text:
\begin{document}

\title {\bf Duality of Weighted Sum Formulas of Alternating Multiple $T$-Values}
\author{
{Ce Xu\thanks{Email: 19020170155420@stu.xmu.edu.cn}}\\[1mm]
\small $*$ School of Mathematics and Statistics, Anhui Normal University,\\ \small Wuhu 241000, P.R. China\\
[5mm]
Dedicated to professor Masanobu Kaneko on the occasion of his 60th birthday}

\date{}
\maketitle \noindent{\bf Abstract} Recently, a new kind of multiple zeta value level two $T({\bf k})$ (which is called multiple $T$-values) was introduced and studied by Kaneko and Tsumura. In this paper, we define a kind of alternating version of multiple $T$-values, and study several duality formulas of weighted sum formulas about alternating multiple $T$-values by using the methods of iterated integral representations and series representations. Some special values of alternating multiple $T$-values can also be obtained.
\\[2mm]
\noindent{\bf Keywords}: Kaneko-Tsumura multiple $T$-values; Alternating multiple $T$-values; Weighted sum formulas; Duality.

\noindent{\bf AMS Subject Classifications (2020):} 11M99; 11M06; 11M32.

\section{Introduction}

We begin with some basic notations. A finite sequence ${\bf k} := (k_1,\ldots, k_r)$ of positive integers is called an index. We
put
\[|{\bf k}|:=k_1+\cdots+k_r,\quad d({\bf k}):=r,\]
and call them the weight and the depth of ${\bf k}$, respectively. If $k_r>1$, ${\bf k}$ is called admissible. Let $I(k,r)$ be the set of all indices of weight $k$ and depth $r$.

For an admissible index ${\bf k}=(k_1,\ldots,k_r)$, the multiple zeta values (abbr. MZVs) are defined by
\begin{align}
\zeta(k_1,k_2,\ldots,k_r):=\sum\limits_{0<m_1<\cdots<m_r } \frac{1}{m_1^{k_1}m_2^{k_2}\cdots m_r^{k_r}}.
\end{align}
When $r=1$ and $k_1=k$, $\z(k)$ stands for Riemann zeta value, which is defined by
\begin{align*}
\z(k):=\su \frac{1}{n^k} \quad (k\in \N\setminus\{1\}).
\end{align*}
The systematic study of multiple zeta values began in the early 1990s with the works of Hoffman \cite{H1992} and Zagier \cite{DZ1994}. After that it has been attracted a lot of research on them in the last three decades (see, for example, the book of Zhao \cite{Z2016}).

So far, surprisingly little work has been done on several variants of multiple zeta values. In recent papers \cite{KTA2018,KTA2019}, Kaneko and Tsumura introduced and studied a new kind of multiple zeta values of level two
\begin{align}
T(k_1,k_2,\ldots,k_r):&=2^r \sum_{0<m_1<\cdots<m_r\atop m_j\equiv j\ {\rm mod}\ 2} \frac{1}{m_1^{k_1}m_2^{k_2}\cdots m_r^{k_r}}\nonumber\\
&=2^r\sum\limits_{0<n_1<\cdots<n_r} \frac{1}{(2n_1-1)^{k_1}(2n_2-2)^{k_2}\cdots (2n_r-r)^{k_r}}\label{a2}
\end{align}
for an admissible index ${\bf k}=(k_1,\ldots,k_r)$, which is called the `multiple $T$-value' (MTV for short). This is in contrast to Hoffman's multiple $t$-value (MtV for short) defined by
\begin{align}
t(k_1,k_2,\ldots,k_r):&=\sum_{0<m_1<\cdots<m_r\atop \forall m_j:\ {\rm odd}} \frac{1}{m_1^{k_1}m_2^{k_2}\cdots m_r^{k_r}},
\end{align}
which was introduced and studied in his recent paper \cite{H2019} as another variant of multiple zeta values of level two.

In (\ref{a2}), we put a bar
on top of $k_j$ if there is a sign $(-1)^{n_j}$ appearing in the denominator on the right. For example,
\begin{align*}
T(k_1,{\overline {k_2}},k_3,{\overline {k_4}}):=2^4\sum\limits_{0<n_1<n_2<n_3<n_4} \frac{(-1)^{n_2+n_4}}{(2n_1-1)^{k_1}(2n_2-2)^{k_2}(2n_3-3)^{k_4}(2n_4-4)^{k_4}}.
\end{align*}
The sums of types above (one of more the $k_j$ barred) are called the alternating multiple $T$-values.
In particular, we let
\begin{align}
\T(k_1,k_2,\ldots,k_r):=2^r\sum\limits_{0<n_1<\cdots<n_r} \frac{(-1)^{n_r}}{(2n_1-1)^{k_1}(2n_2-2)^{k_2}\cdots (2n_r-r)^{k_r}},
\end{align}
for an index ${\bf k}=(k_1,\ldots,k_r)$. In \cite[Corollary 3.4]{WX2020}, the author and Wang proved that the four alternating linear sums
\begin{align*}
&(1-(-1)^{p+q})\su \frac{h^{(p)}_n}{n^q},\quad (1+(-1)^{p+q})\su \frac{{\bar h}^{(p)}_n}{n^q},\\
&(1+(-1)^{p+q})\su \frac{{\bar h}^{(p)}_n}{n^q}(-1)^{n-1},\quad (1-(-1)^{p+q})\su \frac{h^{(p)}_n}{n^q}(-1)^{n-1}
\end{align*}
can be expressed in terms of (alternating) Riemann zeta values and (alternating) $t$-values.
Here $h_n^{(p)}$ stands for odd harmonic number of order $p$ defined by
\begin{align*}
h_n^{(p)}:=\sum\limits_{k=1}^n \frac{1}{(k-1/2)^p},\quad h_n\equiv h_n^{(1)} \quad{\rm and}\quad h_0^{(p)}:=0
\end{align*}
and ${\bar h}_n^{(p)}$ stands for alternating odd harmonic number of order $p$ defined by
\begin{align*}
{ \bar h}_n^{(p)}: = \sum\limits_{k=1}^n {\frac{(-1)^{k-1}}{{{(k-1/2)^p}}}},\quad {\bar h}_n\equiv{\bar h}^{(1)}_n,\quad {\bar h}_0^{(p)}:=0.
\end{align*}
Hence, according to definitions, we know that the two (alternating) double $T$-values
\[T(k_1,k_2)=\frac{1}{2^{k_1+k_2-2}}\su \frac{h_n^{(k_1)}}{n^{k_2}},\quad T(k_1,\overline{k_2})=\frac{1}{2^{k_1+k_2-2}}\su \frac{h_n^{(k_1)}}{n^{k_2}}(-1)^{n-1}\]
can be evaluated by (alternating) Riemann zeta values and (alternating) $t$-values if weight $k_1+k_2$ is odd (Note that the explicit evaluation of $T(k_1,k_2)$ with weight $k_1+k_2$ odd was first found by Kaneko-Tsumura in \cite{KTA2018}). The two (alternating) double $T$-values
\[T(\overline{k_1},k_2)=-\frac{1}{2^{k_1+k_2-2}}\su \frac{{\bar h}_n^{(k_1)}}{n^{k_2}},\quad T(\overline{k_1},\overline{k_2})=-\frac{1}{2^{k_1+k_2-2}}\su \frac{{\bar h}_n^{(k_1)}}{n^{k_2}}(-1)^{n-1}\]
can be evaluated by (alternating) Riemann zeta values and (alternating) $t$-values if weight $k_1+k_2$ is even.
Here the alternating Riemann zeta value ${\bar \z}(k)$ is denoted by
\begin{align}
{\bar \z}(k):=\su \frac{(-1)^{n-1}}{n^k} \quad \text{and} \quad {\bar \z}(0):=\frac{1}{2}\quad (k\in \N),
\end{align}
and the alternating $t$-value ${\bar t}(k)$ is denoted by
\begin{align}
{\bar t}(k):=\su \frac{(-1)^{n-1}}{(n-1/2)^k}\quad (k\in\N).
\end{align}
Clearly, ${\bar \z}(k)=(1-2^{1-k})\z(k)$ if $k>1$ and $\T(k)=T({\bar k})=-{\bar t}(k)/{2^{k-1}}$. Note that from \cite{G1989}, for nonnegative integer $k$, we have the generating function of ${\bar t}(2k+1)$,
\begin{align*}
\frac{\pi}{\cos(\pi s)}=2\sum\limits_{k=0}^{\infty} {\bar t}(2k+1)s^{2k}=\sum\limits_{k=0}^{\infty} \frac{(-1)^kE_{2k}\pi^{2k+1}}{(2k)!}s^{2k},
\end{align*}
where $E_{2k}$ is Euler number. Thus, we compute
\begin{align}
{\bar t}(2k+1)=\frac{(-1)^kE_{2k}\pi^{2k+1}}{2(2k)!}\quad (k\geq 0).
\end{align}

