(1)\) as:

$$\begin{aligned}&\sum _{m+n=p} \sum _{a+b+c=m} \sum _{u+v=n} \frac{1}{a!b!c!u!v!} \\&\quad \times \left( \log \frac{1-t_{1}}{1-t_{2}} \right) ^{a} \left( \log \frac{1-t_{2}}{1-u_{1}} \right) ^{b} \left( \log \frac{1-u_{1}}{1-u_{2}} \right) ^{c} \left( \log \frac{t_{2}}{t_{1}} \right) ^{u} \left( \log \frac{u_{2}}{u_{1}} \right) ^{v}. \end{aligned}$$

The factor

$$\begin{aligned} \frac{1}{a!u!} \left( \log \frac{1-t_{1}}{1-t_{2}} \right) ^{a} \left( \log \frac{t_{2}}{t_{1}} \right) ^{u} \end{aligned}$$

forms a sum and the other factor

$$\begin{aligned} \frac{1}{c!v!} \left( \log \frac{1-u_{1}}{1-u_{2}} \right) ^{c} \left( \log \frac{u_{2}}{u_{1}} \right) ^{v} \end{aligned}$$

forms another sum, so

$$\begin{aligned} \mathbb J(1) = \sum _{m+n=p} \sum _{a+b+c=m} \sum _{u+v=n} \sum _{\begin{array}{c} \mid \varvec{\alpha }\mid =a+u+1 \\ \mid \varvec{\beta }\mid =c+v+1 \end{array}} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{a}+1, \{1\}^{b}, \beta _{0}, \ldots , \beta _{c}+1). \end{aligned}$$

Therefore, we complete the proof. \(\square \)

Proposition 3.2

For any nonnegative integer p, we have

$$\begin{aligned} \mathbb J(2) = \sum _{c+d+m=p} (-1)^{m} \zeta (c+2,\{1\}^{m},d+2). \end{aligned}$$

Proof

On \(D_{2}: 0< u_{1}< u_{2}< t_{1}< t_{2} < 1\), the factor

$$\begin{aligned} \log \frac{1-t_{1}}{1-u_{2}} \end{aligned}$$

is replaced by

$$\begin{aligned} -\log \frac{1-u_{2}}{1-t_{1}} \end{aligned}$$

and the integrand

$$\begin{aligned} \frac{1}{p!} \left( -\log \frac{1-u_{2}}{1-t_{1}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^p \end{aligned}$$

is expanded into

$$\begin{aligned} \sum _{c+d+m=p} \frac{(-1)^{m}}{c!\,d!\,m!} \left( \log \frac{1-u_{2}}{1-t_{1}} \right) ^{m} \left( \log \frac{u_{2}}{u_{1}} \right) ^{c} \left( \log \frac{t_{2}}{t_{1}} \right) ^{d}. \end{aligned}$$

So that

$$\begin{aligned} \mathbb J(2) = \sum _{c+d+m=p} (-1)^m\zeta (c+2,\{1\}^{m},d+2). \end{aligned}$$

\(\square \)

It is noting that we have another expression of \(\mathbb J(2)\) in [7, Theorem 3]:

$$\begin{aligned} \sum _{a+b+m=p}\!\! (-1)^m\zeta (a+2,\{1\}^m,b+2) \!=\!2[(-1)^p-1]\zeta (p+2,\overline{2}) +\left( 1-2^{-p-2}\right) \zeta (p+4). \end{aligned}$$

4 The Integrals \(\mathbb J(3), \mathbb J(4), \mathbb J(5)\), and \(\mathbb J(6)\) Parts

In this section, we will give the evaluations of \(\mathbb J(j)\) for \(3\le j\le 6\). These expressions give us the part of the weighted sum of our main theorem. Our method of proving the following proposition is inspired by the approach used in [21, Theorem 5.3].

Proposition 4.1

Let p be a nonnegative integer. We have

$$\begin{aligned} \mathbb J(3) = \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \sum _{a+b+c=m} W(a,b,c) \end{aligned}$$

with \(\sigma (r) = \alpha _{0} + \alpha _{1} + \cdots + \alpha _{r}\) and

$$\begin{aligned} W(a,b,c) = 2^{\sigma (a+b+1)-\sigma (a)-(b+1)} (1-2^{1-\alpha _{a+b+1}}). \end{aligned}$$
(4.1)

Proof

The integral \(\mathbb J(3)\) is on \(D_3: 0<t_1<u_1<t_2<u_2<1\). Therefore, we replace the integrand of \(\mathbb J(3)\) by

$$\begin{aligned} \frac{1}{p!} \left( \log \frac{1-t_{1}}{1-u_{1}} + \log \frac{1-u_{1}}{1-t_{2}} + \log \frac{1-t_{2}}{1-u_{2}} + \log \frac{u_{1}}{t_{1}} + 2\log \frac{t_{2}}{u_{1}} + \log \frac{u_{2}}{t_{2}} \right) ^{p}. \end{aligned}$$

