多重ゼータ値とその類似　Part２

　
$\mathbf{Lemma}\mathbf{2.1}\mathbf{.}$　　$n\in {\mathbb{Z}}_{\ge 0}$，$\alpha >0$，$\beta \ge 0$，$0\le x<1$に対して

${\displaystyle [2.1.1] {\int }_0^x{t}^{n-1+\alpha }(1-t{)}^{-1+\beta }dt=\frac{(\alpha {)}_n(1-x{)}^{\beta }}{(\alpha +\beta {)}_n}\sum _{n\le m}\frac{(\alpha +\beta {)}_m{x}^{m+\alpha }}{(\alpha {)}_m(m+\alpha )}}$

${\displaystyle [2.1.2] {\int }_0^x{t}^{n+\alpha }(1-t{)}^{-1+\beta }dt=\frac{\alpha +n}{\alpha +\beta +n}\frac{(\alpha {)}_n(1-x{)}^{\beta }}{(\alpha +\beta {)}_n}\sum _{n<m}\frac{(\alpha +\beta {)}_m{x}^{m+\alpha }}{(\alpha {)}_m(m+\alpha )}}$
　
$\mathbf{Proof}\mathbf{.}$

${\displaystyle I_n={\int }_0^x{t}^{n-1+\alpha }(1-t{)}^{-1+\beta }dt}$
　
とおく。

$\begin{array}{rl}{\displaystyle I_{n-1}-I_n}& ={\int }_0^x{t}^{n-2+\alpha }(1-t{)}^{\beta }dt\\ & ={\left[\frac{t^{n-1+\alpha }(1-t{)}^{\beta }}{n-1+\alpha }\right]}_0^x+\frac{\beta }{n-1+\alpha }{\int }_0^x{t}^{n-1+\alpha }(1-t{)}^{-1+\beta }dt\\ & =\frac{x^{n-1+\alpha }(1-x{)}^{\beta }}{n-1+\alpha }+\frac{\beta }{n-1+\alpha }{I}_n\end{array}$
　
整理して，

${\displaystyle \frac{n-1+\alpha +\beta }{n-1+\alpha }{I}_n=-\frac{x^{n-1+\alpha }(1-x{)}^{\beta }}{n-1+\alpha }+I_{n-1}}$
　
これより

$\begin{array}{rl}{\displaystyle \frac{(\alpha +\beta {)}_n}{(\alpha {)}_n}{I}_n}& =I_0-(1-x{)}^{\beta }\sum _{m=0}^{n-1}\frac{(\alpha +\beta {)}_m{x}^{m+\alpha }}{(\alpha {)}_m(m+\alpha )}\\ & =\frac{x^{\alpha }}{\alpha }{}_2{F}_1[\begin{array}{c}\alpha ,1-\beta \\ 1+\alpha \end{array};x]-(1-x{)}^{\beta }\sum _{m=0}^{n-1}\frac{(\alpha +\beta {)}_m{x}^{m+\alpha }}{(\alpha {)}_m(m+\alpha )}\\ & =(1-x{)}^{\beta }\sum _{n\le m}\frac{(\alpha +\beta {)}_m{x}^{m+\alpha }}{(\alpha {)}_m(m+\alpha )}\end{array}$
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$\mathbf{Corollary}\mathbf{2.1}\mathbf{.}$

${\displaystyle [2.2] {\int }_0^x\frac{t^{n-1+\alpha }}{1-t}{\ln }^j\frac{1}{t}dt=\sum _{k=0}^j\frac{j!}{k!}{\ln }^k\frac{1}{x}\sum _{n\le m}\frac{x^{m+\alpha }}{(m+\alpha {)}^{j-k+1}}}$

　
$\mathbf{Proof}\mathbf{.}$
　$[2.1.1]$において，$\beta =0$とし，$\alpha$について$j$回微分すればよい。
$◻$

　
$\mathbf{Definition}\mathbf{.}$

$\begin{array}{rl}{\displaystyle {\int }_a^b{f_1}^{\prime }(t)\circ f_2(t)\circ \cdots \circ f_n(t)}& ={\int }_a^b(f_1(t)-f_1(a))f_2(t)\circ \cdots \circ f_n(t)\\ & ={\int }_{a<t_1<\cdots <t_n<b}{f_1}^{\prime }(t_1)f_2(t_2)\cdots f_n(t_n)d{t}_1\cdots d{t}_n\end{array}$

$\begin{array}{r}{\displaystyle {\int }_a^{b}f(t)\circ \underset{r}{\underbrace{g(t)\circ \cdots \circ g(t)}}\circ h(t)={\int }_a^{b}f(t){(\circ g(t))}^r\circ h(t)}\end{array}$