For multiple zeta values, the classical sum formula is widely known and its variants are enormous (see \cite[Chapter 5]{Z2016}), for example, Zagier \cite{DZ1994} conjectured the following weighted sum formula for MZVs that for an admissible index ${\bf k}$,
\begin{align*}
\sum_{{\bf k}\in I(k,r)}\z(k_1,k_2,\ldots,k_r)=\z(k)\quad (k\geq 2)
\end{align*}
which was first proved by Granville in \cite{G1997}. Some other related results for weighted sums of MZVs may be seen in the works of \cite{GKZ2006,G2016,GL2015,H2017,LQ2019,M2013,N2009} and references therein. For Hoffman's multiple $t$-values, many mathematicians studied the weighed sums formulas problem of MtVs, for example see
\cite{LX2020,SC2012,SJ2017,Z2015}.
In contrast, we know very little about the weighed sum formula for MTVs. For MTVs, only certain formulas are found in depths 2 and 3, for detail see Kaneko and Tsumura \cite[Theorems 3.2 and 3.3]{KTA2019}. Moreover, in \cite{KTA2019}, Kaneko and Tsumura also conjectured an analogue of Machide's formula for MTVs, see \cite[Conjecture 4.6]{KTA2019}. In this paper, we will discuss a kind of weighted sums of alternating MTVs.
For an index ${\bf k}=(k_1,\ldots,k_r)$ and positive integer $m$, we define a weighted sum of alternating MTVs with arguments of weight $k$ and depth $r$ by
\begin{align}
W_m(k,r):=\sum_{{\bf k}\in I(k,r)}\binom{k_r+m-2}{m-1} \T(k_1,\ldots,k_{r-1},k_r+m-1),
\end{align}
if $m=1$, we let $W(k,r):=W_1(k,r)$. Obviously, with the help of \cite[Eq.(3.10)]{WX2020}, by a direct calculation, we arrive at the conclusion that if $k>1$ is an odd, then the weighted sum
$W(k,2)$
can be expressed in terms of (alternating) Riemann zeta values (for detailed, see Theorem \ref{thmc4} in the present paper).

For positive integers $j$ and $p$ with $1\leq j\leq p$, we let
\begin{align}
Z(j,p):=\sum_{k=1}^{p-j}(-2)^k\sum\limits_{i_0<i_1<\cdots<i_{k-1}<i_k,\atop i_0=j,i_k=p} \prod\limits_{l=1}^k {\bar \z}(2i_l-2i_{l-1}),\quad Z(p,p):=1.
\end{align}
It is clear that
$Z(j,p)=a_{j,p}\zeta(2p-2j)\quad (a_{j,p}\in \mathbb{Q})$.
For instance, we have $Z(j,j+1)=-\z(2),\ Z(j,j+2)=3\z(4)/4$ and $Z(j,j+3)=-3\z(6)/16$. In particular, the author and Zhao \cite{XZ2020} shown that
\begin{align}
Z(j,p)=\frac{(-1)^{p-j}\pi^{2p-2j}}{(2p-2j+1)!}.
\end{align}

The purpose of present paper is to evaluate the duality of products of weighted sum $W(k,r)$ and $Z(j,p)$ by using the method of iterated integrals. Our main results are following three theorems.

\begin{thm}\label{thm1} For positive integers $m$ and $p$, we have
\begin{align}\label{a9}
(-1)^m \sum_{j=1}^p W(2j+2m-1,2m)Z(j,p)=(-1)^p \sum_{j=1}^m W(2j+2p-1,2p)Z(j,m).
\end{align}
\end{thm}

\begin{thm}\label{thm2} For positive integers $m$ and $p$, we have
\begin{align}\label{a10}
(-1)^m \sum_{j=1}^p W(2j+2m-3,2m-1)Z(j,p)=(-1)^p \sum_{j=1}^m W(2j+2p-3,2p-1)Z(j,m).
\end{align}
\end{thm}

\begin{thm}\label{thm3} For positive integers $k,m$ and $r$, we have
\begin{align}\label{a11}
&W_m(k+r-1,r)+(-1)^r W_k(m+r-1,r)\nonumber\\
&=\sum_{j=1}^{r-1} (-1)^{j-1} T(\{1\}_{k-2},{\bar 1},\{1\}_{j-1},{\bar 1})T(\{1\}_{m-2},{\bar 1},\{1\}_{r-j-1},{\bar 1}),
\end{align}
where $(\{1\}_{-1},{\bar 1}):=\emptyset$, and $\{l\}_m$ denotes the sequence $\underbrace{l,\ldots,l}_{m \text{\;times}}$.
\end{thm}

Clearly, from Theorems \ref{thm1}, \ref{thm2} and \ref{thm3}, many explicit relations about weighted sums of alternating MTVs can be established. For example, by straightforward calculations, we obtain
\begin{align*}
W(5,4)+W(5,2)=\frac{7}{4}\z(2)\z(3)\quad\text{and}\quad
W(7,5)=\frac{\pi^7}{7680}+\z(2)W(5,3).
\end{align*}

\section{Preliminaries}

In this section we will establish several fundamental formulas for weighed sums $W(k,r)$.

Kaneko and Tsumura \cite{KTA2018,KTA2019} introduced the following a new kind of multiple polylogarithm function
\begin{align}\label{b1}
&{\rm A}(k_1,k_2,\ldots,k_r;z): = 2^r\sum\limits_{1 \le {n_1} <  \cdots  < {n_r}\atop n_j\equiv j\ {\rm mod}\ 2} {\frac{{{z^{{n_r}}}}}{{n_1^{{k_1}}n_2^{{k_2}} \cdots n_r^{{k_r}}}}}\quad (|z|<1),
\end{align}
where $k_1,\ldots,k_r$ are positive integers. If $k_r>1$, we allow $z=1$\ (In \cite{KTA2018}, $2^{-r}{\rm A}(k_1,\ldots,k_r;z)$ is denoted by ${\rm Ath}(k_1,\ldots,k_r;z)$.). When $r=1$, the polylogarithm of level 2 ${\rm A}(k;z)$ was considered by Sasaki in \cite{S2012}.

From definition, it is easy to find that
\begin{align}\label{b2}
\frac{d}{dz}{\mathrm{A}}({{k_1}, \cdots ,k_{r-1},{k_r}}; z)= \left\{ {\begin{array}{*{20}{c}} \frac{1}{z} {\mathrm{A}}({{k_1}, \cdots ,{k_{r-1}},{k_r-1}};z)
   {\ \ (k_r\geq 2),}  \\
   {\frac{2}{1-z^2}{\mathrm{A}}({{k_1}, \cdots ,{k_{r-1}}};z)\;\;\;\ \ \ (k_r = 1).}  \\
\end{array} } \right.
\end{align}
Hence, by (\ref{b2}), we deduce the iterated integral expression
\begin{align}\label{b3}
&{\mathrm{A}}({{k_1}, \cdots,k_{r-1} ,{k_r}};z)=\int\limits_{0}^z \underbrace{\frac{dt}{t}\cdots\frac{dt}{t}}_{k_r-1}\frac{2dt}{1-t^2}\underbrace{\frac{dt}{t}\cdots\frac{dt}{t}}_{k_{r-1}-1}\frac{2dt}{1-t^2}\cdots
\underbrace{\frac{dt}{t}\cdots\frac{dt}{t}}_{k_1-1}\frac{2dt}{1-t^2}\nonumber\\
&=2^r\left\{\prod\limits_{j=1}^r\frac{(-1)^{k_j-1}}{\Gamma(k_j)}\right\}\int\nolimits_{D_r(z)} \frac{\log^{k_1-1}\left(\frac{t_1}{t_2}\right)\cdots \log^{k_{r-1}-1}\left(\frac{t_{r-1}}{t_r}\right)\log^{k_r-1}\left(\frac{t_r}{z}\right)}{(1-t_1^2)\cdots (1-t_{r-1}^2)(1-t_r^2)}dt_1\cdots dt_r,
\end{align}
where $D_{r}(z):=\{(t_1,\ldots,t_r)\mid 0<t_1<\cdots<t_r<z\}\quad (r\in\N).$ Here,
\[\int\limits_{0}^z f_k(t)dtf_{k-1}(t)dt \cdots f_1(t_1)dt:=\int\limits_{D_k(z)} f_k(t_k)f_{k-1}(t_{k-1}) \cdots f_1(t_1)dt_1\cdots dt_{k-1}dt_k.\]
Further, setting $t_j\rightarrow zt_j$ in (\ref{b3}) gives
\begin{align}\label{b4}
&{\rm A}(k_1,\ldots,k_{r-1},k_r;z)\nonumber\\
&=2^rz^r\left\{\prod\limits_{j=1}^r\frac{(-1)^{k_j-1}}{\Gamma(k_j)}\right\}\int\nolimits_{D_r(1)} \frac{\log^{k_1-1}\left(\frac{t_1}{t_2}\right)\cdots \log^{k_{r-1}-1}\left(\frac{t_{r-1}}{t_r}\right)\log^{k_r-1}\left(t_r\right)}{(1-z^2t_1^2)\cdots (1-z^2t_{r-1}^2)(1-z^2t_r^2)}dt_1\cdots dt_r.
\end{align}
Applying the changes of variables $t_j\rightarrow \frac{1-t_{r+1-j}}{1+t_{r+1-j}}$ to (\ref{b4}) yields
\begin{align}\label{b5}
&{\rm A}(k_1,\ldots,k_{r-1},k_r;z)\nonumber\\
&=4^rz^r\left\{\prod\limits_{j=1}^r\frac{(-1)^{k_j-1}}{\Gamma(k_j)}\right\}\int\nolimits_{D_r(1)} \frac{\log^{k_1-1}\left(\frac{(1-t_r)(1+t_{r-1})}{(1+t_r)(1-t_{r-1})} \right)}{(1+t_r)^2-z^2(1-t_r)^2}\cdots  \frac{\log^{k_{r-1}-1}\left(\frac{(1-t_2)(1+t_{1})}{(1+t_2)(1-t_{1})} \right)}{(1+t_2)^2-z^2(1-t_2)^2}\nonumber\\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad \times \frac{\log^{k_{r}-1}\left(\frac{1-t_1}{1+t_1} \right)}{(1+t_1)^2-z^2(1-t_1)^2}dt_1dt_2\cdots dt_r.
\end{align}
As is well-known, we can regard the function ${\rm A}(k_1,k_2,\ldots,k_r;z)$ as a single-valued holomorphic
function in the simply connected domain $\mathbb{C}\setminus \{(-\infty,-1]\cup [1,+\infty)\}$, via the process of iterated integration
starting with ${\rm A}(1;z)=\int_0^z 2dt/{(1-t^2)}$.

