多重ゼータ値とその類似　Part１

　
　
$\mathbf{Definition}\mathbf{.}$

${\displaystyle C_r=2^{-2r}(\genfrac{}{}{0ex}{}{2r}{r}), P_r=\frac{(-1{)}^r(\frac{\pi }{2}{)}^{2r+1}}{(2r+1)!}, Q_r=\frac{(-1{)}^r(\frac{\pi }{2}{)}^{2r}}{(2r)!}}$

$\begin{array}{rl}{\displaystyle }& {\zeta }_n(\mathit{k})=\sum _{0<n_1<\cdots <n_r\le n}{\mathit{n}}^{-\mathit{k}}\\ & t_n^{}(\mathit{k})=\sum _{0\le n_1<\cdots <n_r\le n}{(2\mathit{n}+1)}^{-\mathit{k}}\\ & {\xi }_n(\mathit{k})=\sum _{0<n_1<\cdots <n_r\le n}{(2\mathit{n})}^{-\mathit{k}}\\ & S_n^A(\mathit{k})=\sum _{0<n_1<\cdots <n_r\le n}{C}_{n_1}^{-1}{(2\mathit{n})}^{-\mathit{k}}\\ & S_n^B(\mathit{k})=\sum _{0\le n_1<\cdots <n_r\le n}{C}_{n_1}{(2\mathit{n}+1)}^{-\mathit{k}}\end{array}$

　
$\mathbf{Theorem}\mathbf{1.}$　$m,n\in {\mathbb{Z}}_{\ge 0}$に対して次の式が成り立つ。

${\displaystyle [1.1.1] {\int }_0^{\frac{\pi }{2}}{x}^{2m}{\cos }^{2n}xdx=(-1{)}^m(2m)!C_n\sum _{i=0}^m{P}_{m-i}{\xi }_n^{\star }(\{2{\}}_i)}$

${\displaystyle [1.1.2] {\int }_0^{\frac{\pi }{2}}{x}^{2m}{\cos }^{2n+1}xdx=\frac{(-1{)}^m(2m)!}{(2n+1)C_n}\sum _{i=0}^m{Q}_{m-i}{t}_n^{\star }(\{2{\}}_i)}$

${\displaystyle [1.1.3] {\int }_0^{\frac{\pi }{2}}{x}^{2m+1}{\cos }^{2n}xdx=(-1{)}^{m+1}(2m+1)!C_n(S_n^{A\star }(\{2{\}}_{m+1})+\sum _{i=0}^m{Q}_{m-i+1}{\xi }_n^{\star }(\{2{\}}_i))}$

${\displaystyle [1.1.4] {\int }_0^{\frac{\pi }{2}}{x}^{2m+1}{\cos }^{2n+1}xdx=\frac{(-1{)}^m(2m+1)!}{(2n+1)C_n}(\sum _{i=0}^m{P}_{m-i}{t}_n^{\star }(\{2{\}}_i)-S_n^{B\star }(1,\{2{\}}_m))}$

${\displaystyle [1.1.5] {\int }_0^{\frac{\pi }{2}}{x}^{2m}{\sin }^{2n}xdx=(-1{)}^m(2m)!C_n(\sum _{i=0}^m{P}_{m-i}{\xi }_n^{\star }(\{2{\}}_i)-\sum _{i=1}^m{P}_{m-i}{S}_n^{A\star }(\{2{\}}_i))}$

${\displaystyle [1.1.6] {\int }_0^{\frac{\pi }{2}}{x}^{2m}{\sin }^{2n+1}xdx=\frac{(-1{)}^m(2m)!}{(2n+1)C_n}(t_n^{\star }(\{2{\}}_m)-\sum _{i=0}^m{P}_{m-i}{S}_n^{B\star }(1,\{2{\}}_{i-1}))}$

${\displaystyle [1.1.7] {\int }_0^{\frac{\pi }{2}}{x}^{2m+1}{\sin }^{2n}xdx=(-1{)}^m(2m+1)!C_n(\sum _{i=0}^m{Q}_{m-i}{S}_n^{A\star }(\{2{\}}_{i+1})-\sum _{i=0}^m{Q}_{m-i+1}{\xi }_n^{\star }(\{2{\}}_i))}$

