MZVの和公式の証明

MZVの反復積分表示より${\displaystyle \sum _{I_0(k,r)}\zeta (\mathbf{k})=\sum _{\begin{array}{c}{\epsilon }_2,\dots ,{\epsilon }_{k-1}\in \{0,1\}\\ \text{wt}(\epsilon )=r-1\end{array}}I(1,{\epsilon }_2,\dots ,{\epsilon }_{k-1},0)}$がわかる(${\displaystyle {\omega }_0(t)=\frac{dt}{t},{\omega }_1(t)=\frac{dt}{1-t}}$とし${\epsilon }_1,\dots ,{\epsilon }_k\in \{0,1\}$に対し${\displaystyle I({\epsilon }_1,\dots ,{\epsilon }_k)={\int }_{0<t_1<\cdots <t_k<1}\prod _{i=1}^k{\omega }_i(t_i)}$とおいた)
$$\begin{array}{rl}\sum _{n=1}^{k-1}\sum _{I_0(k,n)}\zeta (\mathbf{k})X^{n-1}& =\sum _{{\epsilon }_2,\dots ,{\epsilon }_{k-1}\in \{0,1\}}I(1,{\epsilon }_2,\dots ,{\epsilon }_k,0)X^{{\epsilon }_2+\cdots +{\epsilon }_{k-1}}\\ & ={\int }_{0<t_1<\cdots <t_k<1}\frac{1}{1-t_1}(\frac{1}{t_2}+\frac{X}{1-t_2})\cdots (\frac{1}{t_{k-1}}+\frac{X}{1-t_{k-1}})\frac{1}{t_k}d{t}_1\cdots d{t}_k\\ & =\frac{1}{(k-2)!}{\int }_{0<t_1<t_k<1}{\left({\int }_{t_1}^{t_k}(\frac{1}{t}+\frac{X}{1-t})dt\right)}^{k-2}\frac{d{t}_1}{1-t_1}\frac{d{t}_k}{t_k}\\ & =\frac{1}{(k-2)!}{\int }_{0<t_1<t_k<1}{(\ln \frac{t_1}{t_k}+X\ln \frac{1-t_k}{1-t_1})}^{k-2}\frac{d{t}_1}{1-t_1}\frac{d{t}_k}{t_k}\end{array}$$
${\displaystyle x=\ln \frac{t_1}{t_k},y=\ln \frac{1-t_k}{1-t_1},\frac{d{t}_1}{1-t_1}\frac{d{t}_k}{t_k}=\frac{dxdy}{e^{x+y}-1}}$と変数変換し$X^{n-1}$で係数比較すると
$$\begin{array}{rl}\sum _{I_0(k,r)}\zeta (\mathbf{k})& ={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{e^{-m(x+y)}}{(k-n-1)!(n-1)!}{x}^{k-n-1}{y}^{n-1}dxdy\\ & =\zeta (k)\end{array}$$