Next, we use (\ref{b5}) to establish four identities involving weighted sum $W(k,r)$, alternating Riemann zeta values and infinite series whose general $n$-terms is a product of $(-1)^n/n$, multiple $T$-harmonic sum and multiple $S$-harmonic sum. First, we need to give the definitions of multiple $T$-harmonic sum and multiple $S$-harmonic sum. For an index ${{\bf k}}:= (k_1,\ldots, k_r)$, let
\begin{align*}
{\bf k}_{2m-1}:=(k_1,k_2,\ldots,k_{2m-1})\quad \text{and}\quad {\bf k}_{2m}:=(k_1,k_2,\ldots,k_{2m}).
\end{align*}
For positive integers $n_1,n_2,\ldots,n_{r}$ and $n$, if $r=2m-1$ is an odd, we define
\begin{align*}
&D_n({{\bf n}_{2m-1}}):=\left\{(n_1,n_2,\ldots,n_{2m-1},n)\mid 0<n_1\leq n_2 <\cdots\leq n_{2m-2}<n_{2m-1}\leq n \right\},\ (n\geq m)\\
&E_n({{\bf n}_{2m-1}}):=\left\{(n_1,n_2,\ldots,n_{2m-1},n)\mid 1\leq n_1<n_2\leq \cdots< n_{2m-2}\leq n_{2m-1}< n \right\},\ (n>m)
\end{align*}
and if $r=2m$ is an even, we define
\begin{align*}
&D_n({{\bf n}_{2m}}):=\left\{(n_1,n_2,\ldots,n_{2m},n)\mid 0<n_1\leq n_2 <\cdots\leq n_{2m-2}<n_{2m-1}\leq n_{2m}<n \right\},\ (n> m)\\
&E_n({{\bf n}_{2m}}):=\left\{(n_1,n_2,\ldots,n_{2m},n)\mid 1\leq n_1<n_2\leq \cdots< n_{2m-2}\leq n_{2m-1}< n_{2m}\leq n \right\},\ (n>m).
\end{align*}

\begin{defn}(\cite{XZ2020})\label{lem1} For positive integer $m$, the multiple $T$-harmonic sums ({\rm MTHSs} for short) and multiple $S$-harmonic sums ({\rm MSHSs} for short) are defined by
\begin{align}
&T_n({{\bf k}_{2m-1}}):= \sum_{D_n({{\bf n}_{2m-1}})} \frac{2^{2m-1}}{(\prod_{j=1}^{m-1} (2n_{2j-1}-1)^{k_{2j-1}}(2n_{2j})^{k_{2j}})(2n_{2m-1}-1)^{k_{2m-1}}},\label{MOT}\\
&T_n({{\bf k}_{2m}}):= \sum_{D_n({{\bf n}_{2m}})} \frac{2^{2m}}{\prod_{j=1}^{m} (2n_{2j-1}-1)^{k_{2j-1}}(2n_{2j})^{k_{2j}}},\label{MET}\\
&S_n({{\bf k}_{2m-1}}):= \sum_{E_n({{\bf n}_{2m-1}})} \frac{2^{2m-1}}{(\prod_{j=1}^{m-1} (2n_{2j-1})^{k_{2j-1}}(2n_{2j}-1)^{k_{2j}})(2n_{2m-1})^{k_{2m-1}}},\label{MOS}\\
&S_n({{\bf k}_{2m}}):= \sum_{E_n({{\bf n}_{2m}})} \frac{2^{2m}}{\prod_{j=1}^{m} (2n_{2j-1})^{k_{2j-1}}(2n_{2j}-1)^{k_{2j}}},\label{MES}
\end{align}
where $T_n({{\bf k}_{2m-1}}):=0$ if $n<m$, and $T_n({{\bf k}_{2m}})=S_n({{\bf k}_{2m-1}})=S_n({{\bf k}_{2m}}):=0$ if $n\leq m$. Moreover, for convenience we let $T_n(\emptyset)=S_n(\emptyset):=1$. We call (\ref{MOT}) and (\ref{MET}) are multiple $T$-harmonic sums, and call (\ref{MOS}) and (\ref{MES}) are multiple $S$-harmonic sums.
\end{defn}

Clearly, according to the definitions of ${\rm A}(k_1,k_2,\ldots,k_r;z)$ and MTHSs, by an elementary calculation, we can find that
\begin{align}
&{\rm A}({\bf k}_{2m-1};z)=2\su \frac{T_n({\bf k}_{2m-2})}{(2n-1)^{k_{2m-1}}}z^{2n-1}
\end{align}
and
\begin{align}
&{\rm A}({\bf k}_{2m};z)=2\su \frac{T_n({\bf k}_{2m-1})}{(2n)^{k_{2m}}}z^{2n}.
\end{align}
From the definitions of alternating MTVs and ${\rm A}(k_1,k_2,\ldots,k_r;z)$, we have
\begin{align}\label{b12}
&{\rm A}({\bf k}_{2m-1};i)=i(-1)^m {\bar T}({\bf k}_{2m-1})
\end{align}
and
\begin{align}\label{b13}
&{\rm A}({\bf k}_{2m};i)=(-1)^m {\bar T}({\bf k}_{2m}),
\end{align}
where $i$ is imaginary unit. Kaneko and Tsumura \cite{KTA2018} gave the following relation
\begin{align}\label{b14}
{\rm A}(\{1\}_r;z)=\frac{1}{r!} \left({\rm A}(1;z)\right)^r=\frac{(-1)^r}{r!}\log^r\left(\frac{1-z}{1+z} \right).
\end{align}
It is clear that $A(1;i)=i\frac{\pi}{2}$. Hence, using (\ref{b12})-(\ref{b14}), a simple calculation yields
\begin{align}\label{b15}
{\bar T}(\{1\}_r)=\frac{(-1)^r}{r!}\left(\frac{\pi}{2}\right)^r.
\end{align}

Next, in order to state our main results, we need to give a lemma.
\begin{lem}(\cite[Theorem 2.1]{XZ2020})\label{lem1} For positive integers $m$ and $n$, the following identities hold
\begin{align}
&\begin{aligned}
\int_{0}^1 t^{2n-2} \log^{2m}\tt dt&= \frac{2(2m)!}{2n-1} \sum_{j=0}^m {\bar \z}(2m-2j)T_n(\{1\}_{2j}),\label{ee}
\end{aligned}\\
&\begin{aligned}
\int_{0}^1 t^{2n-2} \log^{2m-1}\tt dt&= -\frac{2(2m-1)!}{2n-1} \sum_{j=0}^{m-1} {\bar \z}(2m-1-2j)T_n(\{1\}_{2j})\\&\quad-\frac{(2m-1)!}{2n-1} S_n(\{1\}_{2m-1}),\label{eo}
\end{aligned}\\
&\begin{aligned}
\int_{0}^1 t^{2n-1} \log^{2m}\tt dt&=\frac{(2m)!}{n} \sum_{j=0}^{m-1} {\bar \z}(2m-1-2j)T_n(\{1\}_{2j+1})\\&\quad+\frac{(2m)!}{2n} S_n(\{1\}_{2m}),\label{oe}
\end{aligned}\\
&\begin{aligned}
\int_{0}^1 t^{2n-1} \log^{2m-1}\tt dt&= -\frac{(2m-1)!}{n} \sum_{j=0}^{m-1} {\bar \z}(2m-2-2j)T_n(\{1\}_{2j+1}),\label{oo}
\end{aligned}
\end{align}
where ${\bar \zeta}(0)$ should be interpreted as $1/2$ wherever it occurs.
\end{lem}

