多重ゼータ値とその類似　Part４

　
$\mathbf{L}\mathbf{e}\mathbf{m}\mathbf{m}\mathbf{a}\mathbf{4.1}\mathbf{.}$　　$0<x\le 1$に対して

${\displaystyle [4.1] {\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},2}(1-x)=\zeta (a+1)-\sum _{j=0}^a\frac{1}{j!}\sum _{0<n}\frac{x^n}{n^{a-j+1}}{\ln }^j\frac{1}{x}}$

　
$\mathbf{Proof}\mathbf{.}$

$\begin{array}{rl}{\displaystyle {\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},2}(1-x)}& ={\int }_0^{1-x}\frac{1}{1-t}{(\circ \frac{1}{1-t})}^{a-1}\circ \frac{1}{t}\\ & ={\int }_x^1\frac{1}{1-t}{(\circ \frac{1}{t})}^a\\ & =\sum _{0<n}{\int }_x^1{t}^{n-1}{(\circ \frac{1}{t})}^a\\ & =\sum _{0<n}{\int }_x^1\frac{1}{t}\frac{t^n-x^n}{n}{(\circ \frac{1}{t})}^{a-1}\\ & =\sum _{0<n}{\int }_x^1\frac{1}{t}(\frac{t^n-x^n}{n^2}-\frac{x^n}{n}\ln \frac{t}{x}){(\circ \frac{1}{t})}^{a-2}\\ & =\cdots \\ & =\sum _{0<n}(\frac{1}{n^{a+1}}-\sum _{j=0}^a\frac{1}{j!}\frac{x^n}{n^{a-j+1}}{\ln }^j\frac{1}{x})\\ & =\zeta (a+1)-\sum _{j=0}^a\frac{1}{j!}\sum _{0<n}\frac{x^n}{n^{a-j+1}}{\ln }^j\frac{1}{x}\end{array}$
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$\mathbf{C}\mathbf{o}\mathbf{n}\mathbf{j}\mathbf{e}\mathbf{c}\mathbf{t}\mathbf{u}\mathbf{r}\mathbf{e}\mathbf{4.1}\mathbf{.}$　　$a\in {\mathbb{Z}}_{>0},b\in {\mathbb{Z}}_{\ge 0}$に対して

${\displaystyle [4.2] {\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},b+2}(1-x)=\sum _{0<n}(\frac{1}{n^{a+1}}-\sum _{j=0}^a\frac{x^n}{n^{a-j+1}}{\mathrm{L}\mathrm{i}}_{{\{1\}}_j}(1-x))\sum _{k=0}^b(-1{)}^k{\zeta }_{n-1}(\{1{\}}_{b-k}){\mathrm{L}\mathrm{i}}_{{\{1\}}_k}(x)}$

著者は$b=0,1,2$での証明に成功したが，一般の$b$での厳密な証明は未達成。計算方法は確立しているが。

　
$\mathbf{C}\mathbf{o}\mathbf{r}\mathbf{o}\mathbf{l}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{y}\mathbf{4.1}\mathbf{.}$　　$x<1$に対して

${\displaystyle [4.3] \sum _{0<n}\frac{{\zeta }_{n-1}(\{1{\}}_{a-1})}{n^3}\frac{n!}{(1-x{)}_n}=\sum _{0<m\le n}\frac{1}{(m-x)n^{a+1}}-\sum _{j=1}^a\sum _{0<m<n}\frac{1}{m^{a-j+1}(n-x{)}^{j+1}}}$

