# Apéry-Type Series with Summation Indices of Mixed Parities and Colored
Multiple Zeta Values, I

Ce Xua, and Jianqiang Zhaob,
a. School of Mathematics and Statistics, Anhui Normal University, Wuhu 241002,
PRC
b. Department of Mathematics, The Bishop’s School, La Jolla, CA 92037, USA
Email<EMAIL_ADDRESS>corresponding author, ORCID
0000-0002-0059-7420.Email<EMAIL_ADDRESS>ORCID 0000-0003-1407-4230.

Abstract. In this paper, we shall study Aéry-type series in which the central
binomial coefficient appears as part of the summand. Let
$b_{n}=4^{n}/\binom{2n}{n}$. Let $s_{1},\dots,s_{d}$ be positive integers with
$s_{1}\geq 2$. We consider the series

$\displaystyle\displaystyle\sum_{n_{1}>\cdots>n_{d}>0}\displaystyle\frac{b_{n_{1}}}{n_{1}^{s_{1}}\cdots
n_{d}^{s_{d}}}$

and the variants with some or all indices $n_{j}$ replaced by $2n_{j}\pm 1$
and some or all “$>$” replaced by “$\geq$”, provided the series are defined.
We can also replace $b_{n_{1}}$ by its square in the above series when
$s_{1}\geq 3$. The main result is that all such series are $\mathbb{Q}$-linear
combinations of the real and/or the imaginary parts of colored multiple zeta
values of level 4, i.e., multiple polylogarithms evaluated at 4th roots of
unity.

Keywords: Apéry-type series, colored multiple zeta values, mixed parities,
iterated integrals.

AMS Subject Classifications (2020): 11M32, 11B65, 11B37, 44A05, 33B30.

##  1 Introduction

In his celebrated proof of irrationality of ${\zeta}(2)$ and ${\zeta}(3)$ in
1979, Apéry used crucially the following two identities

${\zeta}(2)=3\displaystyle\sum_{n\geq
1}\displaystyle\frac{1}{n^{2}{\binom{2n}{n}}}\quad\text{and}\quad{\zeta}(3)=\displaystyle\frac{5}{2}\displaystyle\sum_{n\geq
1}\displaystyle\frac{(-1)^{n-1}}{n^{3}{\binom{2n}{n}}}.$ (1.1)

Motivated by Apéry’s proof, Leshchiner [14] generalized these to higher weight
Riemann zeta values and some other analogs. However, no irrationality proof
has been found so far for other Riemann zeta values at odd positive integers
greater than 4, although a lot of progress has been made (see e.g., [13, 17,
22]. In particular, in 2020, Lai and Yu [13] proved that for any small
$\varepsilon>0$, the number of irrationals among the following odd zeta
values: $\zeta(3),\zeta(5),\zeta(7),\ldots,\zeta(s)$, is at least
$(c_{0}-\varepsilon)\sqrt{s/\log(s)}$, provided $s$ is a sufficiently large
odd integer with respect to $\varepsilon$, with constant
$c_{0}=1.192507\ldots$.

On the other hand, series generalizing those on the right-hand side of (1.1),
including odd-indexed variations (see Remark 4.2) have appeared in the
calculations of the ${\varepsilon}$-expansions of the Feynman diagrams in
recent years (see, e.g., [8, 9, 10]).

In the meantime, the frequent and sometimes unexpected appearance of colored
multiple zeta values (see (1.2)) in quite a few different branches of
mathematics and physics has attracted the attention of many mathematicians and
physicists alike. These numbers, as vast generalizations of Riemann zeta
values, are all conjectured to be not only irrational but also transcendental.
One naturally wonders if the multiple sums, which we call Apéry-type series,
that generalize those in (1.1) can be related to these numbers. In a series of
papers, we will answer some of these questions. As part I of this series, this
paper concentrates on Apéry-type series such as those defined by (4.43) and
(4.53) in which the central binomial coefficients appear only on the
denominators. We will show that a large class of these series can be expressed
as $\mathbb{Q}$-linear combinations of the real and/or the imaginary parts of
the colored multiple zeta values of level 4, i.e., multiple polylogarithms
evaluated at 4th roots of unity, see Thm. 4.1. Some related results may be
found in [4, 11, 18, 19] and references therein.

###  1.1 Notation.

Let $\mathbb{N}$ be the set of positive integers and
$\mathbb{N}_{0}:=\mathbb{N}\cup\\{0\\}$. A finite sequence
${\boldsymbol{\sl{s}}}:=(s_{1},\ldots,s_{d})\in\mathbb{N}^{d}$ is called a
_composition_. We define the weight and the depth of ${\boldsymbol{\sl{s}}}$
by

$|{\boldsymbol{\sl{s}}}|:=s_{1}+\cdots+s_{d},\quad\text{and}\quad\operatorname*{dp}({\boldsymbol{\sl{s}}}):=d,$

respectively. For any $N$th roots of unity $z_{1},\dotsc,z_{d}$ the _colored
multiple zeta values_ (CMZVs) of level $N$ are defined by

$\operatorname{Li}_{{\boldsymbol{\sl{s}}}}({\boldsymbol{\sl{z}}}):=\displaystyle\sum_{n_{1}>\cdots>n_{d}>0}\displaystyle\frac{z_{1}^{n_{1}}\dots
z_{d}^{n_{d}}}{n_{1}^{s_{1}}\dots n_{d}^{s_{d}}},$ (1.2)

which converge if $(s_{1},z_{1})\neq(1,1)$ (see [16] and [21, Ch. 15]), in
which case we call $({\boldsymbol{\sl{s}}};{\boldsymbol{\sl{z}}})$
_admissible_. The multiple zeta values are CMZVs of level 1, namely,
$\zeta({\boldsymbol{\sl{s}}}):=\operatorname{Li}_{{\boldsymbol{\sl{s}}}}(1_{d})$
where $1_{d}$ is the string of 1’s with $d$ repetitions. Moreover, CMZVs can
be expressed using Chen’s iterated integrals

$\operatorname{Li}_{{\boldsymbol{\sl{s}}}}({\boldsymbol{\sl{z}}})=\displaystyle\int_{0}^{1}{\texttt{a}}^{s_{1}-1}{\texttt{x}}_{\xi_{1}}\cdots{\texttt{a}}^{s_{d}-1}{\texttt{x}}_{\xi_{d}},$
(1.3)

where $\xi_{j}:=\prod_{i=1}^{j}z_{i}^{-1}$, ${\texttt{a}}:=dt/t$ and
${\texttt{x}}_{\xi}:=dt/(\xi-t)$ for any $N$th roots of unity $\xi$, see [21,
Sec. 2.1] for a brief summary of this theory. The theory of iterated integrals
was developed first by K.T. Chen in the 1960’s. It has played important roles
in the study of algebraic topology and algebraic geometry in the past half
century. Its simplest form is

$\displaystyle\displaystyle\int_{0}^{1}f_{1}(t)dtf_{2}(t)dt\cdots f_{p}(t)dt=$
$\displaystyle\,\displaystyle\int_{0}^{1}f_{1}(t)dt\circ
f_{2}(t)dt\circ\cdots\circ f_{p}(t)dt$ $\displaystyle:=$
$\displaystyle\,\displaystyle\int\limits_{1>t_{1}>\cdots>t_{p}>0}f_{1}(t_{1})f_{2}(t_{2})\cdots
f_{p}(t_{p})dt_{1}dt_{2}\cdots dt_{p}.$

One can extend these to iterated integrals over any piecewise smooth path on
the complex plane via pull-backs. We refer the interested reader to Chen’s
original work [6, 7] for more details.

###  1.2 Akhilesh’s result.

In [2, 3] Akhilesh discovered some very important and surprising connections
between MZVs and the following Apéry-type series (which he calls multiple
Apéry-like sums and which are normalized slightly differently here)

${\sigma}({\boldsymbol{\sl{s}}};x):=\displaystyle\sum_{n_{1}>n_{2}>\cdots>n_{d}>0}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{(2x)^{2n_{1}}}{(2n_{1})^{s_{1}}\cdots(2n_{d})^{s_{d}}}.$
(1.4)

His ingenious idea is to study the $n$-tails (and more generally, double
tails) of such series. We reformulate one of his key results as follows to
make it more transparent. Set

$\displaystyle g_{s}(t)=\left\\{\begin{array}[]{ll}\tan t\,dt,&\hbox{if
$s=1$;}\\\ dt\circ(\cot t\,dt)^{s-2}\circ dt,&\hbox{if $s\geq
2$,}\end{array}\right.$ (1.7)

and their non-trigonometric counterpart

$G_{s}(t)=\left\\{\begin{array}[]{ll}{\omega}_{2}&\hbox{if $s=1$;}\\\
{\omega}_{1}{\omega}_{0}^{s-2}{\omega}_{1}&\hbox{if $s\geq
2$,}\end{array}\right.$ (1.8)

where ${\omega}$’s are defined by (2.10). Further, we set $\binom{0}{0}=1$,

$\displaystyle b_{n}(x)=4^{n}{{\binom{2n}{n}}}^{-1}x^{2n}\quad\text{and}\quad
b_{n}=b_{n}(1)=4^{n}{{\binom{2n}{n}}}^{-1}\quad\forall n\geq 0.$

######  Theorem 1.1.

_([3, Thm. 4])_ For all $n\in\mathbb{N}_{0}$,
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$ we have

$\displaystyle{\sigma}({\boldsymbol{\sl{s}}};\sin y)_{n}:=$
$\displaystyle\,\displaystyle\sum_{n_{1}>\cdots>n_{d}>n}\displaystyle\frac{b_{n_{1}}(\sin
y)}{(2n_{1})^{s_{1}}\cdots(2n_{d})^{s_{d}}}=\displaystyle\frac{d}{dy}\displaystyle\int_{0}^{y}g_{s_{1}}\circ\cdots\circ
g_{s_{d}}\circ b_{n}(\sin t)\,dt,$

where $y\in(-\pi/2,\pi/2)$ if $s_{1}=1$ and $y\in[-\pi/2,\pi/2]$ if $s_{1}\geq
2$. Using non-trigonometric 1-forms, we have for all $x\in(-1,1)$

$\displaystyle{\sigma}({\boldsymbol{\sl{s}}};x)_{n}:=$
$\displaystyle\,\displaystyle\sum_{n_{1}>\cdots>n_{d}>n}\displaystyle\frac{b_{n_{1}}(x)}{(2n_{1})^{s_{1}}\cdots(2n_{d})^{s_{d}}}=\sqrt{1-x^{2}}\displaystyle\frac{d}{dx}\displaystyle\int_{0}^{x}G_{s_{1}}\circ\cdots\circ
G_{s_{d}}\circ b_{n}(t){\omega}_{1}.$

We will again call the sum $|{\boldsymbol{\sl{s}}}|:=s_{1}+\dots+s_{d}$ the
weight of the series ${\sigma}({\boldsymbol{\sl{s}}};\sin y)$ and $d$ the
depth.

###  1.3 Main results.

In Thm 2.3 and Thm 3.1 we will show that the tails of the series

$\displaystyle{\tau}^{\star}({\boldsymbol{\sl{s}}};x)_{n}:=$
$\displaystyle\,\displaystyle\sum_{n_{1}\geq\cdots\geq n_{d}\geq
n}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{(2x)^{2n_{1}}}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}},$
$\displaystyle\chi({\boldsymbol{\sl{s}}};x)_{n}:=$
$\displaystyle\,\displaystyle\sum_{n_{1}>\cdots>n_{d}>n}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{(2x)^{2n_{1}}}{(2n_{1}-1)^{s_{1}}\cdots(2n_{d}-1)^{s_{d}}}$

can be written as iterated integrals similar to the one in Thm 1.1. As
corollaries, we show that the series ${\sigma}({\boldsymbol{\sl{s}}};1)_{0}$,
${\tau}^{\star}({\boldsymbol{\sl{s}}};1)_{0}$,
$\chi({\boldsymbol{\sl{s}}};1)_{0}$ and more general similar series defined by
(4.43) and (4.53) with summation indices of any parity pattern can be
expressed as $\mathbb{Q}$-linear combinations of the real and/or the imaginary
parts of CMZVs of level 4.

We also consider other Apéry-type series similar to the above by using the
square of central binomial coefficients such as

$\begin{split}{\sigma}^{(2)}({\boldsymbol{\sl{s}}}):=&\,\displaystyle\sum_{n_{1}>\cdots>n_{d}>0}{\binom{2n_{1}}{n_{1}}}^{-2}\displaystyle\frac{16^{n_{1}}}{(2n_{1})^{s_{1}}\cdots(2n_{d})^{s_{d}}},\\\
{\tau}^{\star,(2)}({\boldsymbol{\sl{s}}}):=&\,\displaystyle\sum_{n_{1}\geq\cdots\geq
n_{d}\geq
0}{\binom{2n_{1}}{n_{1}}}^{-2}\displaystyle\frac{16^{n_{1}}}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}}\end{split}$
(1.9)

and show that they also lie in the $\mathbb{Q}$-vector space of CMZVs of level
4.