Using the iterated integral identity (\ref{b5}) with the help of Lemma \ref{lem1}, we can get the following theorem.
\begin{thm} For positive integers $m$ and $p$,
\begin{align}
&\begin{aligned}\label{b20}
W(2p+2m-1,2m)=2(-1)^m \sum_{j=0}^{p-1} {\bar \z}(2p-2-2j)\su \frac{T_n(\{1\}_{2m-1})T_n(\{1\}_{2j+1})}{n}(-1)^n,
\end{aligned}\\
&\begin{aligned}\label{b21}
W(2p+2m-2,2m)&=2(-1)^m \sum_{j=0}^{p-2} {\bar \z}(2p-3-2j)\su \frac{T_n(\{1\}_{2m-1})T_n(\{1\}_{2j+1})}{n}(-1)^n\\
&\quad+(-1)^m \su \frac{T_n(\{1\}_{2m-1})S_n(\{1\}_{2p-2})}{n}(-1)^n,
\end{aligned}\\
&\begin{aligned}\label{b22}
W(2p+2m-2,2m-1)&=4(-1)^{m-1} \sum_{j=0}^{p-1} {\bar \z}(2p-1-2j)\su \frac{T_n(\{1\}_{2m-2})T_n(\{1\}_{2j})}{2n-1}(-1)^n\\
&\quad+2(-1)^{m-1} \su \frac{T_n(\{1\}_{2m-2})S_n(\{1\}_{2p-1})}{2n-1}(-1)^n,
\end{aligned}\\
&\begin{aligned}\label{b23}
W(2p+2m-3,2m-1)=4(-1)^{m-1} \sum_{j=0}^{p-1} {\bar \z}(2p-2-2j)\su \frac{T_n(\{1\}_{2m-2})T_n(\{1\}_{2j})}{2n-1}(-1)^n.
\end{aligned}
\end{align}
\end{thm}
\pf By using (\ref{b12}) and (\ref{b13}) we know that
\begin{align*}
{\bar T}(k_1,k_2,\ldots,k_r)=i^r {\rm A}(k_1,k_2,\ldots,k_r;i).
\end{align*}
Then summing both sides of above identity and using (\ref{b5}), we have
\begin{align}\label{bb1}
W(k+r-1,r)=\frac{(-1)^{k+r-1}}{(k-1)!}2^r \int_{D_r(1)} \frac{\log^{k-1}\left(\frac{1-t_r}{1+t_r}\right)}{(1+t_1^2)\cdots (1+t_r^2)}dt_1\cdots dt_r.
\end{align}
Expand the $(1+t_j)^{-1}$ into geometric series to deduce that
\begin{align*}
W(k+r-1,r)=\frac{(-1)^{k-1}}{(k-1)!}2^r \sum_{0<m_1<\cdots<m_r} \frac{(-1)^{m_r} \int_0^1 t^{2m_r-r-1}\log^{k-1}\tt dt}{(2m_1-1)(2m_2-2)\cdots (2m_{r-1}-r+1)}.
\end{align*}
By straightforward calculations, if $r=2m$ and $k=2p$, then
\begin{align*}
W(2p+2m-1,2m)=\frac{2(-1)^{m-1}}{(2p-1)!} \su (-1)^n T_n(\{1\}_{2m-1}) \int_0^1 t^{2n-1}\log^{2p-1}\tt dt,
\end{align*}
if $r=2m$ and $k=2p-1$, then
\begin{align*}
W(2p+2m-2,2m)=\frac{2(-1)^{m}}{(2p-2)!} \su (-1)^n T_n(\{1\}_{2m-1}) \int_0^1 t^{2n-1}\log^{2p-2}\tt dt,
\end{align*}
if $r=2m-1$ and $k=2p$, then
\begin{align*}
W(2p+2m-2,2m-1)=\frac{2(-1)^{m}}{(2p-1)!} \su (-1)^n T_n(\{1\}_{2m-2}) \int_0^1 t^{2n-2}\log^{2p-1}\tt dt,
\end{align*}
if $r=2m-1$ and $k=2p-1$, then
\begin{align*}
W(2p+2m-3,2m-1)=\frac{2(-1)^{m-1}}{(2p-2)!} \su (-1)^n T_n(\{1\}_{2m-2}) \int_0^1 t^{2n-2}\log^{2p-2}\tt dt.
\end{align*}
Thus, with the help of formulas (\ref{ee})-(\ref{oo}), we may easily deduce these desired evaluations.\hfill$\square$

Next, we give several special cases. Setting $m=p=1$ in (\ref{b20}) yields
\begin{align}\label{b24}
\su \frac{T_n^2(1)}{n}(-1)^{n-1}={\T}(1,2)+\T(2,1)=\frac{7}{4}\z(3),
\end{align}
where we used these two well-known results
\begin{align*}
\T(1,2)=-\frac{7}{4}\z(3)+\frac{\pi}{4}{\bar t}(2)\quad \text{and}\quad \T(2,1)=\frac{7}{2}\z(3)-\frac{\pi}{4}{\bar t}(2).
\end{align*}
Letting $m=1$ in (\ref{b23}) gives
\begin{align}
\T(2p-1)=2\sum_{j=1}^{p-1} (-1)^j{\bar \z}(2p-2-2j) \T(\{1\}_{2j+1}).
\end{align}

Now, we end this section by a theorem.
\begin{thm}\label{thmb3} For positive integers $r$ and $k$,
\begin{align}
\T(\{1\}_{r-1},k)=\sum_{j=1}^r (-1)^{j-1} \T(\{1\}_{r-j})W(k+j-1,j).
\end{align}
\end{thm}
\pf According to the definitions of alternating MTVs and using (\ref{b5}) with $z=i,\ k_1=\cdots=k_{r-1}=1$ and $k_r=1$, we have
\begin{align}\label{bb2}
\T(\{1\}_{r-1},k)=\frac{(-1)^{k+r-1}}{(k-1)!}2^r \int_{D_r(1)} \frac{\log^{k-1}\left(\frac{1-t_1}{1+t_1}\right)}{(1+t_1^2)\cdots (1+t_r^2)}dt_1\cdots dt_r.
\end{align}
Note that the two integrals
\begin{align*}
\int_{0<t_1<\cdots<t_r<x} \frac{dt_1\cdots dt_r}{(1+t_1^2)\cdots(1+t_r^2)}=\frac{1}{r!} \arctan(x)^r
\end{align*}
and
\begin{align*}
\int_{x<t_1<\cdots<t_r<1} \frac{dt_1\cdots dt_r}{(1+t_1^2)\cdots(1+t_r^2)}=\frac{1}{r!} \left(\frac{\pi}{4}-\arctan(x)\right)^r,
\end{align*}
then the identities (\ref{bb1}) and (\ref{bb2}) can be rewritten in the forms
\begin{align*}
W(k+r-1,r)=\frac{(-1)^{k+r-1}2^r}{(k-1)!(r-1)!} \int_{0}^1 \frac{\log^{k-1}\tt \arctan(t)^{r-1}}{1+t^2}dt
\end{align*}
and
\begin{align*}
\T(\{1\}_{r-1},k)=\frac{(-1)^{k+r-1}2^r}{(k-1)!(r-1)!} \int_{0}^1 \frac{\log^{k-1}\tt \left(\frac{\pi}{4}- \arctan(t)\right)^{r-1}}{1+t^2}dt.
\end{align*}
Thus, combining the two equations above, by an elementary calculation, we prove the desired result.\hfill$\square$

\section{Proofs of Theorems \ref{thm1} and \ref{thm2}}

In this section, we give the proofs of Theorems \ref{thm1} and \ref{thm2}. We need the following a lemma.
\begin{lem}(\cite[Lemma 5.1]{XZ2020})\label{lemc1}  Let $A_{p,q}, B_p, C_p\ (p,q\in \N)$ be any complex sequences. If
\begin{align}\label{e1}
\sum\limits_{j=1}^p A_{j,p}B_j=C_p\quad\text{and}\quad A_{p,p}:=1,
\end{align}
hold, then
\begin{align}\label{e2}
B_p=\sum\limits_{j=1}^p C_j \sum\limits_{k=1}^{p-j} (-1)^k \left\{\sum\limits_{i_0<i_1<\cdots<i_{k-1}<i_k,\atop i_0=j,i_k=p} \prod\limits_{l=1}^k A_{i_{l-1},i_l}\right\},
\end{align}
holds, where $\sum\limits_{k=1}^0 (\cdot):=1$.
\end{lem}

Hence, from (\ref{b20}), (\ref{b23}) and Lemma \ref{lemc1}, we can get the following two theorems.
\begin{thm} For positive integers $m$ and $p$,
\begin{align}\label{c3}
\su \frac{T_n(\{1\}_{2m-1})T_n(\{1\}_{2p-1})}{n}(-1)^n=(-1)^m \sum_{j=1}^p W(2j+2m-1,2m)Z(j,p).
\end{align}
\end{thm}
\pf  Setting
\begin{align*}
&A_{j,p}=2{\bar \z}(2p-2j),\quad B_j=\su \frac{T_n(\{1\}_{2m-1})T_n(\{1\}_{2j-1})}{n}(-1)^n
\end{align*}
and
\begin{align*}
&C_p=(-1)^mW(2p+2m-1,2m)
\end{align*}
in Lemma \ref{lemc1} and using (\ref{b20}) gives the desired formula.\hfill$\square$

