多重ゼータ値とその類似　Part６

　
$\mathbf{P}\mathbf{r}\mathbf{o}\mathbf{p}\mathbf{o}\mathbf{s}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{6.1}\mathbf{.}$

${\displaystyle [6.1] \sum _{0<m\le n}\frac{C_n}{C_m{m}^2{n}^2}={\pi }^2{\ln }^{2}2-\sum _{0<m\le n}\frac{1}{m{n}^3{C}_n}}$

$\mathbf{Proof}\mathbf{.}$
$\begin{array}{rl}{\displaystyle \sum _{0<m\le n}\frac{C_n}{C_m{m}^2{n}^2}}& =\sum _{0<n}\frac{C_n}{n^2}(-\frac{{\pi }^2}{2}+\frac{4}{C_n}{\int }_0^{\frac{\pi }{2}}t{\sin }^{2n}tdt)\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\sum _{0<n}\frac{4}{n^2}{\int }_0^{\frac{\pi }{2}}t{\sin }^{2n}tdt\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\sum _{0<n}\frac{4}{n^2}({[\frac{t^2}{2}{\sin }^{2n}t]}_0^{\frac{\pi }{2}}-n{\int }_0^{\frac{\pi }{2}}{t}^2{\sin }^{2n-1}t\cos tdt)\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\sum _{0<n}\frac{4}{n^2}\frac{{\left(\frac{\pi }{2}\right)}^2}{2}-\sum _{0<n}\frac{4}{n}{\int }_0^{\frac{\pi }{2}}{t}^2{\sin }^{2n-1}t\cos tdt\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\frac{{\pi }^4}{12}-\sum _{0<n}\frac{4}{n}{\int }_0^1{t}^{2n-1}({\sin }^{-1}t{)}^{2}dt\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\frac{{\pi }^4}{12}-\sum _{0<n}\frac{2}{n}{\int }_0^1{t}^{2n-1}\sum _{0<m}\frac{t^{2m}}{m^2{C}_m}dt\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\frac{{\pi }^4}{12}-\sum _{0<m,n}\frac{2}{n{m}^2{C}_m}{\int }_0^1{t}^{2m+2n-1}dt\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n}{n^2}+\frac{{\pi }^4}{12}-\sum _{0<m,n}\frac{1}{n{m}^2(m+n)C_m}\\ & =-\frac{{\pi }^2}{2}(\frac{{\pi }^2}{6}-2{\ln }^{2}2)+\frac{{\pi }^4}{12}-\sum _{0<m,n}\frac{1}{m^3{C}_m}(\frac{1}{n}-\frac{1}{m+n})\\ & ={\pi }^2{\ln }^{2}2-\sum _{0<m\le n}\frac{1}{m{n}^3{C}_n}\end{array}$
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$\mathbf{P}\mathbf{r}\mathbf{o}\mathbf{p}\mathbf{o}\mathbf{s}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{6.2}\mathbf{.}$

${\displaystyle [6.2] \sum _{0<m\le n}\frac{C_n{x}^n}{m^2{n}^2{C}_m}=\frac{{\pi }^2}{2}{\mathrm{L}\mathrm{i}}_2(x)-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n{x}^n}{n^2}-\sum _{0<m,n}\frac{x^n}{m^{2}n(m+n)C_m}}$

$\mathbf{Proof}\mathbf{.}$
$\begin{array}{rl}{\displaystyle \sum _{0<m\le n}\frac{C_n{x}^n}{m^2{n}^2{C}_m}}& =\sum _{0<n}\frac{C_n{x}^n}{n^2}(-\frac{{\pi }^2}{2}+\frac{4}{C_n}{\int }_0^{\frac{\pi }{2}}t{\sin }^{2n}tdt)\\ & =-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n{x}^n}{n^2}+\sum _{0<n}\frac{4x^n}{n^2}({[\frac{t^2}{2}{\sin }^{2n}t]}_0^{\frac{\pi }{2}}-n{\int }_0^{\frac{\pi }{2}}{t}^2{\sin }^{2n-1}t\cos tdt)\\ & =\frac{{\pi }^2}{2}{\mathrm{L}\mathrm{i}}_2(x)-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n{x}^n}{n^2}-\sum _{0<n}\frac{4x^n}{n}{\int }_0^1{t}^{2n-1}({\sin }^{-1}t{)}^{2}dt\\ & =\frac{{\pi }^2}{2}{\mathrm{L}\mathrm{i}}_2(x)-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n{x}^n}{n^2}-\sum _{0<n}\frac{2x^n}{n}{\int }_0^1{t}^{2n-1}\sum _{0<m}\frac{t^{2m}}{m^2{C}_m}dt\\ & =\frac{{\pi }^2}{2}{\mathrm{L}\mathrm{i}}_2(x)-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n{x}^n}{n^2}-\sum _{0<n}\frac{2x^n}{m^{2}n{C}_m}{\int }_0^1{t}^{2m+2n-1}dt\\ & =\frac{{\pi }^2}{2}{\mathrm{L}\mathrm{i}}_2(x)-\frac{{\pi }^2}{2}\sum _{0<n}\frac{C_n{x}^n}{n^2}-\sum _{0<m,n}\frac{x^n}{m^{2}n(m+n)C_m}\end{array}$
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$\mathbf{P}\mathbf{r}\mathbf{o}\mathbf{p}\mathbf{o}\mathbf{s}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{6.3}\mathbf{.}$

${\displaystyle [6.3] \pi \sum _{0\le n}{C}_n^2{x}^n=\sqrt{1-x}\sum _{0\le m<n}\frac{C_m^2}{n{C}_n}{x}^{n-m-1}}$

