# Integral Evaluation of Odd Euler Sums, Multiple $t$-value
$t\left(3,2,\ldots,2\right)$ and Multiple zeta value $\zeta(3,2,\ldots,2)$

Sarth Chavan  ,  Masato Kobayashi  and  Jorge Layja  Euler Circle, Palo Alto,
California 94306<EMAIL_ADDRESS>Department of Engineering
Kanagawa University, 3-27-1 Rokkaku-bashi, Yokohama 221-8686, Japan.
<EMAIL_ADDRESS>Department of Mathematics, Unversidad De Las Amèricas
Puebla, Cholula, Mexico.

###### Abstract.

We construct an analytic approach to evaluate odd Euler sums, multiple zeta
value $\zeta(3,2,\ldots,2)$ and multiple $t$-value
$t\left(3,2,\ldots,2\right)$. Moreover, we also conjecture a closed expression
for multiple $t$-value $t\left(2,\ldots,2,1\right)$.

###### Key words and phrases:

Euler sums; Polylogarithm; Multiple zeta values; Multiple $t$-values; Riemann
zeta function; Inverse sine function; Inverse hyperbolic tangent function.

###### 2020 Mathematics Subject Classification:

Primary: 11M$32$; Secondary: 11M$06$; 40B$05$

##  1\. Introduction and Main Results

###  1.1. Euler sums

The classical _Euler sums_ are infinite series

(1.1)
$H\left(p_{1},p_{2},\ldots,p_{k},q\right)=\sum_{n=1}^{+\infty}\frac{H_{n}^{\left(p_{1}\right)}H_{n}^{\left(p_{2}\right)}\ldots
H_{n}^{(p_{k})}}{n^{q}}$

where $p_{1},p_{2},\ldots,p_{k}\in\mathbb{N}$, $q\geqslant 2$ and
$H_{n}^{\left(p\right)}=\sum_{k=1}^{n}k^{-p}$ is the generalized harmonic
number.

This theory indeed dates back to Euler and Goldbach, while recent research on
this topic has been quite active. For an early introduction and study on the
evaluations of classical Euler sums, the readers may consult in Flajolet and
Salvy’s paper [13], in which they have developed a contour integral
representation approach to the evaluation of Euler sums. For some recent
progress, the readers are referred to [19],[20], [21] and references therein.

In particular, _linear Euler sums_ $H(p,q)$ with $p,q\geqslant 2$ satisfy the
_reflection formula_

$H(p,q)+H(q,p)=\zeta(p)\,\zeta(q)+\zeta(p+q)$

as Vălean has discussed in [18]. As a result of which, in the evaluation of
$H(p,q)$, Riemann zeta values $\zeta(p)$ with $p\geqslant 2$ often show up
with rational coefficients.

###  1.2. Multiple zeta values and multiple $t$-values

_Multiple zeta values_ (MZVs) are real numbers, originally defined by Euler,
that have been much studied in recent years because of their many surprising
properties and the many places they appear in mathematics and mathematical
physics, ranging from periods of mixed Tate motives [4] to values of Feynman
integrals in perturbative quantum field theory.

In fact, multiple zeta values (MZVs) are simply a variant of the Euler sums.
For positive integers $i_{1},i_{2}\ldots,i_{k}$ where $i_{1}\geqslant 2$, the
infinite series

(1.2)
$\zeta\left(i_{1},i_{2},\ldots,i_{k}\right)=\sum_{\begin{subarray}{c}n_{1}>n_{2}>\cdots>n_{k}\geqslant
1\end{subarray}}\frac{1}{n_{1}^{i_{1}}n_{2}^{i_{2}}\cdots n_{k}^{i_{k}}}$

is usually called a _multiple zeta value_. Weight of the index
$\left(i_{1},i_{2},\ldots,i_{k}\right)$ is $i_{1}+i_{2}+\cdots+i_{k}$ and its
depth is $k$. The study of multiple zeta values began in the early 1990s with
the works of Hoffman [9] and Zagier [24], and has attracted a lot of research
in the last two decades. For detailed history and applications, the readers
may consult the book of Zhao [25].

One of the main problems in this topic is to understand the
$\mathbf{Q}$-linear dependence of MZVs. Some MZVs are a combination of single
zeta values while according to Brown, there certainly seems to exist
exceptions such as $\zeta(5,3)$ [4]. We usually call the MZVs
$\zeta(i_{1},i_{2},\ldots,i_{k})$ with $i_{j}\in\\{2,3\\}$ _Hoffman elements_.
In particular, F. Brown [4] proved that every multiple zeta value is a
$\mathbf{Q}$-linear combination of Hoffman elements. Moreover, Zagier [23]
showed explicit relations among Hoffman elements in the form
$\zeta(2,\ldots,2,3,2,\ldots,2)$.

In a recent paper [10], Hoffman introduced and studied an _odd_ variant of the
MZVs:

(1.3)
$t\left(i_{1},i_{2},\ldots,i_{k}\right)=\sum_{\begin{subarray}{c}n_{1}>n_{2}>\cdots>n_{k}\geqslant
1\\\
n_{j}\,\mathrm{odd}\end{subarray}}\frac{1}{n_{1}^{i_{1}}n_{2}^{i_{2}}\ldots
n_{k}^{i_{k}}}=\sum_{n_{1}>n_{2}>\cdots>n_{k}\geqslant
1}\dfrac{1}{\prod_{j=1}^{k}\left(2n_{j}-1\right)^{i_{j}}}$

which are called the _multiple t-values_ (MtVs). As showed in [10, Corollaries
4.1 and 4.2], the MtVs are reducible to a linear combinations of alternating
(or colored) MZVs.

There are indeed many conjectures concerning the arithmetic nature of these
numbers. Let $(d_{n})_{n\geqslant 1}$ be the _Padovan sequence_ defined by
$d_{1}=0,d_{2}=1,d_{3}=1$ and $d_{n}=d_{n-2}+d_{n-3}$ for $n\geqslant 4$ and
let $(f_{n})_{n\geqslant 1}$ be the Fibonacci sequence with $f_{1}=f_{2}=1$,
$f_{n}=f_{n-1}+f_{n-2}$ for $n\geqslant 3$. Let $\mathcal{Z}_{n}$ and
$\mathcal{T}_{n}$ denote the $\mathbf{Q}$-span of all MZVs and MtVs of weight
$n$ respectively. Then, we have the following conjectures by M. Hoffmann [10],
B. Saha [16] and D. Zagier [23].

######  Conjecture 1.1.

  1. (1)

(Zagier [23]) $\dim\mathcal{Z}_{n}=d_{n}$.

  2. (2)

(Hoffman [10]) $\dim\mathcal{T}_{n}=f_{n}$.

  3. (3)

(Hoffman [10]) The following set of weight $n$ MZVs forms a $\mathbf{Q}$-basis
of $\mathcal{Z}_{n}$.

$\\{\zeta(i_{1},i_{2},\ldots,i_{k})\mid
i_{1}+i_{2}+\cdots+i_{k}=n,i_{j}\in\\{2,3\\}\\}.$

  4. (4)

(Saha [16]) The following set of weight $n$ MtVs forms a $\mathbf{Q}$-basis of
$\mathcal{T}_{n}$ for $n\geqslant 2$.

$\\{t\left(a_{1}+1,a_{2},\dots,a_{r}\right)\mid
a_{1}+a_{2}+\cdots+a_{r}=n-1,a_{i}\in\\{1,2\\},1\leqslant r\leqslant n-1\\}.$

We often discuss MZVs and MtVs independently since the algebra of multiple
$t$-values is quite different in some ways from the algebra of multiple zeta
values. Both the duality theorem and the _double shuffle relations_ [11] of
MZVs are missing for MtVs. However, there should be rich connections between
these two theories. In this article, we show that we can compute
$\zeta(3,2,\ldots,2)$ and $t\left(3,2,\ldots,2\right)$ using the same method.

###  1.3. Central binomial sums

_Central binomial sums_ play an important role in number theory. Informally
speaking, it is an infinite series involving $\binom{2n}{n}$. Lehmer [14]
discussed the following two types of such sums:

$\textrm{I}.\,\sum_{n=0}^{+\infty}a_{n}\binom{2n}{n},\quad\textrm{II}.\,\sum_{n=0}^{+\infty}a_{n}\binom{2n}{n}^{-1}.$

Moreover, he also presented connections between such series and Maclaurin
series of $\arcsin\left(z\right)$ and $\arcsin^{2}\left(z\right)$. Some other
examples include

$\sum_{n=1}^{{+\infty}}\frac{1}{n^{2}\binom{2n}{n}}=\frac{1}{3}\,\zeta(2),\quad\sum_{n=1}^{{+\infty}}\frac{(-1)^{n-1}}{n^{3}\binom{2n}{n}}=\frac{2}{5}\,\zeta(3),\quad\sum_{n=1}^{{+\infty}}\frac{1}{n^{4}\binom{2n}{n}}=\frac{17}{36}\,\zeta(4)$

as they arise in the work of Apèry [1] and van der Poorten [17] to prove
irrationality of $\zeta(3)$. Along this line, it is natural to consider an
application of such sums for MZVs and MtVs.

###  1.4. Main results

Our main results are at the crossroads of these three topics. Moreover, our
idea throughout is more analytic; we seek Euler sums, MZVs and MtVs by
_integral evaluation_. This is indeed a powerful method as seen in Doelder [6]
and Vălean [18].