${\displaystyle [1.1.8] {\int }_0^{\frac{\pi }{2}}{x}^{2m+1}{\sin }^{2n+1}xdx=\frac{(-1{)}^m(2m+1)!}{(2n+1)C_n}\sum _{i=0}^m{Q}_{m-i}{S}_n^{B\star }(1,\{2{\}}_i)}$

　
$\mathbf{Proof}\mathbf{.}$
　
いずれも同じような部分積分をし，漸化式を解くことで容易に求まるので，$[1.1.1]$のみ証明を示す。

${\displaystyle I(m,n)={\int }_0^{\frac{\pi }{2}}{x}^{2m}{\cos }^{2n}xdx}$
　
とおく。部分積分し整理すれば，

${\displaystyle 2nI(m,n)=-\frac{(2m-1)2m}{2n}I(m-1,n)+(2n-1)I(m,n-1)}$
　
を得る。両辺$\frac{1}{2n{C}_n}$を掛けると

$\begin{array}{rl}{\displaystyle \frac{I(m,n)}{C_n}}& =-\frac{(2m-1)2m}{(2n{)}^2{C}_n}I(m-1,n)+\frac{I(m,n-1)}{C_{n-1}}\\ & =-\sum _{i=1}^n\frac{(2m-1)2m}{(2i{)}^2{C}_i}I(m-1,i)+\frac{I(m,0)}{C_0}\\ & =-\sum _{i=1}^n\frac{(2m-1)2m}{(2i{)}^2{C}_i}I(m-1,i)+\frac{{\left(\frac{\pi }{2}\right)}^{2m+1}}{2m+1}\end{array}$
　
両辺$\frac{1}{(2m)!}$を掛けると

$\begin{array}{rl}{\displaystyle \frac{I(m,n)}{(2m)!C_n}}& =\frac{{\left(\frac{\pi }{2}\right)}^{2m+1}}{(2m+1)!}-\sum _{i=1}^n\frac{1}{(2i{)}^2}\frac{I(m-1,i)}{(2m-2)!C_i}\\ & =\frac{{\left(\frac{\pi }{2}\right)}^{2m+1}}{(2m+1)!}-\sum _{i=1}^n\frac{1}{(2i{)}^2}(\frac{{\left(\frac{\pi }{2}\right)}^{2m-1}}{(2m-1)!}-\sum _{j=1}^i\frac{1}{(2j{)}^2}\frac{I(m-2,j)}{(2m-4)!C_j})\\ & =\cdots \\ & =(-1{)}^m\sum _{i=0}^m{P}_{m-i}{\xi }_n^{\star }(\{2{\}}_i)\end{array}$
$◻$

　
$\mathbf{Corollary}\mathbf{1.1}\mathbf{.}$

${\displaystyle [1.2.1] \underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^m{P}_{m-i}{\xi }_n^{\star }(\{2{\}}_i)=0 (m\in {\mathbb{Z}}_{>0})}$

${\displaystyle [1.2.2] \underset{n\to \mathrm{\infty }}{lim}\sum _{i=0}^m{Q}_{m-i}{t}_n^{\star }(\{2{\}}_i)=0 (m\in {\mathbb{Z}}_{>0})}$

${\displaystyle [1.2.3] \underset{n\to \mathrm{\infty }}{lim}(S_n^{A\star }(\{2{\}}_{m+1})+\sum _{i=0}^m{Q}_{m-i+1}{\xi }_n^{\star }(\{2{\}}_i))=0 (m\in {\mathbb{Z}}_{\ge 0})}$

${\displaystyle [1.2.4] \underset{n\to \mathrm{\infty }}{lim}(\sum _{i=0}^m{P}_{m-i}{t}_n^{\star }(\{2{\}}_i)-S_n^{B\star }(1,\{2{\}}_m))=0 (m\in {\mathbb{Z}}_{\ge 0})}$