Putting $m=1$ and $p=2$ in (\ref{c3}), we get
\begin{align*}
\su \frac{T_n(1)T_n(1,1,1)}{n}(-1)^n&=\frac{7}{4}\z(2)\z(3)-W(5,2)=\frac{21}{16}\z(2)\z(3)-\frac{31}{16}\z(5),
\end{align*}
where we used the formula (\ref{b24}) and the identity
\[W(5,2)=\frac{7}{16}\z(2)\z(3)+\frac{31}{16}\z(5).\]

\begin{thm} For positive integers $m$ and $p$,
\begin{align}\label{c4}
2\su \frac{T_n(\{1\}_{2m-2})T_n(\{1\}_{2p-2})}{2n-1}(-1)^n=(-1)^{m-1} \sum_{j=1}^p W(2j+2m-3,2m-1)Z(j,p).
\end{align}
\end{thm}
\pf  Setting
\begin{align*}
&A_{j,p}:=2{\bar \z}(2p-2j),\quad B_j:=2\su \frac{T_n(\{1\}_{2m-2})T_n(\{1\}_{2j-2})}{2n-1}(-1)^n
\end{align*}
and
\begin{align*}
C_p:=(-1)^{m-1}W(2p+2m-3,2m-1)
\end{align*}
in Lemma \ref{lemc1} and using (\ref{b23}) yields the desired formula.\hfill$\square$

Letting $m=p=2$ in (\ref{c4}), we have
\begin{align*}
2\su \frac{T^2_n(1,1)}{2n-1}(-1)^n=-W(5,3)+2{\bar \z}(2)\T(1,1,1).
\end{align*}

Applying (\ref{b21}) and (\ref{b22}), we know that for positive integers $m$ and $p$, the sums
\[\su \frac{T_n(\{1\}_{2m-1})S_n(\{1\}_{2p-2})}{n}(-1)^n\quad \text{and}\quad \su \frac{T_n(\{1\}_{2m-2})S_n(\{1\}_{2p-1})}{2n-1}(-1)^n\]
can also be evaluated by weighted sums $W(k,r)$ and $Z(j,p)$.
\\
{\bf Proofs of Theorems \ref{thm1} and \ref{thm2}}. Changing $(m,p)$ to $(p,m)$ in (\ref{c3}) and (\ref{c4}), and using the duality of series on the left hand sides, then
\begin{align*}
\su \frac{T_n(\{1\}_{2m-1})T_n(\{1\}_{2p-1})}{n}(-1)^n&=(-1)^m \sum_{j=1}^p W(2j+2m-1,2m)Z(j,p)\\
&=(-1)^p \sum_{j=1}^m W(2j+2p-1,2p)Z(j,m)
\end{align*}
and
\begin{align*}
2\su \frac{T_n(\{1\}_{2m-2})T_n(\{1\}_{2p-2})}{2n-1}(-1)^n&=(-1)^{m-1} \sum_{j=1}^p W(2j+2m-3,2m-1)Z(j,p)\\
&=(-1)^{p-1} \sum_{j=1}^m W(2j+2p-3,2p-1)Z(j,m).
\end{align*}
Thus, we prove Theorems \ref{thm1} and \ref{thm2}. \hfill$\square$

We get some cases. Setting $m=1$ in (\ref{a9}) and (\ref{a10}) yield
\begin{align}\label{c5}
W(2p+1,2p)=(-1)^{p-1} \sum_{j=1}^p W(2j+1,2)Z(j,p)
\end{align}
and
\begin{align}
\T(\{1\}_{2p-1})=(-1)^{p-1}\sum_{j=1}^p \T(2j-1)Z(j,p).
\end{align}

Now, we give the explicit expression of weighted sums $W(2k+1,2)$ via $t$-values and alternating zeta values.
\begin{thm}\label{thmc4} For positive integer $k$,
\begin{align}\label{c7}
W(2k+1,2)=\frac{1}{2^{2k-1}} \sum_{j=1}^k \t(2j+1){\bar \z}(2k-2j),
\end{align}
where $\t(k):=2^k t(k)=(2^k-1)\z(k)$ if $k>1$ and $\t(1):=0$ if $k=1$.
\end{thm}
\pf From \cite[Eq.(3.10)]{WX2020}, Wang and the author gave the result
\begin{align}\label{c8}
&(1-(-1)^{p+q})\su \frac{h^{(p)}_n}{n^q}(-1)^{n-1}\nonumber\\
=&-(-1)^p(1+(-1)^q)\t(p){\bar \z}(q)+(-1)^p\binom{p+q-1}{p-1}\t(p+q)\nonumber\\
&-(-1)^p\sum_{k=0}^{p-1} ((-1)^k+1)\binom{p+q-k-2}{q-1}{\bar t}(k+1){\bar t}(p+q-k-1)\nonumber\\
&+2(-1)^p \sum_{j=1}^{[q/2]} \binom{p+q-2j-1}{p-1}{\bar \z}(2j)\t(p+q-2j).
\end{align}
According to definition, we have
\begin{align*}
W(2k+1,2)=\sum_{k_1+k_2=2k+1,\atop k_1,k_2\geq 1} \frac{1}{2^{k_1+k_2-2}} \su \frac{h^{(k_1)}_n}{n^{k_2}}(-1)^{n-1}.
\end{align*}
Thus, applying (\ref{c8}) with $p=k_1$ and $q=k_2$, by an elementary calculation, we complete this proof. \hfill$\square$

Hence, using (\ref{c5}) and (\ref{c7}), we obtain the following description.
\begin{cor} For positive integer $p$, the weighted sums $W(2p+1,2p)$ can be expressed in terms of (alternating) Riemann zeta values.
\end{cor}

As two examples, for $p=1$ and $2$, we have
\begin{align*}
W(3,2)=\frac{7}{4}\z(3)\quad \text{and} \quad W(5,4)=\frac{21}{16}\z(2)\z(3)-\frac{31}{16}\z(5).
\end{align*}

\section{Proof of Theorem \ref{thm3}}
In this section we prove Theorem \ref{thm3}. First, we need to give a lemma.
\begin{lem}\label{lem4.1}(\cite[Lemma 2.5]{X2020}) If $f_i\ (i=1,\ldots,m)$ are integrable real functions, the following identity holds:
\begin{align}
 &g\left( {{f_1},{f_2}, \cdots ,{f_m}} \right) + {\left( { - 1} \right)^m}g\left( {{f_m},{f_{m - 1}}, \cdots ,{f_1}} \right) \nonumber\\
 & = \sum\limits_{i = 1}^{m - 1} {{{\left( { - 1} \right)}^{i - 1}}g\left( {{f_i},{f_{i - 1}}, \cdots ,{f_1}} \right)} g\left( {{f_{i + 1}},{f_{i + 2}} \cdots ,{f_m}} \right),
\end{align}
where $g\left( {{f_1},{f_2}, \cdots ,{f_m}} \right)$ is defined by
\[g\left( {{f_1},{f_2}, \cdots ,{f_m}} \right): = \int\limits_{0 < {t_m} <  \cdots <t_2 < {t_1} < 1} {{f_1}\left( {{t_1}} \right){f_2}\left( {{t_2}} \right) \cdots {f_m}\left( {{t_m}} \right)d{t_1}d{t_2} \cdots d{t_m}} .\]
\end{lem}

On the one hand, in (\ref{b5}), replacing $k_r$ by $k_r+m-1$ and letting $z=i$, then summing, we find that
\begin{align}
&W_m(k+r-1,r)=\sum_{{\bf k}\in I(k+r-1,r)} \binom{k_r+m-2}{m-1} \T(k_1,\ldots,k_{r-1},k_r+m-1)\nonumber\\
&=\frac{(-1)^{k+r+m}}{(k-1)!(m-1)!}2^r \int_{D_r(1)} \frac{\log^{m-1}\left(\frac{1-t_1}{1+t_1} \right)\log^{k-1}\left(\frac{1-t_r}{1+t_r} \right)}{(1+t_1^2)\cdots (1+t_r^2)}dt_1\cdots dt_r.
\end{align}
On the other hand, we deduce the identity
\begin{align}\label{d3}
\int_{D_r(1)} \frac{\log^{p}\left(\frac{1-t_1}{1+t_1} \right)}{(1+t_1^2)\cdots (1+t_r^2)}dt_1\cdots dt_r
=\frac{(-1)^{p+r}p!}{2^r} T(\{1\}_{p-1},{\bar 1},\{1\}_{r-1},{\bar 1})
\end{align}
by using (\ref{b1}), (\ref{b14}) and expanding the $(1+t_j^2)^{-1}$ in geometric series with a direct calculation.

Then using Lemma \ref{lem4.1}, by an elementary calculation, we prove the Theorem \ref{thm3}.\hfill$\square$

Letting $m=1$ in Theorem \ref{thm3}, we can get the following a corollary.
\begin{cor} For positive integers $k$ and $r$,
\begin{align}\label{d4}
W(k+r-1,r)+(-1)^r\T(\{1\}_{r-1},k)=\sum_{j=1}^{r-1}(-1)^{j-1}T(\{1\}_{k-2},{\bar 1},\{1\}_{j-1},{\bar 1})\T(\{1\}_{r-j}).
\end{align}
\end{cor}

Setting $r=2$ in (\ref{d4}) yields
\begin{align}\label{d5}
W(k+1,2)+\T(1,k)=T(\{1\}_{k-2},{\bar 1},{\bar 1})\T(1).
\end{align}
where $\T(1)=-{\pi}/{2}$ from (\ref{b15}).