In Section 2 and Section 3, $t\left(3,2,\ldots,2\right)$ and
$\zeta(3,2,\ldots,2)$ are evaluated using the inverse sine integral (2.1) and
the nested sum representations for integer powers of the inverse sine function
[2, Equations (1) and (2)]. In particular, it is shown that
$\zeta(3,2,\ldots,2)$ and $t\left(3,2,\ldots,2\right)$ are rational
polynomials of $\pi$ and odd zeta values $\zeta(2k+1)$ with $k\geqslant 1$.

In Section 4, _multiple mixed value_
$\mu\left(i_{k},i_{k-1},\ldots,i_{1}\right)$ is defined and the Maclaurin
series expansion for integer powers of inverse hyperbolic tangent function [8,
(6.3)] is used to evaluate $\mu\left(2,1,\ldots,1\right)$. Finally, in Section
5, _odd Euler sums_ $\mathcal{O}\left(p,q\right)$ and
$\mathcal{B}\left(p,q\right)$ are defined and a reflection formula for
$\mathcal{O}\left(p,q\right)$ and $\mathcal{B}\left(p,q\right)$ is established
to evaluate $\mathcal{O}\left(q,q\right)$ and $\mathcal{B}\left(q,q\right)$.
Later, an analytic approach is implemented to compute
$\mathcal{O}\left(p,q\right)$ and $\mathcal{B}\left(p,q\right)$ for small
values of $p$ and $q$. Moreover, integral reduction of
$\mathcal{O}\left(p,q\right)$ and $\mathcal{B}\left(p,q\right)$ are also
obtained.

Beyond these specific cases, we hope that the tools used in this study will be
useful to the readers in their endeavour of evaluation of more interesting
series and integrals.

###  1.5. Notation

Let

$\displaystyle\left(2n\right)!!=\prod_{k=0}^{n-1}\left(2n-2k\right)\,\,\,\text{and}\,\,\,\left(2n-1\right)!!=\prod_{k=0}^{n-1}\left(2n-2k-1\right).$

Moreover, for notational convenience, let

$\zeta(3,\underbrace{2,2,\ldots,2}_{n\,\text{times}})=\zeta\left(3,\\{2\\}^{n}\right)\,\,\,\text{and}\,\,\,t\,(3,\underbrace{2,2,\ldots,2}_{n\,\text{times}})=t\left(3,\\{2\\}^{n}\right).$

##  2\. Evaluation of Multiple $t$–value $t\left(3,2,\ldots,2\right)$

Let us first begin with several Definitions and Propositions.

######  Definition 2.1.

Let $\mathbf{R}[[z]]$ denote the set of convergent power series with real
coefficients. For $f(z)\in\mathbf{R}[[z]]$, define
$\mathbb{W}:\mathbf{R}[[z]]\to\mathbf{R}[[z]]$ to be

$\mathbb{W}f(z)=\int_{0}^{1}\frac{f(xz)}{\sqrt{1-x^{2}}}\,\mathrm{d}x.$

It is now helpful to understand the function $\mathbb{W}$ coefficientwise.

######  Proposition 2.2.

Let $f(z)\in\mathbf{R}[[z]]$. Moreover, suppose it can be expressed in the
form

$f(z)=\sum_{k=0}^{{+\infty}}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}\,\mathcal{C}_{2k+1}\,z^{2k+1}.$

Then, the following identity holds.

$\mathbb{W}f(z)=\sum_{k=0}^{{+\infty}}\frac{\mathcal{C}_{2k+1}}{2k+1}\,z^{2k+1}.$

###### Proof.

From Definition 2.1, it follows that

$\mathbb{W}f(z)=\sum_{k=0}^{+\infty}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}\,\mathcal{C}_{2k+1}\,z^{2k+1}{\int_{0}^{1}\frac{x^{2n+1}}{\sqrt{1-x^{2}}}\,\mathrm{d}x}$
$=\sum_{k=0}^{+\infty}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}\frac{\left(2k\right)!!}{\left(2k-1\right)!!}\,\mathcal{C}_{2k+1}\,z^{2k+1}=\sum_{k=0}^{{+\infty}}\frac{\mathcal{C}_{2k+1}}{2k+1}\,z^{2k+1}$

which is the desired result. ∎

######  Proposition 2.3.

Let $f(z)\in\mathbf{R}[[z]]$ and $\alpha\in\mathbf{R}$. If $f(0)=0$, then we
have

$\mathbb{W}\left({\int_{0}^{\alpha}{\frac{f(z)}{z}}\,\mathrm{d}z}\right)=\int_{0}^{1}{\frac{f(\alpha
x)\arccos{\left(x\right)}}{x}}\,\mathrm{d}x.$

###### Proof.

Notice that, if $\alpha=0$, then both sides are 0. Otherwise, we have

$\displaystyle\mathbb{W}\left({\int_{0}^{\alpha}{\frac{f(z)}{z}}\,\mathrm{d}z}\right)$
$\displaystyle=\int_{0}^{1}{\int_{0}^{\alpha
u}{\frac{f(z)}{z}}}\frac{1}{\sqrt{1-u^{2}}}\,\mathrm{d}z\,\mathrm{d}u$
$=\int_{0}^{\alpha}{{\frac{f(z)}{z}}\int_{z/\alpha}^{1}\frac{1}{\sqrt{1-u^{2}}}}\,\mathrm{d}u\,\mathrm{d}z=\int_{0}^{\alpha}{\frac{f(z)}{z}}\arccos\left({\frac{z}{\alpha}}\right)\mathrm{d}z=\int_{0}^{1}{\frac{f(\alpha
x)\arccos{\left(x\right)}}{x}}\,\mathrm{d}x$

as desired. ∎

######  Proposition 2.4 (J.M. Borwein–Chamberland[2, (1.3), (1.4)]).

For $|x|\leqslant 2$ and $N\geqslant 0$

$\frac{1}{\left(2N+1\right)!}\left[\arcsin\left(\dfrac{x}{2}\right)\right]^{2N+1}=\sum_{k=0}^{{+\infty}}\frac{G_{N}(k)}{2^{2k}\left(2k+1\right)}\binom{2k}{k}\left({\frac{x}{2}}\right)^{2k+1}$

where $G_{0}(k)=1$ and

$G_{N}(k)=\sum_{n_{1}=0}^{k-1}\frac{1}{\left(2n_{1}+1\right)^{2}}\sum_{n_{2}=0}^{n_{1}-1}\frac{1}{\left(2n_{2}+1\right)^{2}}\,\,\,\cdots\sum_{n_{N}=0}^{n_{N-1}-1}\frac{1}{\left(2n_{N}+1\right)^{2}}.$

######  Remark 2.5.

Notice that the boundary $|x|=\pm 2$ is inclusive here.

######  Proposition 2.6.

Let

(2.1)
$I(N)=\int_{0}^{1}\dfrac{\arcsin^{N}(z)}{z}\,\mathrm{d}z\quad\left(N\in\mathbb{N}\right)$

Then, the following identities hold.

(2.2)
$I(2N+1)=\dfrac{\left(2N+1\right)!}{2^{2N+1}}\sum_{j=0}^{N}\dfrac{\left(-1\right)^{j}\pi^{2N+1-2j}}{\left(2N+1-2j\right)!}\,\eta\left(2j+1\right)$
(2.3)
$I(2N)=\dfrac{\left(2N\right)!}{2^{2N}}\left(\sum_{j=0}^{N-1}\dfrac{\left(-1\right)^{j}\pi^{2N-2j}}{\left(2N-2j\right)!}\,\eta\left(2j+1\right)+\left(-1\right)^{N}2\left(1-2^{-2N-1}\right)\zeta(2N+1)\right)$

where $\eta\left(m\right)=\left(1-2^{1-m}\right)\zeta(m)$ represents the
Dirichlet eta function with $\eta\left(1\right)=\log(2)$.

###### Proof.

Buhler and Crandall in their very recent article[5, page 280] proved that

$J(n)=\displaystyle\int_{0}^{1/2}{z^{n}\cot\left(\pi z\right)}\,\mathrm{d}z$
$=\frac{n!}{2^{n}}\left({\sum_{1\leqslant k\leqslant
n,\,k\,\textrm{odd}}\frac{\left(-1\right)^{(k-1)/2}\eta\left(k\right)}{\pi^{k}\left(n-k+1\right)!}}\right)+\cos\left(\frac{n\pi}{2}\right)\frac{\left(4n!\right)\left(1-2^{-n-1}\right)}{\left(2\pi\right)^{n+1}}\,\zeta(n+1).$

Note that we corrected the sign $\frac{1+\left(-1\right)^{n}}{2}$ to
$\cos\left(\frac{n\pi}{2}\right)$. Substituting $x=\sin\left(\pi z\right)$
produces

$J(n)=\int_{0}^{1/2}{z^{n}\cot\left(\pi
z\right)}\,\mathrm{d}z=\frac{1}{\pi^{n+1}}\displaystyle\int_{0}^{1}{\frac{\arcsin^{n}(x)}{x}}\,\mathrm{d}z=\frac{I(n)}{\pi^{n+1}}.$

Writing down the cases for $n=2N+1$ and $n=2N$ with $k=2j+1$ yields

$I(2N+1)=\dfrac{\left(2N+1\right)!}{2^{2N+1}}\sum_{j=0}^{N}\dfrac{\left(-1\right)^{j}\pi^{2N+1-2j}}{\left(2N+1-2j\right)!}\,\eta\left(2j+1\right),$
$I(2N)=\dfrac{\left(2N\right)!}{2^{2N}}\left(\sum_{j=0}^{N-1}\dfrac{\left(-1\right)^{j}\pi^{2N-2j}}{\left(2N-2j\right)!}\,\eta\left(2j+1\right)+\left(-1\right)^{N}2\left(1-2^{-2N-1}\right)\zeta(2N+1)\right)$

which is the desired result. ∎

######  Corollary 2.7.

First few values of $I(N)$ are as follows.

