MZVで遊ぼう

[1]
$(1)^{k}\coloneqq \overbrace{1+1+1\cdots +1}^{k個}$の様に定める。すると以下の様に$wt-N+1$は書ける。
\begin{equation} wt-N-1=\overbrace{(1)^{k_{1}-2}+(1)^{k_{2}-1}+\cdots+(1)^{k_{N}-1}}^{N個} \end{equation}
[2]右辺は$1$と$+$がそれぞれ$wt-N-1$,$N-1$個ありそれらを並べる個数分だけ表し方がある。
そしてその表し方の分だけ異なる$MZV$を取る事が出来るので
\begin{equation} d(N,wt)=\begin{pmatrix}wt-2\\N-1\end{pmatrix} \end{equation}
[3]
\begin{eqnarray} d(wt)&=&\sum_{N=1}^{wt-1}\begin{pmatrix}wt-2\\N-1\end{pmatrix}\\ &=&(1+1)^{wt-2}\\ &=&2^{wt-2} \end{eqnarray}

[1]
\begin{eqnarray} \zeta(k_{1},k_{2},...,k_{N})&=&\sum_{n_{1}\gt n_{2}\gt\cdots n_{N}\geq 1}\frac{1}{n_{1}^{k_{1}}n_{2}^{k_{2}}\cdots n_{N}^{k_{N}}}\\ &\lt&\sum_{n_{1}\gt n_{2}}\frac{1}{n_{1}^{2}}(\sum_{n_{3}=1}^{n_{2}}\frac{1}{n_{3}})^{N-1}\\ &=&\sum_{n_{1}\gt n_{2}}\frac{1}{n_{1}^{2}}(H_{n_{2}})^{N-1} \end{eqnarray}
[2]
\begin{eqnarray} 0&\lt&H_{n}-\log{n}\\ &=&\sum_{k=1}^{n}\frac{1}{k}-\sum_{k=1}^{n}\int_{k}^{k+1}\frac{1}{x}dx\\ &=&\sum_{k=1}^{n}\int_{0}^{1}(\frac{1}{k}-\frac{1}{x+k})dx\\ &=&\sum_{k=1}^{n}\int_{0}^{1}(\frac{x}{k(x+k)})dx\\ &\lt&\sum_{k=1}^{n}\frac{1}{k^{2}}\int_{0}^{1}xdx\\ &=&\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k^{2}}\lt\frac{1}{2}\zeta(2)\lt 1 \end{eqnarray}
ゆえに$H_{n}\lt 1+\log{n}$
[3]下記の様な$(0,\infty)$上で定義された関数を考える。
\begin{eqnarray} \forall r\in(0,1]:f(x)\coloneqq\frac{1+\log{x}}{x^{r}} \end{eqnarray}
すると
\begin{eqnarray} \lim_{x\rightarrow\infty}f(x)&=&\lim_{x\rightarrow\infty}\frac{1+\log{x}}{x^{r}}\\ &=&\lim_{x\rightarrow\infty}\frac{\log{x^{r}}}{rx^{r}}\\ &=&\frac{1}{r}\lim_{y\rightarrow\infty}\frac{\log{y}}{y}\\ &=&0 \end{eqnarray}
ゆえに、$r=\frac{1}{2(N-1)}$とすると
\begin{eqnarray} \zeta(k_{1},k_{2},...,k_{N})&\lt&\sum_{n_{1}\gt n_{2}}\frac{1}{n_{1}^{2}}(H_{n_{2}})^{N-1}\\ =\sum_{n_{1}\gt n_{2}}\frac{o(n_{2}^{\frac{1}{2}})}{n_{1}^{2}} &\lt&+\infty \end{eqnarray}

[1]
\begin{eqnarray} x&=&\sum_{m=1}^{\infty}\sum_{n=0}^{\infty}B_{n}\frac{x^{m+n}}{m!n!}\\ &=&\sum_{N=1}^{\infty}x^{N}\sum_{n=0}^{N-1}B_{n}\frac{1}{(N-n)!n!} \end{eqnarray}
[2]係数比較により下記の式が得られる。
\begin{eqnarray} \left\{ \begin{array}{l} B_{0}=1\\ \sum_{n=0}^{N-1}B_{n}\frac{1}{(N-n)!n!}=0 \end{array} \right. \end{eqnarray}
[3]また
\begin{eqnarray} \frac{x}{e^{x}-1}+\frac{x}{2}&=&\frac{x}{2}\frac{e^{x}+1}{e^{x}-1}\\ &=&\frac{x}{2}\frac{e^{\frac{x}{2}}+e^{-\frac{x}{2}}}{e^{\frac{x}{2}}-e^{-\frac{x}{2}}}\\ &=&\frac{x}{2}\coth{\frac{x}{2}}\\ &=&\frac{-x}{2}\coth{\frac{-x}{2}} \end{eqnarray}
より$\frac{x}{2}\coth{\frac{x}{2}}$は遇関数である事が分かったので
\begin{eqnarray} \left\{ \begin{array}{l} B_{0}=1\\ B_{1}=-\frac{1}{2}\\ \frac{1}{(2N+1)!}-\frac{1}{2(2N)!}+\sum_{n=1}^{N}B_{2n}\frac{1}{(2N-2n+1)!(2n)!}=0 \end{array} \right. \end{eqnarray}

[$\Rightarrow$]$f^{(k)}(a)=0\quad (k=0,1,2,...,m-1)$かつ$f^{(m)}(a)=0$が成り立つとする。
(i)$f(a)=0$および因数定理より$\exists g_{0}(x)\in \mathbb{C}[x]\ s.t.\ f(x)=(x-a)g_{0}(x)$が成り立つ。
(ii)$f^{(1)}(x)=(x-a)g^{(1)}_{0}(x)+g_{0}(x)$なので、$f^{(1)}(a)=g_{0}(a)$が得られる。(i)同様に因数定理から$\exists g_{1}(x)\in\mathbb{C}[x]\ s.t.\ g_{0}(x)=(x-a)g_{1}(x)$が得られる。
(iii)これを帰納的に繰り返し$\exists g_{m-1}(x)\in\mathbb{C}[x]\ s.t.\ f(x)=(x-a)^{m}g_{m-1}(x)$が得られる。
(iv)最後に$f^{(m)}(x)=\sum_{k=0}^{m-1}\begin{pmatrix}m\\k\end{pmatrix}\frac{m!}{(m-k)!}(x-a)^{m-k}g_{m-1}^{(m-k)}(x)+g_{m-1}(x)$なので、仮定より$f^{(m)}(a)=g_{m-1}(a)\neq 0$が得られる。この事から因数定理より$g_{m-1}(x)$は$x-a$で割る事は出来ない事が分かる。
(v)以上をまとめると、$\exists g_{m-1}(x)\in \mathbb{C}[x]\ s.t.\ f(x)=(x-a)^{m}g_{m-1}(x)$が得られ$f(x)$が$x=a$で$m$重根を持つ事が示された。
[$\Leftarrow$]$f(x)$が$x=a$で$m$重根を持つとすると$\exists g(x)\in\mathbb{C}[x](g(a)\neq 0)\ s.t.\ f(x)=(x-a)^{m}g(x)$が成り立つ。
ゆえに下記の式が得られるので必要性も示せた。
\begin{eqnarray} \left\{ \begin{array}{l} f^{(k)}(x)=\sum_{l=0}^{k}\begin{pmatrix}k\\l\end{pmatrix}(x-a)^{m-l}g^{(l)}(x)\Rightarrow f^{(k)}(a)=0\\ f^{(m)}(x)=\sum_{k=0}^{m-1}\begin{pmatrix}m\\k\end{pmatrix}\frac{m!}{(m-k)!}(x-a)^{m-k}g^{(m-k)}(x)+g(x)\Rightarrow f^{(m)}(a)=g(a)\neq 0 \end{array} \right. \end{eqnarray}