Next, we give a duality theorem of alternating MTVs and give some weighted sum formulas involving alternating MTVs.
\begin{thm} For integers $p\geq 0$ and $r\geq 1$,
\begin{align}\label{d6}
T(\{1\}_{p-1},{\bar 1},\{1\}_{r-1},{\bar 1})=\T(\{1\}_{r-1},p+1),
\end{align}
where $(\{1\}_{-1},{\bar 1}):=\emptyset$.
\end{thm}
\pf In (\ref{d3}), applying the changes of variables $t_j\rightarrow \frac{1-t_{r+1-j}}{1+t_{r+1-j}}$ to (\ref{b4}) gives
\begin{align*}
\frac{(-1)^{p+r}p!}{2^r} T(\{1\}_{p-1},{\bar 1},\{1\}_{r-1},{\bar 1})&=\int_{D_r(1)} \frac{\log^p(t_r)}{(1+t_1^2)\cdots (1+t_r^2)}dt_1\cdots dt_r\\
&=\frac{(-1)^{p+r}p!}{2^r} \T(\{1\}_{r-1},p+1),
\end{align*}
where in the last step, we expanded the $(1+t_j^2)^{-1}$ in geometric series and used the following well-known identity
$$\int_{0}^1 t^{n-1} \log^p(t)dt=p!\frac{(-1)^p}{n^{p+1}}.$$
Thus, we complete this proof. \hfill$\square$

Letting $r=1$ in (\ref{d6}) yields
\begin{align}\label{d7}
T(\{1\}_{p-1},{\bar 1},{\bar 1})=-\frac{1}{2^p}{\bar t}(p+1).
\end{align}
In particular, if $k=2$, then we have
\[T({\bar 1},{\bar 1})=-\frac{1}{2}{\bar t}(2)=-2G,\]
where $G:=\sum_{n=1}^\infty (-1)^{n-1}/(2n-1)^2$ is Catalan's constant (see \cite[Eq. (7.4.4.183a)]{K1997}).
Hence, substituting (\ref{d7}) into (\ref{d5}) with $p=k-1$, we get the following corollary.
\begin{cor} For positive integer $k$,
\begin{align}\label{d8}
W(k+1,2)=\frac{\pi}{2^k}{\bar t}(k)-\T(1,k).
\end{align}
\end{cor}

\begin{re} The formula (\ref{d8}) can be obtained by Theorem \ref{thmb3} with $r=2$.
\end{re}

Further, applying (\ref{d6}) with $p=k-1$, the formula (\ref{a11}) can be rewritten in the form
\begin{align}\label{d9}
W_m(k+r-1,r)+(-1)^rW_k(m+r-1,r)=\sum_{j=1}^{r-1}(-1)^{j-1}\T(\{1\}_{j-1},k)\T(\{1\}_{r-j-1},m)
\end{align}
and
the formula (\ref{d4}) can be rewritten in the form
\begin{align}\label{d10}
W(k+r-1,r)+(-1)^r\T(\{1\}_{r-1},k)=\sum_{j=1}^{r-1}(-1)^{j-1}\T(\{1\}_{j-1},k)\T(\{1\}_{r-j}).
\end{align}
Clearly, (\ref{d10}) can also be immediately obtained from (\ref{d9}) with $m=1$. Moreover, setting $r=2$ in (\ref{d10}) yields (\ref{d8}).

Now, we end this section by a theorem.
\begin{thm}\label{thmd6} For positive integers $m,k$ and $r$,
\begin{align}\label{d11}
&\sum_{k_1+\cdots+k_r=k+r-1,\atop k_1,\ldots,k_r\geq 1} \binom{k_r+m-2}{m-1}{\rm A}(k_1,\ldots,k_{r-1},k_r+m-1;z)\nonumber\\
&+(-1)^r \sum_{k_1+\cdots+k_r=m+r-1,\atop k_1,\ldots,k_r\geq 1} \binom{k_r+k-2}{k-1}{\rm A}(k_1,\ldots,k_{r-1},k_r+k-1;z)\nonumber\\
&=\sum_{j=1}^{r-1} (-1)^{j-1} {\rm A}(\{1\}_{r-1-j},m;z){\rm A}(\{1\}_{j-1},k;z).
\end{align}
\end{thm}
\pf Replacing $k_r$ by $k_r+m-1$ in (\ref{b5}) and summing, by an elementary calculation, we have
\begin{align}\label{d12}
&\sum_{k_1+\cdots+k_r=k+r-1,\atop k_1,\ldots,k_r\geq 1} \binom{k_r+m-2}{m-1} {\rm A}(k_1,\ldots,k_{r-1},k_r+m-1;z)\nonumber\\
&=\frac{(-1)^{k+m}}{(m-1)!(k-1)!} 4^r z^r \int_{D_r(1)} \frac{\log^{m-1}\left(\frac{1-t_1}{1+t_1} \right)\log^{k-1}\left(\frac{1-t_r}{1+t_r} \right)dt_1\cdots dt_r}{[(1+t_1)^2-z^2(1-t_1)^2]\cdots [(1+t_r)^2-z^2(1-t_r)^2] }.
\end{align}
Letting $k=1$ yields
\begin{align*}
{\rm A}(\{1\}_{r-1},m;z)=\frac{(-1)^{m-1}}{(m-1)!} 4^r z^r \int_{D_r(1)} \frac{\log^{m-1}\left(\frac{1-t_1}{1+t_1} \right)dt_1\cdots dt_r}{[(1+t_1)^2-z^2(1-t_1)^2]\cdots [(1+t_r)^2-z^2(1-t_r)^2] }.
\end{align*}
Finally, using Lemma \ref{lem4.1}, by a direct calculation, we may easily deduce the desired result. \hfill$\square$

It is clear that setting $z=-1$ in (\ref{d11}) gives the identity (\ref{d9}).
\\[5mm]
{\bf Acknowledgments.}  The author expresses his deep gratitude to Professors Masanobu Kaneko and Jianqiang Zhao for valuable discussions and comments.

 {\small

config: arxiv


TITLE: Fourier transform of a locally integrable tempered distribution
QUESTION [3 upvotes]: The following statement is an exercise in a set of lecture notes, and I cannot prove nor find a proof of it.

Let $u \in \mathcal{S}'(\mathbb{R}^n)\cap L^1_{\mathrm{loc}}(\mathbb{R}^n)$ be non-negative, and assume that the Fourier transform $\hat{u}$ of $u$ is in $L^{\infty}(\mathbb{R}^n)$. Then in fact $u \in L^1(\mathbb{R})$, and we have the equality $(2\pi)^{n/2}\|\hat{u}\|_{\infty} = \|u\|_{1}$.

The exercise is supplemented by the hint: Consider the sequence $u_k(x) = u(x)\varphi(x/k)$ for some $\varphi \in C_0^{\infty}(\mathbb{R}^n)$ is a standard bump function with support around zero and $\varphi(0) = 1$.
If we drop the assumption that $u \in L^1_{\mathrm{loc}}(\mathbb{R}^n)$ then the statement has easy counterexamples, so this must be key. I do not see an immediate counterexample if we drop the assumption that $u$ is non-negative.
I have tried to use the function given in the hint to prove the statement. We can see that $u_k$ converges almost everywhere to $u$ as $k \to \infty$. However, without the dominated convergence theorem at my disposal (as we have no bounds by integrable functions), I am unable to use this to relate the norm of $u$ to that of $\hat{u}$. I may require another approach.

REPLY [1 votes]: Let $u_k(x) = u(x) e^{-\pi |x|^2/k^2}$, $\widehat{u_k} = \widehat{u} \ast k^n e^{-\pi |x|^2 k^2}$.  Then since $u_k(x) \ge 0$ and is increasing in $k$ $$\|u\|_{L^1}=\lim_{k \to \infty}\|u_k\|_{L^1} = \lim_{k \to \infty} \widehat{u_k}(0) \le \|\widehat{u}\|_{L^\infty}\lim_{k \to \infty} \|1 \ast k^n e^{-\pi |x|^2 k^2}\|_\infty =\|\widehat{u}\|_{L^\infty}$$


TITLE: Has anything been done on $x^2-y^2-z^2$?
QUESTION [0 upvotes]: The representation of integers by the sums of three squares is a well studied problem. But, has anything been done on $x^2-y^2-z^2$?
Maybe I haven't looked hard enough - although I think I have. A reference or two would be appreciated rather than anyone spending serious time on this. Thank you.

REPLY [3 votes]: You want Modern Elementary Theory of Numbers by Leonard Eugene Dickson. On page 161 we find Theorem 115. Dickson and his students Ross and Oppenheim found all indefinite ternary quadratic forms that are universal. There are four types. Letting $M$ be any nonzero integer and $N$ any odd integer, the forms are equivalent to one of these:
$$ xy - M z^2,  $$
$$  2xy - N z^2, $$
$$  2xy + y^2 - N z^2,  $$
$$  2xy + y^2 - 2N z^2.  $$
Alright, half the Hessian matrix of second partial derivatives of your form is
$$
H =
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & -1 & 0 \\
0 & 0 & -1
\end{array}
\right)
$$
Half the Hessian matrix of second partial derivatives $2xy - z^2$ is
$$
G =
\left(
\begin{array}{rrr}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & -1
\end{array}
\right)
$$
Taking integer matrix
$$
P =
\left(
\begin{array}{rrr}
1 & 1 & 1 \\
1 & 0 & 1 \\
0 & 1 & 1
\end{array}
\right)
$$
we find
$$  P^T H P = G.  $$
If we then take
$$
Q = P^{-1} =
\left(
\begin{array}{rrr}
1 & 0 & -1 \\
1 & -1 & 0 \\
-1 & 1 & 1
\end{array}
\right),
$$
we find
$$  Q^T G Q = H.  $$
Dickson also says the conditions on $ax^2 + b y^2 + c z^2$ to be universal; that is the way you worded your question. Put briefly, there must be integers $u,v,w,$ not all zero, with $au^2 + b v^2 + c w^2=0,$ as well as $abc$ being odd or twice an odd number.


text:
\section{Analysis of Pareto-Optimal (no wind case)}
\label{sec:analysispareto}

In this section, we develop some relations between the Pareto-Optimal
cuve and the case of multiple solutions for a same ``$C_{I}$''
trade-off parameter in the optimal trajectory control problem, recalling
that the Pareto-Optimal is a tradeoff curve between the total flight
time $t_{f}$ and the amount of control energy $\intop u^{T}u$ required
for different values of $C_{I}$.