(2.4) $I(1)=\dfrac{\pi}{2}\,\log\left(2\right),$ (2.5)
$I(2)=\dfrac{\pi^{2}}{4}\,\log\left(2\right)-\dfrac{7}{8}\,\zeta(3),$ (2.6)
$I(3)=\dfrac{\pi^{3}}{8}\,\log\left(2\right)-\dfrac{9\pi}{16}\,\zeta(3),$
(2.7)
$I(4)=\dfrac{\pi^{4}}{16}\,\log\left(2\right)-\dfrac{9\pi^{2}}{16}\,\zeta(3)+\dfrac{93}{32}\,\zeta(5),$
(2.8)
$I(5)=\dfrac{\pi^{5}}{32}\,\log\left(2\right)-\dfrac{15\pi^{3}}{32}\,\zeta(3)+\dfrac{225\pi}{64}\,\zeta(5),$
(2.9)
$I(6)=\dfrac{\pi^{6}}{64}\,\log\left(2\right)-\dfrac{45\pi^{4}}{128}\,\zeta(3)+\dfrac{675\pi^{2}}{128}\,\zeta(5)-\dfrac{5715}{256}\,\zeta(7),$
(2.10)
$I(7)=\dfrac{\pi^{7}}{128}\log\left(2\right)-\dfrac{63\pi^{5}}{256}\,\zeta(3)+\dfrac{1575\pi^{3}}{256}\,\zeta(5)-\dfrac{19845\pi}{512}\,\zeta(7),$
(2.11)
$I(8)=\dfrac{\pi^{8}}{256}\log\left(2\right)-\dfrac{21\pi^{6}}{128}\,\zeta(3)+\dfrac{1575\pi^{4}}{256}\,\zeta(5)-\dfrac{19845\pi^{2}}{256}\,\zeta(7)+\dfrac{160965}{512}\,\zeta(9).$

######  Theorem 2.8.

For $N\geqslant 1$, we have

$t\left(3,\\{2\\}^{N}\right)$
$=\frac{1}{2^{2N+2}}\left[\sum_{j=1}^{N}\frac{\left(-1\right)^{j+1}\left(2j\right)\pi^{2N+2-2j}}{\left(2N+2-2j\right)!}\,\eta(2j+1)+\left(-1\right)^{N}2\left(2N+2\right)\left(1-2^{-2N-3}\right)\zeta(2N+3)\right].$

###### Proof.

From Proposition 2.4, it follows that

$\frac{\arcsin^{2N+1}(x/2)}{(2N+1)!}=\sum_{k>n_{1}>\cdots>n_{N}\geqslant
0}\frac{1}{\left(2k+1\right)\left(2n_{1}+1\right)^{2}\ldots\left(2n_{N}+1\right)^{2}}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}\left({\frac{x}{2}}\right)^{2k+1}$

where we have simply used the identity

$\dfrac{1}{2^{2k}}\dbinom{2k}{k}=\frac{\left(2k-1\right)!!}{\left(2k\right)!!}$

For notational convenience, let $z=x/2$. Therefore we have

$\frac{\arcsin^{2N+1}(z)}{\left(2N+1\right)!}=\sum_{k>n_{1}>\cdots>n_{N}\geqslant
0}\frac{1}{\left(2k+1\right)\left(2n_{1}+1\right)^{2}\ldots\left(2n_{N}+1\right)^{2}}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}\,z^{2k+1}.$

Dividing both sides by $z$ and integrating termwise from $0$ to $\alpha$ with
respect to $z$ produces

$\displaystyle\int_{0}^{\alpha}{\frac{\arcsin^{2N+1}(z)}{z\left(2N+1\right)!}}\,\mathrm{d}z=\sum_{k>n_{1}>\cdots>n_{N}\geqslant
0}\frac{1}{\left(2k+1\right)^{2}\left(2n_{1}+1\right)^{2}\ldots\left(2n_{N}+1\right)^{2}}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}\,\alpha^{2k+1}.$

Next, we use Proposition 2.2 to deduce that

$\mathbb{W}\left({\displaystyle\int_{0}^{\alpha}{\frac{\arcsin^{2N+1}(z)}{z\left(2N+1\right)!}}\,\mathrm{d}z}\right)=\sum_{k>n_{1}>\cdots>n_{N}\geqslant
0}\frac{1}{\left(2k+1\right)^{3}\left(2n_{1}+1\right)^{2}\ldots\left(2n_{N}+1\right)^{2}}\,\alpha^{2k}$

Substituting $\alpha=1$ and combining Proposition 2.3 and 2.6, we finally
conclude that

$t\left(3,\\{2\\}^{N}\right)=\frac{1}{(2N+1)!}\displaystyle\int_{0}^{1}{\frac{\arcsin^{2N+1}(z)\arccos\left(z\right)}{z}}\,\mathrm{d}z=\dfrac{\pi}{2}\left(\frac{I(2N+1)}{\left(2N+1\right)!}\right)-\dfrac{I(2N+2)}{\left(2N+1\right)!}$
$=\frac{1}{2^{2N+2}}\left[\sum_{j=1}^{N}\frac{\left(-1\right)^{j+1}\left(2j\right)\pi^{2N+2-2j}}{\left(2N+2-2j\right)!}\,\eta(2j+1)+\left(-1\right)^{N}2\left(2N+2\right)\left(1-2^{-2N-3}\right)\zeta(2N+3)\right]$

which is the desired result. This completes the proof of Theorem 2.8. ∎

######  Corollary 2.9.

Substituting $N=2,3$ in Theorem 2.8 produces

$t\left(3,2,2\right)=\frac{\pi}{2}\left(\dfrac{I(5)}{120}\right)-\dfrac{I(6)}{120}=\dfrac{\pi^{4}}{1024}\,\zeta(3)-\dfrac{15\pi^{2}}{512}\,\zeta(5)+\dfrac{381}{2048}\,\zeta(7)=0.002109185\ldots$
$t\left(3,2,2,2\right)=\dfrac{\pi^{6}}{122880}\,\zeta(3)-\dfrac{5\pi^{4}}{8192}\,\zeta(5)+\dfrac{189\pi^{2}}{16384}\,\zeta(7)+\dfrac{511}{8192}\,\zeta(9)=0.00005499616\ldots$

##  3\. Evaluation of Multiple Zeta Value $\zeta(3,2,\ldots,2)$

It turns out that we can use the same method as in the previous section to
evaluate multiple zeta value $\zeta(3,2,\ldots,2)$. Although this is merely a
special case of Zagier’s work [23, p.981, Theorem 1], we present our proof
since our approach is quite different.

######  Proposition 3.1 (Borwein–Chamberland[2, (1.1), (1.2)]).

For $|x|\leqslant 2$ and $N\geqslant 1$, we have

$\frac{1}{\left(2N\right)!}\left[\arcsin{\left(\dfrac{x}{2}\right)}\right]^{2N}=\sum_{k=1}^{{+\infty}}\frac{H_{N}(k)}{\binom{2k}{k}}\frac{x^{2k}}{k^{2}}$

where $H_{1}(k)=1/4$ and

$H_{N+1}(k)=\frac{1}{4}\sum_{n_{1}=1}^{k-1}\frac{1}{\left(2n_{1}\right)^{2}}\sum_{n_{2}=1}^{n_{1}-1}\frac{1}{\left(2n_{2}\right)^{2}}\,\,\,\cdots\sum_{n_{N}=1}^{n_{N-1}-1}\frac{1}{\left(2n_{N}\right)^{2}}.$

######  Proposition 3.2.

Let $f(z)\in\mathbf{R}[[z]]$. Moreover, suppose it can be expressed in the
form

$f(z)=\sum_{k=0}^{{+\infty}}{\frac{\left(2k\right)!!}{\left(2k-1\right)!!}}\,\mathcal{C}_{2k}\,z^{2k}.$

Then, the following identity holds.

$\mathbb{W}f(z)=\frac{\pi}{2}\sum_{k=0}^{{+\infty}}\mathcal{C}_{2k}\,z^{2k}.$

###### Proof.

The proof is quite similar to that of Proposition 2.2. Notice that

$\int_{0}^{1}\frac{z^{2k}}{\sqrt{1-z^{2}}}\,\mathrm{d}z=\frac{\pi}{2}\frac{\left(2k-1\right)!!}{\left(2k\right)!!}$

which explains the appearance of $\pi/2$ in $\mathbb{W}f(z)$. ∎

######  Theorem 3.3.

For $N\geqslant 0$, we have

$\zeta\left(3,\\{2\\}^{N}\right)$
$=2\left[\sum_{j=1}^{N}\frac{\left(-1\right)^{j+1}\left(2j\right)\pi^{2N+2-2j}}{\left(2N+3-2j\right)!}\,\eta\left(2j+1\right)-\left(-1\right)^{N}\left[1-\left(1-2^{-2N-2}\right)\left(2N+2\right)\right]\zeta(2N+3)\right].$

###### Proof.

From Proposition 3.1, it follows that

$\frac{1}{\left(2N+2\right)!}\left[\arcsin\left(\dfrac{x}{2}\right)\right]^{2N+2}=\frac{1}{2^{2N+2}}\sum_{k>n_{1}>\cdots>n_{N}}\frac{1}{k^{2}\,n_{1}^{2}\,n_{2}^{2}\ldots
n_{N}^{2}}\frac{\left(2k\right)!!}{\left(2k-1\right)!!}\left({\frac{x}{2}}\right)^{2k}.$