Then, we expand it as:

$$\begin{aligned} \sum _{\begin{array}{c} m+n=p\\ a+b+c=m\\ u+v+w=n \end{array}} \frac{2^{v}}{a!b!c!u!v!w!} \left( \log \frac{1-t_{1}}{1-u_{1}} \right) ^{a}&\left( \log \frac{1-u_{1}}{1-t_{2}} \right) ^{b} \left( \log \frac{1-t_{2}}{1-u_{2}} \right) ^{c}\\&\times \left( \log \frac{u_{1}}{t_{1}} \right) ^{u} \left( \log \frac{t_{2}}{u_{1}} \right) ^{v} \left( \log \frac{u_{2}}{t_{2}} \right) ^{w}. \end{aligned}$$

This yields

$$\begin{aligned} \mathbb J(3) = \sum _{\begin{array}{c} m+n=p\\ a+b+c=m\\ u+v+w=n \end{array}} 2^{v} \sum _{\begin{array}{c} \mid \varvec{\alpha }\mid =a+u+1 \\ \mid \varvec{\beta }\mid =b+v+1 \\ \mid \varvec{\gamma }\mid =c+w+1 \end{array}} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{a}, \beta _{0}, \beta _{1}, \ldots , \beta _{b}+\gamma _{0}, \gamma _{1}, \ldots , \gamma _{c}+1). \end{aligned}$$

We change the variables \(\varvec{\alpha ,\beta ,\gamma }\) to a nonnegative vector variable \(\varvec{e}=(e_0,e_1,\ldots ,e_{m+1})\). Then,

$$\begin{aligned} \mathbb J(3)&=\sum _{\begin{array}{c} m+n=p\\ a+b+c=m\\ \mid \varvec{e}\mid =n \end{array}} \left( \sum _{i=0}^{e_{a+b+1}}2^{e_{a+b+1}-i}\right) 2^{e_{a+1}+e_{a+2}+\cdots +e_{a+b}} \\&\qquad \times \zeta (e_0+1,\ldots ,e_{a+b}+1,e_{a+b+1}+2,e_{a+b+2}+1,\ldots ,e_{m}+1,e_{m+1}+2). \end{aligned}$$

Since

$$\begin{aligned} \sum _{i=0}^{e_{a+b+1}}2^{e_{a+b+1}-i}=2^{e_{a+b+1}+1}-1, \end{aligned}$$

we can change the nonnegative vector variable \(\varvec{e}\) to a new positive variable \(\varvec{\alpha }\) and \(\mathbb J(3)\) becomes

$$\begin{aligned} \sum _{\begin{array}{c} m+n=p\\ a+b+c=m\\ \mid \varvec{\alpha }\mid =m+n+3 \end{array}} 2^{\alpha _{a+1}+\cdots +\alpha _{a+b}-b}\left( 2^{\alpha _{a+b+1}-1}-1\right) \zeta (\alpha _0,\ldots ,\alpha _m,\alpha _{m+1}+1). \end{aligned}$$

We rewrite \(\mathbb J(3)\) as:

$$\begin{aligned} \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \sum _{a+b+c=m} W_{\varvec{\alpha }}(a,b,c) \end{aligned}$$

with \(\sigma (r) = \alpha _{0} + \alpha _{1} + \cdots + \alpha _{r}\) and

$$\begin{aligned} W_{\varvec{\alpha }}(a,b,c) = 2^{\sigma (a+b+1)-\sigma (a)-(b+1)} (1-2^{1-\alpha _{a+b+1}}). \end{aligned}$$

\(\square \)

In the same manner as the previous proposition, we obtain the following:

Proposition 4.2

Notation as introduced above, we have

$$\begin{aligned} \mathbb J(4) = \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \sum _{a+b=m} W_{\varvec{\alpha }}(a,b,0), \end{aligned}$$

where \(W_{\varvec{\alpha }}(a,b,c)\) is defined in Eq. (4.1) and

$$\begin{aligned} W_{\varvec{\alpha }}(a,b,0) = 2^{\sigma (a+b+1)-\sigma (a)-(b+1)} (1-2^{1-\alpha _{a+b+1}}). \end{aligned}$$

Proposition 4.3

Notation as introduced above, then we have

$$\begin{aligned} \mathbb J(5) = \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) W_{\varvec{\alpha }}(0,m,0), \end{aligned}$$

where

$$\begin{aligned} W_{\varvec{\alpha }}(0,m,0) = 2^{\sigma (m+1)-\sigma (0)-(m+1)} (1-2^{1-\alpha _{m+1}}) = 2^{n+2-\sigma (0)} (1-2^{1-\alpha _{m+1}}). \end{aligned}$$

Proposition 4.4

Notation as introduced above, then we have

$$\begin{aligned} \mathbb J(6) = \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \sum _{b+c=m} W_{\varvec{\alpha }}(0,b,c) \end{aligned}$$

with \(\sigma (r) = \alpha _{0} + \alpha _{1} + \cdots + \alpha _{r}\) and

$$\begin{aligned} W_{\varvec{\alpha }}(0,b,c) = 2^{\sigma (b+1)-\sigma (0)-(b+1)} (1-2^{1-\alpha _{b+1}}). \end{aligned}$$

So up to now, our shuffle relation appeared to be the form:

$$\begin{aligned} J_{p}&= \mathbb J(1) + \mathbb J(2) + \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \\&\times \left( \sum _{a+b+c=m} W_{\varvec{\alpha }}(a,b,c) +\sum _{a+b=m}W_{\varvec{\alpha }}(a,b,0) + W_{\varvec{\alpha }}(0,m,0) \right. \\&\left. + \sum _{b+c=m}W_{\varvec{\alpha }}(0,b,c) \right) . \end{aligned}$$