$\mathbf{Proof}\mathbf{.}$
$\begin{array}{rl}{\displaystyle \pi \sum _{0\le n}{C}_n^2{x}^n}& =2{\int }_0^{\frac{\pi }{2}}\frac{d\theta }{\sqrt{1-x{\sin }^2\theta }}\\ & =2{\int }_0^1\frac{dt}{\sqrt{(1-x{t}^2)(1-t^2)}}\\ & ={\int }_0^1\frac{dt}{\sqrt{(1-xt)(1-t)t}}\\ & =\frac{1}{\sqrt{x}}{\int }_0^x\frac{dt}{\sqrt{(1-t)(1-\frac{t}{x})t}}\\ & =\frac{1}{\sqrt{x}}\sum _{0\le m}\frac{C_m}{x^m}{\int }_0^1\frac{t^{m-\frac{1}{2}}}{\sqrt{1-t}}dt\\ & =\frac{1}{\sqrt{x}}\sum _{0\le m}\frac{C_m}{x^m}{C}_m\sum _{m<n}\frac{x^{n-\frac{1}{2}}\sqrt{1-x}}{n{C}_n}\\ & =\sqrt{1-x}\sum _{0\le m<n}\frac{C_m^2}{n{C}_n}{x}^{n-m-1}\end{array}$
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$\mathbf{L}\mathbf{e}\mathbf{m}\mathbf{m}\mathbf{a}\mathbf{6.1}\mathbf{.}$

${\displaystyle [6.4] {\int }_0^{\mathrm{\infty }}\frac{x}{{\cosh }^{2n}x}dx=\frac{\ln 2}{2n{C}_n}+\frac{1}{2}\sum _{k=0}^{n-1}\frac{(-1{)}^k{H}_k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k})}$

$\mathbf{Proof}\mathbf{.}$
$\begin{array}{rl}{\displaystyle {\int }_0^z\frac{x}{{\cosh }^{2n}x}dx}& ={[x\sum _{k=0}^{n-1}\frac{(-1{)}^k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k}){\tanh }^{2k+1}x]}_0^z-{\int }_0^z\sum _{k=0}^{n-1}\frac{(-1{)}^k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k}){\tanh }^{2k+1}xdx\\ & =z\sum _{k=0}^{n-1}\frac{(-1{)}^k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k}){\tanh }^{2k+1}z-\sum _{k=0}^{n-1}\frac{(-1{)}^k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k}){[\ln \cosh x-\sum _{l=1}^k\frac{{\tanh }^{2l}x}{2l}]}_0^z\\ & =\sum _{k=0}^{n-1}\frac{(-1{)}^k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k})(z{\tanh }^{2k+1}z-\ln \cosh z+\sum _{l=1}^k\frac{{\tanh }^{2l}z}{2l})\end{array}$
いま
${\displaystyle \underset{z\to \mathrm{\infty }}{lim}(z{\tanh }^{2k+1}z-\ln \cosh z)=\ln 2}$
ので
${\displaystyle {\int }_0^{\mathrm{\infty }}\frac{x}{{\cosh }^{2n}x}dx=\frac{\ln 2}{2n{C}_n}+\frac{1}{2}\sum _{k=0}^{n-1}\frac{(-1{)}^k{H}_k}{2k+1}(\genfrac{}{}{0ex}{}{n-1}{k})}$
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$\mathbf{P}\mathbf{r}\mathbf{o}\mathbf{p}\mathbf{o}\mathbf{s}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{6.4}\mathbf{.}$　　${\displaystyle -\frac{\pi }{2}\le x\le \frac{\pi }{2}}$に対して

${\displaystyle [6.5] 2x^2\ln 2+\sum _{0<k\le m<n}\frac{(-1{)}^m{\sin }^{2n}x}{k(2m+1)n}(\genfrac{}{}{0ex}{}{n-1}{m})=\sum _{0<n}\frac{(-1{)}^{n-1}}{n^3}(1-\cos 2nx)}$

$\mathbf{Proof}\mathbf{.}$
$\begin{array}{rl}{\displaystyle {\int }_0^{\mathrm{\infty }}t\ln \frac{1}{1-\frac{{\sin }^{2}x}{{\cosh }^{2}t}}dt}& =\sum _{0<n}\frac{{\sin }^{2n}x}{n}{\int }_0^{\mathrm{\infty }}\frac{t}{{\cosh }^{2n}t}dt\\ & =\sum _{0<n}\frac{{\sin }^{2n}x}{n}(\frac{\ln 2}{2n{C}_n}+\frac{1}{2}\sum _{0<k\le m<n}\frac{(-1{)}^{m-1}}{k(2m+1)}(\genfrac{}{}{0ex}{}{n-1}{m}))\\ & =x^2\ln 2+\frac{1}{2}\sum _{0<k\le m<n}\frac{(-1{)}^m{\sin }^{2n}x}{k(2m+1)n}(\genfrac{}{}{0ex}{}{n-1}{m})\end{array}$

$\begin{array}{rl}{\displaystyle {\int }_0^{\mathrm{\infty }}t\ln \frac{1}{1-\frac{{\sin }^{2}x}{{\cosh }^{2}t}}dt}& ={\int }_0^{\mathrm{\infty }}t\ln \frac{(1+e^{-2t}{)}^2}{1+2e^{-2t}\sin 2x+e^{-4t}}dt\\ & ={\int }_0^{\mathrm{\infty }}t\cdot 2\sum _{0<n}\frac{(-1{)}^{n-1}{e}^{-2nt}}{n}(1-\cos 2nx)dt\\ & =\sum _{0<n}\frac{(-1{)}^{n-1}}{2n^3}(1-\cos 2nx)\end{array}$
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