##  2 First variant with odd summation indices

In this section, we consider a variation of the Apéry-type series studied in
[3] by restricting the summation indices to odd numbers only and replacing
strict inequalities among them by non-strict ones. Concerning this we have the
next well-known result. Define

$\displaystyle{\omega}_{0}:=$ $\displaystyle\,\displaystyle\frac{dt}{t},\quad$
$\displaystyle\,{\omega}_{1}:=$
$\displaystyle\,\displaystyle\frac{dt}{\sqrt{1-t^{2}}},\quad$
$\displaystyle\,{\omega}_{2}:=$
$\displaystyle\,\displaystyle\frac{t\,dt}{1-t^{2}},$ (2.10)
$\displaystyle{\omega}_{3}:=$
$\displaystyle\,\displaystyle\frac{dt}{t\sqrt{1-t^{2}}},\quad$
$\displaystyle\,{\omega}_{5}:=$
$\displaystyle\,\displaystyle\frac{t\,dt}{\sqrt{1-t^{2}}},\quad$
$\displaystyle\,{\omega}_{8}:=$
$\displaystyle\,\displaystyle\frac{dt}{1-t^{2}}.$ (2.11)

######  Lemma 2.1.

For all $d\in\mathbb{N}$ and
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$, we have

$\displaystyle\operatorname{ti}_{\boldsymbol{\sl{s}}}(x)_{n}:=\displaystyle\sum_{n_{1}>\cdots>n_{d}\geq
n}\displaystyle\frac{x^{2n_{1}+1}}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}}=\displaystyle\int_{0}^{x}{\omega}_{0}^{s_{1}-1}{\omega}_{2}\cdots{\omega}_{0}^{s_{d-1}-1}{\omega}_{2}{\omega}_{0}^{s_{d}-1}(t^{2n}{\omega}_{8}).$

###### Proof.

This follows easily by direct computation. ∎

For all $n\in\mathbb{N}_{0}$ and
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$ we define

$\displaystyle{\tau}({\boldsymbol{\sl{s}}};x)_{n}:=$
$\displaystyle\,\displaystyle\sum_{n_{1}>\cdots>n_{d}\geq
n}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{(2x)^{2n_{1}}}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}},$
$\displaystyle{\tau}^{\star}({\boldsymbol{\sl{s}}};x)_{n}:=$
$\displaystyle\,\displaystyle\sum_{n_{1}\geq\cdots\geq n_{d}\geq
n}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{(2x)^{2n_{1}}}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}}.$

######  Theorem 2.2.

For all $d\in\mathbb{N}$ and
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$, the tail

$\displaystyle{\tau}({\boldsymbol{\sl{s}}};1/2)_{n}=$
$\displaystyle\,\displaystyle\int_{0}^{1/2}\displaystyle\frac{4t}{\sqrt{1-4t^{2}}}{\omega}_{0}^{s_{1}-1}{\omega}_{2}\cdots{\omega}_{0}^{s_{d-1}-1}{\omega}_{2}{\omega}_{0}^{s_{d}-1}(t^{2n}{\omega}_{8}).$

###### Proof.

Note that

$\displaystyle\displaystyle\int_{0}^{1/2}\displaystyle\frac{4t}{\sqrt{1-4t^{2}}}t^{2n}dt=$
$\displaystyle\,\displaystyle\int_{0}^{1/4}\displaystyle\frac{2t^{n}}{\sqrt{1-4t}}dt=\displaystyle\frac{\Gamma(n+1)^{2}}{\Gamma(2n+2)}={{\binom{2n}{n}}}^{-1}\displaystyle\frac{1}{2n+1}.$

Thus

$\displaystyle\,\displaystyle\sum_{n_{1}>\cdots>n_{d}\geq
n}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{1}{(2n_{1}+1)^{s_{1}}(2n_{2}+1)^{s_{2}}\cdots(2n_{d}+1)^{s_{d}}}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1/2}\displaystyle\frac{4t}{\sqrt{1-4t^{2}}}\displaystyle\sum_{n_{1}>\cdots>n_{d}\geq
n}\displaystyle\frac{t^{2n_{1}}dt}{(2n_{1}+1)^{s_{1}-1}(2n_{2}+1)^{s_{2}}\cdots(2n_{d}+1)^{s_{d}}}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1/2}\displaystyle\frac{4t}{\sqrt{1-4t^{2}}}\bigg{(}\displaystyle\frac{d}{dt}\operatorname{ti}_{\boldsymbol{\sl{s}}}(t)_{n}\bigg{)}dt$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1/2}\displaystyle\frac{4t}{\sqrt{1-4t^{2}}}{\omega}_{0}^{s_{1}-1}{\omega}_{2}\cdots{\omega}_{0}^{s_{d-1}-1}{\omega}_{2}{\omega}_{0}^{s_{d}-1}(t^{2n}{\omega}_{8}),$

as desired. ∎

It turns out that the star version ${\tau}^{\star}$ behaves better. To study
this, we will extend Chen’s iterated integrals by combining 1-forms and
functions as follows. For any $r\in\mathbb{N}$, 1-forms
$f_{1}(t)\,dt,\dots,f_{r+1}(t)\,dt$ and functions $F_{1}(t),\dots,F_{r}(t)$,
we define recursively

$\displaystyle\,\displaystyle\int_{0}^{1}\big{(}f_{1}(t)\,dt+F_{1}(t)\big{)}\circ\cdots\circ\big{(}f_{r}(t)\,dt+F_{r}(t)\big{)}\circ
f_{r+1}(t)\,dt$ $\displaystyle:=$
$\displaystyle\,\displaystyle\int_{0}^{1}\big{(}f_{1}(t)\,dt+F_{1}(t)\big{)}\circ\cdots\circ\big{(}f_{r-1}(t)\,dt+F_{r-1}(t)\big{)}\circ
f_{r}(t)\,dt\circ f_{r+1}(t)\,dt$ $\displaystyle+$
$\displaystyle\,\displaystyle\int_{0}^{1}\big{(}f_{1}(t)\,dt+F_{1}(t)\big{)}\circ\cdots\circ\big{(}f_{r-1}(t)\,dt+F_{r-1}(t)\big{)}\circ\big{(}F_{r}(t)f_{r+1}(t)\big{)}\,dt.$

We now set the 1-forms

$\displaystyle h_{s}(t)=\left\\{\begin{array}[]{ll}2\csc 2t\,dt,&\hbox{if
$s=1$;}\\\ \csc t\,dt\circ(\cot t\,dt)^{s-2}\circ\csc t\,dt,&\hbox{if $s\geq
2$,}\end{array}\right.$ (2.14)

and their non-trigonometric counter part

$\displaystyle
H_{s}(t)=\left\\{\begin{array}[]{ll}{\omega}_{20}:={\omega}_{0}+{\omega}_{2},&\hbox{if
$s=1$;}\\\ {\omega}_{3}{\omega}_{0}^{s-2}{\omega}_{3},&\hbox{if $s\geq
2$.}\end{array}\right.$ (2.17)

######  Theorem 2.3.

For all $n\in\mathbb{N}_{0}$,
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$ we have the tail

$\displaystyle{\tau}^{\star}({\boldsymbol{\sl{s}}};\sin y)_{n}=$
$\displaystyle\,\displaystyle\sum_{n_{1}\geq\cdots\geq n_{d}\geq
n}\displaystyle\frac{b_{n_{1}}(\sin
y)}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}}=\displaystyle\frac{d}{dy}\displaystyle\int_{0}^{y}h_{s_{1}}\circ\cdots
h_{s_{d}}\circ b_{n}(\sin t)\,dt.$ (2.18)

Hence

$\displaystyle{\tau}^{\star}({\boldsymbol{\sl{s}}};x)_{n}=$
$\displaystyle\,\sqrt{1-x^{2}}\displaystyle\frac{d}{dx}\displaystyle\int_{0}^{x}H_{s_{1}}\circ\cdots\circ
H_{s_{d}}\circ b_{n}(t){\omega}_{1}.$

###### Proof.

When $d=s_{1}=1$, the right-hand side of (2.18) is equal to

$\displaystyle\displaystyle\frac{d}{dy}\displaystyle\int_{0}^{y}2\csc
2t\,dtb_{n}(\sin t)\,dt=\sec y\csc y\displaystyle\int_{0}^{y}b_{n}(\sin
t)\,dt.$ (2.19)

Observe that

$\displaystyle{\binom{2k+2}{k+1}}^{-1}\displaystyle\frac{1}{k+1}=\displaystyle\frac{(k+1)!^{2}}{(2k+2)(2k+1)(2k)!}\displaystyle\frac{1}{k+1}=\displaystyle\frac{1}{2}{\binom{2k}{k}}^{-1}\displaystyle\frac{1}{2k+1}.$

By Thm. 1.1 we get

$\displaystyle\displaystyle\sum_{n_{1}\geq
n}{\binom{2n_{1}+2}{n_{1}+1}}^{-1}\displaystyle\frac{(4\sin^{2}y)^{n_{1}+1}}{n_{1}+1}=2\tan
y\displaystyle\int_{0}^{y}b_{n}(\sin t)\,dt.$

Hence

$\displaystyle\displaystyle\sum_{n_{1}\geq
n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}+1}y}{2n_{1}+1}=\sec
y\displaystyle\int_{0}^{y}b_{n}(\sin t)\,dt,$ (2.20)

which is exactly (2.19). Thus the case $d=s_{1}=1$ of the theorem is proved.

Now, repeatedly multiplying (2.20) by $\cot y$ and integrating $s-1$ times, we
get

$\displaystyle\displaystyle\sum_{n_{1}\geq
n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}+1}y}{(2n_{1}+1)^{s}}=\displaystyle\int_{0}^{y}(\cot
t\,dt)^{s-2}\,(\csc t\,dt)\,(b_{n}(\sin t)\,dt).$ (2.21)

Replacing $n$ by $n_{2}$, multiplying by $1/(2n_{2}+1)$ and taking the sum
$\displaystyle\sum_{n_{1}\geq n_{2}\geq n}$, we get

$\displaystyle\displaystyle\sum_{n_{1}\geq n_{2}\geq
n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}+1}y}{(2n_{1}+1)^{s}(2n_{2}+1)}=$
$\displaystyle\,\displaystyle\int_{0}^{y}(\cot t\,dt)^{s-2}(\csc
t\,dt)\displaystyle\sum_{n_{2}\geq n}\displaystyle\frac{b_{n_{2}}(\sin
t)}{2n_{2}+1}\,dt$ $\displaystyle=$
$\displaystyle\,2\displaystyle\int_{0}^{y}(\cot t\,dt)^{s-2}\,(\csc
t\,dt)\,(\csc 2tdt)\,(b_{n}(\sin t)\,dt).$

Multiplying by $1/(2n_{2}+1)^{s_{2}}$ for $s_{2}\geq 2$ we get

$\displaystyle\,\displaystyle\sum_{n_{1}\geq n_{2}\geq
n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}+1}y}{(2n_{1}+1)^{s_{1}}(2n_{2}+1)^{s_{2}}}$
$\displaystyle=$ $\displaystyle\,\displaystyle\int_{0}^{y}(\cot
t\,dt)^{s-2}(\csc t\,dt)\displaystyle\sum_{n_{2}\geq
n}\displaystyle\frac{b_{n_{2}}(\sin t)}{(2n_{2}+1)^{s_{2}}}\,dt$
$\displaystyle=$ $\displaystyle\,\displaystyle\int_{0}^{y}(\cot
t\,dt)^{s_{1}-2}(\csc t\,dt)(\csc t\,dt)(\cot t\,dt)^{s_{2}-2}(\csc
t\,dt)\,(b_{n}(\sin t)\,dt).$

The theorem follows from doing these repeatedly and can be easily proved by
induction. We leave the details to the interested reader. ∎

######  Corollary 2.4.

For all admissible
${\boldsymbol{\sl{s}}}=(s_{1},\ldots,s_{d})\in\mathbb{N}^{d}$ with $s_{1}\geq
2$, we have

$\displaystyle
t^{\star}({\boldsymbol{\sl{s}}})_{n}:=\displaystyle\sum_{n_{1}\geq\cdots\geq
n_{d}\geq
n}\displaystyle\frac{1}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}}=\displaystyle\frac{2}{\pi}\displaystyle\int_{0}^{\pi/2}h_{s_{1}}\circ\cdots\circ
h_{s_{d}}\circ b_{n}(\sin t)\,dt.$

In particular,

$\displaystyle
t^{\star}({\boldsymbol{\sl{s}}}):=\displaystyle\sum_{n_{1}\geq\cdots\geq
n_{d}\geq
0}\displaystyle\frac{1}{(2n_{1}+1)^{s_{1}}\cdots(2n_{d}+1)^{s_{d}}}=\displaystyle\frac{2}{\pi}\displaystyle\int_{0}^{\pi/2}h_{s_{1}}\circ\cdots\circ
h_{s_{d}}\circ\,dt.$

###### Proof.

Integrating (2.18) over $(0,\pi/2)$ and noticing the fact that

$\displaystyle\displaystyle\int_{0}^{\pi/2}(\sin
t)^{2n}dt=\displaystyle\frac{\pi}{2b_{n}},$

we obtain the corollary immediately. ∎

######  Example 2.5.

Let ${\boldsymbol{\sl{s}}}=(2_{d})$ be the string of 2’s with $d$ repetitions.
Then we see that

$\displaystyle
t^{\star}(2_{d})=\displaystyle\frac{2}{\pi}\displaystyle\int_{0}^{\pi/2}\left(\displaystyle\frac{dt}{\sin
t}\right)^{2d}dt=\displaystyle\frac{2}{\pi}i\Big{(}\operatorname{Li}_{2d+1}(-i)-\operatorname{Li}_{2d+1}(i)\Big{)}=\displaystyle\frac{4}{\pi}{\beta}(2d+1),$

where $\beta$ is the Dirichlet beta function

$\displaystyle{\beta}(s)=\displaystyle\sum_{k\geq
0}\displaystyle\frac{(-1)^{k}}{(2k+1)^{s}}.$ (2.22)

Moreover, we have

$\displaystyle
t^{\star}(2_{a},3,2_{b})=\displaystyle\frac{2}{\pi}\displaystyle\int_{0}^{\pi/2}(\csc
tdt)^{2a+1}\circ\cot tdt\circ(\csc tdt)^{2b+1}\circ dt,$

where $a,b\in\mathbb{N}_{0}$.

######  Remark 2.6.

In [15], T. Murakami first proved the analog of Zagier’s 2-3-2 formula of MZVs
for multiple $t$-values $t(2_{a},3,2_{b})$. Thereafter, several other proofs
have appeared in the literature, see for example [12].