We write it in a compact form:

$$\begin{aligned} J_{p}&= \mathbb J(1) + \mathbb J(2) + \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{m}, \alpha _{m+1}+1) \nonumber \\&\times {\sum _{a+b+c=m}}^{*} \Bigl ( W_{\varvec{\alpha }}(a,b,c) + W_{\varvec{\alpha }}(a,b,0) + W_{\varvec{\alpha }}(0,m,0) + W_{\varvec{\alpha }}(0,b,c) \Bigr ). \end{aligned}$$
(4.2)

5 Another Way to Evaluate \(\mathbb J(1)\) and \(\mathbb J(2)\) Parts

The modified Bell polynomials are defined by [2, 8]

$$\begin{aligned} \exp \left\{ \sum _{k=1}^{\infty } \frac{x_{k} z^{k}}{k} \right\} = \sum _{m=0}^{\infty } P_{m}(x_{1},x_{2},\ldots ,x_{m}) z^{m}. \end{aligned}$$

So that \(P_{m}(x_{1},x_{2},\ldots ,x_{m})\) has the expression

$$\begin{aligned} \sum _{k_{1}+2k_{2}+\cdots +mk_{m}=m} \frac{1}{k_{1}! k_{2}! \cdots k_{m}!} \left( \frac{x_{1}}{1} \right) ^{k_{1}} \left( \frac{x_{2}}{2} \right) ^{k_{2}} \cdots \left( \frac{x_{m}}{m} \right) ^{k_{m}}. \end{aligned}$$

In particular, for \(m = 0,1,2,3\),

$$\begin{aligned}&P_{0} = 1, \quad P_{1}(x_{1}) = x_{1}, \quad P_{2}(x_{1},x_{2}) = \frac{1}{2} (x_{1}^{2}+x_{2}), \\&P_{3}(x_{1},x_{2},x_{3}) = \frac{1}{3} (x_{1}^{3}+3x_{1}x_{2}+2x_{3}). \end{aligned}$$

It is well known that

$$\begin{aligned}&P_{n}(\zeta (2), -\zeta (4), \ldots , (-1)^{n+1} \zeta (2n)) = \zeta (\{2\}^{n}) \quad \text {and} \\&P_{n}(\zeta (2), \zeta (4), \ldots , \zeta (2n)) = \zeta ^{\star }(\{2\}^{n}). \end{aligned}$$

In order to give another expression of \(\mathbb J(1)\), we need the following lemma:

Lemma 5.1

[4, Proposition 4.4] For a pair of positive integers \(k_{1}\), \(k_{2}\) with \(k_{1} \le k_{2}\), let

$$\begin{aligned} Q(y) = \frac{\Gamma (k_{1}-y)}{\Gamma (k_{1})} \frac{\Gamma (k_{2}+1)}{\Gamma (k_{2}+1-y)}. \end{aligned}$$

Then, for any nonnegative integer n,

$$\begin{aligned} \frac{Q^{(n)}(0)}{n!} = P_{n}(h_{1},h_{2},\ldots ,h_{n}) = \sum _{k_{1} \le \ell _{1} \le \ell _{2} \le \cdots \le \ell _{n} \le k_{2}} \frac{1}{\ell _{1} \ell _{2} \cdots \ell _{n}} \end{aligned}$$

with \(h_{n} = \sum _{j=k_{1}}^{k_{2}} \frac{1}{j^{n}}\).

Proposition 5.2

Let p be a nonnegative integer and \(\mathbb J(1)\) be equal to

$$\begin{aligned} \sum _{m+n=p} \frac{1}{m!n!} \int _{D_1} \left( \log \frac{1-t_{1}}{1-u_{2}} \right) ^{m} \left( \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{n} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

Then,

$$\begin{aligned} \mathbb J(1)&= \sum _{m+n=p}\sum _{c+d=n} \{ \zeta ^{\star }(c+2,\{1\}^{m},d+2) - \zeta (m+n+4) \} \\&= \left\{ \frac{p(p+3)}{2} + \frac{p+3}{2^{p+2}} \right\} \zeta (p+4). \end{aligned}$$

Proof

Let S(m, n) be the general term inside the summation of \(\mathbb J(1)\). Then,

$$\begin{aligned} \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } S(m,n) x^{m} y^{n}&= \int _{D_1} \left( \frac{1-t_{1}}{1-u_{2}} \right) ^{x} \left( \frac{t_{2}}{t_{1}} \right) ^{y} \left( \frac{u_{2}}{u_{1}} \right) ^{y} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}} \\&\equiv G({x,y}) \end{aligned}$$

is the generating function for the double sequence \(\{S(m,n)\}\) and

$$\begin{aligned} S(m,n) = \frac{1}{m!n!} \left( \frac{\partial }{\partial x} \right) ^{m} \left( \frac{\partial }{\partial y} \right) ^{n} G(x,y) \left. \right| _{x=y=0}. \end{aligned}$$

