リーマンゼータ関数を多重対数関数の和に分解する
$\zeta(s)=\zeta{\qty(\{1\}^{s-2},2)}$の一般化とも言えます。
定義
$\bm s:1\to m \Defarrow \bm s \coloneqq
(s_1,\range s2m)$
$a\vbin<\bm x_{[1\to m]} \Defarrow a< x_1< x_2<\cdots< x_m$
$\displaystyle \prod_{1\to m} \bm n^{-\bm s} \coloneqq
\prod_{k=1}^m n_k^{-s_k}$
$\bm s:1\to m$
$\beginend{align}{
\zeta(\bm s) \acoloneqq
\sum_{0\vbin<\bm n_{[1\to m]}} \prod \bm n^{-\bm s} \\
\mathrm{Li}_{\bm s}(z) \acoloneqq
\sum_{0\vbin<\bm n_{[1\to m]}} z^{n_m}\prod \bm n^{-\bm s}
}$
$\bm\varepsilon\in\{0,1\}:1\to r$
$\beginend{align}{
I_{x,y}(\bm\varepsilon) \acoloneqq
\int_{x\vbin\le\bm t_{[1\to r]}\le y}
\prod \frac{\dd{\bm t}}{\bm\varepsilon+(-1)^{\bm\varepsilon}\bm t} \\
I_{x,y}(\varnothing) \acoloneqq 1 \\
I(\bm\varepsilon) \acoloneqq
I_{0,1}(\bm\varepsilon)
}$
$a\le b$
$a\vbin\le\varnothing\le b\Defarrow\top$
定理
$I_{x,y}(\bm\varepsilon) = I_{1-y,1-x}{\qty(1-\overleftarrow{\bm\varepsilon})}$
$\mathrm{Li}_{\bm s}(z) = I_{0,z}{\qty(\qty{1,\{0\}^{s_k-1}}_{k=1}^m)}$
$\mathrm{Li}_{\{1\}^m}(z) = \dfrac{(-\ln(1-z))^m}{m!}$
$\displaystyle \uhen = \frac1{m!}\qty(\int_0^z \frac{\dd t}{1-t})^m = \int_{0\vbin\le \bm t_{[1\to m]}\le z} \prod \frac{\dd{\bm t}}{1-\bm t} = I_{0,z}{\qty(\{1\}^m)} = \sahen$
$a\le c\le b$
$\bm x:1\to m$
$\displaystyle a\vbin\le \bm x\le b\Longleftrightarrow
\bigvee_{k=0}^m a\vbin\le\bm x_{[\to k]}\le c\vbin\le\bm x_{[k+1\to]}\le b$
$m=3$
$ \beginend{alignat}{4 a\le t_1\le t_2\le t_3\le b \Longleftrightarrow& &&{\color{blue}c}&&\le\ t_1\le &&\ t_2\le &&\ t_3\le b \\ &{}\vee{} &&a &&\le\ t_1\le {\color{blue}c}\le &&\ t_2\le &&\ t_3\le b \\ &{}\vee{} &&a &&\le\ t_1\le &&\ t_2\le {\color{blue}c}\le &&\ t_3\le b \\ &{}\vee{} &&a &&\le\ t_1\le &&\ t_2\le &&\ t_3\le {\color{blue}c}&& }$$\displaystyle I_{x,y}(\bm\varepsilon) = \sum_{k=0}^r I_{x,z}{\qty(\bm\varepsilon_{[\to k]})} I_{z,y}{\qty(\bm\varepsilon_{[k+1\to]})}$
$\beginend{array}{l s\in\Z_{\ge2} \\ \displaystyle \zeta(s) = \mathrm{Li}_{\{1\}^{s-2},2}(1-z) + \sum_{k=0}^{s-1} \frac{(-\ln z)^k}{k!} \mathrm{Li}_{s-k}(z) }$
$\bm\varepsilon \coloneqq \qty(1,\{0\}^{s-1})$
$\beginend{align}{
\sahen &=
I(\bm\varepsilon) =
\sum_{k=0}^s I_{0,z}{\qty(\bm\varepsilon_{[\to k]})}
I_{z,1}{\qty(\bm\varepsilon_{[k+1\to]})} \\&=
\sum_{k=0}^s I_{0,z}{\qty(\bm\varepsilon_{[\to k]})}
I_{0,1-z}{\qty(1-\overleftarrow{\bm\varepsilon_{[k+1\to]}})} \\&=
I_{0,1-z}{\qty(\{1\}^{s-1},0)} +
\sum_{k=1}^s I_{0,z}{\qty(1,\{0\}^{k-1})}
I_{0,1-z}{\qty(\{1\}^{s-k})} \\&=
\mathrm{Li}_{\{1\}^{s-2},2}(1-z) +
\sum_{k=1}^s \mathrm{Li}_k(z)\mathrm{Li}_{\{1\}^{s-k}}(1-z) \\&=
\mathrm{Li}_{\{1\}^{s-2},2}(1-z) +
\sum_{k=1}^s \frac{(-\ln z)^{s-k}}{(s-k)!}
\mathrm{Li}_k(z) \quad\because\textsf{補題3} \\&=
\uhen
}$
$\beginend{array}{l \zeta(2) = \mathrm{Li}_2(z)+\mathrm{Li}_2(1-z)+\ln z\ln(1-z) \\ \zeta(3) = \mathrm{Li}_3(z)+\mathrm{Li}_{1,2}(1-z)- \ln z\,\mathrm{Li}_2(z)-\dfrac{\ln^2z\ln(1-z)}2 \\ z=\frac12 \\ \zeta(2) = 2\mathrm{Li}_2{\qty(\dfrac12)}+\ln^22 \\ \zeta(3) = \mathrm{Li}^\star_{1,2}{\qty(\dfrac12)}+\dfrac{\pi^2\ln2}{12} \quad\because\small \mathrm{Li}_3{\qty(\dfrac12)}+\mathrm{Li}_{1,2}{\qty(\dfrac12)} = \mathrm{Li}^\star_{1,2}{\qty(\dfrac12)} }$