　
$\mathbf{Proof}\mathbf{.}$

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{C_n}{\int }_0^{\frac{\pi }{2}}{x}^m{\cos }^{2n}xdx=0, \underset{n\to \mathrm{\infty }}{lim}(2n+1)C_n{\int }_0^{\frac{\pi }{2}}{x}^m{\cos }^{2n+1}xdx=0}$
　
より明らか。
$◻$

　
$\mathbf{Corollary}\mathbf{1.2}\mathbf{.}$

${\displaystyle [1.3.1] \sum _{i=0}^m{P}_{m-i}\sum _{0<n}\frac{{\xi }_n^{\star }(\{2{\}}_i){\xi }_{n-1}(\{2{\}}_{a-1})}{(2n{)}^2}=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m+1}}{(2a+2m+1)!}}$

${\displaystyle [1.3.2] \sum _{i=0}^m{Q}_{m-i}\sum _{0\le n}\frac{t_n^{\star }(\{2{\}}_i)t_{n-1}(\{2{\}}_{a-1})}{(2n+1{)}^2}=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m}}{(2a+2m)!}}$

${\displaystyle [1.3.3] \sum _{0<n}\frac{{\xi }_{n-1}(\{2{\}}_{a-1})}{(2n{)}^2}(S_n^{A\star }(\{2{\}}_{m+1})+\sum _{i=0}^m{Q}_{m-i+1}{\xi }_n^{\star }(\{2{\}}_i))=\frac{(-1{)}^{m+1}{\left(\frac{\pi }{2}\right)}^{2a+2m+2}}{(2a+2m+2)!}}$

${\displaystyle [1.3.4] \sum _{0\le n}\frac{t_{n-1}(\{2{\}}_{a-1})}{(2n+1{)}^2}(\sum _{i=0}^m{P}_{m-i}{t}_n^{\star }(\{2{\}}_i)-S_n^{B\star }(1,\{2{\}}_m))=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m+1}}{(2a+2m+1)!}}$

${\displaystyle [1.3.5] \sum _{0<n}\frac{{\xi }_{n-1}(\{2{\}}_{a-1})}{(2n{)}^2}(\sum _{i=0}^m{P}_{m-i}{\xi }_n^{\star }(\{2{\}}_i)-\sum _{i=1}^m{P}_{m-i}{S}_n^{A\star }(\{2{\}}_i))=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m+1}}{(2a+2m+1)(2a)!(2m)!}}$

${\displaystyle [1.3.6] \sum _{0\le n}\frac{t_{n-1}(\{2{\}}_{a-1})}{(2n+1{)}^2}(t_n^{\star }(\{2{\}}_m)-\sum _{i=0}^m{P}_{m-i}{S}_n^{B\star }(1,\{2{\}}_{i-1}))=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m}}{(2a+2m)(2a-1)!(2m)!}}$

${\displaystyle [1.3.7] \sum _{0<n}\frac{{\xi }_{n-1}(\{2{\}}_{a-1})}{(2n{)}^2}(\sum _{i=0}^m{Q}_{m-i}{S}_n^{A\star }(\{2{\}}_{i+1})-\sum _{i=0}^m{Q}_{m-i+1}{\xi }_n^{\star }(\{2{\}}_i))=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m+2}}{(2a+2m+2)(2a)!(2m+1)!}}$

${\displaystyle [1.3.8] \sum _{i=0}^m{Q}_{m-i}\sum _{0\le n}\frac{t_{n-1}(\{2{\}}_{a-1})}{(2n+1{)}^2}{S}_n^{B\star }(1,\{2{\}}_i)=\frac{(-1{)}^m{\left(\frac{\pi }{2}\right)}^{2a+2m+1}}{(2a+2m+1)(2a-1)!(2m+1)!}}$

　
$\mathbf{Proof}\mathbf{.}$

${\displaystyle \sum _{0<n}\frac{{\xi }_{n-1}(\{2{\}}_{a-1})}{(2n{)}^2{C}_n}{\sin }^{2n}x=\frac{x^{2a}}{(2a)!}, \sum _{0\le n}\frac{C_n{t}_{n-1}(\{2{\}}_{a-1})}{2n+1}{\sin }^{2n+1}x=\frac{x^{2a-1}}{(2a-1)!}}$
　
より明らか。
$◻$