Beginning with

$$\begin{aligned} \frac{1}{(1-t_{1})^{1-x}} = \sum _{k=1}^{\infty } \frac{\Gamma (k-x)}{\Gamma (k) \Gamma (1-x)} t_{1}^{k-1}, \end{aligned}$$

and then integrating with respect to \(t_{1}\), \(t_{2}\) and \(u_{1}\), we obtain that

$$\begin{aligned} G(x,y) = \sum _{k=1}^{\infty } \sum _{\ell =1}^{\infty } \frac{1}{(k-y)k(k+\ell -y)} \frac{\Gamma (k-x)}{\Gamma (k) \Gamma (1-x)} \int _{0}^{1} (1-u_{2})^{-x} u_{2}^{k+\ell -1} \, du_{2}. \end{aligned}$$

The value of integral is

$$\begin{aligned} \frac{\Gamma (1-x) \Gamma (k+\ell )}{\Gamma (k+\ell +1-x)} \quad \text {or} \quad \frac{1}{k+\ell } \frac{\Gamma (1-x) \Gamma (k+\ell +1)}{\Gamma (k+\ell +1-x)}. \end{aligned}$$

So

$$\begin{aligned} G(x,y) = \sum _{k=1}^{\infty } \sum _{\ell =1}^{\infty } \frac{1}{k(k-y)(k+\ell )(k+\ell -y)} \frac{\Gamma (k-x) \Gamma (k+\ell +1)}{\Gamma (k) \Gamma (k+\ell +1-x)} \end{aligned}$$

after differentiations with respect to x, y for m, n times, and we use Lemma 5.1,

$$\begin{aligned} S(m,n) = \sum _{c+d=n} \left( \sum _{k=1}^{\infty } \sum _{\ell =1}^{\infty } \frac{1}{k^{c+2} (k+\ell )^{d+2}} \sum _{k \le \ell _{1} \le \cdots \le \ell _{m} \le k+\ell } \frac{1}{\ell _{1} \ell _{2} \cdots \ell _{m}}\right) . \end{aligned}$$

The general term of S(m, n) is

$$\begin{aligned} \zeta ^{\star }(c+2,\{1\}^{m},d+2) - \zeta (m+n+4) \end{aligned}$$

and hence

$$\begin{aligned} \mathbb J(1) = \sum _{m+n=p} \sum _{c+d=n} \{ \zeta ^{\star }(c+2,\{1\}^{m},d+2) - \zeta (m+n+4) \}. \end{aligned}$$

We use a result in [7, Theorem 2]:

$$\begin{aligned} \sum _{a+b+m=p}\zeta ^\star (a+2,\{1\}^m,b+2) =\left( p^2+3p+1+\frac{p+3}{2^{p+2}}\right) \zeta (p+4). \end{aligned}$$

Therefore we have

$$\begin{aligned} \mathbb J(1) = \left( \frac{p(p+3)}{2} + \frac{p+3}{2^{p+2}} \right) \zeta (p+4). \end{aligned}$$

\(\square \)

To transform

$$\begin{aligned} \mathbb J(2) = \sum _{c+d+m=p} (-1)^{m}\zeta (c+2,\{1\}^{m},d+2) \end{aligned}$$

into multiple zeta values related to \(\mathbb J(1)\), we need the following reflection formula.

Proposition 5.3

[7, Proposition 4] For an r-tuple \(\varvec{\alpha }=(\alpha _1,\alpha _2,\ldots ,\alpha _r)\) of positive integers with \(\alpha _1\ge 2\), \(\alpha _r\ge 2\), we have

$$\begin{aligned} \zeta (\alpha _1,\ldots ,\alpha _r)+(-1)^r\zeta ^\star (\alpha _r,\ldots ,\alpha _1) =\sum ^{r-1}_{k=1}(-1)^{k+1}\zeta ^\star (\alpha _k,\ldots ,\alpha _1) \zeta (\alpha _{k+1},\ldots ,\alpha _r). \end{aligned}$$
(5.1)

Proposition 5.4

Let p be a nonnegative integer and

$$\begin{aligned} \mathbb J(2) = \sum _{c+d+m=p} (-1)^{m} \zeta (c+2,\{1\}^{m},d+2). \end{aligned}$$

Then, we have

$$\begin{aligned} \mathbb J(2) = J_{p} - \sum _{c+d+m=p} \zeta ^{\star }(d+2,\{1\}^{m},c+2). \end{aligned}$$

Proof

By the reflection formula (see Proposition 5.3), we know that \(\zeta (c+2,\{1\}^{m},d+2)\) is equal to

$$\begin{aligned} \sum _{a+b=m} (-1)^{a} \zeta ^{\star }(\{1\}^{a},c+2) \zeta (\{1\}^{b},d+2) + (-1)^{m+1} \zeta ^{\star }(d+2,\{1\}^{m},c+2), \end{aligned}$$

so that

$$\begin{aligned} \mathbb J(2)&= \sum _{m+n=p}\left( (-1)^{m} \sum _{a+b=m} (-1)^{a} \sum _{c+d=n} \zeta ^{\star }(\{1\}^{a},c+2) \zeta (\{1\}^{b},d+2) \right. \\&\left. - \sum _{c+d=n} \zeta ^{\star }(d+2,\{1\}^{m},c+2) \right) \\&= \!\!\!\!\sum _{a+b+c+d=p} (-1)^{b} \zeta ^{\star }(\{1\}^{a},c+2) \zeta (\{1\}^{b},d+2) - \!\!\!\!\sum _{c+d+m=n} \zeta ^{\star }(d+2,\{1\}^{m},c+2). \end{aligned}$$