######  Proposition 2.7.

For every positive integer $d$, we have

$\displaystyle{\tau}^{\star}(1_{d};\sin y)=$ $\displaystyle\,2\csc
2y\operatorname{{{Im}}}\operatorname{Li}_{d}(i\tan y),$
$\displaystyle{\tau}^{\star}(2_{d};\sin y)=$ $\displaystyle\,2\csc
y\operatorname{{{Im}}}\operatorname{Li}_{2d}\big{(}i\tan(y/2)\big{)}.$

###### Proof.

By Thm. 2.3 we obtain

$\displaystyle{\tau}^{\star}(1_{d};\sin y)=$ $\displaystyle\,2\csc
2y\displaystyle\int_{0}^{y}\left(\displaystyle\frac{2dt}{\sin
2t}\right)^{d-1}dt=\displaystyle\frac{2\csc
2y}{(d-1)!}\displaystyle\int_{0}^{y}\left(\displaystyle\int_{t}^{y}\displaystyle\frac{2dx}{\sin
2x}\right)^{d-1}dt$ $\displaystyle=$ $\displaystyle\,\displaystyle\frac{2\csc
2y}{(d-1)!}\displaystyle\int_{0}^{y}\log^{d-1}\left|\displaystyle\frac{\csc
2t+\cot 2t}{\csc 2y+\cot 2y}\right|dt$ $\displaystyle=$
$\displaystyle\,\displaystyle\frac{2\csc
2y}{(d-1)!}\displaystyle\int_{0}^{y}\log^{d-1}\big{|}\tan y\cot t\big{|}\,dt.$
(2.23)

By routine differentiation and induction it can be proved easily that

$\displaystyle\displaystyle\frac{1}{(d-1)!}\displaystyle\int_{0}^{y}\log^{d-1}\big{|}\tan
y\cot t\big{|}\,dt=\operatorname{{{Im}}}\operatorname{Li}_{d}(i\tan
y)=\displaystyle\frac{i}{2}\Big{(}\operatorname{Li}_{d}(-i\tan
y)-\operatorname{Li}_{d}(i\tan y)\Big{)}$

for all $d\geq 1$. To see that both sides $\to 0$ as $y\to 0$ we can use
(2.23). Similarly,

$\displaystyle{\tau}^{\star}(2_{d};\sin y)=$ $\displaystyle\,\csc
y\displaystyle\int_{0}^{y}\left(\displaystyle\frac{dt}{\sin
t}\right)^{2d-1}dt=\displaystyle\frac{\csc
y}{(2d-1)!}\displaystyle\int_{0}^{y}\left(\displaystyle\int_{t}^{y}\displaystyle\frac{dx}{\sin
x}\right)^{2d-1}dt$ $\displaystyle=$ $\displaystyle\,\displaystyle\frac{\csc
y}{(2d-1)!}\displaystyle\int_{0}^{y}\log^{2d-1}\big{|}\tan y/2\cot
t/2\big{|}\,dt=2\csc
y\operatorname{{{Im}}}\operatorname{Li}_{2d}\big{(}i\tan(y/2)\big{)}.$

This completes the proof of the proposition. ∎

######  Example 2.8.

Specializing at $y=\pi/4$ and $\pi/2$ in the two identities of Prop. 2.7
respectively, we see that

$\displaystyle\displaystyle\sum_{n_{1}\geq\cdots\geq n_{d}\geq
0}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{2^{n_{1}}}{(2n_{1}+1)\cdots(2n_{d}+1)}=$
$\displaystyle\,2\operatorname{{{Im}}}\operatorname{Li}_{d}(i)=2{\beta}(d),$
$\displaystyle\displaystyle\sum_{n_{1}\geq\cdots\geq n_{d}\geq
0}{\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{4^{n_{1}}}{(2n_{1}+1)^{2}\cdots(2n_{d}+1)^{2}}=$
$\displaystyle\,2\operatorname{{{Im}}}\operatorname{Li}_{2d}(i)=2{\beta}(2d),$

where ${\beta}$ is the Dirichlet beta function defined by (2.22).

##  3 Second variant with odd summation indices

In this section, we consider another variation of the Apéry-type series (1.4)
by restricting the summation indices to odd numbers only and keeping the
strict inequalities among them. These series do not behave as well as the
first variant studied in the last section but are still of interest.

Define the 1-forms

$\displaystyle{\kappa}_{s}(t)=\left\\{\begin{array}[]{ll}\sin t\,dt\,\csc
t\,dt+\tan t\,dt,&\hbox{if $s=1$;}\\\ \sin t\,dt(\cot t\,dt+1)(\cot
t\,dt)^{s-2}\csc t\,dt,\phantom{\displaystyle\frac{1}{2}}&\hbox{if $s\geq
2$,}\end{array}\right.$ (3.26)

and their non-trigonometric counterpart

$\displaystyle
K_{s}(t)=\left\\{\begin{array}[]{ll}{\omega}_{5}{\omega}_{3}+{\omega}_{2},&\hbox{if
$s=1$;}\\\
{\omega}_{5}({\omega}_{0}+1){\omega}_{0}^{s-2}{\omega}_{3},\phantom{\displaystyle\frac{1}{2}}&\hbox{if
$s\geq 2$.}\end{array}\right.$ (3.29)

######  Theorem 3.1.

For all $n\in\mathbb{N}_{0}$ and
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$ the tail

$\displaystyle\chi({\boldsymbol{\sl{s}}};\sin
y)_{n}:=\displaystyle\sum_{n_{1}>\cdots>n_{d}>n}\displaystyle\frac{b_{n_{1}}(\sin
y)}{(2n_{1}-1)^{s_{1}}\cdots(2n_{d}-1)^{s_{d}}}=\displaystyle\frac{d}{dy}\displaystyle\int_{0}^{y}{\kappa}_{s_{1}}\circ\cdots\circ{\kappa}_{s_{d}}\circ
b_{n}(\sin t)\,dt.$

In the above $y\in[-\pi/2,\pi/2]$ if $s_{1}>1$ and $y\in(-\pi/2,\pi/2)$ if
$s_{1}=1$. Using non-trigonometric 1-forms, for $x\in(-1,1)$ we have

$\displaystyle\chi({\boldsymbol{\sl{s}}};x)_{n}:=\displaystyle\sum_{n_{1}>\cdots>n_{d}>n}\displaystyle\frac{b_{n_{1}}(x)}{(2n_{1}-1)^{s_{1}}\cdots(2n_{d}-1)^{s_{d}}}=\sqrt{1-x^{2}}\displaystyle\frac{d}{dx}\displaystyle\int_{0}^{x}K_{s_{1}}\circ\cdots\circ
K_{s_{d}}\circ b_{n}(t){\omega}_{1}.$

###### Proof.

With $s=2$ the identity (2.21) yields that

$\displaystyle\displaystyle\sum_{n_{1}>n}{\binom{2n_{1}-2}{n_{1}-1}}^{-1}\displaystyle\frac{4^{n_{1}-1}\sin^{2n_{1}-1}y}{(2n_{1}-1)^{2}}=\displaystyle\int_{0}^{y}(\csc
t\,dt)b_{n}(\sin t)\,dt.$

Noting that

${\binom{2n_{1}-2}{n_{1}-1}}^{-1}={\binom{2n_{1}}{n_{1}}}^{-1}\displaystyle\frac{2(2n_{1}-1)}{n_{1}},$

we have

$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(\sin
y)}{n_{1}(2n_{1}-1)}=2\sin y\displaystyle\int_{0}^{y}(\csc t\,dt)b_{n}(\sin
t)\,dt.$

Differentiating, we obtain

$\begin{split}\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}-1}y\cos
y}{2n_{1}-1}=&\,\cos y\displaystyle\int_{0}^{y}(\csc t\,dt)b_{n}(\sin
t)\,dt+\displaystyle\int_{0}^{y}b_{n}(\sin t)\,dt.\end{split}$ (3.30)

Multiplying (3.30) by $\tan y$ we get

$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(\sin
y)}{2n_{1}-1}=$ $\displaystyle\,\sin y\displaystyle\int_{0}^{y}(\csc
t\,dt+\sec y)b_{n}(\sin t)\,dt.$ (3.31)

Dividing (3.30) by $\sin y$ and integrating

$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}-1}y}{(2n_{1}-1)^{2}}=$
$\displaystyle\,\displaystyle\int_{0}^{y}(\cot t\,dt)(\csc t\,dt)b_{n}(\sin
t)\,dt+\displaystyle\int_{0}^{y}(\csc t\,dt)b_{n}(\sin t)\,dt.$

Repeatedly multiplying by $\cot y$ and integrating, we see that for all $s\geq
2$

$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}\sin^{2n_{1}-1}y}{(2n_{1}-1)^{s}}=$
$\displaystyle\,\displaystyle\int_{0}^{y}(\cot t\,dt)^{s-1}(\csc
t\,dt)b_{n}(\sin t)\,dt+\displaystyle\int_{0}^{y}(\cot t\,dt)^{s-2}(\csc
t\,dt)b_{n}(\sin t)\,dt.$

Hence if $s\geq 2$ then

$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(\sin
y)}{(2n_{1}-1)^{s}}=$ $\displaystyle\,\sin y\displaystyle\int_{0}^{y}(\cot
t\,dt+1)(\cot t\,dt)^{s-2}(\csc t\,dt)b_{n}(\sin t)\,dt.$ (3.32)

The theorem now follows from repeatedly applying (3.31) or (3.32) at each
depth. This concludes the proof of the theorem. ∎

##  4 Variant with summation indices of mixed parities

By combining Thm. 1.1, Thm. 2.3, and Thm. 3.1 we obtain the following result
easily.

######  Theorem 4.1.

Suppose $d\in\mathbb{N}$ and
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$. Let
$y\in(-\pi/2,\pi/2)$ if $s_{1}=1$ and $y\in[-\pi/2,\pi/2]$ if $s_{1}\geq 2$.
Set ${\lambda}_{2n,s}(t)=g_{s}(t)$, ${\lambda}_{2n+1,s}(t)=h_{s}(t)$ and
${\lambda}_{2n-1,s}(t)={\kappa}_{s}(t)$ which are defined by (1.7), (2.14) and
(3.26), respectively. Then for any $l_{1}(n),\dots,l_{d}(n)=2n,2n\pm 1$ we
have the tails

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
n}\displaystyle\frac{b_{n_{1}}(\sin y)}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=\displaystyle\frac{d}{dy}\displaystyle\int_{0}^{y}{\lambda}_{l_{1},s_{1}}\circ\cdots{\lambda}_{l_{d},s_{d}}\circ
b_{n}(\sin t)\,dt,$ (4.33)

where “$\underset{j}{\succ}$” is “$\geq$” if $l_{j}(n)=2n+1$ and is “$>$”
otherwise. Using non-trigonometric 1-forms, we get for all $x\in(-1,1)$

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
n}\displaystyle\frac{b_{n_{1}}(x)}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=\sqrt{1-x^{2}}\displaystyle\frac{d}{dx}\displaystyle\int_{0}^{x}{\Lambda}_{l_{1},s_{1}}\circ\cdots{\Lambda}_{l_{d},s_{d}}\circ
b_{n}(t){\omega}_{1},$ (4.34)

where ${\Lambda}_{2n,s}(t)=G_{s}(t)$, ${\Lambda}_{2n+1,s}(t)=H_{s}(t)$ and
${\Lambda}_{2n-1,s}(t)=K_{s}(t)$ which are defined by (1.8), (2.17) and
(3.29), respectively.

###### Proof.

Define

$\displaystyle f_{1}(t):=1,\quad
f_{2}(t):=\displaystyle\frac{t}{\sqrt{1-t^{2}}},\quad
f_{3}(t)=\displaystyle\frac{1}{t},\quad
f_{20}(t):=\displaystyle\frac{1}{t\sqrt{1-t^{2}}},\quad f_{5}(t)=t.$

By Thm. 1.1, Thm. 2.3, and Thm. 3.1

$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(x)}{2n_{1}}=$
$\displaystyle\,f_{2}(x)\displaystyle\int_{0}^{x}b_{n}(t){\omega}_{1},$ (4.35)
$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(x)}{(2n_{1})^{s}}=$
$\displaystyle\,f_{1}(x)\displaystyle\int_{0}^{x}{\omega}_{0}^{s-2}{\omega}_{1}\,b_{n}(t){\omega}_{1}\quad\forall
s\geq 2,$ (4.36) $\displaystyle\displaystyle\sum_{n_{1}\geq
n}\displaystyle\frac{b_{n_{1}}(x)}{2n_{1}+1}=$
$\displaystyle\,f_{20}(x)\displaystyle\int_{0}^{x}b_{n}(t){\omega}_{1},$
(4.37) $\displaystyle\displaystyle\sum_{n_{1}\geq
n}\displaystyle\frac{b_{n_{1}}(x)}{(2n_{1}+1)^{s}}=$
$\displaystyle\,f_{3}(x)\displaystyle\int_{0}^{x}{\omega}_{0}^{s-2}{\omega}_{3}\,b_{n}(t){\omega}_{1}\quad\forall
s\geq 2,$ (4.38)
$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(x)}{2n_{1}-1}=$
$\displaystyle\,f_{5}(x)\displaystyle\int_{0}^{x}{\omega}_{3}\,b_{n}(t){\omega}_{1}+f_{2}(x)\displaystyle\int_{0}^{x}b_{n}(t){\omega}_{1},$
(4.39)
$\displaystyle\displaystyle\sum_{n_{1}>n}\displaystyle\frac{b_{n_{1}}(x)}{(2n_{1}-1)^{s}}=$
$\displaystyle\,f_{5}(x)\displaystyle\int_{0}^{x}({\omega}_{0}+1){\omega}_{0}^{s-2}{\omega}_{3}\,b_{n}(t){\omega}_{1}\quad\forall
s\geq 2.$ (4.40)

For convenience, we call the right-hand side of (4.35) and (4.36) (resp.
(4.37) and (4.38), resp. (4.39) and (4.40)) a ${\sigma}$-block (resp.
${\tau}^{\star}$-block, resp. $\chi$-block). In (4.34), each $s_{j}$
corresponds to (a variation of) such a block. Observing that
$f_{j}(t){\omega}_{1}={\omega}_{j}$ we find that after starting with a block
in (4.35)-(4.40), all the middle blocks should be modified as follows: (i)
change $f_{j}(x)$ to ${\omega}_{j}$, (ii) remove the integral sign, and (iii)
drop the 1-form $b_{n}(t){\omega}_{1}$. Repeating this until the end block,
for which only operations (i) and (ii) are required.