For notational convenience, let $z=x/2$, therefore we have

$\frac{\arcsin^{2N+2}(z)}{\left(2N+2\right)!}=\frac{1}{2^{2N+2}}\sum_{k>n_{1}>\cdots>n_{N}}\frac{1}{k^{2}\,n_{1}^{2}\,n_{2}^{2}\ldots
n_{N}^{2}}\frac{\left(2k\right)!!}{\left(2k-1\right)!!}\,z^{2k}.$

Next, we divide both sides by $z$ and integrate them termwise from 0 to
$\alpha$ to get

$\displaystyle\int_{0}^{\alpha}{\frac{\arcsin^{2N+2}(z)}{z\left(2N+2\right)!}}\,\mathrm{d}z=\frac{1}{2^{2N+3}}\sum_{k>n_{1}>\cdots>n_{N}}\frac{1}{k^{3}\,n_{1}^{2}\,n_{2}^{2}\ldots
n_{N}^{2}}\frac{\left(2k\right)!!}{\left(2k-1\right)!!}\,\alpha^{2k}.$

Next, we use Proposition 3.2 to deduce that

$\mathbb{W}\left({\displaystyle\int_{0}^{\alpha}{\frac{\arcsin^{2N+2}(z)}{z\left(2N+2\right)!}}\,\mathrm{d}z}\right)=\frac{\pi}{2^{2N+4}}\sum_{k>n_{1}>\cdots>n_{N}}\frac{1}{k^{3}\,n_{1}^{2}\,n_{2}^{2}\ldots
n_{N}^{2}}\,\alpha^{2k}.$

Substituting $\alpha=1$ and combining Proposition 2.3, Proposition 2.6 finally
produces

$\zeta\left(3,\\{2\\}^{N}\right)=\frac{2^{2N+4}}{\left(2N+2\right)!}\left[\dfrac{I(2N+2)}{2}-\dfrac{I(2N+3)}{\pi}\right]$
$=2\left[\sum_{j=1}^{N}\frac{\left(-1\right)^{j+1}\left(2j\right)\pi^{2N+2-2j}}{\left(2N+3-2j\right)!}\,\eta\left(2j+1\right)-\left(-1\right)^{N}\left[1-\left(1-2^{-2N-2}\right)\left(2N+2\right)\right]\zeta(2N+3)\right]$

as desired. This completes the proof of Theorem 3.3. ∎

######  Corollary 3.4.

Substituting $N=1,2,3$ in Throem 3.3 produces

$\zeta(3,2)=\dfrac{\pi^{2}}{2}\,\zeta(3)-\dfrac{11}{2}\,\zeta(5)=0.22881039\ldots,$
$\zeta(3,2,2)=\dfrac{\pi^{4}}{40}\,\zeta(3)-\dfrac{5\pi^{2}}{4}\,\zeta(5)+\dfrac{157}{16}\,\zeta(7)=0.02912562\ldots,$
$\zeta(3,2,2,2)=\dfrac{\pi^{6}}{1680}\,\zeta(3)-\dfrac{\pi^{4}}{16}\,\zeta(5)+\dfrac{63\pi^{2}}{32}\,\zeta(7)-\dfrac{223}{16}\,\zeta(9)=0.00252145\ldots.$

We have also verified these identities numerically at the computational
website EZ-Face [7].

######  Corollary 3.5.

Moreover

$\zeta\left(3,\\{2\\}^{N}\right)\in\mathbf{Q}\left[\pi,\zeta(3),\zeta(5),\dots,\zeta(2N+3)\right]_{2N+3}$

where $\mathbf{Q}\left[\pi,\zeta(3),\zeta(5),\ldots,\zeta(2N+3)\right]_{2N+3}$
is the set of all elements of degree $2N+3$ in the rational polynomial ring in
$\pi,\zeta(3),\zeta(5),\zeta(7),\ldots,\zeta(2N+3)$ with grading
$\deg\left(\pi\right)=1$ and $\deg\left(\zeta(2j+1)\right)=2j+1$. Indeed, the
same is true for multiple $t$-value $t\left(3,\\{2\\}^{N}\right)$.

##  4\. Evaluation of Multiple Mixed Value $\mu\left(2,1,\ldots,1\right)$

######  Definition 4.1.

For positive integers $i_{1},i_{2}\ldots,i_{k}$, define

$\mu\left(i_{k},i_{k-1},\ldots,i_{1}\right)=\sum_{\begin{subarray}{c}n_{k}>n_{k-1}>\cdots>n_{1}\\\
n_{j}\,\,\equiv\,\,j\,\,\textrm{(mod
$2$)}\end{subarray}}\dfrac{1}{n_{k}^{i_{k}}n_{k-1}^{i_{k-1}}\ldots
n_{1}^{i_{1}}}.$

Notice the little reversal of indices. If $i_{k}\geqslant 2$, this infinite
sum is certainly convergent since it is a partial sum of
$\zeta(i_{k},i_{k-1},\ldots,i_{1})$. Let us call this sum a _multiple mixed
value_.

It is not so immediate to evaluate such sums. However, quite recently,
Guo–Lim–Qi in their preprint [8], announced that they found Maclaurin series
expansion for integer powers of inverse hyperbolic tangent function which is
helpful in the evaluation $\mu\left(2,1,\ldots,1\right)$.

######  Proposition 4.2 (Guo–Lim–Qi [8, (6.3)]).

For $N\geqslant 0$, $\ell_{N}=k$ and $|z|<1$, we have

(4.1)
$\frac{\textnormal{arctanh}^{N}(z)}{N!}=\sum_{k=0}^{+\infty}\left(\prod_{m=1}^{N-1}\sum_{\ell_{m}=0}^{\ell_{m+1}}\dfrac{1}{2\ell_{m}+m}\right)\dfrac{z^{2k+N}}{2k+N}=\sum_{\begin{subarray}{c}n_{N}>n_{N-1}>\cdots>n_{1}\\\
n_{j}\,\,\equiv\,\,j\,\,\textnormal{(mod
$2$)}\end{subarray}}\dfrac{z^{n_{N}}}{n_{N}n_{N-1}\ldots n_{1}}.$

######  Theorem 4.3.

For $N\geqslant 1$, the following identity holds

$\mu\left(2,\\{1\\}^{N-1}\right)=\dfrac{2^{N+1}-1}{2^{2N}}\,\zeta(N+1).$

###### Proof.

Integrating equation (4.1) with respect to $z$ from $0$ to $1$ produces

$\mu\left(2,\\{1\\}^{N-1}\right)=\dfrac{1}{N!}\int_{0}^{1}\dfrac{\textrm{arctanh}^{N}(z)}{z}\,\mathrm{d}z.$

Let us call this integral $K(N)$. We can indeed evaluate $K(N)$ by simply
using the identity

$\mathrm{arctanh}\left(z\right)=-\dfrac{1}{2}\log\left(\dfrac{1-z}{1+z}\right).$

We have

$K(N)=\int_{0}^{1}\dfrac{\mathrm{arctanh}^{N}(z)}{z}\,\mathrm{d}z=\dfrac{\left(-1\right)^{N}}{2^{N}}\int_{0}^{1}\dfrac{1}{z}\log^{N}\left(\dfrac{1-z}{1+z}\right)\mathrm{d}z=\dfrac{\left(-1\right)^{N}}{2^{N-1}}\int_{0}^{1}\dfrac{\log^{N}(z)}{1-z^{2}}\,\mathrm{d}z$

$=\dfrac{\left(-1\right)^{N}}{2^{N-1}}\sum_{k=0}^{+\infty}\int_{0}^{1}z^{2k}\log^{N}(z)\mathrm{d}z=\dfrac{\left(-1\right)^{N}}{2^{N-1}}\sum_{k=0}^{+\infty}\dfrac{\left(-1\right)^{N}N!}{\left(2k+1\right)^{N+1}}=\dfrac{N!\left[2^{N+1}-1\right]}{2^{2N}}\,\zeta(N+1).$

Therefore, putting all things together produces the desired result. ∎

######  Example 4.4.

Substituting $N=1,2,3$ in Theorem 4.3 produces

$\mu\left(2,1\right)=\frac{7}{16}\,\zeta(3),\,\,\,\mu\left(2,1,1\right)=\frac{15}{64}\,\zeta(4)=\dfrac{\pi^{4}}{384},\,\,\,\mu\left(2,1,1,1\right)=\frac{31}{256}\,\zeta(5).$

Indeed, $K(2N)$ appears to be a part (in fact, the boundary term) of $I(2N)$,
that is

(4.2)
$I(2N)=\dfrac{\left(2N\right)!}{2^{2N}}\left(\sum_{j=0}^{N-1}\dfrac{\left(-1\right)^{j}\pi^{2N-2j}}{\left(2N-2j\right)!}\,\eta\left(2j+1\right)\right)+\left(-1\right)^{N}K(2N).$

Notice that Hoffman [10, Appendix] highlights the relations

$t\left(2,1\right)=t\left(2\right)\log(2)-\frac{1}{2}\,t\left(3\right),$
$t\left(2,2,1\right)=\frac{1}{8}\,t\left(5\right)-\frac{1}{14}\,t\left(2\right)t\left(3\right)+\frac{1}{4}\,t\left(4\right)\log(2),$
$t\left(2,2,2,1\right)=-\frac{1}{32}\,t\left(7\right)-\frac{3}{56}\,t\left(3\right)t\left(4\right)+\frac{15}{248}\,t\left(2\right)t\left(5\right)+\frac{1}{48}\,t\left(6\right)\log(2).$

With equations (2.5), (2.7), (2.9) and using
$t\left(i\right)=\left(1-2^{-i}\right)\zeta(i)$, we observe that

$t\left(2,1\right)=\frac{1}{2}\left({\frac{\pi^{2}}{4}\,\log(2)-\frac{7}{8}\,\zeta(3)}\right)=\frac{I(2)}{2!},$
$t\left(2,2,1\right)=\frac{1}{24}\left({\frac{\pi^{4}}{16}\,\log(2)-\frac{9\pi^{2}}{16}\,\zeta(3)+\frac{93}{32}\,\zeta(5)}\right)=\frac{I(4)}{4!},$
$t\left(2,2,2,1\right)=\frac{1}{720}\left({\dfrac{\pi^{6}}{64}\,\log(2)-\dfrac{45\pi^{4}}{128}\,\zeta(3)+\dfrac{675\pi^{2}}{128}\,\zeta(5)-\dfrac{5715}{256}\,\zeta(7)}\right)=\frac{I(6)}{6!}.$

This pattern seems to continue further. However, we have not figured out why
yet.