The first sum can be rewritten as:

$$\begin{aligned} \sum _{m+n=p} \sum _{b+d=m} (-1)^{b} \zeta (\{1\}^{b},d+2) \sum _{a+c=n} \zeta ^{\star }(\{1\}^{a},c+2), \end{aligned}$$

which is precisely equal to \(J_{p}\) expressed as:

$$\begin{aligned}&\sum _{m+n=p} \frac{1}{m!n!} \int _{E_{2}} \left( -\log \frac{1}{1-t_{1}} + \log \frac{t_{2}}{t_{1}} \right) ^{m} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \\&\times \int _{E_{2}} \left( \log \frac{1}{1-u_{2}} + \log \frac{u_{2}}{u_{1}} \right) ^{n} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

\(\square \)

According to previous propositions concerning \(\mathbb J(1)\) and \(\mathbb J(2)\), we have the following

Corollary 5.5

Notation as above, then we have

$$\begin{aligned} \mathbb J(1) + \mathbb J(2) = J_{p} - \frac{(p+1)(p+2)}{2} \zeta (p+4). \end{aligned}$$

Now we combine Eq. (4.2) and we conclude our main theorem.

6 An Application of \(\mathbb J(3)+\mathbb J(4)+\mathbb J(5)+\mathbb J(6)\)

Let us begin with another integral

$$\begin{aligned} \mathbb I(p) =\frac{1}{p!}\int _{E_2\times E_2}\left( \log \frac{1-t_1}{1-t_2} +\log \frac{t_2}{t_1}+\log \frac{u_2}{u_1}\right) ^p \frac{dt_1dt_2}{(1-t_1)t_2}\frac{du_1du_2}{(1-u_1)u_2}, \end{aligned}$$

where p is a nonnegative integer. Since

$$\begin{aligned}&\frac{1}{m!}\int _{0<t_1<t_2<1} \left( \log \frac{1-t_1}{1-t_2}+\log \frac{t_2}{t_1}\right) ^m \frac{dt_1}{1-t_1}\frac{dt_2}{t_2}\\&=\zeta ^\star (\{1\}^m,2)=(m+1)\zeta (m+2), \end{aligned}$$

the integral \(\mathbb I(p)\) can be written as:

$$\begin{aligned} \sum _{m+n=p}(m+1)\zeta (m+2)\zeta (n+2)\quad \text {or}\quad \frac{p+2}{2}\sum _{m+n=p}\zeta (m+2)\zeta (n+2). \end{aligned}$$

On the other hand, we decomposed the integral \(\mathbb I(p)\) into six parts in a similar way to \(\mathbb J_p\):

$$\begin{aligned} \mathbb I(p)=\sum ^6_{j=1}\mathbb I(p;j), \end{aligned}$$

where

$$\begin{aligned} \mathbb I(p;j)=\frac{1}{p!}\int _{D_j} \left( \log \frac{1-t_1}{1-t_2} +\log \frac{t_2}{t_1}+\log \frac{u_2}{u_1}\right) ^p \frac{dt_1dt_2}{(1-t_1)t_2}\frac{du_1du_2}{(1-u_1)u_2}. \end{aligned}$$

The first integral \(\mathbb I(p;1)\) is on \(D_1: 0<t_1<t_2<u_1<u_2<1\), and we have

$$\begin{aligned} \mathbb I(p;1)=\sum _{a+b+c=p} \sum _{\mid \varvec{\alpha }\mid =a+b+1}\zeta (\alpha _0,\ldots ,\alpha _{a-1},\alpha _a+1,c+2). \end{aligned}$$

The integral \(\mathbb I(p;2)\) is

$$\begin{aligned} \mathbb I(p;2)=\sum _{a+b+c=p} \sum _{\mid \varvec{\beta }\mid =b+c+1}\zeta (a+2,\beta _0,\ldots ,\beta _{b-1},\beta _b+1). \end{aligned}$$

The next four parts have corresponding equality with the parts \(\mathbb J(i)\), for \(3\le i\le 6\), as

$$\begin{aligned} \mathbb I(p;3)=\mathbb J(4),\quad \mathbb J(p;4)=\mathbb J(3),\quad \mathbb J(p;5)=\mathbb J(6),\quad \mathbb J(p;6)=\mathbb J(5). \end{aligned}$$

Therefore, we get

$$\begin{aligned} \sum _{j=3}^6\mathbb I(p;j)=\sum ^6_{j=3}\mathbb J(j)=\frac{(p+1)(p+2)}{2}\zeta (p+4). \end{aligned}$$

This gives us the following theorem.

Theorem 6.1

Let p be an nonnegative integer. Then,

$$\begin{aligned}&\sum _{a+b+c=p} \left( \sum _{\mid \varvec{\alpha }\mid =a+b+1}\zeta (\alpha _0,\ldots ,\alpha _{a-1},\alpha _a+1,c+2)\right. \\&\left. +\sum _{\mid \varvec{\beta }\mid =b+c+1}\zeta (a+2,\beta _0,\ldots ,\beta _{b-1},\beta _b+1)\right) \\&\qquad \qquad =\frac{p+2}{2}\sum _{m+n=p}\zeta (m+2)\zeta (n+2)-\frac{(p+1)(p+2)}{2}\zeta (p+4). \end{aligned}$$