厳密な証明ではありません。
[1]$\sin{\pi x}$は$x=n\in\mathbb{Z}$で1位の零点を持つ。
\begin{eqnarray} \left\{ \begin{array}{l} \sin{\pi n}=0\quad(n\in\mathbb{Z})\\ \pi\cos{\pi n}=(-1)^{n}\pi\neq 0\quad(n\in\mathbb{Z}) \end{array} \right. \end{eqnarray}
補題2より$x=n\in\mathbb{Z}$で一位の零点を持つ事が分かる。👈本当は補題2が適用できるのか確認する必要がある。
[2][1]から
\begin{eqnarray} \sin{\pi x}&=&\lim_{N\rightarrow\infty}C_{N}\prod_{n=-N}^{N}(x-n)\\ &=&x\lim_{N\rightarrow\infty}C_{N}\prod_{n=1}^{N}(x^{2}-n^{2}) \end{eqnarray}
実際にはこの乗積が$N\rightarrow\infty$で発散しない様にしないといけない。
それらを考慮すると
\begin{equation} \sin{\pi x}=x\lim_{N\rightarrow\infty}\{C_{N}\prod_{n=1}^{N}n^{2}\}\prod_{n=1}^{N}(\frac{x^{2}}{n^{2}}-1) \end{equation}
両辺を$x$で割り$x\rightarrow 0$の極限を取ると下記の式が得られるので$C_{N}=\frac{(-1)^{N}}{\prod_{n=1}^{N}n^{2}}$
\begin{equation} 1=\lim_{N\rightarrow\infty}\{C_{N}\prod_{n=1}^{N}n^{2}\}(-1)^{N} \end{equation}
これを用いると
\begin{equation} \sin{\pi x}=x\prod_{n=1}^{\infty}(1-\frac{x^{2}}{n^{2}}) \end{equation}

[1]
\begin{eqnarray} x\coth{\frac{x}{2}}=\sum_{n=1}^{\infty}B_{2n}\frac{1}{(2n)!}x^{2n} \end{eqnarray}
[2]
\begin{equation} 2x\coth{x}=\sum_{n=1}^{\infty}B_{2n}\frac{2^{2n}}{(2n)!}x^{2n} \end{equation}
[3]
\begin{eqnarray} 2ix\coth{(ix)}&=&2ix\frac{\cos{x}}{2i\sin{x}}\\ &=&x\cot{x}\\ &=&\sum_{n=1}^{\infty}(-1)^{n}B_{2n}\frac{2^{2n}}{(2n)!}x^{2n} \end{eqnarray}
[4]$cot{x}$は定理3より$x=\pi n\in\mathbb{Z}$で一位の極を持つ。その留数を求めると下記の様に計算できる。👈ただし収束性に関して十分に注意せよ。
\begin{eqnarray} \lim_{x\rightarrow \pi n}(x-\pi n)\cot{x}&=&\lim_{x\rightarrow \pi n}(x-\pi n)\frac{\cos{x}}{\sin{x}}\\ &=&\lim_{x\rightarrow \pi n}\frac{x-\pi n}{\sin{(x-\pi n)}\cos{\pi n}}\cos{x}\\ &=&1 \end{eqnarray}
[5]
\begin{eqnarray} \cot{x}&=&\frac{1}{x}+\lim_{N\rightarrow\infty}\sum_{n=1}^{N}(\frac{1}{x+\pi n}+\frac{1}{x-\pi n})\\ &=&\frac{1}{x}+2\sum_{n=1}^{\infty}\frac{x}{x^{2}-\pi^{2}n^{2}} \end{eqnarray}
[6]
\begin{eqnarray} x\cot{x}&=&1+2\sum_{n=1}^{\infty}\frac{x^{2}}{x^{2}-\pi^{2}n^{2}}\\ &=&1-\frac{2}{\pi^{2}}\sum_{n=1}^{\infty}\frac{x^{2}}{n^{2}}\sum_{m=0}^{\infty}\frac{x^{2m}}{\pi^{2m}n^{2m}}\\ &=&1-\sum_{m=0}^{\infty}\frac{2x^{2m+2}}{\pi^{2m+2}}\sum_{n=1}^{\infty}\frac{1}{n^{2m+2}}\\ &=&1-\sum_{m=0}^{\infty}\frac{2x^{2m+2}}{\pi^{2m+2}}\zeta(2m+2)\\ &=&1-\sum_{m=1}^{\infty}\frac{2x^{2m}}{\pi^{2m}}\zeta(2m) \end{eqnarray}
[7]最後に$x^{2n}$の係数を比較して証明完了

\begin{eqnarray} \zeta(k_{1})\zeta(k_{2})&=&\sum_{n_{1},n_{2}\geq 1}\frac{1}{n_{1}^{k_{1}}n_{2}^{k_{2}}}\\ &=&(\sum_{n_{1}\gt n_{2}\geq 1}+\sum_{n_{2}\gt n_{1}\geq 1}+\sum_{n_{1}=n_{2}\geq 1})\frac{1}{n_{1}^{k_{1}}n_{2}^{k_{2}}}\\ &=&\zeta(k_{1},k_{2})+\zeta(k_{2},k_{1})+\zeta(k_{1}+k_{2}) \end{eqnarray}

\begin{eqnarray} \zeta(2,1)&=&\sum_{n_{1}\gt n_{2}\geq 1}\frac{1}{n_{1}^{2}n_{2}}\\ &=&\sum_{n_{1},n_{2}\geq 1}\frac{1}{(n_{1}+n_{2})^{2}n_{2}}\\ &=&\sum_{n_{1},n_{2}\geq 1}\frac{1}{n_{1}(n_{1}+n_{2})}(\frac{1}{n_{2}}-\frac{1}{n_{1}+n_{2}})\\ &=&\sum_{n_{1}, n_{2}\geq 1}\frac{1}{n_{1}n_{2}(n_{1}+n_{2})}-\zeta(2,1)\\ &=&\sum_{n_{1}, n_{2}\geq 1}\frac{1}{n_{1}^{2}}(\frac{1}{n_{2}}-\frac{1}{n_{1}+n_{2}})-\zeta(2,1)\\ &=&\sum_{n_{1}\geq 1}\frac{1}{n_{1}^{2}}\sum_{n_{2}=1}^{n_{1}}\frac{1}{n_{3}}-\zeta(2,1)\\ &=&\zeta(3) \end{eqnarray}

以下$\{0\}^{n}=(\overbrace{0,0,...,0}^{n個})$の様に定める。
そして次の計算を行っていく。
\begin{eqnarray} (-1)^{N}G(\{0\}^{k_{1}-1},1,\{0\}^{k_{2}-1},0,...,\{0\}^{k_{N}-1},1;x)=(-1)^{N}\overbrace{\int_{0}^{x}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{x_{2}}\cdots\int_{0}^{x_{k-2}}dx_{k-1}}^{k_{1}-1個}\int_{0}^{x_{k-1}}\frac{dx_{k}}{x_{k}-1}G(\{0\}^{k_{2}-1},0,...,\{0\}^{k_{N}-1},1;x_{k}) \end{eqnarray}
[1]$N=1$の場合
\begin{eqnarray} -G(\{0\}^{k_{1}-1},1;x)&=&-\overbrace{\int_{0}^{x}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{x_{2}}\cdots\int_{0}^{x_{k-2}}dx_{k-1}}^{k_{1}-1個}\int_{0}^{x_{k-1}}\frac{dx_{k}}{x_{k}-1}\\ &=&\sum_{n=1}^{\infty}\frac{x^{n}}{n^{k_{1}}}\\ &=&Li_{k_{1}}(x) \end{eqnarray}
[2]そこで$1,2,...,N-1$まで$(-1)^{N-1}G(\{0\}^{k_{1}-1},1,\{0\}^{k_{2}-1},0,...,\{0\}^{k_{N-1}-1},1;x)=Li_{k_{1},k_{2},...,k_{N-1}}(x)$が成り立つとする。
[3]
\begin{eqnarray} (-1)^{N}G(\{0\}^{k_{1}-1},1,\{0\}^{k_{2}-1},0,...,\{0\}^{k_{N}-1},1;x)&=&-\overbrace{\int_{0}^{x}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{x_{2}}\cdots\int_{0}^{x_{k-2}}dx_{k-1}}^{k_{1}-1個}\int_{0}^{x_{k-1}}\frac{dx_{k}}{x_{k}-1}(-1)^{N-1}G(\{0\}^{k_{2}-1},0,...,\{0\}^{k_{N}-1},1;x_{k})\\ &=&-\overbrace{\int_{0}^{x}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{x_{2}}\cdots\int_{0}^{x_{k-2}}dx_{k-1}}^{k_{1}-1個}\int_{0}^{x_{k-1}}\frac{dx_{k}}{x_{k}-1}(-1)^{N-1}Li_{k_{2},k_{3},...,k_{N}}(x_{k})\\ &=&\sum_{n_{2},n_{3},\cdots,n_{N}\geq 1}\frac{1}{(n_{2}+n_{3}+\cdots+n_{N})^{k_{2}}(n_{2}+n_{3}+\cdots+n_{N})^{k_{3}}\cdots n_{N}^{k_{N}}}(-1)\overbrace{\int_{0}^{x}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{x_{2}}\cdots\int_{0}^{x_{k-2}}dx_{k-1}}^{k_{1}-1個}\int_{0}^{x_{k-1}}\frac{x_{k}^{n_{2}+n_{3}+\cdots+n_{N}}dx_{k}}{x_{k}-1}\\ &=&\sum_{n_{1},n_{2},...,n_{N}\geq 1}\frac{x^{n_{1}+n_{2}+\cdots+n_{N}}}{(n_{1}+n_{2}+\cdots+n_{N})^{k_{1}}(n_{2}+n_{3}+\cdots+n_{N})^{k_{2}}\cdots n_{N}^{k_{N}}}\\ &=&Li_{k_{1},k_{2},...,k_{N}}(\overbrace{x,x,...,x}^{N}) \end{eqnarray}