This concludes the constructive proof of the theorem. ∎

######  Remark 4.2.

We note that Apéry-type series with indices of mixed parity already appeared
implicitly in [9, (1.1)]. Indeed, using their notation one need to consider,
for e.g., the following series:

$\displaystyle\displaystyle\sum_{j=1}^{\infty}\displaystyle\frac{1}{\binom{2j}{j}}\displaystyle\frac{u^{j}}{j^{c}}S_{a}(2j-1)=$
$\displaystyle\,2^{c}\displaystyle\sum_{j=1}^{\infty}\displaystyle\frac{1}{\binom{2j}{j}}\displaystyle\frac{u^{j}}{(2j)^{c}}\left(\displaystyle\sum_{k=0}^{j-1}\displaystyle\frac{1}{(2k+1)^{a}}+\displaystyle\sum_{k=1}^{j-1}\displaystyle\frac{1}{(2k)^{a}}\right)$
$\displaystyle=$ $\displaystyle\,2^{c}\displaystyle\sum_{j>k\geq
0}\displaystyle\frac{1}{\binom{2j}{j}}\displaystyle\frac{u^{j}}{(2j)^{c}(2k+1)^{a}}+2^{c}\displaystyle\sum_{j>k>0}\displaystyle\frac{1}{\binom{2j}{j}}\displaystyle\frac{u^{j}}{(2j)^{c}(2k)^{a}}.$

######  Example 4.3.

By composing (4.36) and (4.37) we see that

$\displaystyle\displaystyle\sum_{n_{1}\geq
n_{2}>0}\displaystyle\frac{b_{n_{1}}(x)}{(2n_{1}+1)(2n_{2})^{2}}=\displaystyle\frac{1}{x\sqrt{1-x^{2}}}\displaystyle\int_{0}^{x}{\omega}_{1}^{3}.$

Taking $x=1/2,\sqrt{3}/2$ we see immediately that

$\displaystyle\displaystyle\sum_{n_{1}\geq
n_{2}>0}\binom{2n_{1}}{n_{1}}^{-1}\displaystyle\frac{1}{(2n_{1}+1)(2n_{2})^{2}}=\displaystyle\frac{4}{\sqrt{3}}\displaystyle\frac{(\sin^{-1}(1/2))^{3}}{3!}=\displaystyle\frac{\pi^{3}}{4\cdot
81\sqrt{3}},$ $\displaystyle\displaystyle\sum_{n_{1}\geq
n_{2}>0}\binom{2n_{1}}{n_{1}}^{-1}\displaystyle\frac{3^{n}}{(2n_{1}+1)(2n_{2})^{2}}=\displaystyle\frac{4}{\sqrt{3}}\displaystyle\frac{(\sin^{-1}(\sqrt{3}/2))^{3}}{3!}=\displaystyle\frac{2\pi^{3}}{81\sqrt{3}}.$

These are consistent with the first two identities at the beginning of [1].
Many other evaluations in the loc. cit. can be verified using similar ideas by
repeatedly applying (4.35)–(4.40). Taking $x=\sqrt{2}/2$ we also get

$\displaystyle\displaystyle\sum_{n_{1}\geq
n_{2}>0}\binom{2n_{1}}{n_{1}}^{-1}\displaystyle\frac{2^{n}}{(2n_{1}+1)(2n_{2})^{2}}=\displaystyle\frac{4}{\sqrt{3}}\displaystyle\frac{(\sin^{-1}(\sqrt{2}/2))^{3}}{3!}=\displaystyle\frac{\pi^{3}}{96\sqrt{3}}.$

By specializing at $y=\pi/2$ (or taking limit as $x\to 1^{-}$) we obtain the
following theorem, which helps answer two questions at the end of [20]
affirmatively in Cor. 5.1.

######  Theorem 4.4.

Suppose $d\in\mathbb{N}$,
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$ and $s_{1}\geq
2$. Set $\delta(l)=0$ if $l(n)=2n$ and $\delta(l)=1$ if $l(n)=2n\pm 1$.

  1. (a)

Suppose $l_{1}(n),\dots,l_{d}(n)=2n,2n+1$. Then we have

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\ \cdots\
\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in
i^{\delta(l_{1})}\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|}^{4},$

where “$\underset{j}{\succ}$” is “$\geq$” if $l_{j}(n)=2n+1$ and is “$>$”
otherwise.

  2. (b)

Suppose $l_{1}(n),\dots,l_{d}(n)=2n,2n\pm 1$. If for all $l_{j}(n)=2n-1$
($j\geq 2$) we have $l_{j-1}(n)\neq 2n$, then we have

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\ \cdots\
\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in
i^{\delta(l_{1})}\Big{(}\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|}^{4}+\nu(l_{1})\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|+1}^{4}\Big{)},$
(4.41)

where $\nu(l)=1$ if $l(n)=2n-1$ and $\nu(l)=0$ otherwise. In particular, if
$l_{j}(n)\neq 2n$ for all $j$ then (4.41) holds.

  3. (c)

More generally, $l_{1}(n),\dots,l_{d}(n)=2n,2n\pm 1$ then we have

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\ \cdots\
\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in\mathsf{CMZV}_{\leq|{\boldsymbol{\sl{s}}}|+\nu(l_{1})}^{4}\otimes\mathbb{Q}[i].$
(4.42)

  4. (d)

Moreover, the claim in (4.42) still holds if one changes any of the strict
inequalities $n_{j}>n_{j+1}$ to $n_{j}\geq n_{j+1}$ and vice versa, provided
the series is defined. Here we set $n_{d+1}=0$. In particular, if
$l_{1}(n),\dots,l_{d}(n)=2n,2n\pm 1$ then

$\displaystyle\displaystyle\sum_{n_{1}\succ n_{2}\succ\cdots\ \succ
n_{d}\succ\ 0}\displaystyle\frac{b_{n_{1}}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in\mathsf{CMZV}_{\leq|{\boldsymbol{\sl{s}}}|+\nu(l_{1})}^{4}\otimes\mathbb{Q}[i],$
(4.43)

where “$\succ$” can be either “$\geq$” or “$>$”, provided the series is
defined.

######  Remark 4.5.

Let $q=\displaystyle\max\\{j:l_{j}(n)\neq 2n+1\\}$. The series is defined if
and only if “$\underset{q}{\succ}$” is “$>$”.

###### Proof.

Put ${\texttt{x}}_{\xi}=dt/(\xi-t)$ for any $\xi\in\mathbb{C}$ and
${\texttt{d}}_{\xi,\xi^{\prime}}={\texttt{x}}_{\xi}-{\texttt{x}}_{\xi^{\prime}}$.
First, we observe that under the change of variables

$\displaystyle t\to\sin^{-1}t\quad\text{then}\quad
t\to\displaystyle\frac{1-t^{2}}{1+t^{2}},$ (4.44)

we have

$\displaystyle\cot t\,dt\to$
$\displaystyle\,{\omega}_{0}={\texttt{a}}:=\displaystyle\frac{dt}{t}\to{\texttt{y}},\quad$
$\displaystyle\csc t\,dt\to$
$\displaystyle\,{\omega}_{3}:=\displaystyle\frac{dt}{t\sqrt{1-t^{2}}}\to{\texttt{d}}_{-1,1},$
(4.45) $\displaystyle dt\to$
$\displaystyle\,{\omega}_{1}:=\displaystyle\frac{dt}{\sqrt{1-t^{2}}}\to
i{\texttt{d}}_{-i,i},\quad$ $\displaystyle\sec t\csc t\,dt\to$
$\displaystyle\,{\omega}_{20}:=\displaystyle\frac{dt}{t(1-t^{2})}\to{\texttt{y}}+{\texttt{z}},$
(4.46) $\displaystyle\tan t\,dt\to$
$\displaystyle\,{\omega}_{2}:=\displaystyle\frac{t\,dt}{1-t^{2}}\to{\texttt{z}},\quad$
$\displaystyle\sec t\,dt\to$
$\displaystyle\,{\omega}_{8}:=\displaystyle\frac{dt}{1-t^{2}}\to-{\texttt{a}},$
(4.47)

where
${\texttt{y}}={\texttt{x}}_{-i}+{\texttt{x}}_{i}-{\texttt{x}}_{-1}-{\texttt{x}}_{1}$
and ${\texttt{z}}=-{\texttt{a}}-{\texttt{x}}_{-i}-{\texttt{x}}_{i}$.
Furthermore, we notice

$\displaystyle\sin t\,dt\to{\omega}_{5}$
$\displaystyle\,=\displaystyle\frac{t\,dt}{\sqrt{1-t^{2}}}\to\displaystyle\frac{dt}{(i-t)^{2}}+\displaystyle\frac{dt}{(i+t)^{2}}.$
(4.48)

Let

$\displaystyle{\mathsf{O}}:=\mathbb{Q}\left\langle{\omega}_{j}:0\leq j\leq
3\text{ or }j=5\right\rangle.$

By repeatedly using the six cases (4.35)–(4.40) we see that every sum in
(4.42) can be expressed as $\mathbb{Q}$-linear combinations of the following
form

$\displaystyle\displaystyle\int_{0}^{1}{\alpha}_{1}\dots\alpha_{m}$ (4.49)

with $m\leq|{\boldsymbol{\sl{s}}}|$ and ${\alpha}_{j}\in{\mathsf{O}}$.

(a) In this case ${\omega}_{5}$ never appears and the weight in (4.35)–(4.38)
is always the same as the number of 1-forms appearing on the right-hand side
so that there is no weight drop. Thus we only need to consider the number
${\omega}_{1}$’s appearing in (4.34). From (4.45)–(4.47) we see that only
${\omega}_{1}$ produces $i$ after the change of variables
$t\to(1-t^{2})/(1+t^{2})$.

Among the four cases (4.35)–(4.38) only iteration (4.36) affects the number of
${\omega}_{1}$’s in (4.42), by adding two ${\omega}_{1}$’s. Then ending block
$b_{n}(t){\omega}_{1}={\omega}_{1}$ when $n=0$. So we need to consider the
starting 1-form inside the iterated integral of (4.34), which is chopped off
after taking the derivative $d/dx$ and then multiplied by $\sqrt{1-x^{2}}$.
This 1-form is ${\omega}_{1}$ if and only if it is a ${\sigma}$-block with
$s_{1}\geq 2$. Hence, after taking $d/dx$ we find that on the right-hand side
of (4.34) the total number of ${\omega}_{1}$’s is even for a starting
${\sigma}$-block (i.e., $l_{1}(n)=2n$) and the number is odd for a starting
${\tau}^{\star}$-block (i.e., $l_{1}(n)=2n+1$). The claim of (a) is thus
proved.

(b) We first claim that we may reduce this case to the case where
$l_{j}(n)=2n-1$ appears only when $j=1$. We will prove this by induction on
the depth. We have nothing to do when the depth is 1. In general, we change
the index $n_{j}\to n_{j}+1$ for all $j\geq 2$ such that $l_{j}(n)=2n-1$. Then
we need to consider the following three possible blocks in front of $j$-th
block. Setting $k=n_{j-1},m=n_{j},r=s_{j-1},s=s_{j}$, we see that

$\displaystyle\displaystyle\sum_{k>m}\displaystyle\frac{1}{(2k-1)^{r}(2m-1)^{s}}\Longrightarrow\displaystyle\sum_{k>m}\displaystyle\frac{1}{(2k+1)^{r}(2m+1)^{s}}=\displaystyle\sum_{k\geq
m}\displaystyle\frac{1}{(2k+1)^{r}(2m+1)^{s}}-\displaystyle\frac{1}{(2k+1)^{r+s}},$
$\displaystyle\displaystyle\sum_{k\geq
m}\displaystyle\frac{1}{(2k+1)^{r}(2m-1)^{s}}\Longrightarrow\displaystyle\sum_{k\geq
m+1}\displaystyle\frac{1}{(2k+1)^{r}(2m+1)^{s}}$ $\displaystyle\hskip
142.26378pt=\displaystyle\sum_{k\geq
m}\displaystyle\frac{1}{(2k+1)^{r}(2m+1)^{s}}-\displaystyle\frac{1}{(2k+1)^{r+s}}.$

For the possible ${\tau}^{\star}$-block after the $j$-th block, setting
$k=n_{j+1},m=n_{j},r=s_{j+1},s=s_{j}$ we have

$\displaystyle\displaystyle\sum_{k>m}\displaystyle\frac{1}{(2m-1)^{s}(2k+1)^{r}}\Longrightarrow\displaystyle\sum_{k+1>m}\displaystyle\frac{1}{(2m+1)^{s}(2k+1)^{r}}=\displaystyle\sum_{k\geq
m}\displaystyle\frac{1}{(2m+1)^{s}(2k+1)^{r}}$

which looks in good shape. We also need to consider the possible
${\sigma}$-blocks after the $j$-th block: (setting
$k=n_{j+1},m=n_{j},r=s_{j+1},s=s_{j}$)

$\displaystyle\displaystyle\sum_{k>m}\displaystyle\frac{1}{(2k-1)^{s}(2m)^{r}}\Longrightarrow\displaystyle\sum_{k+1>m}\displaystyle\frac{1}{(2k+1)^{s}(2m)^{r}}=\displaystyle\sum_{k\geq
m}\displaystyle\frac{1}{(2k+1)^{s}(2m)^{r}},$

which looks in good shape, too.