Here, we also give another proof using a duality theorem due to Ohno [19]. Given any nonnegative integer a, we know that the dual of \(\zeta (\{1\}^a,2,2)\) is \(\zeta (2,a+2)\) and the dual of \(\zeta (2,\{1\}^a,2)\) is \(\zeta (a+2,2)\), respectively. Then, for a nonnegative integer m, we have

$$\begin{aligned} \sum _{\mid \varvec{c}\mid =m}\zeta (c_1+1,\ldots ,c_a+1,c_{a+1}+2,c_{a+2}+2)&=\sum _{\mid \varvec{d}\mid =m}\zeta (d_1+2,d_2+a+2),\\ \sum _{\mid \varvec{c}\mid =m}\zeta (c_1+2,c_2+1,\ldots ,c_{a+1}+1,c_{a+2}+2)&=\sum _{\mid \varvec{d}\mid =m}\zeta (d_1+a+2,d_2+2). \end{aligned}$$

We substitute the above formulas in \(\mathbb I(p;1)+\mathbb I(p;2)\), and then, the desired result will be obtained.

7 Another Expression of \(\mathbb J(3)\)

The sum of multiple zeta values

$$\begin{aligned} \mathbb J(3) = \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =p+3} \zeta (\alpha _{0},\alpha _{1},\ldots ,\alpha _{m},\alpha _{m+1}+1) \sum _{a+b+c=m} W(a,b,c) \end{aligned}$$

came from a joint of three sums

$$\begin{aligned}&\sum _{m+n=p} \sum _{a+b+c=m} \sum _{u+v+w=n} 2^{v} \\&\times \sum _{\begin{array}{c} \mid \varvec{\alpha }\mid =a+u+1 \\ \mid \varvec{\beta }\mid =b+v+1 \\ \mid \varvec{\gamma }\mid =c+w+1 \end{array}} \zeta (\alpha _{0}, \alpha _{1}, \ldots , \alpha _{a}, \beta _{0}, \beta _{1}, \ldots , \beta _{b}+\gamma _{0}, \ldots , \gamma _{c}+1) \end{aligned}$$

and it has the integral representation

$$\begin{aligned} \frac{1}{p!} \int _{D_{3}} \left( \log \frac{1-t_{1}}{1-u_{2}} + \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{p} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}, \end{aligned}$$

which is equal to

$$\begin{aligned} \sum _{m+n=p} \frac{1}{m!n!} \int _{D_{3}} \left( \log \frac{1-t_{1}}{1-u_{2}} \right) ^{m} \left( \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{t_{1}} \right) ^{n} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

Let T(m, n) be the general term in the above sum. Here, we are going to find the value of T(m, n) through its generating function.

Proposition 7.1

For integers \(m,n \ge 0\), let

$$\begin{aligned} T(m,n) = \frac{1}{m!n!} \int _{D_{3}} \left( \log \frac{1-t_{1}}{1-u_{2}} \right) ^{m} \left( \log \frac{t_{2}}{t_{1}} + \log \frac{u_{2}}{u_{1}} \right) ^{n} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

Then,

$$\begin{aligned} T(m,n) = \sum _{\mid \varvec{\alpha }\mid =n+3} \{ \zeta ^{\star }(\alpha _{1},\{1\}^{m},\alpha _{2}+1) - \zeta (m+n+4) \} (2^{\alpha _{2}-1}-1). \end{aligned}$$

Proof

The generating function for the double sequence T(m, n) is

$$\begin{aligned} \sum _{m=0}^{\infty } \sum _{n=0}^{\infty } T(m,n) x^{m} y^{n}&\equiv G(x,y) \\&= \int _{D_{3}} \left( \frac{1-t_{1}}{1-u_{2}} \right) ^{x} \left( \frac{t_{2}}{t_{1}} \right) ^{y} \left( \frac{u_{2}}{u_{1}} \right) ^{y} \frac{dt_{1} dt_{2}}{(1-t_{1})t_{2}} \frac{du_{1} du_{2}}{(1-u_{1})u_{2}}. \end{aligned}$$

Like the case of \(\mathbb J(1)\), G(x, y) can be evaluated as:

$$\begin{aligned} \sum _{k=1}^{\infty } \sum _{\ell =1}^{\infty } \frac{1}{(k-y)(k+\ell -2y)(k+\ell -y)(k+\ell )} \frac{\Gamma (k-x)}{\Gamma (k)} \frac{\Gamma (k+\ell +1)}{\Gamma (k+\ell +1-x)}{.} \end{aligned}$$

Hence,

$$\begin{aligned} T(m,n)&= \left. \frac{1}{m\!n\!} \left( \frac{\partial }{\partial x} \right) ^{m} \left( \frac{\partial }{\partial y} \right) ^{n} G(x,y) \right| _{x=y=0} \\&= \sum _{a+b+c=n} \sum _{k=1}^{\infty } \sum _{\ell =1}^{\infty } \frac{2^{b}}{k^{a+1} (k+\ell )^{b+c+3}} \sum _{k \le k_{1} \le k_{2} \le \cdots \le k_{m} \le k+\ell } \frac{1}{k_{1} k_{2} \cdots k_{m}}. \end{aligned}$$

It is equal to

$$\begin{aligned} \sum _{\mid \varvec{\alpha }\mid =n+3} \{ \zeta ^{\star }(\alpha _{1},\{1\}^{m},\alpha _{2}+1) - \zeta (m+n+4) \} (2^{\alpha _{2}-1}-1), \end{aligned}$$

as asserted. \(\square \)