有理数か終端が$y$である様な$x,y$を並べた項の$Q-$線形結合全体が$\mathfrak{h}^{1}$である事を考えれば明らか。

帰納法を用いて示す。
まず、語$w_{1},w_{2}$の$y$に関する次数を$d_{1},d_{2}$とする。
[1]$(d_{1},d_{2})=(1,1)$の場合:
この場合は$w_{1}=z_{k_{1}},w_{2}=z_{l_{1}}$となる様な自然数$k_{1},l_{1}$を取る事が出来る。
これを用いて計算を行う。
\begin{eqnarray} w_{1}\ast w_{2}&=&(z_{k_{1}}1\ast z_{l_{1}}1)\\ &=&z_{k_{1}}(1\ast z_{l_{1}}1)+z_{l_{1}}(z_{k_{1}}1\ast 1)+z_{k_{1}+l_{1}}(1\ast 1)\\ &=&z_{k_{1}}z_{l_{1}}+z_{l_{1}}z_{k_{1}}+z_{k_{1}+l_{1}}\\ &=&z_{l_{1}}z_{k_{1}}+z_{k_{1}}z_{l_{1}}+z_{l_{1}+k_{1}}\\ &=&z_{l_{1}}(1\ast z_{k_{1}})+z_{k_{1}}(z_{l_{1}}1\ast 1)+z_{l_{1}+k_{1}}(1\ast 1)\\ &=&w_{2}\ast w_{1} \end{eqnarray}
[2]$(d_{1},d_{2})\neq(1,1)$の場合
(A)$d_{1}$を固定し$d_{2}+1$の場合を考える。
\begin{eqnarray} \left\{ \begin{array}{l} w_{1}=z_{k_{1}}z_{k_{2}}\cdots z_{k_{d_{2}}}\\ w_{2}=z_{l_{1}}z_{l_{2}}\cdots z_{l_{d_{2}}}z_{l_{d_{2}+1}} \end{array} \right. \end{eqnarray}
\begin{eqnarray} w_{1}\ast w_{2}&=&z_{k_{1}}(z_{k_{2}}\cdots z_{k_{d_{2}}}\ast z_{l_{1}}z_{l_{2}}\cdots z_{l_{d_{2}}}z_{l_{d_{2}+1}})+z_{l_{1}}(z_{k_{1}}z_{k_{2}}\cdots z_{k_{d_{2}}}\ast z_{l_{2}}\cdots z_{l_{d_{2}}}z_{l_{d_{2}+1}} )+(可換な項)\\ &=&z_{k_{1}}z_{k_{2}}(z_{k_{3}}\cdots z_{k_{d_{2}}}\ast z_{l_{1}}z_{l_{2}}\cdots z_{l_{d_{2}}}z_{l_{d_{2}+1}})+z_{l_{1}}z_{l_{2}}(z_{k_{1}}z_{k_{2}}\cdots z_{k_{d_{2}}}\ast z_{l_{3}}\cdots z_{l_{d_{2}}}z_{l_{d_{2}+1}}+(可換な項)\\ &=&z_{1}z_{2}+z_{2}z_{1}+(可換な項)\\ &=&z_{2}z_{1}+z_{1}z_{2}+(可換な項)\\ &=&w_{2}\ast w_{1} \end{eqnarray}
(B)$d_{2}$を固定し$d_{1}+1$の場合も同様の計算をすればいい。
[3]最終的に帰納法の仮定より証明完了。

帰納法を用いて示す。
まず、語$w_{1},w_{2},w_{3}$の$y$に関する次数を$d_{1},d_{2},d_{3}$とする。
[1]$(d_{1},d_{2},d_{3})=(1,1,1)$の場合:
\begin{eqnarray} (w_{k}\ast w_{l})\ast w_{m}&=&(z_{k}\ast z_{l})\ast z_{m}\\ &=&\{z_{k}z_{l}+z_{l}z_{k}+z_{k+l}\}\ast z_{m}\\ &=&z_{k}z_{l}\ast z_{m}1+z_{l}z_{k}\ast z_{m}1+z_{k+l}\ast z_{m}\\ &=&z_{k}(z_{l}\ast z_{m}1)+z_{m}(z_{k}z_{l}\ast 1)+z_{k+m}(z_{l}\ast 1)+z_{l}(z_{k}\ast z_{m}1)+z_{m}(z_{l}z_{k}\ast 1)+z_{l+m}z_{k}+z_{k+l}z_{m}+z_{m}z_{k+l}+z_{k+l+m}\\ &=&z_{k}z_{l}z_{m}+z_{k}z_{m}z_{l}+z_{k}z_{l+m}+z_{m}z_{k}z_{l}+z_{k+m}z_{l}+z_{l}z_{k}z_{m}+z_{l}z_{m}z_{k}+z_{l}z_{k+m}+z_{m}z_{l}z_{k}+z_{l+m}z_{k}+z_{k+l}z_{m}+z_{m}z_{k+l}+z_{k+l+m} \end{eqnarray}
\begin{eqnarray} w_{k}\ast(w_{l}\ast w_{m})&=&z_{k}\ast (z_{l}\ast z_{m})\\ &=&z_{k}\ast(z_{l}z_{m}+z_{m}z_{l}+z_{l+m})\\ &=&z_{k}1\ast z_{l}z_{m}+z_{k}1\ast z_{m}z_{l}+z_{k}1\ast z_{l+m}1\\ &=&z_{k}(1\ast z_{l}z_{m})+z_{l}(z_{k}1\ast z_{m})+z_{k+l}(1\ast z_{m})+z_{k}(1\ast z_{m}z_{l})+z_{m}(z_{k}1\ast z_{l})+z_{k+m}(1\ast z_{l})+z_{k}(1\ast z_{l+m}1)+z_{l+m}(z_{k}1\ast 1)+z_{k+l+m}(1\ast 1)\\ &=&z_{k}z_{l}z_{m}+z_{l}z_{k}z_{m}+z_{l}z_{m}z_{k}+z_{l}z_{k+m}+z_{k+l}z_{m}+z_{k}z_{m}z_{l}+z_{m}z_{k}z_{l}+z_{m}z_{l}z_{k}+z_{m}z_{k+l}+z_{k+m}z_{l}+z_{k}z_{l+m}+z_{l+m}z_{k}+z_{k+l+m} \end{eqnarray}
[2]$(d_{1},d_{2},d_{3})\neq(1,1,1)$まで成り立つと仮定
\begin{eqnarray} \left\{ \begin{array}{l} w_{1}=z_{k_{1}}w_{1}^{'}\\ w_{2}=z_{l_{1}}w_{2}^{'}\\ w_{3}=z_{m_{1}}w_{3}^{'} \end{array} \right. \end{eqnarray}
$d_{2},d_{3}$を固定して$d_{1}+1$とした場合
\begin{eqnarray} (w_{1}\ast w_{2})\ast w_{3}&=&\{z_{k_{1}}(w_{1}^{'}\ast z_{l_{1}}w_{2}^{'})+z_{l_{1}}(z_{k_{1}}w_{1}^{'}\ast w_{2}^{'})+z_{k_{1}+l_{1}}w_{1}^{'}\ast w_{2}^{'}\}\ast z_{m_{1}}w_{3}^{'}\\ &=&\color{red}z_{k_{1}}\{(w_{1}^{'}\ast z_{l_{1}}w_{2}^{'})\ast z_{m_{1}}w_{3}^{'}\}\color{black}+z_{l_{1}}(z_{k_{1}}w_{1}^{'}\ast w_{2}^{'})\ast z_{m_{1}}w_{3}^{'}\color{red}+z_{k_{1}+l_{1}}\{(w_{1}^{'}\ast w_{2}^{'})\ast z_{m_{1}}w_{3}^{'}\}👈赤色部分は結合法則が成り立つ事が帰納法の仮定で既に分かっている。以下結合法則が成立している部分は省略する。\\ &=&z_{l_{1}}(z_{k_{1}}w_{1}^{'}\ast w_{2}^{'})\ast z_{m_{1}}w_{3}^{'}+(結合律が成り立つ事が分かっている部分)\\ &=&z_{l_{1}}z_{l_{2}}(z_{k_{1}}w_{1}^{'}\ast w_{2}^{''})\ast z_{m_{1}}w_{3}^{'}+(結合律が成り立つ事が分かっている部分)\\ &=&z_{l_{1}}z_{l_{2}}\cdots z_{l_{d_{2}}}(z_{k_{1}}w_{1}^{'}\ast 1)\ast z_{m_{1}}w_{3}^{'}+(結合律が成り立つ事が分かっている部分)\\ &=&\color{blue}z_{l_{1}}z_{l_{2}}\cdots z_{l_{d_{2}}}\{z_{k_{1}}w_{1}^{'}\ast (1\ast z_{m_{1}}w_{3}^{'})\}\color{black}+(結合律が成り立つ事が分かっている部分)\color{green}👈出てくるすべての項が結合律を満たす項で出来ている事が分かる。後は逆向きに構成しなおしていけばいい。\\ &=&w_{1}\ast(w_{2}\ast w_{3}) \end{eqnarray}
他の$d_{2},d_{3}$に対しても同様に計算すればいい。