To summarize, the above shows that if no ${\sigma}$-$\chi$-block chain appears
then we see that no weight drops can occur in the decomposed sums.

Furthermore, from the above, we can assume the $\chi$-block appears only as
the first block, if it ever does. By the explicit iterated integral
expressions of the ${\sigma}$\- and ${\tau}^{\star}$-blocks (4.35)–(4.38), we
see that if $\chi$-block does not appear then all the CMZVs involved are of
the same weight. If a $\chi$-block appears at the beginning then (4.40) shows
that the weight can increase by one for some CMZVs and the counting of
${\omega}_{1}$ is the same as the case with a starting ${\tau}^{\star}$-block.
This completes the proof of (b).

(c) As the proof of (b), we first claim that we may reduce the general case to
the case where $l_{j}(n)=2n-1$ appears, if it ever does, then $j=1$. We will
prove this by induction on the depth. We have nothing to do when the depth is
1. In general, we change the index $n_{j}\to n_{j}+1$ for all $j\geq 2$ such
that $l_{j}(n)=2n-1$. Then we need to consider the following possible block
${\sigma}$ in front of $j$-th block since the other two possibilities have
been already handled by case (b): (setting
$k=n_{j-1},m=n_{j},r=s_{j-1},s=s_{j}$)

$\displaystyle\displaystyle\sum_{k>m}\displaystyle\frac{1}{(2k)^{r}(2m-1)^{s}}\Longrightarrow\displaystyle\sum_{k>m+1}\displaystyle\frac{1}{(2k)^{r}(2m+1)^{s}}=\displaystyle\sum_{k>m}\displaystyle\frac{1}{(2k)^{r}(2m+1)^{s}}-\displaystyle\frac{1}{(2k)^{r}(2k-1)^{s}}.$

Thus by partial fractions we may decompose the second term above as pure
powers of either $2k$ or $2k-1$, thus reducing the depth by 1. We point out
that this is also the reason why the weight may drop due to the partial
fractions when ${\sigma}$-$\chi$-block chain appears.

Thus we will assume the $\chi$-block appears only as the first block. Then
this case is proved again by the explicit formula (4.40).

(d) Observe that for any fixed $n$,

$\displaystyle\displaystyle\sum_{m\geq n}\displaystyle\frac{1}{m^{s}(2n\pm
1)^{t}}=$ $\displaystyle\,\displaystyle\frac{1}{n^{s}(2n\pm
1)^{t}}+\displaystyle\sum_{m>n}\displaystyle\frac{1}{m^{s}(2n\pm 1)^{t}},$
$\displaystyle\displaystyle\sum_{m>n}\displaystyle\frac{1}{(2m+1)^{t}n^{s}}=$
$\displaystyle\,-\displaystyle\frac{1}{n^{s}(2n+1)^{t}}+\displaystyle\sum_{m\geq
n}\displaystyle\frac{1}{(2m+1)^{t}n^{s}},$
$\displaystyle\displaystyle\sum_{m>n}\displaystyle\frac{1}{(2m+1)^{t}(2n-1)^{s}}=$
$\displaystyle\,-\displaystyle\frac{1}{(2n+1)^{t}(2n-1)^{s}}+\displaystyle\sum_{m\geq
n}\displaystyle\frac{1}{(2m+1)^{t}(2n-1)^{s}},$
$\displaystyle\displaystyle\sum_{m\geq
n}\displaystyle\frac{1}{(2m-1)^{t}n^{s}}=$
$\displaystyle\,-\displaystyle\frac{1}{(2n-1)^{t}n^{s}}+\displaystyle\sum_{m>n}\displaystyle\frac{1}{(2m+1)^{t}n^{s}},$
$\displaystyle\displaystyle\sum_{m\geq
n}\displaystyle\frac{1}{(2m-1)^{t}(2n+1)^{s}}=$
$\displaystyle\,\displaystyle\frac{1}{(2n-1)^{t}(2n+1)^{s}}+\displaystyle\sum_{m>n}\displaystyle\frac{1}{(2m+1)^{t}(2n+1)^{s}}.$

By partial fraction decomposition we can reduce all the first terms on the
right-hand side of the above to a $\mathbb{Q}$-linear combination of single
powers of either $1/n$ or $1/(2n\pm 1)$. The only complication is that when
$j=1$ non-admissible terms may appear, which requires us to use the shuffle
regularization. This is similar to the proof of [20, Thm. 9.6] using [20,
Lemma. 9.4 and Lemma 9.5]. We leave the details to the interested reader who
may also refer to Example 5.10 (especially the computation for $S_{2}$ on page
5.10) for the detailed steps to carry this out explicitly. ∎

######  Corollary 4.6.

Let $d\in\mathbb{N}$ and
${\boldsymbol{\sl{s}}}=(s_{1},\dots,s_{d})\in\mathbb{N}^{d}$. Let
${\boldsymbol{\sl{l}}}=(l_{1},\dots,l_{d})$ where $l_{j}(n)=2n$ or $2n\pm 1$
for all $j=1,\dots,d$. Let “$\succ$” denote either “$\geq$” or “$>$”. Then for
every real algebraic $x$ such that $0<x\leq 1$ the value

$\displaystyle\Xi({\boldsymbol{\sl{s}}},{\boldsymbol{\sl{l}}};x):=\displaystyle\sum_{n_{1}\succ
n_{2}\succ\cdots\ \succ n_{d}\succ\
0}\displaystyle\frac{b_{n_{1}}(x)}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}},$

if it exists, can be expressed as a $\mathbb{Q}[i,x,\sqrt{1-x^{2}}]$-linear
combination of the multiple polylogarithms evaluated at algebraic points.

###### Proof.

By the proof of Thm. 4.4, up to factors of $x$ and $\sqrt{1-x^{2}}$ in front
(which can be seen more easily from (4.35)–(4.40)),
$\Xi({\boldsymbol{\sl{s}}},{\boldsymbol{\sl{l}}};x)$ can be expressed as a
$\mathbb{Q}$-linear combinations of

$\displaystyle\displaystyle\int_{0}^{x}\Big{[}{\omega}_{j}:j=0,1,2,3\Big{]}_{\ell}$
(4.50)

where $\Big{[}{\omega}_{j}:j=0,1,2,3\Big{]}_{|{\boldsymbol{\sl{s}}}|}$ is an
iteration of 1-forms of length $\ell\leq|{\boldsymbol{\sl{s}}}|+1$. Here the
1-form ${\omega}_{5}$ is not needed since the proof of Thm. 4.4 shows that if
$l_{j}(n)=2n-1$ then we may assume $j=1$ (i.e., $\chi$-block only appears at
the beginning). Therefore, after applying the change of variables
$t\to\displaystyle\frac{1-t^{2}}{1+t^{2}}$, by (4.45)–(4.47) we see that
(4.50) is transformed to a $\mathbb{Q}[i]$-linear combination of iterated
integrals of the form

$\displaystyle\displaystyle\int_{{\lambda}(x)}^{1}{\alpha}_{1}\dots\alpha_{\ell},.$

where ${\lambda}(x)=\sqrt{\displaystyle\frac{1-x}{1+x}},\
{\alpha}_{j}\in\\{{\texttt{x}}_{0},{\texttt{x}}_{\mu}:\mu^{8}=1\\}.$ Note
${\lambda}(x)\to x$ under the change of variables
$t\to\displaystyle\frac{1-t^{2}}{1+t^{2}}$. If $x\neq 1$, to convert this to
multiple polylogs we generally need to use the regularization process. Thus,
for an arbitrarily small ${\varepsilon}>0$ we write

$\displaystyle\displaystyle\int_{{\lambda}(x)}^{1}{\alpha}_{1}\dots\alpha_{\ell}=$
$\displaystyle\,\displaystyle\sum_{j=0}^{\ell}\displaystyle\int_{{\lambda}(x)}^{\varepsilon}{\alpha}_{1}{\alpha}_{2}\dots\alpha_{j}\displaystyle\int_{\varepsilon}^{1}{\alpha}_{j+1}\dots\alpha_{\ell}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{j=0}^{\ell}(-1)^{j}\displaystyle\int_{\varepsilon}^{{\lambda}(x)}{\alpha}_{j}\dots\alpha_{2}{\alpha}_{1}\displaystyle\int_{\varepsilon}^{1}{\alpha}_{j+1}\dots\alpha_{\ell}$
$\displaystyle=$
$\displaystyle\,\displaystyle\sum_{j=0}^{\ell}(-1)^{j}\displaystyle\int_{{\varepsilon}/{\lambda}(x)}^{1}{\alpha}^{\prime}_{j}\dots\alpha^{\prime}_{2}{\alpha}^{\prime}_{1}\displaystyle\int_{\varepsilon}^{1}{\alpha}_{j+1}\dots\alpha_{\ell}$

where ${\alpha}^{\prime}_{k}={\texttt{x}}_{\xi/{\lambda}(x)}$ if
${\alpha}_{k}={\texttt{x}}_{\xi}$ where $\xi=0$ or $\xi^{8}=1$. Note that
$\xi/{\lambda}(x)$ is still algebraic. By the usual regularization procedure
we see that the last expression can be written as a polynomial
$P(\log({\varepsilon}))$ plus $O({\varepsilon}\log^{\ell}({\varepsilon}))$,
such that all the coefficients of $P$ are $\mathbb{Q}$-linear combination of
the multiple polylogarithms evaluated at algebraic points. Here we have used
the fact that
$\log({\lambda}(x))=\tfrac{1}{2}\big{(}\operatorname{Li}_{1}(-x)-\operatorname{Li}_{1}(x)\big{)}$.
Finally, taking ${\varepsilon}\to 0$ yields the corollary at once. ∎

######  Theorem 4.7.

Keep notation as in Thm. 4.4. Assume $s_{1}\geq 3$.

  1. (a)

Let $l_{1}(n),\dots,l_{d}(n)=2n,2n\pm 1$. If $l_{1}(n)\neq 2n-1$ and for all
$l_{j}(n)=2n-1$ ($j\geq 2$) we have $l_{j-1}(n)\neq 2n$, then

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\ \cdots\
\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|}^{4}.$ (4.51)

In particular, if $l_{j}(n)\neq 2n$ for all $j$ then (4.51) holds. If
$l_{1}(n)=2n-1$ and for all $l_{j}(n)=2n-1$ ($j\geq 2$) we have
$l_{j-1}(n)\neq 2n$, then

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\ \cdots\
\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|}^{4}+\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|+1}^{4}+\mathsf{CMZV}_{|{\boldsymbol{\sl{s}}}|+2}^{4}.$

  2. (b)

More generally, if $l_{1}(n),\dots,l_{d}(n)=2n,2n\pm 1$ then we have

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\ \cdots\
\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in\mathsf{CMZV}_{\leq|{\boldsymbol{\sl{s}}}|+2\nu(l_{1})}^{4}\otimes\mathbb{Q}[i].$
(4.52)

  3. (c)

Moreover, the claim in (b) still holds if one changes any of the strict
inequalities $n_{j}>n_{j+1}$ to $n_{j}\geq n_{j+1}$ in (4.52) and vice versa,
provided the series is defined. In particular,

$\displaystyle\displaystyle\sum_{n_{1}\ \succ\ \cdots\ \succ\ n_{d}\succ\
0}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}\in\mathsf{CMZV}_{\leq|{\boldsymbol{\sl{s}}}|+2\nu(l_{1})}^{4}\otimes\mathbb{Q}[i],$
(4.53)

where “$\succ$” can be either “$\geq$” or “$>$”, provided the series is
defined.

###### Proof.

The key observation is that

$\displaystyle\displaystyle\int_{0}^{1}\displaystyle\frac{x^{2n+1}}{\sqrt{1-x^{2}}}\,dx=\displaystyle\int_{0}^{\pi/2}\sin^{2n+1}t\,dt=B\Big{(}n+1,\displaystyle\frac{1}{2}\Big{)}=\displaystyle\frac{b_{n}}{2n+1}.$

(a) When $l_{1}(n)=2n+1$ by (4.34) we see that the sum

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}(x)/\sqrt{1-x^{2}}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=\displaystyle\frac{1}{x\sqrt{1-x^{2}}}\displaystyle\int_{0}^{x}{\omega}_{0}^{s_{1}-2}{\omega}_{3}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ{\omega}_{1}$

when $s_{1}\geq 2$. Thus multiplying by $x$ on both sides and integrating over
$(0,1)$ we get

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}+1}\cdots
l_{d}(n_{d})^{s_{d}}}=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{1}{\omega}_{0}^{s_{1}-2}{\omega}_{3}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ{\omega}_{1}.$

The claim follows immediately since there are even number of ${\omega}_{1}$’s
in this case.

(b) If $l_{1}(n)=2n$ then we see that

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}(x)}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{0}^{s-2}{\omega}_{1}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ{\omega}_{1}$

when $s_{1}\geq 2$. Then we can divide by $x\sqrt{1-x^{2}}$ and integrate over
$(0,1)$ to get

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}+1}\cdots
l_{d}(n_{d})^{s_{d}}}=$ $\displaystyle\,\displaystyle\sum_{n_{1}\
\underset{1}{\succ}\ \dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
0}\displaystyle\frac{b_{n_{1}}b_{n_{1}-1}}{(2n_{1}-1)l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{3}{\omega}_{0}^{s_{1}-2}{\omega}_{1}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ{\omega}_{1}$

since

$\displaystyle\frac{b_{n_{1}-1}}{2n_{1}-1}=\displaystyle\frac{b_{n_{1}}}{2n_{1}}.$
(4.54)

The theorem holds as well in this case as the number of ${\omega}_{1}$’s is
still even.