　
$\mathbf{Proof}\mathbf{.}$

$\begin{array}{rl}{\displaystyle \sum _{0<n}\frac{{\zeta }_{n-1}(\{1{\}}_{a-1})}{n^3}\frac{n!}{(1-x{)}_n}}& ={\int }_0^1\frac{t^{-x}}{1-t}{\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},2}(1-t)dt\\ & ={\int }_0^1\frac{t^{-x}}{1-t}\sum _{0<n}(\frac{1-t^n}{n^{a+1}}-\sum _{j=1}^a\frac{1}{j!}\frac{t^n}{n^{a-j+1}}{\ln }^j\frac{1}{t})dt\\ & =\sum _{0<n}(\frac{1}{n^{a+1}}{\int }_0^1\frac{t^{-x}(1-t^n)}{1-t}dt-\sum _{j=1}^a\frac{1}{j!n^{a-j+1}}{\int }_0^1\frac{t^{n-x}}{1-t}{\ln }^j\frac{1}{t})\\ & =\sum _{0<n}(\frac{1}{n^{a+1}}\sum _{m=1}^n{\int }_0^1{t}^{m-1+x}dt-\sum _{j=1}^a\frac{1}{j!n^{a-j+1}}\sum _{n<m}{\int }_0^1{t}^{m-1-x}{\ln }^j\frac{1}{t}dt)\\ & =\sum _{0<n}(\frac{1}{n^{a+1}}\sum _{m=1}^n\frac{1}{m-x}-\sum _{j=1}^a\frac{1}{j!n^{a-j+1}}\sum _{n<m}\frac{j!}{(m-x{)}^{j+1}})\\ & =\sum _{0<m\le n}\frac{1}{(m-x)n^{a+1}}-\sum _{j=1}^a\sum _{0<m<n}\frac{1}{m^{a-j+1}(n-x{)}^{j+1}}\end{array}$
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$\mathbf{C}\mathbf{o}\mathbf{r}\mathbf{o}\mathbf{l}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{y}\mathbf{4.2}\mathbf{.}$

${\displaystyle [4.4] \sum _{0<n}\frac{{\zeta }_{n-1}(\{1{\}}_{a-1}){\zeta }_n^{\star }(\{1{\}}_b)}{n^3}={\zeta }^{\star }(b+1,a+1)-\sum _{j=1}^a(\genfrac{}{}{0ex}{}{j+b-1}{b})\zeta (a-j+1,j+b+1)}$

　
$\mathbf{Proof}\mathbf{.}$
　$[4.3]$において，$x$について冪級数展開して係数を比較する。
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$\mathbf{T}\mathbf{h}\mathbf{e}\mathbf{o}\mathbf{r}\mathbf{e}\mathbf{m}\mathbf{4.1}\mathbf{.}$

${\displaystyle [4.5] {\mathrm{L}\mathrm{i}}_{\{1{\}}_{a-1},b+1}(1-x)=\sum _{j=0}^{b-1}(-1{)}^j\zeta (\{1{\}}_{a-1},b-j+1){\mathrm{L}\mathrm{i}}_{\{1{\}}_j}(x)+(-1{)}^b\sum _{0\le n_1\le \cdots \le n_b\le a}{\mathrm{L}\mathrm{i}}_{\{1{\}}_{n_1}}(1-x){\mathrm{L}\mathrm{i}}_{a-n_b+1,n_b-n_{b-1}+1,\cdots ,n_2-n_1+1}(x)}$

　
$\mathbf{Proof}\mathbf{.}$

${\displaystyle {\int }_0^x\frac{1}{1-t}{\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},2}(1-t)dt=\zeta (\{1{\}}_{a-1},3)-{\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},3}(1-x)}$

$\begin{array}{rl}{\displaystyle {\int }_0^x\frac{1}{1-t}{\mathrm{L}\mathrm{i}}_{{\{1\}}_{a-1},2}(1-t)dt}& ={\int }_0^x\frac{1}{1-t}(\zeta (a+1)-\sum _{j=0}^a\frac{1}{j!}\sum _{0<n}\frac{t^n}{n^{a-j+1}}{\ln }^j\frac{1}{t})dt\\ & =\zeta (a+1)\ln \frac{1}{1-x}-\sum _{j=0}^a\frac{1}{j!}\sum _{0<n}\frac{1}{n^{a-j+1}}{\int }_0^x\frac{t^n}{1-t}{\ln }^j\frac{1}{t}dt\\ & =\zeta (\{1{\}}_{a-1},2)\ln \frac{1}{1-x}-\sum _{j=0}^a\frac{1}{j!}\sum _{0<n}\frac{1}{n^{a-j+1}}\sum _{k=0}^j\frac{j!}{k!}{\ln }^k\frac{1}{x}\sum _{n<m}\frac{x^m}{m^{j-k+1}}\\ & =\zeta (\{1{\}}_{a-1},2)\ln \frac{1}{1-x}-\sum _{0\le k\le j\le a}\frac{1}{k!}{\ln }^k\frac{1}{x}\sum _{0<n<m}\frac{x^m}{n^{a-j+1}{m}^{j-k+1}}\end{array}$

${\displaystyle {\int }_0^x\frac{1}{1-t}dt}$を繰り返せばよい。
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