[1]
\begin{eqnarray} \left\{ \begin{array}{l} w_{1}=z_{k_{1}}\\ w_{2}=z_{l_{1}} \end{array} \right. \end{eqnarray}
\begin{eqnarray} Z(w_{1}\ast w_{2})&=&Z(z_{k_{1}}z_{l_{1}}+z_{l_{1}}z_{k_{1}}+z_{k_{1}+l_{1}})\\ &=&Z(z_{k_{1}}z_{l_{1}})+Z(z_{l_{1}}z_{k_{1}})+Z(z_{k_{1}+l_{1}})\\ &=&\zeta(k_{1},l_{1})+\zeta(l_{1},k_{1})+\zeta(k_{1}+l_{1})\\ &=&\zeta(k_{1})\zeta(l_{1})\\ &=&Z(k_{1})Z(l_{1}) \end{eqnarray}
[2]
\begin{eqnarray} \left\{ \begin{array}{l} w_{1}=z_{k_{1}}z_{k_{2}}\cdots z_{k_{M}}\\ w_{2}=z_{l_{1}}z_{l_{2}}\cdots z_{l_{N}}\\ \end{array} \right. \end{eqnarray}
\begin{eqnarray} Z(w_{1}\ast w_{2})&=&Z(z_{k_{1}}(z_{k_{2}}\cdots z_{k_{M}}\ast z_{l_{1}}z_{l_{2}}\cdots z_{l_{N}}))+Z(z_{l_{1}}(z_{k_{2}}\cdots z_{k_{M}}\ast z_{l_{1}}z_{l_{2}}\cdots z_{l_{N}}))+Z(z_{k_{1}}z_{l_{1}}(z_{k_{2}}\cdots z_{k_{M}}\ast z_{l_{2}}\cdots z_{l_{N}}))👈Zの線形性を用いている。 \end{eqnarray}
[3]次に
\begin{eqnarray} \left\{ \begin{array}{l} S(X,Y)\coloneqq\{(\square_{0} x_{1}\square_{1}\cdots\square_{|X-1|}x_{N}\square_{|X|})|\square_{k}=\left\{\begin{array}{l}\varnothing\\y_{k_{1}},y_{k_{2}},...,y_{k_{j_{k}}}\quad(k_{1}\lt k_{2}\lt\cdots \lt k_{j_{k}})\end{array}\right.,m(\square_{k})=\left\{\begin{array}{l}0\quad(\square_{k}=\varnothing)\\k_{1}\quad(other wise)\end{array}\right.,M(\square_{k})=\left\{\begin{array}{l}|Y|\quad(\square_{k}=\varnothing)\\k_{j_{k}}\quad(other wise)\end{array}\right.,M(\square_{k})\lt m(\square_{k+1}),\square_{0}\cup\square_{1}\cdots\square_{|X|}=Y\}\\ T(S(X,Y))\coloneqq\{z_{1}\diamond_{1}z_{2}\diamond_{2}\cdots\diamond_{|Z|-1}z_{|Z|}|\diamond_{k}=\left\{\begin{array}{l}\gt\quad\\\geq\end{array}\right.ただし、x_{k}\diamond_{k} x_{k+1}またはy_{k}\diamond y_{k+1}ならば\diamond=\gt\}\\ \overline{M}_{k}=(m_{k},m_{k+1},...,m_{M})\\ \overline{N}_{k}=(n_{k},n_{k+1},...,n_{N}) \end{array} \right. \end{eqnarray}
の様に定める。
\begin{eqnarray} Z(w_{1})Z(w_{2})&=&\sum_{m_{1}\gt m_{2}\gt\cdots \gt m_{M}\geq 1}\frac{1}{m_{1}^{k_{1}}m_{2}^{k_{2}}\cdots m_{M}^{k_{M}}}\sum_{n_{1}\gt n_{2}\gt\cdots \gt n_{N}\geq 1}\frac{1}{n_{1}^{l_{1}}n_{2}^{l_{2}}\cdots n_{N}^{l_{N}}}\\ &=&(\sum_{m_{1}\gt T(S(M_{2},N_{1}))}+\sum_{n_{1}\geq T(S(M_{1},N_{2}))}+\sum_{m_{1}=n_{1}\geq T(S(M_{2},N_{2}))})\frac{1}{m_{1}^{k_{1}}m_{2}^{k_{2}}\cdots m_{M}^{k_{M}}n_{1}^{l_{1}}n_{2}^{l_{2}}\cdots n_{N}^{l_{N}}}\\ &=&Z(z_{k_{1}}w_{1}^{'}\ast w_{2})+Z(z_{l_{1}}w_{1}\ast w_{2}^{'})+Z(z_{k_{1}+l_{1}}w_{1}^{'}\ast w_{2}^{'})\\ &=&Z(w_{1}\ast w_{2}) \end{eqnarray}

\begin{eqnarray} \zeta(k,l)&=&G(\{0\}^{k-1},1,\{0\}^{l-1},1;1)\\ &=&\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{x_{2}}\cdots\int_{0}^{x_{k-1}}\frac{dx_{k}}{1-x_{k}}\int_{0}^{x_{k}}\frac{dx_{k+1}}{x_{k+1}}\int_{0}^{x_{k+1}}\frac{dx_{k+2}}{x_{k+2}}\cdots\int_{0}^{x^{k+l-1}}\frac{dx_{k+l}}{1-x_{k+l}}\\ &=&\int_{0}^{1}\frac{dx_{k+l}}{1-x_{k+l}}\int_{x_{k+l}}^{1}\frac{dx_{k+l-1}}{x_{k+l-1}}\cdots\int_{x_{k+1}}^{1}\frac{dx_{k}}{1-x_{k}}\int_{x_{k}}^{1}\frac{dx_{k-1}}{x_{k-1}}\int_{x_{k-1}}^{1}\frac{dx_{k-2}}{x_{k-2}}\cdots\int_{x_{2}}^{1}\frac{dx_{1}}{x_{1}}\\ &=&\int_{0}^{1}\frac{dy_{k+l}}{y_{k+l}}\int_{0}^{y_{k+l}}\frac{dy_{k+l-1}}{1-y_{k+l-1}}\cdots\int_{0}^{y_{k+1}}\frac{dy_{k}}{y_{k}}\int_{0}^{y_{k}}\frac{dy_{k-1}}{1-y_{k-1}}\int_{0}^{y_{k-1}}\frac{dy_{k-2}}{1-y_{k-2}}\cdots\int_{0}^{y_{2}}\frac{dy_{1}}{1-y_{1}}\\ &=&G(0,\{1\}^{l-1},0,\{1\}^{k-1})\\ &=&\left\{ \begin{array}{l} \zeta(3,\{1\}^{k-2})\quad(l=1)\\ \zeta(2,\{1\}^{l-1},2,\{1\}^{k-2})\quad(l=2,3,...) \end{array} \right. \end{eqnarray}