(c) If $l_{1}(n)=2n-1$ then since $s_{1}\geq 3$ by (4.34) we have

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
n}\displaystyle\frac{b_{n_{1}}(x)}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=x\displaystyle\int_{0}^{x}({\omega}_{0}+1){\omega}_{0}^{s-2}{\omega}_{3}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ
b_{n}(t){\omega}_{1}.$

We first differentiate this to get

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
n}\displaystyle\frac{2n_{1}b_{n_{1}}x^{2n_{1}-1}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=$
$\displaystyle\,\displaystyle\int_{0}^{x}({\omega}_{0}+1)^{2}{\omega}_{0}^{s-3}{\omega}_{3}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ
b_{n}(t){\omega}_{1}.$

As in the $l_{1}(n)=2n$ case, we can divide by $\sqrt{1-x^{2}}$ and integrate
over $(0,1)$ to get

$\displaystyle\displaystyle\sum_{n_{1}\ \underset{1}{\succ}\
\dots\underset{d-1}{\succ}n_{d}\ \underset{d}{\succ}\
n}\displaystyle\frac{b_{n_{1}}^{2}}{l_{1}(n_{1})^{s_{1}}\cdots
l_{d}(n_{d})^{s_{d}}}=\displaystyle\int_{0}^{1}{\omega}_{1}({\omega}_{0}+1)^{2}{\omega}_{0}^{s_{1}-3}{\omega}_{3}\circ{\Lambda}_{l_{2},s_{2}}(t)\circ\cdots{\Lambda}_{l_{d},s_{d}}(t)\circ
b_{n}(t){\omega}_{1}$

by using (4.54) again. This completes the proof of the theorem. ∎

##  5 A corollary and some examples

In this last section, we will apply our main theorems to compute a few typical
Apéry type series to illustrate the power of our method. We can also see how
the regularization process is needed in some of the examples.

First, we can answer affirmatively a few questions we posted at the end of
[20]. For ${\boldsymbol{\sl{k}}}\in\mathbb{N}^{d}$ and
${\boldsymbol{\sl{l}}}\in\mathbb{N}^{e}$ we define

$\displaystyle\zeta_{n}({\boldsymbol{\sl{k}}}):=$
$\displaystyle\,\displaystyle\sum_{n\geq
m_{1}>\dots>m_{d}>0}\displaystyle\frac{1}{m_{1}^{k_{1}}\cdots m_{d}^{k_{d}}},$
$\displaystyle t_{n}({\boldsymbol{\sl{l}}}):=$
$\displaystyle\,\displaystyle\sum_{n\geq
r_{1}>\dots>r_{e}>0}\displaystyle\frac{1}{(2r_{1}-1)^{l_{1}}\cdots(2r_{e}-1)^{l_{e}}}.$

######  Corollary 5.1.

For all $m\in\mathbb{N}$, $p\in\mathbb{N}_{\geq 2}$, $q\in\mathbb{N}_{\geq
3}$, and all compositions of positive integers ${\boldsymbol{\sl{k}}}$ and
${\boldsymbol{\sl{l}}}$ (including the cases ${\boldsymbol{\sl{k}}}=\emptyset$
or ${\boldsymbol{\sl{l}}}=\emptyset$), we have

$\displaystyle{\rm(a)}$ $\displaystyle\ \
\displaystyle\sum_{n=1}^{\infty}b_{n}\displaystyle\frac{\zeta_{n}({\boldsymbol{\sl{k}}})t_{n}({\boldsymbol{\sl{l}}})}{n^{p}}\in\mathsf{CMZV}_{|{\boldsymbol{\sl{k}}}|+|{\boldsymbol{\sl{l}}}|+p}^{4},$
$\displaystyle\quad{\rm(b)}$ $\displaystyle\ \
\displaystyle\sum_{n=1}^{\infty}b_{n}^{2}\displaystyle\frac{\zeta_{n}({\boldsymbol{\sl{k}}})t_{n}({\boldsymbol{\sl{l}}})}{n^{q}}\in\mathsf{CMZV}^{4}_{|{\boldsymbol{\sl{k}}}|+|{\boldsymbol{\sl{l}}}|+q},$
$\displaystyle{\rm(c)}$ $\displaystyle\ \
\displaystyle\sum_{n=0}^{\infty}b_{n}\displaystyle\frac{\zeta_{n}({\boldsymbol{\sl{k}}})t_{n}({\boldsymbol{\sl{l}}})}{(2n+1)^{p}}\in
i\mathsf{CMZV}^{4}_{|{\boldsymbol{\sl{k}}}|+|{\boldsymbol{\sl{l}}}|+p},$
$\displaystyle\quad{\rm(d)}$ $\displaystyle\ \
\displaystyle\sum_{n=0}^{\infty}b_{n}^{2}\displaystyle\frac{\zeta_{n}({\boldsymbol{\sl{k}}})t_{n}({\boldsymbol{\sl{l}}})}{(2n+1)^{q}}\in\mathsf{CMZV}^{4}_{|{\boldsymbol{\sl{k}}}|+|{\boldsymbol{\sl{l}}}|+q}.$

###### Proof.

Write

$\displaystyle\zeta_{n}({\boldsymbol{\sl{k}}})=\displaystyle\sum_{n\geq
m_{1}>\dots>m_{d}>0}\displaystyle\frac{1}{m_{1}^{k_{1}}\cdots
m_{d}^{k_{d}}},\quad
t_{n}({\boldsymbol{\sl{l}}})=\displaystyle\sum_{n>r_{1}>\dots>r_{e}\geq
0}\displaystyle\frac{1}{(2r_{1}+1)^{l_{1}}\cdots(2r_{e}+1)^{l_{e}}}.$

We only need to note the following facts: (i) for any summation index $m$ for
$\zeta_{n}({\boldsymbol{\sl{k}}})$ and summation index $r$ for
$t_{n}({\boldsymbol{\sl{l}}})$ there are only two possibilities: $m>r$ or
$r\geq m$; (ii) we can re-write

$\displaystyle\displaystyle\sum_{n>r_{1}}\displaystyle\frac{1}{(2n+1)^{q}(2r_{1}+1)^{l_{1}}}=\displaystyle\sum_{n\geq
r_{1}}\displaystyle\frac{1}{(2n+1)^{q}(2r_{1}+1)^{l_{1}}}-\displaystyle\frac{1}{(2n+1)^{q+l_{1}}}$

and obtain similar identities when $n$ and $r_{1}$ are replaced by $r_{j}$ and
$r_{j+1}$. Therefore, we see that (a) and (c) are special cases of Thm.
4.4(a). (b) and (d) are special cases of Thm. 4.7(a). ∎

In the following we will compute a series of examples using our main theorems.

######  Example 5.2.

When depth $d=1$, by (4.38) we see that for all $x\in[-1,1]$

$\displaystyle\displaystyle\sum_{n\geq
0}\displaystyle\frac{b_{n}(x)}{(2n+1)^{m+2}}=$
$\displaystyle\,\displaystyle\frac{1}{x}\displaystyle\int_{0}^{x}{\omega}_{0}^{m}{\omega}_{3}{\omega}_{1}$
(5.55)

for all $m\geq 0$. Applying $t\to\displaystyle\frac{1-t^{2}}{1+t^{2}}$ to
(5.55) we have

$\displaystyle\displaystyle\sum_{n\geq
0}\displaystyle\frac{b_{n}(x)}{(2n+1)^{m+2}}=$
$\displaystyle\,\displaystyle\frac{i(-1)^{m}}{x}\displaystyle\int_{{\lambda}(x)}^{1}({\texttt{x}}_{i}-{\texttt{x}}_{-i})({\texttt{x}}_{1}-{\texttt{x}}_{-1}){\texttt{y}}^{m},$

where ${\lambda}(x)=\sqrt{\displaystyle\frac{1-x}{1+x}}$ and
${\texttt{y}}={\texttt{x}}_{-i}+{\texttt{x}}_{i}-{\texttt{x}}_{-1}-{\texttt{x}}_{1}$
as defined in (4.45). Taking $x=1$ and applying Au’s Mathematica package [4]
we get

$\displaystyle\displaystyle\sum_{n\geq
0}\displaystyle\frac{b_{n}}{(2n+1)^{2}}=$
$\displaystyle\,2\,\operatorname{{{Im}}}(\operatorname{Li}_{1,1}(i,-i)+\operatorname{Li}_{1,1}(-i,-i))=2G\approx
1.83193119,$ (5.56) $\displaystyle\displaystyle\sum_{n\geq
0}\displaystyle\frac{b_{n}}{(2n+1)^{3}}=$
$\displaystyle\,2\,\operatorname{{{Im}}}\Big{(}\operatorname{Li}_{1_{3}}(-i,i,i)+\operatorname{Li}_{1_{3}}(-i,i,-i)-\operatorname{Li}_{1_{3}}(-i,i,-1)-\operatorname{Li}_{1_{3}}(-i,i,1)$
$\displaystyle-$
$\displaystyle\,\operatorname{Li}_{1_{3}}(-i,-i,-i)-\operatorname{Li}_{1_{3}}(-i,-i,i)+\operatorname{Li}_{1_{3}}(-i,-i,-1)+\operatorname{Li}_{1_{3}}(-i,-i,1)\Big{)}$
$\displaystyle=$
$\displaystyle-\displaystyle\frac{\pi^{3}}{32}-\displaystyle\frac{1}{8}\pi\log^{2}2+4{\rm
Im}\operatorname{Li}_{3}\left(\displaystyle\frac{1+i}{2}\right)\approx
1.122690025,$

where $G={\beta}(2)$ is Catalan’s constant. This sum appears in [5, Example
2.12], too.

######  Example 5.3.

As an application of Cor. 4.6, we now compute

$\displaystyle\displaystyle\sum_{n\geq
0}\displaystyle\frac{1}{{\binom{2n}{n}}(2n+1)^{2}}=\displaystyle\sum_{n\geq
0}\displaystyle\frac{b_{n}(1/2)}{(2n+1)^{2}}.$

From the previous example we see that

$\displaystyle\,\displaystyle\sum_{n\geq
0}\displaystyle\frac{1}{{\binom{2n}{n}}(2n+1)^{2}}==\displaystyle\int_{\tfrac{1}{\sqrt{3}}}^{1}({\texttt{x}}_{i}-{\texttt{x}}_{-i})({\texttt{x}}_{1}-{\texttt{x}}_{-1})$
$\displaystyle=$
$\displaystyle\,2i\left(\displaystyle\int_{\tfrac{1}{\sqrt{3}}}^{0}({\texttt{x}}_{i}-{\texttt{x}}_{-i})({\texttt{x}}_{1}-{\texttt{x}}_{-1})+\displaystyle\int_{0}^{1}({\texttt{x}}_{i}-{\texttt{x}}_{-i})\displaystyle\int_{1/\sqrt{3}}^{0}({\texttt{x}}_{1}-{\texttt{x}}_{-1})+\displaystyle\int_{0}^{1}({\texttt{x}}_{i}-{\texttt{x}}_{-i})({\texttt{x}}_{1}-{\texttt{x}}_{-1})\right)$
$\displaystyle=$
$\displaystyle\,2i\left(\displaystyle\int_{0}^{\tfrac{1}{\sqrt{3}}}({\texttt{x}}_{1}-{\texttt{x}}_{-1})({\texttt{x}}_{i}-{\texttt{x}}_{-i})-\displaystyle\int_{0}^{1}({\texttt{x}}_{i}-{\texttt{x}}_{-i})\displaystyle\int_{0}^{\tfrac{1}{\sqrt{3}}}({\texttt{x}}_{1}-{\texttt{x}}_{-1})\right)+4G\quad(\text{by
\eqref{eq-Catalan}})$ $\displaystyle=$
$\displaystyle\,4\operatorname{{{Im}}}\Big{(}\operatorname{Li}_{1,1}\Big{(}\displaystyle\frac{-1}{\sqrt{3}},i\Big{)}-\operatorname{Li}_{1,1}\Big{(}\displaystyle\frac{1}{\sqrt{3}},-i\Big{)}\Big{)}-\pi\log(2+\sqrt{3})+4G\approx
1.063459833.$

######  Example 5.4.

As an easy example of Thm. 4.4(a), by (4.36) and (4.37) we have

$\displaystyle\displaystyle\sum_{n_{1}>n_{2}\geq
0}\displaystyle\frac{b_{n_{1}}}{n_{1}^{2}(2n_{2}+1)}=$
$\displaystyle\,4\displaystyle\int_{0}^{\pi/2}dt\circ(\csc t\sec tdt)\circ dt$
$\displaystyle=$
$\displaystyle\,4\displaystyle\int_{0}^{1}{\omega}_{1}\circ{\omega}_{20}\circ{\omega}_{1}\qquad(\text{by
$t\to\sin^{-1}t$})$ $\displaystyle=$
$\displaystyle\,-4\displaystyle\int_{0}^{1}({\texttt{x}}_{-i}-{\texttt{x}}_{i})\circ({\texttt{x}}_{0}+{\texttt{x}}_{-1}+{\texttt{x}}_{1})\circ({\texttt{x}}_{-i}-{\texttt{x}}_{i})=7\zeta(3)$

by the change of variables $t\to(1-t^{2})/(1+t^{2})$ then using Au’s package
[4].