${\displaystyle {\int }_a^{b}f(t)({\int }_a^t{g}_1(u)\circ \cdots \circ g_n(u))dt={\int }_a^b{g}_1(t)\circ \cdots \circ g_n(t)\circ f(t)}$

　
$\mathbf{Corollary}\mathbf{2.1}\mathbf{.}$

${\displaystyle [2.3.1] {\int }_0^x{t}^{n-1+\alpha }(1-t{)}^{-1+\beta }{(\circ \frac{1}{t(1-t)})}^{a-1}=\frac{(\alpha {)}_n}{(\alpha +\beta {)}_n}\sum _{n\le n_1\le \cdots \le n_a}\frac{x^{n_a+\alpha }(1-x{)}^{\beta }}{(n_1+\alpha )\cdots (n_a+\alpha )}\frac{(\alpha +\beta {)}_{n_a}}{(\alpha {)}_{n_a}}}$

${\displaystyle [2.3.2] {\int }_0^x{t}^{n-1+\alpha }(1-t{)}^{-1+\beta }{(\circ \frac{1}{1-t})}^{a-1}=\frac{(\alpha {)}_n}{(\alpha +\beta {)}_n}\sum _{n\le n_1<\cdots <n_a}\frac{x^{n_a+\alpha }(1-x{)}^{\beta }}{(n_1+\alpha +\beta )\cdots (n_{a-1}+\alpha +\beta )(n_a+\alpha )}\frac{(\alpha +\beta {)}_{n_a}}{(\alpha {)}_{n_a}}}$

　
$\mathbf{Proof}\mathbf{.}$
　$a$に関する数学的帰納法により容易にわかる。
$◻$

　
$\mathbf{Corollary}\mathbf{2.2}\mathbf{.}$　　$0\le x<1$に対して

${\displaystyle [2.4.1] {\int }_0^x\frac{t^{n-1}}{1-t}dt=\sum _{n<m}\frac{x^m}{m}}$

${\displaystyle [2.4.2] {\int }_0^x\frac{t^n}{1-t}dt=\sum _{n\le m}\frac{x^m}{m}}$

${\displaystyle [2.4.3] {\int }_0^x\frac{t^{n-\frac{1}{2}}}{1-t}dt=\sum _{n<m}\frac{x^{m-\frac{1}{2}}}{m-\frac{1}{2}}=\sum _{n\le m}\frac{x^{m+\frac{1}{2}}}{m+\frac{1}{2}}}$

${\displaystyle [2.4.4] {\int }_0^x\frac{t^{n+\frac{1}{2}}}{1-t}dt=\sum _{n<m}\frac{x^{m+\frac{1}{2}}}{m+\frac{1}{2}}}$

${\displaystyle [2.4.5] {\int }_0^x\frac{t^{n-1}}{\sqrt{1-t}}dt=\frac{1}{n{C}_n}\sum _{n\le m}{C}_m{x}^m\sqrt{1-x}}$

${\displaystyle [2.4.6] {\int }_0^x\frac{t^n}{\sqrt{1-t}}dt=\frac{1}{(n+\frac{1}{2})C_n}\sum _{n<m}{C}_m{x}^m\sqrt{1-x}}$

${\displaystyle [2.4.7] {\int }_0^x\frac{t^{n-\frac{1}{2}}}{\sqrt{1-t}}dt=C_n\sum _{n<m}\frac{x^{m-\frac{1}{2}}\sqrt{1-x}}{m{C}_m}=C_n\sum _{n\le m}\frac{x^{m+\frac{1}{2}}\sqrt{1-x}}{(m+\frac{1}{2})C_m}}$

${\displaystyle [2.4.8] {\int }_0^x\frac{t^{n+\frac{1}{2}}}{\sqrt{1-t}}dt=C_{n+1}\sum _{n<m}\frac{x^{m+\frac{1}{2}}\sqrt{1-x}}{(m+\frac{1}{2})C_m}}$

　
$\mathbf{Proof}\mathbf{.}$
　いずれも$\mathbf{Lemma}\mathbf{2.1}$の特殊な場合である。適当な$\alpha ,\beta$を代入すればよい。
$◻$

　
$\mathbf{Theorem}\mathbf{2.1}\mathbf{.}$　　$n\in {\mathbb{Z}}_{\ge 0}$，$a\in {\mathbb{Z}}_{>0}$に対して

${\displaystyle [2.5.1] {\int }_0^x\frac{t^{2n}({\sin }^{-1}x-{\sin }^{-1}t{)}^{2a-1}}{\sqrt{1-t^2}}dt=(2a-1)!C_n\sum _{n<n_1<\cdots <n_a}\frac{x^{2n_a}}{(2n_1{)}^2\cdots (2n_a{)}^2{C}_{n_a}}}$