[1]
\begin{eqnarray} \zeta(2)^{2}&=&G(0,1;1)^{2}\\ &=&\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{1-x_{2}}\int_{0}^{1}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{4}}{1-x_{4}}\\ &=&\int_{1\geq x_{1}\geq x_{2}\geq 0\land 1\geq x_{3}\geq x_{4}\geq 0}\frac{dx_{1}}{x_{1}}\frac{dx_{2}}{1-x_{2}}\frac{dx_{3}}{x_{3}}\frac{dx_{4}}{1-x_{4}} \end{eqnarray}
[2]
\begin{eqnarray} \left\{ \begin{array}{l} S(X,Y)\coloneqq\{(\square_{0},x_{1},\square_{1},...,x_{|X|},\square_{|X|})|\square_{k}=\left\{\begin{array}{l}\varnothing\\y_{k_{1}},y_{k_{2}},...,y_{k_{j_{k}}}\quad(k_{1}\lt k_{2}\lt\cdots \lt k_{j_{k}})\end{array}\right.,m(\square_{k})=\left\{\begin{array}{l}0\quad(\square_{k}=\varnothing)\\k_{1}\end{array}\right.,M(\square_{k})=\left\{\begin{array}{l}|Y|\quad(\square_{k}=\varnothing)\\k_{j_{k}}\end{array}\right.,\square_{0}\cup\square_{1}\cup\cdots\square_{|X|}=Y\}\\ T(S(X,Y))\coloneqq\{1\geq z_{1}\geq z_{2},...\geq z_{|Z|}\geq 0|(z_{1},z_{2},...,z_{|Z|})\in S(X,Y)\} \end{array} \right. \end{eqnarray}
の様に置く。
特に$X=(x_{1},x_{2}),Y=(x_{3},x_{4})$とすると
\begin{eqnarray} T(S(X,Y))=&\{&1\geq x_{3}\geq x_{4}\geq x_{1}\geq x_{2}\geq 0\\ &,&1\geq x_{1}\geq x_{3}\geq x_{4}\geq x_{2}\geq 0\\ &,&1\geq x_{1}\geq x_{2}\geq x_{3}\geq x_{4}\geq 0\\ &,&1\geq x_{3}\geq x_{1}\geq x_{4}\geq x_{2}\geq 0\\ &,&1\geq x_{3}\geq x_{1}\geq x_{2}\geq x_{4}\geq 0\\ &,&1\geq x_{1}\geq x_{3}\geq x_{2}\geq x_{4}\geq 0 \} \end{eqnarray}
以下$\int_{v\in T(S(X,Y))}=\sum_{v\in T(S(X,Y))}\int_{v}$の様に記号を定めると
\begin{eqnarray} \zeta(2)^{2}&=&\int_{v\in T(S(X,Y))}\frac{dx_{1}}{x_{1}}\frac{dx_{2}}{1-x_{2}}\frac{dx_{3}}{x_{3}}\frac{dx_{4}}{1-x_{4}}\\ &=&\int_{0}^{1}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{4}}{1-x_{4}}\int_{0}^{x_{4}}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{1-x_{2}}\\ &+&\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{4}}{1-x_{4}}\int_{0}^{x_{4}}\frac{dx_{2}}{1-x_{2}}\\ &+&\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{1-x_{2}}\int_{0}^{x_{2}}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{4}}{1-x_{4}}\\ &+&\int_{0}^{1}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{4}}{1-x_{4}}\int_{0}^{x_{4}}\frac{dx_{2}}{1-x_{2}}\\ &+&\int_{0}^{1}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{1-x_{2}}\int_{0}^{x_{2}}\frac{dx_{4}}{1-x_{4}}\\ &+&\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{3}}{x_{3}}\int_{0}^{x_{3}}\frac{dx_{2}}{1-x_{2}}\int_{0}^{x_{2}}\frac{dx_{4}}{1-x_{4}}\\ &=&\zeta(2,2)+\zeta(3,1)+\zeta(2,2)+\zeta(3,1)+\zeta(3,1)+\zeta(3,1)\\ &=&2\zeta(2,2)+4\zeta(3,1) \end{eqnarray}

証明は、調和積と同じなので省略。

\begin{eqnarray} z_{2}\smallsmile z_{2}&=&xy\smallsmile xy\\ &=&x(y\smallsmile xy)+x(xy\smallsmile y)\\ &=&x\{y(1\smallsmile xy)+x(y1\smallsmile y)\}+x\{x(y\smallsmile y1)+y(xy\smallsmile 1)\}\\ &=&xyxy+2x^{2}y^{2}+2x^{2}y^{2}+xyxy\\ &=&2 xyxy+4x^{2}y^{2} \end{eqnarray}
\begin{eqnarray} Z(z_{2}\smallsmile z_{2})&=&2Z(xyxy)+4Z(x^{2}y^{2})\\ &=&2\zeta(2,2)+4\zeta(3,1)\\ &=&Z(z_{2})^{2}\\ &=&\zeta(2)^{2} \end{eqnarray}

これは$Z$が$\ast-$積そして$\smallsmile-$積に対して同型なので
\begin{eqnarray} Z(w_{1}\ast w_{2})&=&Z(w_{1})Z(w_{2})\\ &=&Z(w_{1}\smallsmile z_{2}) \end{eqnarray}

[1]
$f(x)=\log{(x+\alpha-1)}$とおく。するとEuler-Maclaurinの定理より
\begin{eqnarray} \log{(\alpha)_{N}}&=&\sum_{n=1}^{N}\log{(n+\alpha-1)}\\ &=&\int_{1}^{N}\log{(x+\alpha-1)}+\frac{f(N)+f(1)}{2}+\frac{f^{'}(N)+f^{'}(1)}{12}-\frac{f^{(3)}(N)-f^{(3)}(1)}{720}+\\ &=&(N+\alpha-1)\log{(N+\alpha-1)}-N+1+\frac{\log{(N+\alpha-1)+\log{\alpha}}}{2}+\frac{\frac{1}{N+\alpha-1}-\frac{1}{\alpha}}{12}-\frac{\frac{2}{(N+\alpha-1)^{3}}-\frac{1}{\alpha^{3}}}{720}+\cdots\\ &=&\log{C}+\log{(N+\alpha-1)^{N+\alpha-1}}-N+\frac{1}{2}\log{(N+\alpha-1)}+O(\frac{1}{N})\\ &=&\log{C(N+\alpha-1)^{N+\alpha-\frac{1}{2}}e^{-N}}+O(\frac{1}{N}) \end{eqnarray}
[2]
\begin{equation} (\alpha)_{N}=C(N+\alpha-1)^{N+\alpha-\frac{1}{2}}e^{-N+O(\frac{1}{N})} \end{equation}