######  Example 5.5.

For a pure $\chi$-sum, by (4.40) we have

$\displaystyle\displaystyle\sum_{n>0}\displaystyle\frac{b_{n}}{(2n-1)^{2}}=$
$\displaystyle\,\displaystyle\int_{0}^{1}({\omega}_{0}+1){\omega}_{3}{\omega}_{1}=i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}(1-{\texttt{y}})$
$\displaystyle=$
$\displaystyle\,2G-\displaystyle\frac{1}{32}\pi^{3}+4\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}-\displaystyle\frac{1}{8}\pi\log^{2}2\approx
2.954621213,$

where we see the weight can increase by one as predicted by Thm. 4.4(b).
Similarly,

$\displaystyle\displaystyle\sum_{n>0}\displaystyle\frac{b_{n}}{(2n-1)^{3}}=$
$\displaystyle\,\displaystyle\int_{0}^{1}({\omega}_{0}+1){\omega}_{0}{\omega}_{3}{\omega}_{1}=i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}{\texttt{y}}({\texttt{y}}-1)$
$\displaystyle=$
$\displaystyle\,-4{\beta}(4)+\displaystyle\frac{1}{96}\bigg{(}2\operatorname{{{Im}}}\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}+4\pi\log^{3}2+3\pi^{3}\log
2$ $\displaystyle\,\hskip
85.35826pt-12\pi\log^{2}2-3\pi^{3}-\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}\bigg{)}\approx
2.1543060048.$

######  Example 5.6.

For a sum of mixed parities as examples of Thm. 4.4(b), by (4.37) and (4.40)
we have

$\displaystyle\displaystyle\sum_{n_{1}>n_{2}\geq
0}\displaystyle\frac{b_{n_{1}}}{(2n_{1}-1)^{2}(2n_{2}+1)}=$
$\displaystyle\,\displaystyle\int_{0}^{1}({\omega}_{0}+1){\omega}_{3}{\omega}_{20}{\omega}_{1}=i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}({\texttt{a}}-{\texttt{x}}_{-1}-{\texttt{x}}_{1}){\texttt{d}}_{-1,1}({\texttt{y}}-1)$
$\displaystyle=$
$\displaystyle\,14{\beta}(4)-16\operatorname{{{Im}}}\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}-\displaystyle\frac{1}{12}\pi\log^{3}2-\displaystyle\frac{3}{16}\pi^{3}\log
2$ $\displaystyle\,\hskip
56.9055pt+\displaystyle\frac{1}{8}\pi\log^{2}2+\displaystyle\frac{5}{32}\pi^{3}-4\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}\approx
3.937040753.$

So we see the weight can increase by one with a starting $\chi$-block.

######  Example 5.7.

For a sum of mixed parities without ${\sigma}$-block but with a starting
${\tau}^{\star}$-block, by (4.38) and (4.39) we have

$\displaystyle\,\displaystyle\sum_{n_{1}\geq
n_{2}>0}\displaystyle\frac{b_{n_{1}}}{(2n_{1}+1)^{2}(2n_{2}-1)}=\displaystyle\int_{0}^{1}{\omega}_{3}({\omega}_{5}{\omega}_{3}+{\omega}_{2}){\omega}_{1}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{3}{\omega}_{20}{\omega}_{1}-\displaystyle\int_{0}^{1}{\omega}_{0}{\omega}_{3}{\omega}_{1}\qquad(\text{since
}{\omega}_{5}=-d\sqrt{1-t^{2}},\ \sqrt{1-t^{2}}{\omega}_{3}={\omega}_{0})$
$\displaystyle=$
$\displaystyle\,-i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}({\texttt{y}}+{\texttt{z}}){\texttt{d}}_{-1,1}+i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}{\texttt{y}}$
$\displaystyle=$
$\displaystyle\,\displaystyle\frac{3}{16}\pi^{3}-8\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}+\displaystyle\frac{1}{4}\pi\log^{2}2\approx
1.630404535576,$

where
${\texttt{y}}+{\texttt{z}}=-{\texttt{a}}-{\texttt{x}}_{-1}-{\texttt{x}}_{1}$.
Thus the weight is unchanged as predicted by Thm. 4.4(b).

######  Example 5.8.

For a sum of mixed parities with a starting ${\sigma}$-block, followed by a
${\tau}$-block then a $\chi$-block, by (4.36), (4.37) and (4.40) we get

$\displaystyle\,\displaystyle\sum_{n_{1}>n_{2}\geq
n_{3}>0}\displaystyle\frac{b_{n_{1}}}{(2n_{1})^{2}(2n_{2}+1)(2n_{3}-1)^{2}}=\displaystyle\int_{0}^{1}{\omega}_{1}{\omega}_{20}{\omega}_{5}({\omega}_{1}+1){\omega}_{3}{\omega}_{1}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{1}\big{(}{\omega}_{20}{\omega}_{3}-{\omega}_{3}{\omega}_{0}\big{)}{\omega}_{3}{\omega}_{1}\qquad(\text{since
}{\omega}_{5}=-d\sqrt{1-t^{2}},\ \sqrt{1-t^{2}}{\omega}_{3}={\omega}_{0})$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}{\texttt{d}}_{-1,1}({\texttt{y}}+{\texttt{z}}){\texttt{d}}_{-i,i}-\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}{\texttt{y}}{\texttt{d}}_{-1,1}{\texttt{d}}_{-i,i}$
$\displaystyle=$
$\displaystyle\,\displaystyle\frac{G}{4}\Big{(}\displaystyle\frac{\pi^{3}}{4}-32\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}+\pi\log^{2}2\Big{)}-\displaystyle\frac{15}{2}\bigg{(}\operatorname{Li}_{5}\Big{(}\displaystyle\frac{1}{2}\Big{)}+\log
2\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1}{2}\Big{)}\bigg{)}-\displaystyle\frac{1}{4}\log^{5}2$
$\displaystyle\,+6\pi{\beta}(4)+24\operatorname{{{Re}}}\operatorname{Li}_{3,1,1}(1,1,I)+\displaystyle\frac{1}{384}\bigg{(}80\pi^{2}\log^{3}2-15\pi^{4}\log
2-87\pi^{2}\zeta(3)-2250\zeta(5)\bigg{)}$ $\displaystyle\,\approx
0.98658158829.$

So weight is unchanged in every step as predicted by Thm. 4.4(b).

######  Example 5.9.

For a sum of mixed parities with ${\sigma}$-block followed by a $\chi$-block,
by (4.36) and (4.39) we obtain

$\displaystyle\,\displaystyle\sum_{n_{1}>n_{2}>0}\displaystyle\frac{b_{n_{1}}}{(2n_{1})^{2}(2n_{2}-1)}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{\pi/2}dt\,dt\Big{(}\sin t\,dt\csc
t\,dt+\tan t\,dt\Big{)}\,dt$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{\pi/2}dt\,dt\Big{(}d(-\cos t)\,\csc
t\,dt+\tan t\,dt\Big{)}\,dt$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{\pi/2}dt\,\cos t\,dt\csc
t\,dt+dt\Big{(}\cot t\,dt+\tan t\,dt\Big{)}\,dt$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{\pi/2}dt\,(\sin t-1)\csc t\,dt+dt\,\csc
t\sec t\,dt\,dt$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{\pi/2}dt\,dt-\csc t\,dt+dt\,\csc t\sec
t\,dt\,dt$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{1}{\omega}_{1}-{\omega}_{3}{\omega}_{1}+{\omega}_{1}{\omega}_{20}{\omega}_{1}$
$\displaystyle=$
$\displaystyle\,-i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}+\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}({\texttt{y}}+tz){\texttt{d}}_{-i,i}-{\texttt{d}}_{-i,i}^{2}$
$\displaystyle=$
$\displaystyle\,\displaystyle\frac{1}{8}\pi^{2}-2G+\displaystyle\frac{7}{4}\zeta(3)\approx
1.5053689423.$

Note that not only the weight is a mix of 2 and 3, but this is a mix of both
real and imaginary parts of some CMZVs of level 4. The main complication is
brought in by the 1-form $\sin t\,dt$ (corresponding to ${\omega}_{5}$)
appearing in the $\chi$-block, which moves to the front after integration by
parts if the block in front is a ${\sigma}$-block but disappears if the block
in front is a ${\tau}^{\star}$-block.

######  Example 5.10.

We apply the idea of proof of Thm. 4.1 to the sum

$S:=\displaystyle\sum_{n_{1}\geq n_{2}\geq n_{3}\geq
1}\displaystyle\frac{b_{n_{1}}}{n_{1}^{2}(2n_{2}+1)(2n_{3}-1)}.$

First, we break the sum in two sub-sums $S=S_{1}+S_{2}$ where

$\displaystyle S_{1}=$ $\displaystyle\,\displaystyle\sum_{n_{1}>n_{2}\geq
n_{3}\geq 1}\displaystyle\frac{b_{n_{1}}}{n_{1}^{2}(2n_{2}+1)(2n_{3}-1)},$
$\displaystyle S_{2}=$ $\displaystyle\,\displaystyle\sum_{n_{1}\geq n_{3}\geq
1}\displaystyle\frac{b_{n_{1}}}{n_{1}^{2}(2n_{1}+1)(2n_{3}-1)}.$

Then by (4.36), (4.37) and (4.39) we see that

$\displaystyle S_{1}=$
$\displaystyle\,4\displaystyle\int_{0}^{1}{\omega}_{1}\displaystyle\sum_{n_{2}\geq
n_{3}>0}\displaystyle\frac{b_{n_{2}}(t)}{(2n_{2}+1)(2n_{3}-1)}{\omega}_{1}$
$\displaystyle=$
$\displaystyle\,4\displaystyle\int_{0}^{1}{\omega}_{1}{\omega}_{20}\displaystyle\sum_{n_{3}>0}\displaystyle\frac{b_{n_{3}}(t)}{2n_{3}-1}{\omega}_{1}$
$\displaystyle=$
$\displaystyle\,4\displaystyle\int_{0}^{1}{\omega}_{1}({\omega}_{0}+{\omega}_{2}){\omega}_{5}{\omega}_{3}{\omega}_{1}+4\displaystyle\int_{0}^{1}{\omega}_{1}({\omega}_{0}+{\omega}_{2}){\omega}_{2}{\omega}_{1}.$

By the change of variables $t\to\displaystyle\frac{1-t^{2}}{1+t^{2}}$ we
obtain

$\displaystyle S_{1}=$
$\displaystyle\,-4A-4\bar{A}-4\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}({\texttt{a}}+{\texttt{x}}_{-i}+{\texttt{x}}_{i})({\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1}){\texttt{d}}_{-i,i},$

where $\bar{A}$ is the complex conjugation of

$\displaystyle
A=\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\displaystyle\frac{dt}{(i-t)^{2}}({\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1}){\texttt{d}}_{-i,i}.$

Integration by parts yields

$\displaystyle A=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\displaystyle\frac{{\texttt{x}}_{-1}-{\texttt{x}}_{1}}{i-t}({\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1}){\texttt{d}}_{-i,i}-\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\displaystyle\frac{{\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1}}{i-t}{\texttt{d}}_{-i,i}.$

Explicitly, for all fourth roots of unity $\xi\neq i$

$\displaystyle\displaystyle\frac{{\texttt{a}}}{i-t}=i({\texttt{x}}_{i}-{\texttt{a}}),\
\displaystyle\frac{{\texttt{x}}_{\xi}}{i-t}=\displaystyle\frac{{\texttt{d}}_{i,\xi}}{\xi-i}.$
(5.57)

We obtain

$\displaystyle A=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\left(\displaystyle\frac{1}{-1-i}{\texttt{d}}_{i,-1}-\displaystyle\frac{1}{1-i}{\texttt{d}}_{i,1}\right)({\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1}){\texttt{d}}_{-i,i}$
$\displaystyle-$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\left(-i({\texttt{x}}_{i}+{\texttt{a}})-\displaystyle\frac{1}{1+i}{\texttt{d}}_{i,-1}+\displaystyle\frac{1}{1-i}{\texttt{d}}_{i,1}\right){\texttt{d}}_{-i,i}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\left(-{\texttt{x}}_{i}+\displaystyle\frac{1-i}{2}{\texttt{x}}_{-1}+\displaystyle\frac{1+i}{2}{\texttt{x}}_{1}\right)({\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1}){\texttt{d}}_{-i,i}$
$\displaystyle-$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\left(-i{\texttt{a}}+\displaystyle\frac{1-i}{2}{\texttt{x}}_{-1}-\displaystyle\frac{1+i}{2}{\texttt{x}}_{1}\right){\texttt{d}}_{-i,i}.$

Therefore using Au’s package [4] we find that

$\displaystyle
S_{1}=-4\displaystyle\int_{0}^{1}\Big{(}{\texttt{d}}_{-i,i}({\texttt{a}}+{\texttt{x}}_{-1}+{\texttt{x}}_{1})^{2}{\texttt{d}}_{-i,i}-{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}^{2}{\texttt{d}}_{-i,i}\Big{)}=8G^{2}\approx
6.71194375752575.$

Now we turn to $S_{2}$. Set

$\displaystyle S_{2}(x)=$ $\displaystyle\,\displaystyle\sum_{n_{1}\geq
n_{3}\geq 1}\displaystyle\frac{b_{n_{1}}(x)}{n_{1}^{2}(2n_{1}+1)(2n_{3}-1)}.$