Corollary 7.2

For any nonnegative integer p, we have

$$\begin{aligned} \mathbb J(3) = \sum _{m+n=p} \sum _{\mid \varvec{\alpha }\mid =n+3} \Bigl \{ \zeta ^{\star }(\alpha _{1},\{1\}^{m},\alpha _{2}+1) - \zeta (p+4) \Bigr \} (2^{\alpha _{2}-1}-1). \end{aligned}$$

Remark 7.3

We have [7, Theorem 1] a duality theorem:

$$\begin{aligned} \sum _{\mid \varvec{\alpha }\mid =n+3} \zeta ^{\star }(\alpha _{1},\{1\}^{m},\alpha _{2}+1) = (m+n+3) \zeta (m+n+4). \end{aligned}$$

However, it is still unknown for the weighted sum

$$\begin{aligned} \sum _{\mid \varvec{\alpha }\mid =n+3} \zeta ^{\star }(\alpha _{1},\{1\}^{m},\alpha _{2}+1) 2^{\alpha _{2}}. \end{aligned}$$

However, our weighted sum formula did provide an approximate evaluation of weighted sum of zeta-star values of such kind.

In the next section, we will try give some weighted alternating Euler sums.

8 Weighted Alternating Euler Sum Formulas

The weighted alternating Euler sum formula

$$\begin{aligned} \sum _{m+n=p}2^n\zeta (m+1,\overline{n+1}) =\frac{p+1}{2}\zeta (p+2)+\zeta (p+1,\overline{1})+\zeta (\overline{p+2}) \end{aligned}$$

was obtained in [6, Eq. (5.3)]. Now we produce a more general formula which it covers the above.

Theorem 8.1

For any nonnegative integers p, q and a real number \(\lambda \), we have

$$\begin{aligned}&\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) \lambda ^{n-q}\zeta (\overline{m+1})\zeta (\overline{n+1}) \nonumber \\&=\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) (\lambda +1)^{n-q}\zeta (m+1,\overline{n+1})\nonumber \\& +\sum _{\begin{array}{c} m+n=p\\ a+b=q \end{array}}\left( {\begin{array}{c}m\\ a\end{array}}\right) \left( {\begin{array}{c}n\\ b\end{array}}\right) \lambda ^{m-a}(\lambda +1)^{n-b}\zeta (m+1,\overline{n+1}). \end{aligned}$$
(8.1)

Proof

It is known that [6, Eq. (2.3)]:

$$\begin{aligned} \zeta (\overline{p+1})=\frac{-1}{p!}\int ^1_0\left( \log \frac{1}{t}\right) ^p\frac{dt}{1+t}. \end{aligned}$$

We evaluate the following sum:

$$\begin{aligned} S(p)=\sum _{m+n=p}\lambda ^n\zeta (\overline{m+1})\zeta (\overline{n+1}) \end{aligned}$$

as an integral form

$$\begin{aligned} \frac{1}{p!}\int _{\begin{array}{c} 0<t_1<t_2<1\\ 0<t_2<t_1<1 \end{array}} \left( \log \frac{1}{t_1}+\lambda \log \frac{1}{t_2}\right) ^p \frac{dt_1dt_2}{(1+t_1)(1+t_2)}. \end{aligned}$$

This integral can be decomposed into two parts:

$$\begin{aligned} \mathbb K_1&= \frac{1}{p!}\int _{0<t_1<t_2<1} \left( \log \frac{1}{t_1}+\lambda \log \frac{1}{t_2}\right) ^p \frac{dt_1dt_2}{(1+t_1)(1+t_2)},\text { and}\\ \mathbb K_2&= \frac{1}{p!}\int _{0<t_2<t_1<1} \left( \log \frac{1}{t_1}+\lambda \log \frac{1}{t_2}\right) ^p \frac{dt_1dt_2}{(1+t_1)(1+t_2)}. \end{aligned}$$

It can be seen that

$$\begin{aligned} \mathbb K_1=\sum _{a+b=p}(\lambda +1)^b\zeta (a+1,\overline{b+1})\text { and } \mathbb K_2=\sum _{a+b=p}\lambda ^a(\lambda +1)^b\zeta (a+1,\overline{b+1}). \end{aligned}$$

Therefore, we have

$$\begin{aligned} \sum _{m+n=p}\lambda ^n\zeta (\overline{m+1})\zeta (\overline{n+1}) =\sum _{m+n=p}\zeta (m+1,\overline{n+1})\Bigl [ (\lambda +1)^n(\lambda ^m+1)\Bigr ]. \end{aligned}$$
(8.2)

We differential q times with \(\lambda \) in the above equation, and we obtain the final identity. \(\square \)

We list some examples. Let \(\lambda =0\) in Eq. (8.1). The following identity is obtained.

$$\begin{aligned}&\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) \zeta (m+1,\overline{n+1})\\&=\zeta (\overline{q+1})\zeta (\overline{p+1-q}) -\sum _{a+b=q}\left( {\begin{array}{c}p-a\\ b\end{array}}\right) \zeta (a+1,\overline{p+1-a}). \end{aligned}$$