[1]$|\frac{(\vb*{\alpha}_{1})_{n_{1}}(\vb*{\alpha_{2}})_{n_{2}}\cdots(\vb*{\alpha}_{N})_{n_{N}}}{(\vb*{\beta}_{1})_{n_{1}}(\vb*{\beta}_{2})_{n_{2}}\cdots(\vb*{\beta}_{L})_{n_{N}}}|$が$n_{1},n_{2},...,n_{N}$が$o(n_{i}^{r})\quad(0\leq r\lt 1,i=1,2,...,N)$となる条件を探す。
\begin{eqnarray} |\frac{(\vb*{\alpha}_{1})_{n_{1}}(\vb*{\alpha_{2}})_{n_{2}}\cdots(\vb*{\alpha}_{N})_{n_{N}}}{(\vb*{\beta}_{1})_{n_{1}}(\vb*{\beta}_{2})_{n_{2}}\cdots(\vb*{\beta}_{L})_{n_{N}}}|&=&|\prod_{i=1}^{N}\frac{\prod_{j=1}^{s_{i}}(a_{ij})_{n_{i}}}{\prod_{j=1}^{t_{i}}(b_{ij})_{n_{i}}}|\\ &=&|\prod_{i=1}^{N}\frac{\prod_{j=1}^{s_{i}}C(N+\alpha_{ij}-1)^{n_{i}+\alpha_{ij}-\frac{1}{2}}e^{-n_{i}}}{\prod_{j=1}^{t_{i}}C(N+\beta_{ij}-1)^{n_{i}+\beta_{ij}-\frac{1}{2}}e^{-n_{i}}}|\\ &=&C^{\sum_{i}^{N}(s_{i}-t_{i})}e^{-\sum_{i=1}^{N}(s_{i}-t_{i})n_{i}}\prod_{i=1}^{N}\frac{\prod_{j=1}^{s_{i}}|(N+\alpha_{ij}-1)^{n_{i}+\alpha_{ij}-\frac{1}{2}}|}{\prod_{j=1}^{t_{i}}|(N+\beta_{ij}-1)^{n_{i}+\beta_{ij}-\frac{1}{2}}|}\\ &=&C^{\sum_{i}^{N}(s_{i}-t_{i})}e^{-\sum_{i=1}^{N}(s_{i}-t_{i})n_{i}}\prod_{i=1}^{N}\frac{\prod_{j=1}^{s_{i}}|N+\alpha_{ij}-1|^{n_{i}+Re(\alpha_{ij})-\frac{1}{2}}}{\prod_{j=1}^{t_{i}}|N+\beta_{ij}-1|^{n_{i}+Re(\beta_{ij})-\frac{1}{2}}} \end{eqnarray}
[2]
\begin{eqnarray} \log{\prod_{i=1}^{N}\frac{\prod_{j=1}^{s_{i}}|N+\alpha_{ij}-1|^{n_{i}+Re(\alpha_{ij})-\frac{1}{2}}}{\prod_{j=1}^{t_{i}}|N+\beta_{ij}-1|^{n_{i}+Re(\beta_{ij})-\frac{1}{2}}}}&\sim&\sum_{i=1}^{N}\sum_{j=1}^{s_{i}}\{n_{i}+Re(\alpha_{ij}-\frac{1}{2})\}\log{|N+\alpha_{ij}-1|}-\sum_{i=1}^{N}\sum_{j=1}^{t_{i}}\{n_{i}+Re(\beta_{ij}-\frac{1}{2})\}\log{|N+\beta_{ij}-1|} \end{eqnarray}
[3]$n_{1},n_{2},...,n_{N}$に依存する部分だけが問題なのでその部分だけを抜き出して考えると
\begin{equation} \sum_{i=1}^{N}n_{i}\{-(s_{i}-t_{i})+\sum_{j=1}^{s_{i}}\log{|N+\alpha_{ij}-1|}-\sum_{j=1}^{t_{i}}\log{|N+\beta_{ij}-1|}\}\leq 0 \end{equation}
[4]上記不等式を整理すると与えられた不等式が得られる。
\begin{eqnarray} \color{red}s_{i}-t_{j}\geq\sum_{j=1}^{s_{i}}\log{|N+\alpha_{ij}-1|}-\sum_{j=1}^{t_{i}}\log{|N+\beta_{ij}-1|} \end{eqnarray}
[5]以上の計算結果を用いると、$|\frac{(\vb*{\alpha}_{1})_{n_{1}-1}(\vb*{\alpha_{2}})_{n_{2}-1}\cdots(\vb*{\alpha}_{N})_{n_{N}-1}}{(\vb*{\beta}_{1})_{n_{1}-1}(\vb*{\beta}_{2})_{n_{2}-1}\cdots(\vb*{\beta}_{L})_{n_{N}-1}}|$は上に有界なので$\exists M\in\mathbb{R}(0\lt M)\ s.t.\ |\frac{(\vb*{\alpha}_{1})_{n_{1}-1}(\vb*{\alpha_{2}})_{n_{2}-1}\cdots(\vb*{\alpha}_{N})_{n_{N}-1}}{(\vb*{\beta}_{1})_{n_{1}-1}(\vb*{\beta}_{2})_{n_{2}-1}\cdots(\vb*{\beta}_{L})_{n_{N}-1}}|\lt M$が成り立つ。これを用いて計算すると一般化されたMZVは絶対収束する事が示された。
\begin{eqnarray} |\vb*{\zeta}(\begin{pmatrix}\vb*{\alpha}_{1}&\vb*{\alpha}_{2}&\cdots&\vb*{\alpha}_{K}\\\vb*{\beta}_{1}&\vb*{\beta}_{2}&\cdots&\vb*{\beta}_{L}\end{pmatrix};k_{1},k_{2},...,k_{N})|&\leq&\sum_{n_{1}\gt n_{2}\gt\cdots\gt n_{N}\geq 1}|\frac{(\vb*{\alpha}_{1})_{n_{1}-1}(\vb*{\alpha_{2}})_{n_{2}-1}\cdots(\vb*{\alpha}_{N})_{n_{N}}-1}{(\vb*{\beta}_{1})_{n_{1}-1}(\vb*{\beta}_{2})_{n_{2}-1}\cdots(\vb*{\beta}_{L})_{n_{N}-1}}|\frac{1}{n_{1}^{k_{1}}n_{2}^{k_{2}}\cdots n_{N}^{k_{N}}}\\ &\leq&M\sum_{n_{1}\gt n_{2}\gt\cdots\gt n_{N}\geq 1}\frac{1}{n_{1}^{k_{1}}n_{2}^{k_{2}}\cdots n_{N}^{k_{N}}}\\ &\lt&M\zeta(2,\overbrace{1,1,...,1}^{N-1個}) \end{eqnarray}

\begin{eqnarray} \zeta^{2}(\begin{pmatrix}\vb*{\alpha}_{1}\\\vb*{\beta}_{1}\end{pmatrix},2)&=&\sum_{n_{1}\geq 1}\frac{(\vb*{\alpha}_{1})_{n_{1}-1}}{(\vb*{\beta}_{1})_{n_{1}-1}}\frac{1}{n_{1}^{2}}\sum_{n_{1}\geq 1}\frac{(\vb*{\alpha}_{1})_{n_{2}-1}}{(\vb*{\beta}_{1})_{n_{2}-1}}\frac{1}{n_{2}^{2}}\\ &=&(\sum_{n_{1}\gt n_{2}\geq 1}+\sum_{n_{2}\gt n_{1}\geq 1}+\sum_{n_{1}=n_{2}=1})\frac{(\vb*{\alpha}_{1})_{n_{1}-1}}{(\vb*{\beta}_{1})_{n_{1}-1}}\frac{1}{n_{1}^{2}}\frac{(\vb*{\alpha}_{1})_{n_{2}-1}}{(\vb*{\beta}_{1})_{n_{2}-1}}\frac{1}{n_{2}^{2}}\\ &=&2\zeta(\begin{pmatrix}\vb*{\alpha_{1}}&\vb*{\alpha_{1}}\\\vb*{\beta_{1}}&\vb*{\beta_{1}}\end{pmatrix},2,2)+\zeta(\begin{pmatrix}\vb*{\alpha_{1}}\sharp\vb*{\alpha}_{1}\\\vb*{\beta}_{1}\sharp\vb*{\beta}_{1}\end{pmatrix},4) \end{eqnarray}

\begin{eqnarray} \zeta(\begin{pmatrix}\vb*{\alpha}_{1}\\\vb*{\beta}_{1}\end{pmatrix},2)\zeta(\begin{pmatrix}\vb*{\alpha}_{2}\\\vb*{\beta}_{2}\end{pmatrix},2)&=&\sum_{n_{1}\geq 1}\frac{(\vb*{\alpha}_{1})_{n_{1}-1}}{(\vb*{\beta}_{1})_{n_{1}-1}}\frac{1}{n_{1}^{2}}\sum_{n_{1}\geq 1}\frac{(\vb*{\alpha}_{2})_{n_{2}-1}}{(\vb*{\beta}_{2})_{n_{2}-1}}\frac{1}{n_{2}^{2}}\\ &=&(\sum_{n_{1}\gt n_{2}\geq 1}+\sum_{n_{2}\gt n_{1}\geq 1}+\sum_{n_{2}= n_{1}\geq 1})\frac{(\vb*{\alpha}_{1})_{n_{1}-1}}{(\vb*{\beta}_{1})_{n_{1}-1}}\frac{1}{n_{1}^{2}}\frac{(\vb*{\alpha}_{2})_{n_{2}-1}}{(\vb*{\beta}_{2})_{n_{2}-1}}\frac{1}{n_{2}^{2}}\\ &=&\zeta(\begin{pmatrix}\vb*{\alpha}_{1}&\vb*{\alpha}_{2}\\\vb*{\beta}_{1}&\vb*{\beta}_{2}\end{pmatrix},2,2)+\zeta(\begin{pmatrix}\vb*{\alpha}_{2}&\vb*{\alpha}_{1}\\\vb*{\beta}_{2}&\vb*{\beta}_{1}\end{pmatrix},2,2)+\zeta(\begin{pmatrix}\vb*{\alpha_{1}}\sharp\vb*{\alpha}_{2}\\\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\end{pmatrix},4) \end{eqnarray}