By partial fraction

$\displaystyle S_{2}(x)=$ $\displaystyle\,\displaystyle\sum_{m\geq
n>0}\left(\displaystyle\frac{4b_{m}(x)}{(2m+1)(2n-1)}-\displaystyle\frac{2b_{m}(x)}{m(2n-1)}+\displaystyle\frac{b_{m}(x)}{m^{2}(2n-1)}\right)$
$\displaystyle=$ $\displaystyle\,\displaystyle\sum_{m\geq
n>0}\displaystyle\frac{4b_{m}(x)}{(2m+1)(2n-1)}-\displaystyle\sum_{m>n>0}\left(\displaystyle\frac{2b_{m}(x)}{m(2n-1)}-\displaystyle\frac{b_{m}(x)}{m^{2}(2n-1)}\right)-\displaystyle\sum_{n>0}\left(\displaystyle\frac{2b_{n}(x)}{n(2n-1)}-\displaystyle\frac{b_{n}(x)}{n^{2}(2n-1)}\right)$
$\displaystyle=$
$\displaystyle\,4\displaystyle\sum_{n>0}\left(f_{20}(x)\displaystyle\int_{0}^{x}\displaystyle\frac{b_{n}(t)}{2n-1}{\omega}_{1}-f_{2}(x)\displaystyle\int_{0}^{x}\displaystyle\frac{b_{n}(t)}{2n-1}{\omega}_{1}+\displaystyle\int_{0}^{x}{\omega}_{1}\displaystyle\frac{b_{n}(t)}{2n-1}{\omega}_{1}-\displaystyle\frac{b_{n}(x)}{4n^{2}}\right)$
$\displaystyle=$
$\displaystyle\,4\big{(}f_{20}(x)-f_{2}(x)\big{)}\displaystyle\int_{0}^{x}({\omega}_{5}{\omega}_{3}{\omega}_{1}+{\omega}_{2}{\omega}_{1})+4\displaystyle\int_{0}^{x}{\omega}_{1}({\omega}_{5}{\omega}_{3}{\omega}_{1}+{\omega}_{2}{\omega}_{1})-4\displaystyle\int_{0}^{x}{\omega}_{1}{\omega}_{1}$

by (4.35)–(4.39). Note that

$\displaystyle\displaystyle\lim_{x\to 1^{-}}\displaystyle\int_{0}^{1}$
$\displaystyle\,{\omega}_{5}{\omega}_{3}{\omega}_{1}=-i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\bigg{(}\displaystyle\frac{dt}{(i-t)^{2}}+\displaystyle\frac{dt}{(i+t)^{2}}\bigg{)}=2\,\operatorname{{{Re}}}\left(-i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\displaystyle\frac{{\texttt{d}}_{-1,1}}{i-t}\right)$
$\displaystyle=$
$\displaystyle\,2\,\operatorname{{{Im}}}\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\left(\displaystyle\frac{1}{-1-i}{\texttt{d}}_{i,-1}-\displaystyle\frac{1}{1-i}{\texttt{d}}_{i,1}\right)=\operatorname{{{Im}}}\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\left(-2{\texttt{x}}_{i}+(1-i){\texttt{x}}_{-1}+(1+i){\texttt{x}}_{1}\right),$

which is a finite value in $i\mathsf{CMZV}_{2}^{4}$. Further, setting
${\lambda}(x)=\sqrt{\displaystyle\frac{1-x}{1+x}}$ we get

$\displaystyle\displaystyle\int_{0}^{x}{\omega}_{2}{\omega}_{1}=$
$\displaystyle\,-i\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{d}}_{-i,i}({\texttt{a}}+{\texttt{x}}_{-i}+{\texttt{x}}_{i}).$

We only need to take care of

$\displaystyle\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{x}}_{-i}{\texttt{a}}=$
$\displaystyle\,\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{x}}_{-i}\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{a}}-\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{a}}{\texttt{x}}_{-i}=-\log\bigg{(}\displaystyle\frac{i+{\lambda}(x)}{i+1}\bigg{)}\log{\lambda}(x)-\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{a}}{\texttt{x}}_{-i}$

of which the last term $\to-\operatorname{Li}_{2}(i)$ as $x\to 1^{-}$. Hence

$\displaystyle\displaystyle\lim_{x\to
1^{-}}(f_{20}(x)-f_{2}(x))\displaystyle\int_{{\lambda}(x)}^{1}{\texttt{x}}_{-i}{\texttt{a}}=$
$\displaystyle\,-\log\bigg{(}\displaystyle\frac{i}{i+1}\bigg{)}\displaystyle\lim_{x\to
1^{-}}\sqrt{1-x^{2}}\log\sqrt{\displaystyle\frac{1-x}{1+x}}$ $\displaystyle=$
$\displaystyle\,-\log\bigg{(}\displaystyle\frac{i}{i+1}\bigg{)}\displaystyle\frac{\sqrt{2}}{2}\displaystyle\lim_{{\varepsilon}\to
0^{+}}\sqrt{{\varepsilon}}\log({\varepsilon}/2)=0.$

Thus

$\displaystyle S_{2}=$ $\displaystyle\,\displaystyle\lim_{x\to
1^{-}}S_{2}(x)=4\displaystyle\int_{0}^{1}{\omega}_{1}({\omega}_{5}{\omega}_{3}{\omega}_{1}+{\omega}_{2}{\omega}_{1})-4\displaystyle\int_{0}^{1}{\omega}_{1}{\omega}_{1}$
$\displaystyle=$
$\displaystyle\,4\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\bigg{(}{\texttt{z}}-{\texttt{d}}_{-1,1}\Big{(}\displaystyle\frac{dt}{(i-t)^{2}}+\displaystyle\frac{dt}{(i+t)^{2}}\Big{)}\bigg{)}{\texttt{d}}_{-i,i}-2\left(\displaystyle\int_{0}^{1}{\omega}_{1}\right)^{2}$
$\displaystyle=$
$\displaystyle\,-4\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}({\texttt{a}}+{\texttt{x}}_{-i}+{\texttt{x}}_{i}){\texttt{d}}_{-i,i}-2B-2\bar{B}-\displaystyle\frac{\pi^{2}}{2},$

where

$\displaystyle B=$
$\displaystyle\,2\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\displaystyle\frac{{\texttt{d}}_{-1,1}}{i-t}{\texttt{d}}_{-i,i}-2\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\displaystyle\frac{{\texttt{d}}_{-i,i}}{i-t}$
$\displaystyle=$
$\displaystyle\,2\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\left(\displaystyle\frac{1}{-1-i}{\texttt{d}}_{i,-1}-\displaystyle\frac{1}{1-i}{\texttt{d}}_{i,1}\right){\texttt{d}}_{-i,i}-\,2\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\left(\displaystyle\frac{1}{-2i}({\texttt{x}}_{i}-{\texttt{x}}_{-i})-\displaystyle\frac{1}{i-t}-i\right)$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\big{(}{\texttt{x}}_{-1}+{\texttt{x}}_{1}-i{\texttt{d}}_{-1,1}-2{\texttt{x}}_{i}\big{)}{\texttt{d}}_{-i,i}-i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}{\texttt{d}}_{-i,i}+2\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\displaystyle\frac{{\texttt{d}}_{-1,1}}{i-t}+2i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}$
$\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\big{(}{\texttt{x}}_{-1}+{\texttt{x}}_{1}-i{\texttt{d}}_{-1,1}-2{\texttt{x}}_{i}\big{)}({\texttt{d}}_{-i,i}+1)-i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}{\texttt{d}}_{-i,i}+2i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}.$

Hence

$\displaystyle
B+\bar{B}=2\left(\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}\big{(}{\texttt{x}}_{-1}+{\texttt{x}}_{1}-{\texttt{x}}_{-i}-{\texttt{x}}_{i}\big{)}{\texttt{d}}_{-i,i}+\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-i,i}+i\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}\right).$

We can see that $S_{2}=7\zeta(3)-8G\approx 1.0866735685$ by Au’s package [4]
and therefore

$\displaystyle\sum_{n_{1}\geq n_{2}\geq n_{3}\geq
1}\displaystyle\frac{b(n_{1})}{n_{1}^{2}(2n_{2}+1)(2n_{3}-1)}=7\zeta(3)+8G^{2}-8G\approx
7.79861732643.$

In the next three examples, we consider some Apéry-type series which involve
the square of the central binomial coefficients.

######  Example 5.11.

Since $\chi$-block does not appear in this example we see that the weight of
the CMZVs is the same as the weight of the series, as predicted by Thm.
4.7(a):

$\displaystyle\displaystyle\sum_{n_{1}\geq
n_{2}>0}\displaystyle\frac{b_{n_{1}}^{2}}{(2n_{1}+1)^{4}(2n_{2})}=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{1}{\omega}_{0}{\omega}_{3}{\omega}_{2}{\omega}_{1}=\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{z}}{\texttt{d}}_{-1,1}{\texttt{y}}{\texttt{d}}_{-i,i}=W_{5}\approx
0.04433915814,$

where

$\displaystyle W_{5}=$ $\displaystyle\,5\log
2\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1}{2}\Big{)}+21\operatorname{Li}_{5}\Big{(}\displaystyle\frac{1}{2}\Big{)}+\pi\bigg{(}16\operatorname{{{Im}}}\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}-17{\beta}(4)+8\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}\log
2\bigg{)}$ $\displaystyle\,+\displaystyle\frac{379}{2880}\pi^{4}\log
2+\displaystyle\frac{1}{30}\log^{5}2-16\operatorname{{{Re}}}\operatorname{Li}_{3,1,1}(1,1,i)-\displaystyle\frac{1}{192}\pi^{2}\Big{(}16\log^{3}2-29\zeta(3)\Big{)}-\displaystyle\frac{27}{4}\zeta(5).$

######  Example 5.12.

The sum next has weight 4 but due to the ${\sigma}$-$\chi$-block chain we need
to use CMZVs of weight of both 3 and 4 to express it:

$\displaystyle\displaystyle\sum_{n_{1}>n_{2}>0}$
$\displaystyle\,\displaystyle\frac{b_{n_{1}}^{2}}{(2n_{1})^{3}(2n_{2}-1)}=\displaystyle\int_{0}^{1}{\omega}_{3}{\omega}_{1}({\omega}_{5}{\omega}_{3}+{\omega}_{2}){\omega}_{1}=\displaystyle\int_{0}^{1}{\omega}_{3}({\omega}_{1}{\omega}_{0}-dt{\omega}_{3}){\omega}_{1}+{\omega}_{3}{\omega}_{1}{\omega}_{2}{\omega}_{1}$
$\displaystyle\,\ \hskip 56.9055pt\qquad(\text{since
}{\omega}_{5}=-d\sqrt{1-t^{2}},\ \sqrt{1-t^{2}}{\omega}_{3}={\omega}_{0},\
\sqrt{1-t^{2}}{\omega}_{1}=dt)$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}({\omega}_{3}{\omega}_{1}-{\omega}_{1}{\omega}_{3}){\omega}_{1}+{\omega}_{3}{\omega}_{1}{\omega}_{0}{\omega}_{1}+{\omega}_{3}{\omega}_{1}{\omega}_{2}{\omega}_{1}\qquad(\text{since
}t{\omega}_{3}={\omega}_{1})$ $\displaystyle=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}({\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}-{\texttt{d}}_{-1,1}{\texttt{d}}_{-i,i})-{\texttt{d}}_{-i,i}({\texttt{y}}+{\texttt{z}}){\texttt{d}}_{-i,i}{\texttt{d}}_{-1,1}$
$\displaystyle=$ $\displaystyle\,2G^{2}-G\pi\log
2+\displaystyle\frac{1}{64}\pi\bigg{(}3\pi^{3}-128\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}+4\pi\log^{2}2\bigg{)}+2G\pi-\displaystyle\frac{21}{4}\zeta(3)$
$\displaystyle\approx$ $\displaystyle\,0.40829155182.$

######  Example 5.13.

The series in this example does not have a ${\sigma}$-$\chi$-block chain so
that there is no weight drop. But the first block is a $\chi$-block so the
weight can increase by two as predicted by Thm. 4.7(b):

$\displaystyle\displaystyle\sum_{n_{1}>n_{2}>0}\displaystyle\frac{b_{n_{1}}^{2}}{(2n_{1}-1)^{3}(2n_{2})}=$
$\displaystyle\,\displaystyle\int_{0}^{1}{\omega}_{1}({\omega}_{0}+1)^{2}{\omega}_{3}{\omega}_{2}{\omega}_{1}$
$\displaystyle=$
$\displaystyle\,-\displaystyle\int_{0}^{1}{\texttt{d}}_{-i,i}{\texttt{z}}{\texttt{d}}_{-1,1}({\texttt{y}}-1)^{2}{\texttt{d}}_{-i,i}=W_{6}+2W_{5}+W_{4}\approx
0.38530528471,$

where $W_{5}$ is defined in Example 5.11 and

$\displaystyle W_{4}=$
$\displaystyle\,-2G^{2}-\displaystyle\frac{49}{720}\pi^{4}+2\pi\operatorname{{{Im}}}\operatorname{Li}_{3}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}-\displaystyle\frac{11}{48}\pi^{2}\log^{2}2+\displaystyle\frac{1}{6}\log^{4}2+G\pi\log^{2}2+4\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1}{2}\Big{)},$
$\displaystyle W_{6}=$
$\displaystyle\,68\operatorname{Li}_{6}\Big{(}\displaystyle\frac{1}{2}\Big{)}-\displaystyle\frac{7655}{27648}\pi^{6}+\displaystyle\frac{61}{2}\pi\operatorname{{{Im}}}\operatorname{Li}_{4,1}(i,1)-\displaystyle\frac{41}{2}\pi\operatorname{{{Im}}}\operatorname{Li}_{4,1}(i,-1)+96\pi\operatorname{{{Im}}}\operatorname{Li}_{5}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}$
$\displaystyle\,-19\pi{\beta}(4)\log
2+32\pi\operatorname{{{Im}}}\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1+i}{2}\Big{)}\log
2-\displaystyle\frac{181}{2880}\pi^{4}\log^{2}2-\displaystyle\frac{1}{96}\pi^{2}\log^{4}2+\displaystyle\frac{1}{90}\log^{6}2$
$\displaystyle\,-\displaystyle\frac{169}{4}\zeta(\bar{5},1)-\displaystyle\frac{5}{12}\pi^{2}\operatorname{Li}_{4}\Big{(}\displaystyle\frac{1}{2}\Big{)}+10\log
2\operatorname{Li}_{5}\Big{(}\displaystyle\frac{1}{2}\Big{)}-24G{\beta}(4)-6\pi^{2}\log
2\zeta(3)$
$\displaystyle\,-24\operatorname{{{Re}}}\operatorname{Li}_{4,2}(-1,i)+64\operatorname{{{Re}}}\operatorname{Li}_{3,1,1,1}(1,1,1,i)+\displaystyle\frac{8195}{128}\zeta(3)^{2}+\displaystyle\frac{2821}{32}\log
2\zeta(5).$

Acknowledgement. Ce Xu is supported by the National Natural Science Foundation
of China [Grant No. 12101008], the Natural Science Foundation of Anhui
Province [Grant No. 2108085QA01] and the University Natural Science Research
Project of Anhui Province [Grant No. KJ2020A0057]. Jianqiang Zhao is supported
by the Jacobs Prize from The Bishop’s School.

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