######  Conjecture 4.5.

Let $N\in\mathbb{N}$, then the following identity holds.

$t\left(\\{2\\}^{N},1\right)=\dfrac{I(2N)}{\left(2N\right)!}=\dfrac{1}{2^{2N}}\left(\sum_{j=0}^{N-1}\dfrac{\left(-1\right)^{j}\pi^{2N-2j}}{\left(2N-2j\right)!}\,\eta\left(2j+1\right)+\left(-1\right)^{N}2\left(1-2^{-2N-1}\right)\zeta(2N+1)\right).$

Notice that $t\left(\\{2\\}^{N},1\right)$ is a MtV appearing in Saha’s
conjecture [16].

##  5\. Odd Euler Sums

###  5.1. Definitions

Let us begin with several definitions and notation. Set

$\mathcal{O}_{n}\left(p\right)=\sum_{k=1}^{n}\frac{1}{(2k-1)^{p}}\,\,\,\,\text{and}\,\,\,\,\mathcal{B}_{n}\left(p\right)=\sum_{k=1}^{n}\frac{(-1)^{k}}{(2k-1)^{p}}.$

Notice that

$\mathcal{O}\left(p\right)=\mathcal{O}_{\infty}\left(p\right)=\left(1-2^{-p}\right)\zeta(p)\,\,\,\text{and}\,\,\,\mathcal{B}\left(p\right)=\mathcal{B}_{\infty}\left(p\right)=\beta\left(p\right)$

where $\beta$ denotes the _Dirichlet beta function_. In particular,
$\beta\left(2\right)=\mathcal{G}$ (the Catalan constant).

######  Definition 5.1.

Call each of

$\mathcal{O}\left(p,q\right)=\sum_{n=1}^{+\infty}\frac{\mathcal{O}_{n}\left(p\right)}{(2n-1)^{q}}\,\,\,\,\text{and}\,\,\,\,\mathcal{B}\left(p,q\right)=\sum_{n=1}^{+\infty}\frac{(-1)^{n}\,\mathcal{B}_{n}\left(p\right)}{(2n-1)^{q}}$

an _odd Euler sum_.

###  5.2. Reflection formula for Odd Euler Sums and Evaluation of
$\mathcal{O}\left(q,q\right)$ and $\mathcal{B}\left(q,q\right)$

Euler sums $\\{H(p,q)\\}$ satisfy reflection formula. Here is a natural analog
for odd Euler sums.

######  Proposition 5.2.

The following identities hold.

(5.1)
$\mathcal{O}\left(p,q\right)+\mathcal{O}\left(q,p\right)=\mathcal{O}\left(p\right)\mathcal{O}\left(q\right)+\mathcal{O}\left(p+q\right),$
(5.2)
$\mathcal{B}\left(p,q\right)+\mathcal{B}\left(p,q\right)=\beta\left(p\right)\beta\left(q\right)+\mathcal{B}\left(p+q\right).$

###### Proof.

Since the double sum

$\sum_{k}\sum_{n}\frac{1}{\left(2k-1\right)^{p}\left(2n-1\right)^{q}}$

is absolutely convergent, we can switch the order of summation to get

$\mathcal{O}\left(p,q\right)=\sum_{n=1}^{{+\infty}}\left({\sum_{k=1}^{n}\frac{1}{\left(2k-1\right)^{p}}}\right)\frac{1}{(2n-1)^{q}}=\sum_{k=1}^{{+\infty}}\sum_{n=k}^{{+\infty}}\frac{1}{\left(2n-1\right)^{q}}\frac{1}{\left(2k-1\right)^{p}}$

$=\sum_{k=1}^{{+\infty}}\left({\sum_{n=1}^{{+\infty}}-\sum_{n=1}^{k-1}}\right)\frac{1}{\left(2n-1\right)^{q}}\frac{1}{\left(2k-1\right)^{p}}=\sum_{k=1}^{{+\infty}}\frac{\left[\mathcal{O}\left(q\right)-\mathcal{O}_{k-1}\left(q\right)\right]}{\left(2k-1\right)^{p}}$

$=\mathcal{O}\left(p\right)\mathcal{O}\left(q\right)-\sum_{k=1}^{{+\infty}}\left({\mathcal{O}_{k}\left(q\right)-\frac{1}{\left(2k-1\right)^{q}}}\right)\frac{1}{\left(2k-1\right)^{p}}=\mathcal{O}\left(p\right)\mathcal{O}\left(q\right)-\mathcal{O}\left(q,p\right)+\mathcal{O}\left(p,q\right).$

Inserting the signs $(-1)^{k-1}$ and $(-1)^{n-1}$ appropriately, we can show
the other one (5.2). ∎

######  Corollary 5.3.

The reflection formula enables us to evaluate odd Euler sums for $p=q$.

(5.3)
$\mathcal{O}\left(q,q\right)=\dfrac{1}{2}\left[\left(1-2^{-2q}\right)\zeta(2q)+\left(1-2^{-q}\right)^{2}\zeta(q)\,\zeta(q)\right],$
(5.4)
$\mathcal{B}\left(q,q\right)=\dfrac{1}{2}\left[\left(1-2^{-2q}\right)\zeta(2q)+\beta\left(q\right)\beta\left(q\right)\right].$

######  Example 5.4.

The following identities hold.

$\mathcal{O}\left(2,2\right)=\frac{1}{2}\left({\frac{15}{16}\,\zeta(4)+\frac{9}{16}\,\zeta^{2}(2)}\right)=\frac{5\pi^{4}}{384},\,\,\,\mathcal{B}\left(3,3\right)=\frac{1}{2}\left({\frac{31}{32}\,\zeta(6)+\beta^{2}\left(3\right)}\right)=\dfrac{1937\pi^{6}}{1935360}.$

It is thus natural to ask whether it is possible to evaluate
$\mathcal{O}\left(p,q\right)$ or $\mathcal{B}\left(p,q\right)$ for $p\neq q$.
We spend some time discussing this question in the next two subsections.

###  5.3. Evaluation of $\mathcal{O}\left(p,q\right)$

We begin by evaluating $\mathcal{O}\left(2,3\right)$ by transforming
$\mathcal{O}_{n}\left(p\right)$ to an integral and then transforming the
initial double sum $\mathcal{O}\left(p,q\right)$ to a double integral followed
by its reduction to single integrals involving the logarithm and polylogarithm
function.

Next, we show that our approach can be generalized to obtain an integral
reduction for $\mathcal{O}\left(p,q\right)$. Let’s start with a variant of
$\mathcal{O}\left(2,2\right)$:

$\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}\sum_{k=0}^{n-1}\dfrac{1}{\left(2k+1\right)^{2}}=-\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}\int_{0}^{1}\left[\sum_{k=0}^{n-1}z^{2k}\right]\log\left(z\right)\mathrm{d}z$

$=-\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}\int_{0}^{1}\dfrac{\log\left(z\right)}{1-z^{2}}\,\mathrm{d}z+\int_{0}^{1}\left[\sum_{n=1}^{+\infty}\dfrac{z^{2n}}{\left(2n+1\right)^{2}}\right]\dfrac{\log\left(z\right)}{1-z^{2}}\,\mathrm{d}z$

$=\dfrac{3}{4}\,\zeta(2)\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}-\int_{0}^{1}\left[\int_{0}^{1}\sum_{n=1}^{+\infty}\left(xz\right)^{2n}\log\left(x\right)\mathrm{d}x\right]\dfrac{\log\left(z\right)}{1-z^{2}}\,\mathrm{d}z$

$=\dfrac{3}{4}\,\zeta(2)\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}-\int_{0}^{1}x^{2}\log\left(x\right)\int_{0}^{1}\dfrac{z^{2}\log\left(z\right)}{\left(1-x^{2}z^{2}\right)\left(1-z^{2}\right)}\,\mathrm{d}z\,\mathrm{d}x$

$=\dfrac{3}{4}\zeta(2)\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}-\int_{0}^{1}\dfrac{x\log(x)\mathrm{Li}_{2}(x)}{1-x^{2}}\mathrm{d}x+\dfrac{1}{4}\int_{0}^{1}\dfrac{x\log(x)\mathrm{Li}_{2}(x^{2})}{1-x^{2}}\mathrm{d}x+\dfrac{\pi^{2}}{8}\int_{0}^{1}\dfrac{x^{2}\log(x)}{1-x^{2}}\mathrm{d}x$
$=\dfrac{3}{4}\zeta(2)\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}-\int_{0}^{1}\dfrac{x\log(x)\mathrm{Li}_{2}(x)}{1-x^{2}}\mathrm{d}x+\dfrac{1}{4}\int_{0}^{1}\dfrac{x\log(x)\mathrm{Li}_{2}(x^{2})}{1-x^{2}}\mathrm{d}x-\dfrac{3}{4}\zeta(2)\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n+1\right)^{2}}$