[1]調和積は可換
まず一般化されたMZVの定義より$\zeta(\begin{pmatrix}\vb*{\alpha}\sharp\vb*{\gamma}\\\vb*{\beta}\sharp\vb*{\delta}\end{pmatrix},k)=\zeta(\begin{pmatrix}\vb*{\beta}\sharp\vb*{\alpha}\\\vb*{\delta}\sharp\vb*{\beta}\end{pmatrix},k)$である事に注意する。
つまり$\sharp$は可換ではないが、一般化されたMZVでは可換に振る舞うという事である。
以上の事を用いる。
[1A]以下$z_{1},z_{2}\in\mathfrak{y}$の$y$の次数$d_{1},d_{2}$により帰納法を用いて示す。
$(d_{1},d_{2})=(1,1)$の場合：$z_{1}=x^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}y,z_{2}=x^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}y$と書けるので以下の様に計算できる。
\begin{eqnarray} x^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}y\ast x^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}y&=&x^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}yx^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}y+x^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}yx^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}y+x_{\vb*{\alpha}\sharp\vb*{\gamma},\vb*{\beta}\sharp\vb*{\delta}}^{k_{1}+k_{2}-1}y\\ &=&x^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}yx^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}y+x^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}yx^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}y+x_{\vb*{\gamma}\sharp\vb*{\alpha},\vb*{\delta}\sharp\vb*{\beta}}^{k_{1}+k_{2}-1}y\\ &=&x^{k_{2}-1}_{\vb*{\beta},\vb*{\gamma}}y\ast x^{k_{1}-1}_{\vb*{\alpha},\vb*{\beta}}y \end{eqnarray}
[1-B]
$(d_{1},d_{2})$の場合まで成り立つと仮定する。そして$(d_{1}+1,d_{2})$でも成り立つ事を示す。
まず、$z_{1}=\prod_{i=1}^{d_{1}+1}z_{\alpha_{i},\beta_{i}}^{k_{2}-1},z_{2}=\prod_{i=1}^{d_{2}}z_{\gamma_{i},\delta_{i}}^{k_{2}-1}$の様に置く。
すると下記の様に計算できる。
\begin{eqnarray} z_{1}\ast z_{2}&=&z_{\vb*{\alpha}_{1},\beta_{1}}^{k_{1}-1}(\prod_{i=2}^{d_{1}+1}z_{\alpha_{i},\beta_{i}}^{k_{2}-1},z_{2})+z_{\gamma_{1},\delta_{1}}^{k_{1}-1}(z_{1}\ast \prod_{i=2}^{d_{2}}z_{\gamma_{i},\delta_{i}}^{k_{2}-1})+(可換な項)\\ &=&z_{\vb*{\alpha}_{1},\beta_{1}}^{k_{1}-1}z_{\vb*{\alpha}_{2},\beta_{2}}^{k_{2}-1}(\prod_{i=3}^{d_{1}+1}z_{\alpha_{i},\beta_{i}}^{k_{2}-1},z_{2})+z_{\gamma_{1},\delta_{1}}^{k_{1}-1}z_{\gamma_{2},\delta_{2}}^{k_{2}-1}(z_{1}\ast\prod_{i=3}^{d_{2}}x^{k_{i}-1}_{\vb*{\gamma_{i}},\vb*{\delta_{i}}}y)+(可換な項)\\ &=&\cdots\\ &=&z_{1}z_{2}+z_{2}z_{1}+(可換な項)\\ \end{eqnarray}
[1-C]$(d_{1},d_{2}+1)$の場合も同様の計算を行えばよいので可換性に関しては証明完了。
[2]結合律：
[2-A]$z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}},z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}},z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}$
\begin{eqnarray} z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}\ast (z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}})&=&z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}\ast(z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}})\\ &=&z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}1\ast z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}1\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}+z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{2}+k_{3}}\\ &=&z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}(z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}1\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}})+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}}^{k_{1}+k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{3}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}(z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}1\ast z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}})+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}+z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{2}+k_{3}}\\ &=&z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{3}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}}^{k_{1}+k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}}^{k_{1}+k_{2}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}+z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{2}+k_{3}} \end{eqnarray}
\begin{eqnarray} (z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}\ast z_{\vb*{\alpha}_{2}\ast\vb*{\beta}_{2}}^{k_{2}})\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}&=&(z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\alpha}_{2}}^{k_{1}+k_{2}})\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}1\\ &=&z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}(z_{\vb*{\alpha}_{2}\ast\vb*{\beta}_{2}}^{k_{2}}\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}1)+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}(z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}\ast 1)+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}(z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}\ast z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}1)+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}(z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}\ast 1)+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{1}}+z_{\vb*{\alpha}_{2}\sharp \vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp \vb*{\beta}_{3}}^{k_{2}+k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\alpha}_{2}}^{k_{1}+k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\alpha}_{2}}^{k_{1}+k_{2}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{2}+k_{3}}\\ &=&z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{2}+k_{3}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{3}}z_{\vb*{\alpha_{2}},\vb*{\beta}_{2}}^{k_{2}}+z_{\vb*{\alpha_{2}},\vb*{\beta_{2}}}^{k_{2}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha_{2}},\vb*{\beta_{2}}}^{k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha_{2}},\vb*{\beta_{2}}}^{k_{2}}z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{3}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{2}\sharp \vb*{\alpha}_{3},\vb*{\beta}_{2}\sharp \vb*{\beta}_{3}}^{k_{2}+k_{3}}z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\alpha}_{2}}^{k_{1}+k_{2}}z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}+z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2},\vb*{\beta}_{1}\sharp\vb*{\alpha}_{2}}^{k_{1}+k_{2}}+z_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2}\sharp\vb*{\alpha}_{3},\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\sharp\vb*{\beta}_{3}}^{k_{1}+k_{2}+k_{3}} \end{eqnarray}
[2-B]$(d_{1},d_{2},d_{3})=(1,1,1)$の場合は正しい事が分かったので、$(d_{1},d_{2},d_{3})\neq(1,1,1)$まで正しいと仮定して通常のMZV同様の計算を行えばよい。
[3]まず、$z_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}},z_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}},z_{\vb*{\alpha}_{3},\vb*{\beta}_{3}}^{k_{3}}$の場合は上記実験結果から成り立つ事が分かる。
後はMZV同様帰納法を用いて証明できるので省略する。

\begin{eqnarray} G(f_{1},f_{2},...,f_{N};x)&=&\int_{0}^{x}dx_{1}f_{1}(x_{1})\int_{0}^{x_{1}}dx_{2}f_{2}(x_{2})\cdots\int_{0}^{x_{N-1}}dx_{N-1}f_{1}(x_{N})dx_{N}\\ &=&\int_{0}^{1}dx_{N}f(x_{N})\int_{x_{N}}^{1}dx_{N-1}f_{N-1}(x_{N-1})\cdots\int_{x_{2}}^{1}dx_{1}f_{1}(x_{1})\\ &=&\int_{0}^{1}dy_{N}f(1-y_{N})\int_{0}^{y_{N}}dy_{N-1}f_{N-1}(1-y_{N-1})\cdots\int_{0}^{y_{2}}dy_{1}f_{1}(1-y_{1})\\ &=&G(f_{N}\circ(1-x),f_{N-1}\circ(1-x),...,f_{1}\circ(1-x);x) \end{eqnarray}