${\displaystyle \sum _{k=r+1}^{\mathrm{\infty }}\sum _{\mathbf{k}\in I_0(k,r)}\zeta (\mathbf{k})Y^{k-r-1}}$という母関数を考える
この母関数は$|Y|<1$で絶対収束する(詳しくは参考文献を参照)
${\displaystyle \zeta (2)\sum _{k=r+1}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{k-2}{r-1})Y^{k-r-1}=\frac{\zeta (2)}{(1-Y{)}^r}}$
$$\begin{array}{rl}\sum _{k=r+1}^{\mathrm{\infty }}\sum _{\mathbf{k}\in I_0(k,r)}\zeta (\mathbf{k})Y^{k-r-1}& =\sum _{0<n_1<\cdots <n_r}\sum _{\mathbf{k}\in I_0(r)}\frac{Y^{\text{wt}(\mathbf{k})-r-1}}{{\mathbf{n}}^{\mathbf{k}}}\\ & =\sum _{0<n_1<\cdots <n_r}\sum _{\mathbf{k}\in I_0(r)}\frac{Y^{k_1-1}}{n_1^{k_1}}\frac{Y^{k_2-1}}{n_2^{k_2}}\cdots \frac{Y^{k_r-2}}{n_r^{k_r}}\\ & =\sum _{0<n_1<\cdots <n_r}\frac{1}{n_1-Y}\frac{1}{n_2-Y}\cdots \frac{1}{n_r(n_r-Y)}\end{array}$$
一方で
$$\begin{array}{rl}\sum _{k=r+1}^{\mathrm{\infty }}\zeta (k)Y^{k-r-1}& =\sum _{k=r+1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{Y^{k-r-1}}{m^k}\\ & =\sum _{m=1}^{\mathrm{\infty }}\frac{1}{m^r(m-Y)}\end{array}$$
また、${\displaystyle \sum _{0<r<k}\sum _{\mathbf{k}\in I_0(k,r)}\zeta (\mathbf{k})X^{r-1}{Y}^{k-r-1}}$を考える
($|X|<|1-Y|$で収束する)
$$\begin{array}{rl}\sum _{0<r<k}\sum _{\mathbf{k}\in I_0(k,r)}\zeta (\mathbf{k})X^{r-1}{Y}^{k-r-1}& =\sum _{r=1}^{\mathrm{\infty }}(\sum _{k=r+1}^{\mathrm{\infty }}\sum _{\mathbf{k}\in I_0(k,r)}\zeta (\mathbf{k})Y^{k-r-1})X^{r-1}\\ & =\sum _{r=1}^{\mathrm{\infty }}\frac{1}{(r-1)!}{\int }_{0<t_1<t_2<1}\frac{t_2^{Y-1}}{t_1^Y(1-t_1)}{(X\ln \frac{1-t_1}{1-t_2})}^{r-1}d{t}_{1}d{t}_2\dots (\ast )\\ & ={\int }_{0<t_1<t_2<1}\frac{t_2^{Y-1}}{t_1^Y(1-t_1)}\exp (X\ln \frac{1-t_1}{1-t_2})d{t}_{1}d{t}_2\\ & ={\int }_{0<t_1<t_2<1}\frac{t_2^{Y-1}}{t_1^Y(1-t_1)}{\left(\frac{1-t_1}{1-t_2}\right)}^{X}d{t}_{1}d{t}_2\end{array}$$
ここで${\displaystyle u=\frac{t_2}{t_1},v=\frac{1-t_1}{1-t_2}}$と変数変換すると
${\displaystyle RHS=\sum _{m=1}^{\mathrm{\infty }}\frac{1}{(m-X)(m-Y)}}$
上より
$$\begin{array}{rl}\sum _{0<r<k}\zeta (k)X^{r-1}{Y}^{k-r-1}& =\sum _{n=1}^{\mathrm{\infty }}(\sum _{k=r+1}^{\mathrm{\infty }}\zeta (k)Y^{k-r-1})X^{r-1}\\ & =\sum _{r=1}^{\mathrm{\infty }}\sum _{m=1}^{\mathrm{\infty }}\frac{X^{r-1}}{m^r(m-Y)}\\ & =\sum _{m=1}^{\mathrm{\infty }}\frac{1}{(m-X)(m-Y)}\end{array}$$
以上より${\displaystyle \sum _{0<r<k}\sum _{\mathbf{k}\in I_0(k,r)}\zeta (\mathbf{k})X^{r-1}{Y}^{k-r-1}=\sum _{0<r<k}\zeta (k)X^{r-1}{Y}^{k-r-1}}$を係数比較すれば得る

$s\ge 1,n\ge s,k\ge n+s$、${\displaystyle \gamma (x,y,z;t)=\sum _{k,n,s}\sum _{\mathbf{k}\in I_0(k,r,s)}{\text{Li}}_{\mathbf{k}}(t)x^{k-n-s}{y}^{n-s}{z}^{2s-2}}$
という母関数を考える
(ここで$I_0(k,r,s)$は重さ$k$深さ$r$高さ$s$な収束インデックス全体の集合を表す)
この母関数は${\displaystyle t(1-t)\frac{d^2\gamma }{d{t}^2}+((1-t)(1-x)-yt)\frac{d\gamma }{dt}+(xy-z^2)\gamma =1\dots (1)}$
という超幾何微分方程式を満たす
(多重ポリログの微分方程式より)
$z^2=xy$としたときの$x^{k-n-1}{y}^{n-1}$の係数は
${\displaystyle \sum _s\sum _{\mathbf{k}\in I_0(k,r,s)}{\text{Li}}_{\mathbf{k}}(t)=\sum _{\mathbf{k}\in I_0(k,r)}{\text{Li}}_{\mathbf{k}}(t)}$
(${\displaystyle I_0(k,r)=\bigcup _s{I}_0(k,n,s)}$)
$(1)$において$z^2=xy$とおいたもの($\overline{\gamma }$とおく)を解く
階数を下げるために$u={\overline{\gamma }}^{\prime }$とする
${\displaystyle t(1-t)u^{\prime }+((1-t)(1-x)-yt)u=1}$
この方程式の解は${\displaystyle \overline{\gamma }(x,y;t)=\frac{1}{1-x}{\int }_0^t(1-w{)}^{-y}{}_2{F}_1[\begin{array}{c}1-x,1-y\\ 2-x\end{array};w]dw}$となる
($t=0$で正則である必要があるため)
Abelの連続性定理より
$$\begin{array}{rl}\overline{\gamma }(x,y;1)& =\frac{1}{1-x}{\int }_0^1(1-w{)}^{-y}{}_2{F}_1[\begin{array}{c}1-x,1-y\\ 2-x\end{array};w]dw\\ & =\sum _{m=0}^{\mathrm{\infty }}\frac{1}{(m+1-x)(m+1-y)}\end{array}$$
このとき$x^{k-r-1}{y}^{r-1}$で係数比較することで得る