A typical example of a Pareto-Optimal is the red curve in Figure 1
below, obtained for the case $x_{g0}=-4,\,v_{xg0}=-12,\,v_{yg0}=0$
and null wind condition ($W_{x}=0,\,W_{y}=0).$ This plot shows the
conflict that appears when one searches for the minimum travel time
$t_{f}$ and minimum energy at the same time.

\begin{figure}
\noindent \begin{centering}
\includegraphics[scale=0.5]{figs/Paretos}
\par\end{centering}
\noindent \centering{}\caption{{Pareto-Optimal curves for ($x_{0}=4,\,v_{x0}=-12,\,v_{y0}=0)$ and
no wind, when there are 3 optimal solutions for $t_{f}$. }}
\end{figure}

We can see from this plot that there are three main branches, corresponding
to the three positive real roots. The first branch (in blue) corresponds
to the lower positive real root that exhibits no reversal movement
(0<K<3), with $C_{I}$ allowed to vary in the range from 0 to infinty,
and with high energy consumption. The second branch (in black) corresponds
to the middle positive real root (with 3<K<6) and $C_{I}$ allowed
to vary in the range $0<C_{I}<15.$ The third branch (in red) is related
to the bigger positive real root with $C_{I}$ in the range $0<C_{I}<18$
and $t_{f}$ varying from $6t_{r}$ to infinty, and with low energy
consumption.

We will show, in the sequel, that all the critical values of time,
energy and $C_{I}$ in this figure can be derived analytically. An
interesting related result for ``LQR minimum-time problem'' showing
the same three critical points of K=0, K=3 and K=6 in a root locus
plot can be found in the reference (Verriest and Lewis, 1991), although
there is no analysis for the critical $C_{I}$ trade-off parameter.

We will show that the critical point of $C_{I}=18=\left(\dfrac{v_{xg0}^{4}}{72x_{g0}^{2}}\right)$
corresponding to the optimal final time $t_{f}=6t_{r}=6\left(-\dfrac{x_{g0}}{v_{xg0}}\right)$
in this figure is associated to one bifurcation point (see figure
7) when the roots of the quartic polynomial are changing from the
condition of ``3 positives and 1 negative real roots'' to ``two
complex roots, one positive root and one negative root''.

\begin{figure}
\noindent \begin{centering}
\includegraphics[scale=0.46]{figs/Paretos_CIb}
\par\end{centering}
\noindent \centering{}\caption{Roots of optimal polynomial for the same case, and for 3 different
$C_{I}$ parameters. }
\end{figure}

We present, in the sequel, the analytical developments for the derivation
of these ``critical'' values.

Firstly, we recall the quartic polynomial equation for the optimal
$t_{f}$ (Carvalho, 2018), with $v_{yg0}=0$ and $W_{x}=0$ is reduced
to
\[
C_{I}t_{f}^{4}-2(v_{xg0}^{2})t_{f}^{2}-12(v_{xg0}x_{g0})t_{f}-18(x_{g0}^{2})=0
\]

As well as the equations of the optimal velocity and optimal acceleration
\[
\begin{array}{c}
v_{xg}(t)=\dfrac{J_{x}t^{2}}{2}-J_{v_{x0}}t+v_{xg0}\\
u(t)=-J_{x}t+J_{v_{x0}}
\end{array}
\]

where the costates are given by
\[
J_{x}=\dfrac{6(2x_{g0}+v_{xg0}t_{f})}{t_{f}^{3}},\,\,\,\,\,\,\,\,\,\,J_{vx0}=\dfrac{2(3x_{g0}+2v_{xg0}t_{f})}{t_{f}^{2}}
\]

As presented in (Carvalho, 2018), an alternative equation for the
optimal acceleration is given by
\begin{equation}
u(t)=-J_{vx}=\sqrt{2C_{I}+2J_{x}v_{x}(t)}\label{eq:u_opt}
\end{equation}

As we know that the final velocity $v_{xg}(t_{f})$ should be null,
we can look for an explicit equation for the roots of the quadratic
function $v_{xg}(t)$ as
\[
t_{v1,v2}=\dfrac{J_{vx0}\pm\sqrt{J_{vx0}^{2}-2J_{x}v_{0}}}{J_{x}}
\]

Where we used the subscript ``v'' to distinguish the roots of the
quadratic velocity polynomial from the roots of the quartic polynomial
(although some of the roots of the quartic polynomial will also be
roots of the quadratic velocity polynomial). From the acceleration
equation in (\ref{eq:u_opt}) we have that $J_{vx0}^{2}=u_{0}^{2}$
and $2C_{I}=J_{vx0}^{2}-2J_{x}v_{xg0}$, and thus the two roots of
the velocity equation can be written as
\[
t_{v1,v2}=\dfrac{J_{vx0}\pm\sqrt{2C_{I}}}{J_{x}}
\]

Look that the middle point between these two roots occur in the time
instant $t=\dfrac{J_{vx0}}{J_{x}}$ that is exactly the point where
the acceleration is null.

As we know that in the critical point of the bifurcation we have the
critical value of K=6, and with the relations obtained previously
for the gains of the quartic polynomial in the bifurcation point as

\[
\left\{ \begin{array}{ccc}
k_{0}=K^{2}\dfrac{x_{g0}^{2}}{v_{x0}^{2}} & \,\,\,\,\,\, & k_{1}=\dfrac{18v_{xg0}^{2}}{C_{I}K^{2}}\end{array}\right.
\]

then we have a simple relation for the ``critical'' $C_{I}$ as
\[
C_{I}=\dfrac{v_{xg0}^{4}}{72x_{g0}^{2}}
\]

Now, with this critical value of $C_{I}$ and knowing that $t_{f}=6t_{r}=6\left(-\dfrac{x_{0}}{v_{0}}\right)$,
we can check for the second root of the quadratic polynomial of the
velocity equation:
\[
t_{v1,v2}=\dfrac{J_{vx0}\pm\sqrt{2C_{I}}}{J_{x}}=\dfrac{J_{vx0}\pm\sqrt{\dfrac{v_{0}^{4}}{36x_{0}^{2}}}}{J_{x}}=\dfrac{J_{vx0}\pm\dfrac{v_{0}^{2}}{6x_{0}}}{J_{x}}
\]

Or
\[
t_{v1,v2}=\dfrac{\dfrac{2(3x_{g0}+2v_{xg0}t_{f})}{t_{f}^{2}}\pm\dfrac{v_{0}^{2}}{6x_{0}}}{\dfrac{6(2x_{g0}+v_{xg0}t_{f})}{t_{f}^{3}}}=\dfrac{2(3x_{g0}+2v_{xg0}t_{f})t_{f}\pm\dfrac{v_{0}^{2}}{6x_{0}}t_{f}^{3}}{6(2x_{g0}+v_{xg0}t_{f})}
\]

Or
\[
t_{v1,v2}=\dfrac{2(3x_{g0}+12v_{xg0}t_{r})t_{f}\pm\dfrac{v_{0}^{2}}{6x_{0}}t_{f}^{3}}{6(2x_{g0}+6v_{xg0}t_{r})}=\dfrac{2(-9x_{g0})t_{f}\pm\dfrac{v_{0}^{2}}{6x_{0}}t_{f}^{3}}{6(-4x_{g0})}
\]

Or

\[
t_{v1,v2}=\dfrac{2(-9x_{g0})t_{f}\pm\dfrac{v_{0}^{2}}{6x_{0}}t_{f}^{3}}{6(-4x_{g0})}=\dfrac{2(-9x_{g0})(6t_{r})\pm\dfrac{v_{0}^{2}}{6x_{0}}(6t_{r})^{3}}{6(-4x_{g0})}
\]

Or finally

\[
t_{v1,v2}=\dfrac{2(-9x_{g0})(6t_{r})\pm\dfrac{v_{0}^{2}}{6x_{0}}(6t_{r})^{3}}{6(-4x_{g0})}=4.5t_{r}\pm1.5t_{r}
\]