\begin{eqnarray} G(\frac{1}{x},\frac{1}{\sqrt{1-x}};x)&=&\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}\frac{dx_{2}}{\sqrt{1-x_{2}}}\\ &=&\sum_{n=0}^{\infty}\frac{\begin{pmatrix}2n\\n\end{pmatrix}}{2^{n}}\int_{0}^{1}\frac{dx_{1}}{x_{1}}\int_{0}^{x_{1}}dx_{2}x_{2}^{n}\\ &=&\sum_{n=0}^{\infty}\frac{\begin{pmatrix}2n\\n\end{pmatrix}}{2^{2n}}\frac{1}{(n+1)^{2}}\\ &=&G(\frac{1}{\sqrt{x}},\frac{1}{1-x};x)\\ &=&\int_{0}^{1}\frac{dx_{2}}{\sqrt{x_{2}}}\int_{0}^{x_{2}}\frac{dx_{1}}{1-x_{1}}\\ &=&\sum_{n=0}^{\infty}\frac{1}{n+1}\int_{0}^{1}dx_{2}x_{2}^{n+\frac{1}{2}}\\ &=&\sum_{n=0}^{\infty}\frac{1}{(n+1)(n+\frac{3}{2})}\\ &=&\sum_{n=1}^{\infty}\frac{1}{n(n+\frac{1}{2})}\\ &=&2\sum_{n=1}^{\infty}(\frac{1}{n}-\frac{2}{2n+1})\\ &=&2\lim_{N\rightarrow\infty}\{2+2H_{N}-2\log{N}-2H_{2N}+2\log{(2N)}-\frac{2}{2N+1}+2\log{N}-2\log{(2N)}\}\\ &=&4(1-\log{2}) \end{eqnarray}

MZVの場合と同様の計算を行うだけなので省略。

MZVと同様なので省略。

\begin{eqnarray} \sum_{n=0}^{\infty}\frac{x^{n}}{\begin{pmatrix}2n\\n\end{pmatrix}}&=&\sum_{n=0}^{\infty}x^{n}\frac{\Gamma(n+1)\Gamma(n+1)}{\Gamma(2n+1)}\\ &=&1+\sum_{n=1}^{\infty}x^{n}nB(n,n+1)\\ &=&1+\sum_{n=0}^{\infty}x^{n}n\int_{0}^{1}t^{n-1}(1-t)^{n}dt\\ &=&1+\int_{0}^{1}\frac{xt(1-t)dt}{t\{1-xt(1-t)\}^{2}}\\ &=&1+x\int_{0}^{1}\frac{(1-t)dt}{\{x(t-\frac{1}{2})^{2}+1-\frac{x}{4}\}^{2}}\\ &=&1+\frac{1}{x}\int_{0}^{1}\frac{(1-t)dt}{\{(t-\frac{1}{2})^{2}+\frac{4-x}{4x}\}^{2}}\\ &=&1+\frac{1}{x}(\frac{4x}{4-x})^{2}\sqrt{\frac{4-x}{4x}}\int_{-\sqrt{\frac{x}{4-x}}}^{\sqrt{\frac{x}{4-x}}}\frac{\frac{1}{2}-\sqrt{\frac{4-x}{x}}u}{(u^{2}+1)^{2}}du\quad(u=\tan{\theta})\\ &=&1+\frac{1}{x}(\frac{4x}{4-x})^{\frac{3}{2}}\int_{0}^{\arctan{\sqrt{\frac{x}{4-x}}}}\cos^{2}{\theta}d\theta\\ &=&1+\frac{1}{x}(\frac{4x}{4-x})^{\frac{3}{2}}\frac{1}{2}\{\arctan{\sqrt{\frac{x}{4-x}}}+\frac{1}{2}\sin{(2\arctan{\sqrt{\frac{x}{4-x}}})}\}\\ &=&1+\frac{1}{x}(\frac{4x}{4-x})^{\frac{3}{2}}\frac{1}{2}\{\arctan{\sqrt{\frac{x}{4-x}}}+\frac{\sqrt{\frac{x}{4-x}}}{1+\frac{x}{4-x}}\}\\ &=&1+\frac{1}{2x}(\frac{4x}{4-x})^{\frac{3}{2}}\arctan{\sqrt{\frac{x}{4-x}}}+\frac{\sqrt{x(4-x)}}{8x}(\frac{4x}{4-x})^{\frac{3}{2}}\\ &=&1+\frac{1}{2x}(\frac{4x}{4-x})^{\frac{3}{2}}\arctan{\sqrt{\frac{x}{4-x}}}+\frac{x}{4-x}\\ &=&\frac{4}{4-x}+\frac{1}{2x}(\frac{4x}{4-x})^{\frac{3}{2}}\arctan{\sqrt{\frac{x}{4-x}}} \end{eqnarray}

\begin{eqnarray} \sum_{n=1}^{\infty}\frac{x^{n}}{n\begin{pmatrix}2n\\n\end{pmatrix}}&=&\sum_{n=1}^{\infty}x^{n}\frac{\Gamma(n)\Gamma(n+1)}{\Gamma(2n+1)}\\ &=&\sum_{n=1}^{\infty}x^{n}\int_{0}^{1}t^{n-1}(1-t)^{n}dt\\ &=&\int_{0}^{1}\frac{xt(1-t)dt}{t\{1-xt(1-t)\}}\\ &=&\int_{0}^{1}\frac{x(1-t)dt}{\{x(t-\frac{1}{2})^{2}+\frac{4-x}{4}\}}\\ &=&\int_{0}^{1}\frac{(1-t)dt}{\{(t-\frac{1}{2})^{2}+\frac{4-x}{4x}\}}\\ &=&\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{(\frac{1}{2}-u)du}{(u^{2}+\frac{4-x}{4x})}\\ &=&\frac{1}{2}\int_{-\frac{1}{2}}^{\frac{1}{2}}\frac{du}{u^{2}+\frac{4-x}{4x}}\\ &=&\sqrt{\frac{4x}{4-x}}\int_{0}^{\arctan{\sqrt{\frac{x}{4-x}}}}d\theta\\ &=&\sqrt{\frac{4x}{4-x}}\arctan{\sqrt{\frac{x}{4-x}}} \end{eqnarray}

[1]調和積
\begin{eqnarray} x_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}y\ast x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y&=&x_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}yx_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y+x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}yx_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}y+x_{\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{1},\vb*{\beta}_{1}\sharp\vb*{\beta}_{1}}^{k_{1}+k_{2}-1}y \end{eqnarray}
[2]シャッフル積
\begin{eqnarray} x_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}y\smallsmile x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y&=&x_{\vb*{\alpha}_{1},\vb*{\beta}}^{k_{1}-1}(y1\smallsmile x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y)+x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}(x_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}y\smallsmile y1)\\ &=&x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}y(1\smallsmile x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y)+x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}(y1\smallsmile y1)+x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}(y\smallsmile y1)+x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y(x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}y\smallsmile 1)\\ &=&x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}yx_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y+2x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y^{2}+2x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}x_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}y^{2}+x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}yx_{\vb*{\alpha}_{1},\vb*{\beta}_{1}}^{k_{1}-1}y \end{eqnarray}
[3]$Y(x_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}y\ast x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y)=Y(x_{\vb*{\alpha_{1}},\vb*{\beta}_{1}}^{k_{1}-1}y\smallsmile x_{\vb*{\alpha}_{2},\vb*{\beta}_{2}}^{k_{2}-1}y)$を用いると
\begin{eqnarray} \zeta(\begin{pmatrix}\vb*{\alpha}_{1}\sharp\vb*{\alpha}_{2}\\\vb*{\beta}_{1}\sharp\vb*{\beta}_{2}\end{pmatrix},k_{1}+k_{2})=2\{\zeta(\begin{pmatrix}\vb*{\alpha}_{1}&\vb*{\alpha}_{2}\\\vb*{\beta}_{1}&\vb*{\beta}_{2}\end{pmatrix},k_{1},k_{2})+\{\zeta(\begin{pmatrix}\vb*{\alpha}_{2}&\vb*{\alpha}_{1}\\\vb*{\beta}_{2}&\vb*{\beta}_{1}\end{pmatrix},k_{2},k_{1})\} \end{eqnarray}