${\displaystyle [2.5.2] {\int }_0^x\frac{t^{2n}({\sin }^{-1}x-{\sin }^{-1}t{)}^{2a-2}}{\sqrt{1-t^2}}dt=(2a-2)!C_n\sum _{n<n_1<\cdots <n_a}\frac{x^{2n_a-1}\sqrt{1-x^2}}{(2n_1{)}^2\cdots (2n_{a-1}{)}^2(2n_a)C_{n_a}}}$

${\displaystyle [2.5.3] {\int }_0^x\frac{t^{2n+1}({\sin }^{-1}x-{\sin }^{-1}t{)}^{2a-1}}{\sqrt{1-t^2}}dt=\frac{(2a-1)!}{(2n+1)C_n}\sum _{n<n_1<\cdots <n_a}\frac{C_{n_a}{x}^{2n_a+1}}{(2n_1+1{)}^2\cdots (2n_{a-1}+1{)}^2(2n_a+1)}}$

${\displaystyle [2.5.4] {\int }_0^x\frac{t^{2n+1}({\sin }^{-1}x-{\sin }^{-1}t{)}^{2a-2}}{\sqrt{1-t^2}}dt=\frac{(2a-2)!}{(2n+1)C_n}\sum _{n<n_1<\cdots <n_a}\frac{C_{n_a}{x}^{2n_a}\sqrt{1-x^2}}{(2n_1+1{)}^2\cdots (2n_{a-1}+1{)}^2}}$

　
$\mathbf{Proof}\mathbf{.}$
　いずれも計算の原理は同じなので，ここでは$[2.5.1]$のみ証明する。

$\begin{array}{rl}{\displaystyle {\int }_0^x\frac{t^{2n}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-1}}& =C_n\sum _{n<n_1}\frac{1}{2n_1{C}_{n_1}}{\int }_0^x\frac{t^{2n_1-1}\sqrt{1-t^2}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-2}\\ & =C_n\sum _{n<n_1}\frac{1}{(2n_1{)}^2{C}_{n_1}}{\int }_0^x\frac{t^{2n_1}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-3}\\ & =C_n\sum _{n<n_1<n_2}\frac{C_{n_1}}{(2n_1{)}^2{C}_{n_1}(2n_2)}{\int }_0^x\frac{t^{2n_2-1}\sqrt{1-t^2}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-4}\\ & =C_n\sum _{n<n_1<n_2}\frac{1}{(2n_1{)}^2(2n_2{)}^2}{\int }_0^x\frac{t^{2n_2}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-5}\\ & =\cdots \\ & =C_n\sum _{n<n_1<\cdots <n_a}\frac{x^{2n_a}}{(2n_1{)}^2\cdots (2n_a{)}^2{C}_{n_a}}\end{array}$

$\begin{array}{rl}{\displaystyle {\int }_0^x\frac{t^{2n}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-1}}& =\frac{1}{1!}{\int }_0^x\frac{t^{2n}({\sin }^{-1}x-{\sin }^{-1}t)}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-2}\\ & =\frac{1}{2!}{\int }_0^x\frac{t^{2n}({\sin }^{-1}x-{\sin }^{-1}t{)}^2}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-3}\\ & =\cdots \\ & =\frac{1}{(2a-1)!}{\int }_0^x\frac{t^{2n}({\sin }^{-1}x-{\sin }^{-1}t{)}^{2a-1}}{\sqrt{1-t^2}}dt\end{array}$
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$\mathbf{Theorem}\mathbf{2.2}\mathbf{.}$

${\displaystyle [2.6] \frac{1}{(2a-1)!(2b)!}{\int }_0^x\frac{(x-t{)}^{2b}}{\tan t}{\int }_0^t(t-u{)}^{2a-1}({\sin }^{-1}u{)}^{2n}dudt=C_n\sum _{n<m_1<\cdots <m_a<n_1<\cdots <n_b}\frac{{\sin }^{2n_b}x}{(2m_1{)}^2\cdots (2m_{a-1}{)}^2(2n_a{)}^3(2n_1{)}^2\cdots (2n_b{)}^2{C}_{n_b}} }$
　
$\mathbf{Proof}\mathbf{.}$