Which confirms that for the critical $C_{I}$, the two roots of the
quadratic velocity equation (for $t_{f}=t_{4}$ coming from the quartic
polynomial) are $t_{v1}=3t_{r}$ and $t_{v2}=6t_{r}.$ And as we have
seen previously that the notable value of $t_{f}=3t_{r}$ is an optimal
solution for this problem only when $C_{I}=0,$ then we conclude that
$t_{v2}=6t_{r},$ that is indeed a double root ($t_{3}=t_{4})$ of
the quartic polynomial with critical $C_{I}$, is the only possible
solution from these two roots for the quadratic polynomial. The other
optimal solution of the quartic polynomial, in this critical $C_{I}$
case, comes from ($t_{2})$. The figure below illustrate these relations
on the roots for the critical case ($C_{I}=C_{I,critic})$, while
the other figure in the sequel illustrates the general case ($C_{I}<C_{I,critic}).$

\begin{figure}
\noindent \begin{centering}
\includegraphics[scale=0.52]{figs/Pareto3_CI_18b}
\par\end{centering}
\noindent \centering{}\caption{{Optimal acceleration, velocity and position responses for $C_{I}=18$.
The 3 positive real solutions of the time polynomial degenerate to
2 solutions, as $t_{3}=t_{4}$ is a double root. Note that the ``yellow''
root is a root of the quadratic velocity polynomial for $t_{f}=6t_{r}$
but it is not a root of the optimal quartic polynomial in $t_{f.}$}}
\end{figure}


text:
{\bf Problem.} The numbers $a,$ $b,$ $c,$ $d$ are equal to 1, 2, 3, 4, in some order.  Find the largest possible value of
\[ab + bc + cd + da.\]
{\bf Level.} Level 3
{\bf Type.} Intermediate Algebra
{\bf Solution.} We can factor $ab + bc + cd + da$ as $(a + c)(b + d).$  Then by AM-GM,
\[(a + c)(b + d) \le \frac{[(a + c) + (b + d)]^2}{4} = \frac{10^2}{4} = 25.\]Equality occurs when $a = 1,$ $b = 2,$ $c = 4,$ and $d = 3,$ so the largest possible value is $\boxed{25}.$

set_name: MATH


TITLE: Showing two matrices have the same minimal polynomial
QUESTION [0 upvotes]: Show that the matrices  $\begin{pmatrix}-1&0&0&0\\0&-1&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$ and $\begin{pmatrix}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$ have the same minimal polynomials.

Is it correct to say the characteristic polynomials of the two are ($x-2$)($x+1$$)^3$ and ($x-2$$)^3$($x+1$) respectivly and the minimal polynomials are ($x-2)(x+1$) since that's the irreducible factorization of both?

REPLY [0 votes]: Let $A=\begin{pmatrix}-1&0&0&0\\0&-1&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$ and $B=\begin{pmatrix}2&0&0&0\\0&2&0&0\\0&0&2&0\\0&0&0&-1\end{pmatrix}$.
we know that the dimension of the eigen space: $dim(E_\lambda)=n-rank(A-\lambda I)$, for a $n\times n$ matrix $A$. So $dim(E_{-1})=4-rank(A+ I)=3$. So the number of Jordan block corresponding to the eigenvalue $=3$. Since algebraic multiplicity of $-1$  is $3$, so highest order Jordan block of $-1$ is $1$. Similarly for the eigenvalue $2$ highest order Jordan block is $1$. So the minimal polynomial of $A$ is $m_A(t)=(t+1)(t-2)$.
Also for the matrix $B$, $dim(E_{-2})=4-rank(A+ 2I)=3$ and by same reason  $m_B(t)=(t+1)(t-2)$. Although their characteristic polynomial are different.


TITLE: How to solve $(\partial_x + i\partial_y)u - c(\partial_x+i\partial_y)au=0$?
QUESTION [0 upvotes]: The following equation,
$$(\partial_x + i\partial_y)u - c(\partial_x+i\partial_y)au=0$$
($a=a(x,y)$ and $\partial_x=\frac{\partial}{\partial x}$)
with solution,
$$u=\exp(ca)f(x+iy)$$
where $f$ and $g$ are arbitrary entire functions, a is some scalar function and $c$ is a scalar.
How can I derive the solution ?

REPLY [0 votes]: $\newcommand{\+}{^{\dagger}}
 \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\down}{\downarrow}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\isdiv}{\,\left.\right\vert\,}
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}
 \newcommand{\wt}[1]{\widetilde{#1}}$
$\ds{\pars{\partial_{x} + \ic\partial_{y}}{\rm u}\pars{x,y}
     -c\pars{\partial_{x} + \ic\partial_{y}}{\rm a}\pars{x,y}{\rm u}\pars{x,y}
     = 0:\ {\large ?}}$

Your notation is confused. For the second term, I'll assume $\ds{\pars{\partial_{x} + \ic\partial_{y}}}$ is 'applied' just to $\ds{\rm a}$. It means I understand $\quad\ds{\pars{\partial_{x} + \ic\partial_{y}}{\rm au}}\quad$
as $\ds{\bracks{\pars{\partial_{x} + \ic\partial_{y}}{\rm a}}{\rm u}}$

In that case, we can rewrite the equation as:
$$
\fermi\pars{x,y}{\rm u}\pars{x,y}=0\,,\qquad\qquad
\fermi\pars{x,y} \equiv
\bracks{1 - \pars{\partial_{x} + \ic\partial_{y}}{\rm a}\pars{x,y}}
$$
The solution will be $\ds{{\rm u}\pars{x,y} = 0}$ unless $\ds{\fermi\pars{x,y} = 0}$.

Otherwise,
\begin{align}
0&=\pars{\partial_{x} + \ic\partial_{y}}{\rm u}\pars{x,y}
-\bracks{c\pars{\partial_{x} + \ic\partial_{y}}{\rm a}\pars{x,y}}{\rm u}\pars{x,y}
\\[3mm]&-c\,{\rm a}\pars{x,y}\pars{\partial_{x} + \ic\partial_{y}}{\rm u}\pars{x,y}
\end{align}
The equation becomes:
\begin{align}
&\bracks{\partial_{x} + \ic\partial_{y}
+\phi\pars{x,y}}{\rm u}\pars{x,y} = 0
\\[3mm]&\mbox{where}\quad\phi\pars{x,y} \equiv
{c\pars{\partial_{x} + \ic\partial_{y}}{\rm a}\pars{x,y} \over c\,{\rm a}\pars{x,y} - 1}
\end{align}


TITLE: Multivariable calculus, change of variables, expression of derivative operators
QUESTION [2 upvotes]: I have the following relations that express a change of variables
$$\begin{cases}
x=ae^\theta\cos(\varphi) \\
y=ae^\theta\sin(\varphi)
\end{cases}
$$
And I want to use it to transform the operator:
$$L=x^2 \frac {\partial^2}{\partial x^2}-2xy \frac{\partial^2}{\partial x\partial y}+y^2 \frac{\partial^2}{\partial y^2}$$
The derivatives become rather complicated and I also have trouble defining the order of regularity of the change of variables.
I would love an explicit solution as to how $$\frac{\partial}{\partial x}, \frac{\partial}{\partial y}$$
are expressed.

REPLY [1 votes]: I'm using subscripts for more compact notation
$$ f_\theta = f_x x_\theta + f_y y_\theta = ae^\theta\big(\cos\phi f_x + \sin\phi f_y\big)  $$
$$ f_\phi = f_x x_\phi + f_y y_\phi = ae^\theta \big(-\sin\phi f_x + \cos\phi f_y\big) $$
Cross-multiply by the trig functions to eliminate one of $f_x$ or $f_y$
$$ \cos\phi f_\theta - \sin\phi f_\phi = ae^\theta f_x $$
$$ \sin\phi f_\theta + \cos\phi f_\phi = ae^\theta f_y $$
Therefore
$$ f_x = \frac{e^{-\theta}}{a}\left(\cos\phi f_\theta - \sin\phi f_\phi\right) $$
$$ f_y = \frac{e^{-\theta}}{a}\left(\sin\phi f_\theta + \cos\phi f_\phi \right) $$
For the second derivatives, just apply the same operators again
$$ \begin{align} f_{xx} &= \frac{e^{-\theta}}{a}\left(\cos\phi\ \partial_\theta f_x - \sin\phi \ \partial_\phi f_x \right) \\
&= \frac{e^{-\theta}}{a}\cos\phi\left(-\frac{e^{-\theta}}{a}(\cos\phi f_\theta - \sin\phi f_\phi) + \frac{e^{-\theta}}{a}(\cos\phi f_{\theta\theta} - \sin\phi f_{\theta\phi}) \right) \\
&\quad - \frac{e^{-2\theta}}{a^2}\sin\phi \left(-\sin\phi f_\theta - \cos\phi f_\phi + \cos\phi f_{\theta\phi} - \sin\phi f_{\phi\phi} \right) \\
&= \frac{e^{-2\theta}}{a^2}\left(\cos^2\phi f_{\theta\theta} + \sin^2\phi f_{\phi\phi} - \sin2\phi f_{\theta\phi} - \cos2\phi f_\theta \right)
\end{align} $$
and so on...
As for the mapping, this is called the log-polar coordinate system. Curves of constant $\theta$ are circles in the $xy$ plane, with the scale being logarithmic ($\theta = 0, \ln 2, \ln 3, \dots$ have radii $a, 2a, 3a, \dots$). The only exception is the origin, which corresponds to $\theta = -\infty$, making it a singular point.
Curves of constant $\phi$ are rays starting from (but not including) the origin, similar to regular polar coordinates.