Then,

$$\begin{aligned} \sum _{m+n=p}\zeta (m+1,\overline{n+1})&=\zeta (\overline{1})\zeta (\overline{p+1})-\zeta (1,\overline{p+1}) \quad \text {(for } q=0){.}\\ \sum _{m+n=p}n\zeta (m+1,\overline{n+1})&=\zeta (\overline{2})\zeta (\overline{p}) -p\zeta (1,\overline{p+1})-\zeta (2,\overline{p}) \quad \text {(for } q=1){.}\\ \sum _{m+n=p}\left( {\begin{array}{c}n\\ 2\end{array}}\right) \zeta (m+1,\overline{n+1})&= \zeta (\overline{3})\zeta (\overline{p-1})-\left( {\begin{array}{c}p\\ 2\end{array}}\right) \zeta (1,\overline{p+1})\\&\qquad -(p-1)\zeta (2,\overline{p})-\zeta (3,\overline{p-1}) \quad \text {(for } q=2){.} \end{aligned}$$

It is note that the first identity in the above formulas can be found in [22, Eq. (2.15)].

Let \(\lambda =-1\) in Eq. (8.1). Then,

$$\begin{aligned}&\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) (-1)^{m}\zeta (\overline{m+1})\zeta (\overline{n+1})\\&=(-1)^{p+q}\zeta (p+1-q,\overline{q+1}) +\sum _{a+b=q}\left( {\begin{array}{c}p-b\\ a\end{array}}\right) \zeta (p+1-b,\overline{b+1}). \end{aligned}$$

The \(q=0\) and \(q=1\) cases are

$$\begin{aligned} \sum _{m+n=p}(-1)^m\zeta (\overline{m+1})\zeta (\overline{n+1})&= \left[ 1+(-1)^p\right] \zeta (p+1,\overline{1}),\\ \sum _{m+n=p}(-1)^mn\zeta (\overline{m+1})\zeta (\overline{n+1})&= \left[ 1+(-1)^{p+1}\right] \zeta (p,\overline{2})+p\zeta (p+1,\overline{1}). \end{aligned}$$

Let \(\lambda =1\) in Eq. (8.1), we have

$$\begin{aligned}&\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) \zeta (\overline{m+1})\zeta (\overline{n+1})\\&=\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) 2^{n-q}\zeta (m+1,\overline{n+1}) +\sum _{\begin{array}{c} m+n=p\\ a+b=q \end{array}}\left( {\begin{array}{c}m\\ a\end{array}}\right) \left( {\begin{array}{c}n\\ b\end{array}}\right) 2^{n-b}\zeta (m+1,\overline{n+1}). \end{aligned}$$

We set \(q=0\) in the above identity; we will get [6, Eq. (2.14)]

$$\begin{aligned} \sum _{m+n=p}\zeta (\overline{m+1})\zeta (\overline{n+1}) =\sum _{m+n=p}2^{n+1}\zeta (m+1,\overline{n+1}). \end{aligned}$$

The well-known double-shuffle relation [22, Eq. (2.2)]

$$\begin{aligned} \zeta (\overline{r})\zeta (\overline{s}) =\zeta (\overline{r},\overline{s})+\zeta (\overline{s},\overline{r})+\zeta (r+s), \end{aligned}$$
(8.3)

where \(r\ge 1\), \(s\ge 1\). We use the above relation and the sum formula [22, Eq. (2.14)]

$$\begin{aligned} \sum _{m+n=p}\zeta (\overline{m+1},\overline{n+1}) =\zeta (\overline{1})\zeta (p+1)-\zeta (\overline{1},p+1), \end{aligned}$$

we derive the following

$$\begin{aligned} \sum _{m+n=p}2^n\zeta (m+1,\overline{n+1}) =\frac{p+1}{2}\zeta (p+2)+\zeta (p+1,\overline{1})+\zeta (\overline{p+2}). \end{aligned}$$

This identity was obtained in [6, Eq. (5.3)]. We indicate this formula in the first paragraph of this section.

Let \(\lambda =-2\) in Eq. (8.1), we have

$$\begin{aligned} \sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) (-1)^m2^{n-q}&\zeta (\overline{m+1})\zeta (\overline{n+1}) =\sum _{m+n=p}\left( {\begin{array}{c}n\\ q\end{array}}\right) (-1)^{m}\zeta (m+1,\overline{n+1})\\&\qquad +\sum _{\begin{array}{c} m+n=p\\ a+b=q \end{array}}\left( {\begin{array}{c}m\\ a\end{array}}\right) \left( {\begin{array}{c}n\\ b\end{array}}\right) 2^{m-a}\zeta (m+1,\overline{n+1}). \end{aligned}$$

We set \(q=0\) in the above identity; we have

$$\begin{aligned} \sum _{m+n=p}(-1)^m2^n\zeta (\overline{m+1})\zeta (\overline{n+1}) =\sum _{m+n=p}\Bigl [2^{m}+(-1)^m\Bigr ]\zeta (m+1,\overline{n+1}). \end{aligned}$$

Using the double-shuffle relation Eq (8.3), we have

$$\begin{aligned} \frac{(-2)^{p+1}-1}{3}\zeta (p+2)&=\sum _{m+n=p}[(-2)^n+(-2)^m]\zeta (\overline{m+1},\overline{n+1})\\&\qquad -\sum _{m+n=p}(-1)^n[1+(-2)^m]\zeta (m+1,\overline{n+1}). \end{aligned}$$