$\begin{array}{rl}{\displaystyle {\int }_0^x\frac{t^{2n}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-1}\circ \frac{1}{t}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2b}}& =C_n\sum _{n<m_1<\cdots <m_a}\frac{1}{(2m_1{)}^2\cdots (2m_a{)}^2{C}_{m_a}}{\int }_0^x\frac{t^{2m_a}}{t}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2b}\\ & =C_n\sum _{n<m_1<\cdots <m_a}\frac{1}{(2m_1{)}^2\cdots (2m_{a-1}{)}^2(2m_a{)}^3{C}_{m_a}}{\int }_0^x\frac{t^{2m_a}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2b-1} \\ & =C_n\sum _{n<m_1<\cdots <m_a<n_1<\cdots <n_b}\frac{x^{2n_b}}{(2m_1{)}^2\cdots (2m_{a-1}{)}^2(2n_a{)}^3(2n_1{)}^2\cdots (2n_b{)}^2{C}_{n_b}}\end{array}$

$\begin{array}{rl}{\displaystyle {\int }_0^x\frac{t^{2n}}{\sqrt{1-t^2}}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2a-1}\circ \frac{1}{t}{(\circ \frac{1}{\sqrt{1-t^2}})}^{2b}}& ={\int }_0^x\frac{1}{\sqrt{1-s^2}}{(\circ \frac{1}{\sqrt{1-s^2}})}^{2b-1}{\int }_0^s\frac{1}{t}{\int }_0^t\frac{u^{2n}}{\sqrt{1-u^2}}{(\circ \frac{1}{\sqrt{1-u^2}})}^{2a-1}\\ & =\frac{1}{(2b-1)!(2a-1)!}{\int }_0^x\frac{({\sin }^{-1}x-{\sin }^{-1}s{)}^{2b-1}}{\sqrt{1-s^2}}{\int }_0^s\frac{1}{t}{\int }_0^t\frac{u^{2n}({\sin }^{-1}t-{\sin }^{-1}u{)}^{2a-1}}{\sqrt{1-u^2}}dudtds \\ & =\frac{1}{(2a-1)!(2b)!}{\int }_0^x\frac{({\sin }^{-1}x-{\sin }^{-1}t{)}^{2b}}{t}{\int }_0^t\frac{u^{2n}({\sin }^{-1}t-{\sin }^{-1}u{)}^{2a-1}}{\sqrt{1-u^2}}dudt\end{array}$

あとは$x$を$\sin x$に置き換えればよい。
$◻$

　
$\mathbf{Definition}\mathbf{.}$

${\displaystyle S_n^C(\mathit{k})=\sum _{0<n_1<\cdots <n_r\le n}(2\mathit{n}{)}^{-\mathit{k}}{C}_{n_r}^{-1}}$

${\displaystyle S_n^D(\mathit{k})=\sum _{0\le n_1<\cdots <n_r\le n}(2\mathit{n}+1{)}^{-\mathit{k}}{C}_{n_r}}$

${\displaystyle S_n^C(\mathit{k};x)=\sum _{0<n_1<\cdots <n_r\le n}(2\mathit{n}{)}^{-\mathit{k}}{C}_{n_r}^{-1}{\sin }^{2n_r}x}$

${\displaystyle S_n^D(\mathit{k};x)=\sum _{0\le n_1<\cdots <n_r\le n}(2\mathit{n}+1{)}^{-\mathit{k}}{C}_{n_r}{\sin }^{2n_r+1}x}$

　
$\mathbf{Theorem}\mathbf{2.3}\mathbf{.}$

${\displaystyle [2.7] S_{\mathrm{\infty }}^C(\{2{\}}_{a-1},3,\{2{\}}_b;x)=\frac{1}{(2a)!(2b)!}{\int }_0^x\frac{t^{2a}(x-t{)}^{2b}}{\tan t}dt}$

　
$\mathbf{Proof}\mathbf{.}$
　$[2.6]$において，$n=0$とすればよい。
$◻$

　
$\mathbf{Corollary}\mathbf{2.3}\mathbf{.}$

${\displaystyle [2.8] S_{\mathrm{\infty }}^C(\{2{\}}_{a-1},3,\{2{\}}_b)=\frac{1}{(2a)!(2b)!}{\int }_0^{\frac{\pi }{2}}\frac{t^{2a}{(\frac{\pi }{2}-t)}^{2b}}{\tan t}dt}$

　
$\mathbf{Proof}\mathbf{.}$
　$[2.7]$において，$x=\frac{\pi }{2}$とすればよい。
$◻$

　
$\mathbf{Theorem}\mathbf{2.4}\mathbf{.}$

${\displaystyle [2.9] S_{\mathrm{\infty }}^D(\{2{\}}_{a-1},3,\{2{\}}_{b-1},1;x)=\frac{1}{(2a-1)!(2b)!}{\int }_0^x\frac{t^{2a-1}(x-t{)}^{2b}}{\tan t}dt}$

　
$\mathbf{Proof}\mathbf{.}$
　$[2.5.3]$について，$\mathbf{Theorem}\mathbf{2.2},\mathbf{2.3}$と同様にすればよい。
$◻$