$=\dfrac{1}{4}\int_{0}^{1}\dfrac{x\log(x)\mathrm{Li}_{2}(x^{2})}{1-x^{2}}\mathrm{d}x-\int_{0}^{1}\dfrac{x\log(x)\mathrm{Li}_{2}(x)}{1-x^{2}}\mathrm{d}x=\dfrac{1}{2}\int_{0}^{1}\dfrac{\log(x)\mathrm{Li}_{2}(x)}{1+x}\mathrm{d}x-\dfrac{7}{16}\int_{0}^{1}\dfrac{\log(x)\mathrm{Li}_{2}(x)}{1-x}\mathrm{d}x$
(5.5)
$=\dfrac{1}{2}\left[-\dfrac{3}{16}\,\zeta(4)\right]-\dfrac{7}{16}\left[-\dfrac{3}{4}\,\zeta(4)\right]=-\dfrac{3}{32}\,\zeta(4)+\dfrac{21}{64}\,\zeta(4)=\dfrac{15}{64}\,\zeta(4)=\frac{\pi^{4}}{384}.$

Similarly, we have

$\mathcal{O}\left(2,3\right)=\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n-1\right)^{3}}\sum_{k=1}^{n}\dfrac{1}{\left(2k-1\right)^{2}}=-\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n-1\right)^{3}}\int_{0}^{1}\left[\sum_{k=1}^{n}z^{2k}\right]\dfrac{\log\left(z\right)}{z^{2}}\,\mathrm{d}z$

$=-\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}(x)}{x^{2}}\int_{0}^{1}\left[\sum_{n=1}^{+\infty}x^{2n}\left(1-z^{2n}\right)\right]\dfrac{\log\left(z\right)}{1-z^{2}}\,\mathrm{d}z\,\mathrm{d}x$

$=-\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}(x)}{1-x^{2}}\int_{0}^{1}\dfrac{\log\left(z\right)}{1-x^{2}z^{2}}\,\mathrm{d}z\,\mathrm{d}x=\dfrac{1}{4}\int_{0}^{1}\dfrac{\log^{2}(x)\,\mathrm{Li}_{2}\left(x\right)}{x\left(1-x^{2}\right)}\mathrm{d}x-\dfrac{1}{4}\int_{0}^{1}\dfrac{\log^{2}(x)\,\mathrm{Li}_{2}\left(-x\right)}{x\left(1-x^{2}\right)}\mathrm{d}x$
(5.6)
$=\dfrac{1}{4}\left(\dfrac{11}{16}\,\zeta(5)+\dfrac{3}{4}\,\zeta(2)\,\zeta(3)\right)-\dfrac{1}{4}\left(-\dfrac{5}{4}\,\zeta(5)-\dfrac{3}{8}\,\zeta(2)\,\zeta(3)\right)=\dfrac{31}{64}\,\zeta(5)+\dfrac{9}{32}\,\zeta(3)\,\zeta(2).$

This approach can be easily generalized to the following.

######  Theorem 5.5.

Let $p,q\in\mathbb{N}$ and $\mathrm{Li}_{n}(z)$ represent the polylogarithm,
then we have

$\mathcal{O}\left(p,q\right)=\dfrac{\left(-1\right)^{q}}{2\left(q-1\right)!}\left[\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(-x\right)}{x\left(1-x^{2}\right)}\,\mathrm{d}x-\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(x\right)}{x\left(1-x^{2}\right)}\,\mathrm{d}x\right].$

###### Proof.

As earlier, we begin by transforming $\mathcal{O}_{n}\left(p\right)$ to an
integral

$\mathcal{O}\left(p,q\right)=\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n-1\right)^{q}}\sum_{k=1}^{n}\dfrac{1}{\left(2k-1\right)^{p}}=\sum_{n=1}^{+\infty}\dfrac{1}{\left(2n-1\right)^{q}}\dfrac{\left(-1\right)^{p-1}}{\left(p-1\right)!}\int_{0}^{1}\left[\sum_{k=1}^{n}z^{2k}\right]\dfrac{\log^{p-1}(z)}{z^{2}}\,\mathrm{d}z$

$=\dfrac{\left(-1\right)^{q-1}}{\left(q-1\right)!}\dfrac{\left(-1\right)^{p-1}}{\left(p-1\right)!}\int_{0}^{1}\dfrac{\log^{q-1}(x)}{x^{2}}\int_{0}^{1}\left[\sum_{n=1}^{+\infty}x^{2n}\left(1-z^{2n}\right)\right]\dfrac{\log^{p-1}(z)}{1-z^{2}}\,\mathrm{d}z\,\mathrm{d}x$

$=\dfrac{\left(-1\right)^{q-1}}{\left(q-1\right)!}\dfrac{\left(-1\right)^{p-1}}{\left(p-1\right)!}\int_{0}^{1}\dfrac{\log^{q-1}(x)}{1-x^{2}}\int_{0}^{1}\dfrac{\log^{p-1}(z)}{1-x^{2}z^{2}}\,\mathrm{d}z\,\mathrm{d}x$

$=\dfrac{\left(-1\right)^{q-1}}{2\left(q-1\right)!}\int_{0}^{1}\dfrac{\log^{q-1}(x)}{1-x^{2}}\left[\dfrac{\left[\mathrm{Li}_{p}\left(x\right)-\mathrm{Li}_{p}\left(-x\right)\right]}{x}\right]\mathrm{d}x$

$=\dfrac{\left(-1\right)^{q}}{2\left(q-1\right)!}\left[\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(-x\right)}{x\left(1-x^{2}\right)}\,\mathrm{d}x-\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(x\right)}{x\left(1-x^{2}\right)}\,\mathrm{d}x\right]$

as desired. ∎

######  Remark 5.6.

Note that Piscos Mathematica Package produces the integral evaluations

$\int_{0}^{1}\dfrac{\log^{3}\left(z\right)\mathrm{Li}_{3}\left(z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=-\dfrac{3\pi^{4}}{64}\,\zeta(3)+\dfrac{5\pi^{2}}{16}\,\zeta(5)-\dfrac{489}{128}\,\zeta(7),$

$\int_{0}^{1}\dfrac{\log^{3}\left(z\right)\mathrm{Li}_{3}\left(-z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=\dfrac{3\pi^{4}}{64}\,\zeta(3)-\dfrac{5\pi^{2}}{32}\,\zeta(5)+\dfrac{273}{128}\,\zeta(7),$

$\int_{0}^{1}\dfrac{\log^{4}\left(z\right)\mathrm{Li}_{4}\left(z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=\dfrac{\pi^{4}}{24}\,\zeta(5)+\dfrac{35\pi^{2}}{32}\,\zeta(7)+\dfrac{579}{64}\,\zeta(9),$

$\int_{0}^{1}\dfrac{\log^{4}\left(z\right)\mathrm{Li}_{4}\left(-z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=-\dfrac{7\pi^{4}}{192}\,\zeta(5)-\dfrac{35\pi^{2}}{64}\,\zeta(7)-\dfrac{477}{32}\,\zeta(9),$

$\int_{0}^{1}\dfrac{\log^{5}\left(z\right)\mathrm{Li}_{5}\left(z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=-\dfrac{15\pi^{6}}{128}\,\zeta(5)+\dfrac{7\pi^{4}}{32}\,\zeta(7)+\dfrac{315\pi^{2}}{64}\,\zeta(9)-\dfrac{18825}{256}\,\zeta(11),$

$\int_{0}^{1}\dfrac{\log^{5}\left(z\right)\mathrm{Li}_{5}\left(-z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=\dfrac{15\pi^{6}}{128}\,\zeta(5)-\dfrac{49\pi^{4}}{256}\,\zeta(7)-\dfrac{315\pi^{2}}{128}\,\zeta(9)+\dfrac{1485}{32}\,\zeta(11),$

$\int_{0}^{1}\dfrac{\log^{6}\left(z\right)\mathrm{Li}_{6}\left(z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=\dfrac{\pi^{6}}{24}\,\zeta(7)+\dfrac{21\pi^{4}}{16}\,\zeta(9)+\dfrac{3465\pi^{2}}{128}\,\zeta(11)+\dfrac{72855}{256}\,\zeta(13),$

$\int_{0}^{1}\dfrac{\log^{6}\left(z\right)\mathrm{Li}_{6}\left(-z\right)}{z\left(1-z^{2}\right)}\,\mathrm{d}z=-\dfrac{31\pi^{6}}{768}\,\zeta(7)-\dfrac{147\pi^{4}}{128}\,\zeta(9)-\dfrac{3465\pi^{2}}{256}\,\zeta(11)-\dfrac{222885}{512}\,\zeta(13).$

Combining these integral evaluations with Theorem 5.5 yields

$\mathcal{O}\left(3,4\right)=\dfrac{\pi^{4}}{128}\,\zeta(3)-\dfrac{5\pi^{2}}{128}\,\zeta(5)+\dfrac{127}{256}\,\zeta(7),$

$\mathcal{O}\left(4,5\right)=\dfrac{5\pi^{4}}{3072}\,\zeta(5)+\dfrac{105\pi^{2}}{3072}\,\zeta(7)+\dfrac{511}{1024}\,\zeta(9),$