　
$\mathbf{Corollary}\mathbf{1.3}\mathbf{.}$

${\displaystyle [1.4.1] \sum _{i=0}^m{P}_{m-i}\sum _{0<n}\frac{C_n^2{\xi }_n^{\star }(\{2{\}}_i)}{2n}=\frac{(-1{)}^m}{(2m)!}{\int }_0^{\frac{\pi }{2}}{x}^{2m}\ln \frac{2}{1+\sin x}dx}$

${\displaystyle [1.4.1] \sum _{i=0}^m{Q}_{m-i}\sum _{0\le n}\frac{t_n^{\star }(\{2{\}}_i)}{(2n+1{)}^2{C}_n^2}=\frac{(-1{)}^m}{(2m)!}{\int }_0^{\frac{\pi }{2}}\frac{x^{2m}(\frac{\pi }{2}-x)}{\sin x}dx}$

　
$\mathbf{Proof}\mathbf{.}$

${\displaystyle \sum _{0<n}\frac{C_n}{2n}{x}^n=\ln \frac{2}{1+\sqrt{1-x}}, \sum _{0\le n}\frac{x^{2n+1}}{(2n+1)C_n}=\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}}$
　
より明らか。
$◻$
　
$\mathbf{Remarks}\mathbf{.}$
　複素解析などにより，$\mathbf{Corollary}\mathbf{1.3}$の右辺の積分は計算できるが，ここでは省く。
　いずれにしても，都合がいい$\mathrm{M}\mathrm{a}\mathrm{c}\mathrm{l}\mathrm{a}\mathrm{u}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{x}\mathrm{p}\mathrm{a}\mathrm{n}\mathrm{s}\mathrm{i}\mathrm{o}\mathrm{n}$を適応することで，積分を計算できるものにした。
　著者は$\mathbf{Theorem}\mathbf{1}$をもとに様々な結果を得た。導出過程は省くが，結果だけいくつか列挙する。

　
$\mathbf{Corollary}\mathbf{1.4}\mathbf{.}$　$E_r:\mathrm{E}\mathrm{u}\mathrm{l}\mathrm{e}\mathrm{r}\mathrm{n}\mathrm{u}\mathrm{m}\mathrm{b}\mathrm{e}\mathrm{r}$

${\displaystyle [1.5.1] \sum _{0\le h}{\xi }_n^{\star }(\{2{\}}_h)=\frac{1}{(2n+1)C_n^2}}$

${\displaystyle [1.5.2] \sum _{0<h}{S}_n^{A\star }(\{2{\}}_h)=\frac{1}{(2n+1)C_n}(\frac{1}{C_n}-1)}$

${\displaystyle [1.5.3] \sum _{0<n}\frac{C_n{\zeta }_n^{\star }(\{2{\}}_{a-1})}{n^2}=-4{\zeta }^{\star }(2a-1,\bar{1})-2\zeta (2a)}$

${\displaystyle [1.5.4] \sum _{0\le n}\frac{S_n^{B\star }(1,\{2{\}}_{a-1})}{(2n+1{)}^2{C}_n}=2t(2a+1)}$

${\displaystyle [1.5.5] \sum _{0\le n}\frac{t_n^{\star }(\{2{\}}_{a-1})}{(2n+1{)}^3{C}_n^2}=4at(2a+1)-\pi \sum _{i=0}^{a-1}{E}_{2i}{Q}_i\beta (2a-2i)}$

${\displaystyle [1.5.6] \sum _{0\le n}\frac{E_{2a}{Q}_a-t_n^{\star }(\{2{\}}_a)}{(2n+1{)}^2{C}_n^2}=2(2a+1)\beta (2a+2)-2E_{2a}{Q}_a-\pi \sum _{i=0}^{a-1}{E}_{2i}{Q}_{i}t(2a-2i+1)}$