$\mathcal{O}\left(5,6\right)=\dfrac{\pi^{6}}{1024}\,\zeta(5)-\dfrac{7\pi^{4}}{4096}\,\zeta(7)-\dfrac{63\pi^{2}}{2048}\,\zeta(9)+\dfrac{2047}{4096}\,\zeta(11),$
$\mathcal{O}\left(6,7\right)=\dfrac{7\pi^{6}}{122880}\,\zeta(7)+\dfrac{7\pi^{4}}{4096}\,\zeta(9)+\dfrac{231\pi^{2}}{8192}\,\zeta(11)+\dfrac{8191}{16384}\,\zeta(13).$

Consequently, using the reflection formula we have

$\mathcal{O}\left(4,3\right)=\dfrac{\pi^{4}}{728}\,\zeta(3)+\dfrac{5\pi^{2}}{128}\,\zeta(5)+\dfrac{127}{256}\,\zeta(7),$

$\mathcal{O}\left(5,4\right)=\dfrac{13\pi^{4}}{1536}\,\zeta(5)-\dfrac{105\pi^{2}}{3072}\,\zeta(7)+\dfrac{511}{1024}\,\zeta(9),$

$\mathcal{O}\left(6,5\right)=\dfrac{\pi^{6}}{30720}\,\zeta(5)+\dfrac{7\pi^{4}}{4096}\,\zeta(7)+\dfrac{63\pi^{2}}{2048}\,\zeta(9)+\dfrac{2047}{4096}\,\zeta(11),$

$\mathcal{O}\left(7,6\right)=\dfrac{\pi^{6}}{1024}\,\zeta(7)-\dfrac{7\pi^{4}}{4096}\,\zeta(9)-\dfrac{231\pi^{2}}{8192}\,\zeta(11)+\dfrac{8191}{16384}\,\zeta(13).$

However, note that in these specific values of $\mathcal{O}\left(p,q\right)$,
no simple pattern can be discerned.

###  5.4. Evaluation of $\mathcal{B}\left(p,q\right)$

We begin by evaluating $\mathcal{B}\left(2,3\right)$ the same way we evaluated
$\mathcal{O}\left(2,3\right)$ in the previous subsection. Next, we show that
our approach can be generalized to obtain an integral reduction for
$\mathcal{B}\left(p,q\right)$. Let’s begin with $\mathcal{B}\left(2,3\right)$:

$\mathcal{B}\left(2,3\right)=\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{3}}\sum_{k=1}^{n}\dfrac{\left(-1\right)^{k-1}}{\left(2k-1\right)^{2}}=\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{3}}\int_{0}^{1}\left[\sum_{k=1}^{n}\left(-z^{2}\right)^{k}\right]\dfrac{\log\left(z\right)}{z^{2}}\,\mathrm{d}z$

$=\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{3}}\int_{0}^{1}\dfrac{\left(-z^{2}\right)^{n}\log\left(z\right)}{1+z^{2}}\,\mathrm{d}z-\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)^{3}}\int_{0}^{1}\dfrac{\log\left(z\right)}{1+z^{2}}\,\mathrm{d}z$

$=-\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)}{x^{2}}\int_{0}^{1}\left[\sum_{n=1}^{+\infty}\left(-x^{2}\right)^{n}\left(-z^{2}\right)^{n}\right]\dfrac{\log\left(z\right)}{1+z^{2}}\,\mathrm{d}z\,\mathrm{d}x+\dfrac{\mathcal{G}\pi^{3}}{32}$

$=-\dfrac{\mathcal{G}}{2}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)}{1+x^{2}}\,\mathrm{d}x-\dfrac{1}{8}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x^{2}\right)}{x\left(1+x^{2}\right)}+\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x+\dfrac{\mathcal{G}\pi^{3}}{32}$

$=\dfrac{\mathcal{G}\pi^{3}}{32}-\dfrac{1}{8}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x^{2}\right)}{x\left(1+x^{2}\right)}+\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x+\dfrac{\mathcal{G}\pi^{3}}{32}=\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x$

$-\dfrac{1}{8}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x^{2}\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x=\dfrac{1}{2}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x-\dfrac{1}{64}\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x\right)}\,\mathrm{d}x$

Some routine manipulations produce

(5.7)
$\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x\right)}\,\mathrm{d}x=\dfrac{83}{8}\,\zeta(5)-\dfrac{9}{2}\,\zeta(2)\,\zeta(3).$

Next, we have

$\int_{0}^{1}\dfrac{\log^{2}\left(x\right)\mathrm{Li}_{2}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x=-\int_{0}^{1}\int_{0}^{1}\dfrac{\log\left(z\right)\log^{2}\left(x\right)}{\left(1-xz\right)\left(1+x^{2}\right)}\,\mathrm{d}x\,\mathrm{d}z$

$=-2\int_{0}^{1}\dfrac{z\log\left(z\right)\mathrm{Li}_{2}\left(z\right)}{1+z^{2}}\,\mathrm{d}z-\dfrac{3}{16}\,\zeta(3)\int_{0}^{1}\dfrac{z\log\left(z\right)}{1+z^{2}}\,\mathrm{d}z-\dfrac{\pi^{3}}{16}\int_{0}^{1}\dfrac{\log\left(z\right)}{1+z^{2}}\,\mathrm{d}z$

$=\int_{0}^{1}\dfrac{\log\left(1+z^{2}\right)\mathrm{Li}_{3}\left(z\right)}{z}\,\mathrm{d}z-\int_{0}^{1}\dfrac{\log\left(z\right)\log\left(1+z^{2}\right)\mathrm{Li}_{2}\left(z\right)}{z}\,\mathrm{d}z+\dfrac{3}{128}\,\zeta(3)\,\zeta(2)+\dfrac{\mathcal{G}\pi^{3}}{16}$

$=\dfrac{1}{8}\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n}H_{2n}}{n^{4}}+\dfrac{1}{4}\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n}H^{\left(2\right)}_{2n}}{n^{3}}+\dfrac{1}{4}\,\zeta(2)\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n+1}}{n^{3}}+\dfrac{1}{2}\,\zeta(3)\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n+1}}{n^{2}}+\dfrac{\mathcal{G}\pi^{3}}{16}$
(5.8)
$+\dfrac{3}{128}\,\zeta(3)\,\zeta(2)=\dfrac{\mathcal{G}\pi^{3}}{16}-\dfrac{9}{16}\,\zeta(2)\,\zeta(3)+\dfrac{331}{256}\,\zeta(5)=\dfrac{\mathcal{G}\pi^{3}}{16}-\dfrac{3\pi^{2}}{32}\,\zeta(3)+\dfrac{331}{256}\,\zeta(5)$

where the arising non-linear alternating harmonic series are evaluated by C.
Vălean in [15]:

$\sum_{n=1}^{+\infty}\frac{\left(-1\right)^{n-1}H_{2n}^{(2)}}{n^{3}}=\frac{61\pi^{2}}{192}\,\zeta(3)+\frac{1973}{128}\zeta(5)+\frac{\pi^{5}}{16}-\frac{\pi}{128}\,\psi^{(3)}\left(\frac{1}{4}\right)$
$\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n-1}H_{2n}}{n^{4}}=-\frac{\pi^{2}}{3}\,\zeta(3)-\frac{437}{64}\,\zeta(5)-\frac{\pi^{5}}{24}+\frac{\pi}{192}\,\psi^{(3)}\left(\frac{1}{4}\right)$

Putting all things together produces

$\mathcal{B}\left(2,3\right)=\dfrac{331}{512}\,\zeta(5)-\dfrac{83}{512}\,\zeta(5)+\dfrac{3\pi^{2}}{256}\,\zeta(3)-\dfrac{3\pi^{2}}{64}\,\zeta(3)+\dfrac{\mathcal{G}\pi^{3}}{32}=\dfrac{31}{64}\,\zeta(5)-\dfrac{9\pi^{2}}{256}\,\zeta(3)+\dfrac{\mathcal{G}\pi^{3}}{32}$

where $\mathcal{G}$ is the Catalan constant. This approach can be easily
generalized to the following.

######  Theorem 5.7.

Let $p,q\in\mathbb{N}$ and $\mathrm{Li}_{n}(z)$ represent the polylogarithm,
then we have

$\mathcal{B}\left(p,q\right)=\dfrac{\left(-1\right)^{q}}{2\left(q-1\right)!}\left[\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(-x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x-\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x\right].$

###### Proof.

The proof is quite similar to that of Theorem 5.5, we have

$\mathcal{B}\left(p,q\right)=\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n}}{\left(2n-1\right)^{q}}\sum_{k=1}^{n}\dfrac{\left(-1\right)^{k}}{\left(2k-1\right)^{p}}=\sum_{n=1}^{+\infty}\dfrac{\left(-1\right)^{n}}{\left(2n-1\right)^{q}}\dfrac{\left(-1\right)^{p-1}}{\left(p-1\right)!}\int_{0}^{1}\left[\sum_{k=1}^{n}\left(-z^{2}\right)^{k}\right]\dfrac{\log^{p-1}\left(z\right)}{z^{2}}\,\mathrm{d}z$

$=\dfrac{\left(-1\right)^{q}}{\left(q-1\right)!}\dfrac{\left(-1\right)^{p}}{\left(p-1\right)!}\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)}{x^{2}}\int_{0}^{1}\left[\sum_{n=1}^{+\infty}\left(-x^{2}\right)^{n}\left(\left(-z^{2}\right)^{n}-1\right)\right]\dfrac{\log^{p-1}\left(z\right)}{1+z^{2}}\,\mathrm{d}z\,\mathrm{d}x$

$=\dfrac{\left(-1\right)^{q}}{\left(q-1\right)!}\dfrac{\left(-1\right)^{p}}{\left(p-1\right)!}\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)}{1+x^{2}}\int_{0}^{1}\dfrac{\log^{p-1}\left(z\right)}{1-x^{2}z^{2}}\,\mathrm{d}z\,\mathrm{d}x$
$=\dfrac{\left(-1\right)^{q-1}}{2\left(q-1\right)!}\int_{0}^{1}\dfrac{\log^{q-1}(x)}{1+x^{2}}\left[\dfrac{\left[\mathrm{Li}_{p}\left(x\right)-\mathrm{Li}_{p}\left(-x\right)\right]}{x}\right]\mathrm{d}x$

$=\dfrac{\left(-1\right)^{q}}{2\left(q-1\right)!}\left[\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(-x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x-\int_{0}^{1}\dfrac{\log^{q-1}\left(x\right)\mathrm{Li}_{p}\left(x\right)}{x\left(1+x^{2}\right)}\,\mathrm{d}x\right]$

as desired. ∎

##  6\. Further Research and Acknowledgements

Several paths have not been explored yet and will be the subject of future
work. Here, we record some ideas for our future research.
(1) We can regard our inverse sine integral as a _log-sine integral_ :

$I(n)=\int_{0}^{1}\frac{\arcsin^{n}(z)}{z}\,\mathrm{d}z=-n\int_{0}^{1}\frac{\log(z)\arcsin^{n-1}(z)}{\sqrt{1-z^{2}}}\,\mathrm{d}z=-n\int_{0}^{\pi/2}{z^{n-1}\log(\sin(z))}\,\mathrm{d}z.$

We refer the reader to Borwein–Broadhurst–Kamnitzer [3] and Williams-Yue [22]
for relations of such log-sine integrals with central binomial series. Notice
that they discussed

$\int_{0}^{\pi/3}\log^{\alpha}\left({2\sin\left(\frac{\vartheta}{2}\right)}\right)\vartheta^{\beta}\,\mathrm{d}\vartheta\quad\left(\alpha,\beta\geqslant
0\right)$

which is not quite same to ours but it should be possible to relate $I(n)$
with other central binomial series. This may serve as a subject of future
work.
(2) Together with multiple zeta values and Euler sums evaluated in this
article, we can compute other MZVs. For example, we may show that

(6.1) $\zeta(3,1,1)=2\,\zeta(5)-\zeta(2)\,\zeta(3).$

Indeed, Vălean [18, page 303, problem 4.44] shows that

$\sum_{n=1}^{+\infty}\dfrac{H_{n}}{n}\left(\zeta(3)-1-\dfrac{1}{2^{3}}-\dfrac{1}{3^{3}}-\cdots-\dfrac{1}{n^{3}}\right)=2\,\zeta(2)\,\zeta(3)-\dfrac{7}{2}\,\zeta(5)$

where $H_{n}$ is the $n$-th Harmonic number. Notice that the left–hand side is
equal to

$\sum_{m>n\geqslant
k}\dfrac{1}{m^{3}nk}=\sum_{m>n>k}\dfrac{1}{m^{3}nk}+\sum_{m>n=k}\dfrac{1}{m^{3}nk}=\zeta(3,1,1)+\zeta(3,2)$

Hence, with $\zeta(3,2)$ we just found above (3.4), we deduce that

$\zeta(3,1,1)=2\,\zeta(2)\,\zeta(3)-3\,\zeta(2)\,\zeta(3)-\dfrac{7}{2}\,\zeta(5)+\dfrac{11}{2}\,\zeta(5)=2\thinspace\zeta(5)-\zeta(2)\,\zeta(3).$

(3) For a multi-index

$\mathbf{i}=(a_{1}+1,\underbrace{1,\dots,1}_{b_{1}-1},a_{2}+1,\underbrace{1,\dots,1}_{b_{2}-1},\dots,a_{k}+1,\underbrace{1,\dots,1}_{b_{k}-1}),$

with $a_{i},b_{i}\geqslant 1$, we define its _dual_ to be

$\mathbf{i}^{\dagger}=(b_{k}+1,\underbrace{1,\ldots,1}_{a_{k}-1},b_{k-1}+1,\underbrace{1,\dots,1}_{a_{k-1}-1},\dots,b_{1}+1,\underbrace{1,\dots,1}_{a_{1}-1}).$

Duality formula for MZVs claims that
$\zeta(\mathbf{i})=\zeta(\mathbf{i}^{\dagger})$ for all such indices.

In particular, $\zeta(2,\\{1\\}^{n-1})=\zeta(n+1)$. After the preparation of
this manuscript, we found out that Kaneko and Tsumura [12] introduced a
_multiple $T$-value_

$T\left(i_{k},i_{k-1},\ldots,i_{1}\right)=2^{k}\sum_{\begin{subarray}{c}n_{k}>n_{k-1}>\cdots>n_{1}\\\
n_{j}\,\,\equiv\,\,j\,\,\textrm{(mod
$2$)}\end{subarray}}\dfrac{1}{n_{k}^{i_{k}}n_{k-1}^{i_{k-1}}\ldots
n_{1}^{i_{1}}}$

with the factor $2^{k}$ for normalization (we changed the sum convention to
ours). They further showed that exactly the same duality formula hold for
multiple $T$-values. In particular, $T(2,\\{1\\}^{n-1})=T(n+1)$. This is
equivalent to Theorem 4.3, thus, we gave another proof of a special case of
$T$-duality by integration of powers of inverse hyperbolic tangent function.
(4) These duality formulas come from _iterated integral expressions_ for
multiple zeta values with two kinds of integrals:

$\int\frac{\mathrm{d}z}{1-z}\,\,\,\text{and}\,\,\,\int\frac{\mathrm{d}z}{z}.$

For example, the famous Euler-Goldbach relation $\zeta(2,1)=\zeta(3)$ is
nothing but

$\int_{0}^{1}\frac{\mathrm{d}x_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{\mathrm{d}x_{2}}{1-x_{2}}\int_{0}^{x_{2}}\frac{\mathrm{d}x_{1}}{1-x_{1}}=\int_{0}^{1}\frac{\mathrm{d}z_{3}}{z_{3}}\int_{0}^{z_{3}}\frac{\mathrm{d}z_{2}}{z_{2}}\int_{0}^{z_{2}}\frac{\mathrm{d}z_{1}}{1-z_{1}}.$

Indeed, it should be possible to bring this idea of iterated integrals into
the study of multiple $t$-values, multiple zeta values and central binomial
series with more integrals in the form

$\int\frac{\mathrm{d}z}{1-z^{2}},\int\frac{\mathrm{d}z}{z\left(1-z^{2}\right)}\,\,\,\text{and}\,\,\,\int\frac{\mathrm{d}z}{\sqrt{1-z^{2}}}.$

Definition 2.1, Proposition 2.2 and Proposition 3.2 implicitly highlight this
little idea.

Notice that there exist iterated integral expressions for integer powers of
inverse sine function and inverse hyperbolic tangent function:

$\frac{\arcsin^{n}(z)}{n!}=\int_{0}^{z}\frac{\mathrm{d}z_{1}}{\sqrt{1-z_{1}^{2}}}\int_{0}^{z_{1}}\frac{\mathrm{d}z_{2}}{\sqrt{1-z_{2}^{2}}}\cdots\int_{0}^{z_{n-1}}\frac{\mathrm{d}z_{n}}{\sqrt{1-z_{n}^{2}}}$

as remarked in [2, (4.1)]. Quite similarly, we can show that

$\frac{\text{arctanh}^{n}(z)}{n!}=\int_{0}^{z}\frac{\mathrm{d}z_{1}}{{1-z_{1}^{2}}}\int_{0}^{z_{1}}\frac{\mathrm{d}z_{2}}{{1-z_{2}^{2}}}\cdots\int_{0}^{z_{n-1}}\frac{\mathrm{d}z_{n}}{{1-z_{n}^{2}}}.$

Thus, as observed above, the equality

$t\left(2,1\right)=\frac{I(2)}{2!}=\dfrac{1}{2}\int_{0}^{1}\frac{\arcsin^{2}(z)}{z}\,\mathrm{d}z$

implies the relation of iterated integrals

$\int_{0}^{1}\frac{\mathrm{d}x_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{\mathrm{d}x_{2}}{x_{2}\left(1-x_{2}^{2}\right)}\int_{0}^{x_{2}}\frac{\mathrm{d}x_{1}}{1-x_{1}^{2}}=\int_{0}^{1}\frac{\mathrm{d}z_{3}}{z_{3}}\int_{0}^{z_{3}}\frac{\mathrm{d}z_{2}}{\sqrt{1-z_{2}^{2}}}\int_{0}^{z_{2}}\frac{\mathrm{d}z_{1}}{\sqrt{1-z_{1}^{2}}}$

which is quite remarkable and not so obvious at a first glance. This can be
indeed extended and we wish to study more about such iterated integral
relations at some another opportunity.
(5) It remains an open problem to compute a closed expression for the
integrals

$\int_{0}^{1}\dfrac{\log^{q-1}\left(z\right)\mathrm{Li}_{p}\left(z\right)}{z\left(1+z^{2}\right)}\,\,\,\text{and}\,\,\,\int_{0}^{1}\dfrac{\log^{q-1}\left(z\right)\mathrm{Li}_{p}\left(z\right)}{z\left(1-z^{2}\right)}$

as it would allow us to compute $\mathcal{O}\left(p,q\right)$ (Theorem 5.5).

The first author would like to thank Christophe Vignat for his guidance and
support throughout the completion of this work. The second author would like
to thank Satomi Abe, Yuko Takada and Michihito Tobe for sincerely supporting
his research.

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