ラマヌジャンの公式集：級数の関係式

$$\begin{array}{rl}{\mathrm{Li}}_2(1-z)+{\mathrm{Li}}_2(1-\frac{1}{z})& =-\frac{1}{2}(\log z{)}^2\\ {\mathrm{Li}}_2(1-z)+{\mathrm{Li}}_2(-\frac{1}{z})& =-\frac{{\pi }^2}{2}-\frac{1}{2}(\log z{)}^2\\ {\mathrm{Li}}_2(z)+{\mathrm{Li}}_2(1-z)& =\frac{{\pi }^2}{6}-\log \left(z\right)\log (1-z)\\ {\mathrm{Li}}_2(z)+{\mathrm{Li}}_2(-z)& =\frac{1}{2}{\mathrm{Li}}_2(z^2)\\ {\chi }_2(z)+{\chi }_2\left(\frac{1-z}{1+z}\right)& =\frac{{\pi }^2}{8}-\frac{1}{2}\log \left(z\right)\log \left(\frac{1-z}{1+z}\right)\\ {\mathrm{Li}}_2\left(\frac{z}{1-w}\right)+{\mathrm{Li}}_2\left(\frac{w}{1-z}\right)& ={\mathrm{Li}}_2(z)+{\mathrm{Li}}_2(w)+{\mathrm{Li}}_2\left(\frac{zw}{(1-z)(1-w)}\right)+\log (1-z)\log (1-w)\\ {\mathrm{Li}}_2(e^{-z})& =\frac{{\pi }^2}{6}+z\log z-z+\sum _{n=1}^{\mathrm{\infty }}\frac{B_n}{(n+1)!n}{z}^{n+1}\\ {\mathrm{Li}}_2(1-e^{-z})& =\sum _{n=0}^{\mathrm{\infty }}\frac{B_n}{(n+1)!}{z}^{n+1}\end{array}$$

$$\begin{array}{rl}{\mathrm{Li}}_2\left(\frac{1}{2}\right)& =\frac{{\pi }^2}{12}-\frac{1}{2}(\log 2{)}^2\\ {\mathrm{Li}}_2\left(\frac{\sqrt{5}-1}{2}\right)& =\frac{{\pi }^2}{10}-{\log }^2\left(\frac{\sqrt{5}-1}{2}\right)\\ {\mathrm{Li}}_2\left(\frac{3-\sqrt{5}}{2}\right)& =\frac{{\pi }^2}{15}-{\log }^2\left(\frac{\sqrt{5}-1}{2}\right)\\ {\chi }_2(\sqrt{2}-1)& =\frac{{\pi }^2}{16}-\frac{1}{4}{\log }^2(\sqrt{2}-1)\\ {\chi }_2\left(\frac{\sqrt{5}-1}{2}\right)& =\frac{{\pi }^2}{12}-\frac{3}{4}{\log }^2\left(\frac{\sqrt{5}-1}{2}\right)\\ {\chi }_2(\sqrt{5}-2)& =\frac{{\pi }^2}{24}-\frac{3}{4}{\log }^2\left(\frac{\sqrt{5}-1}{2}\right)\end{array}$$

$$\begin{array}{rl}{\mathrm{Li}}_3(z)+{\mathrm{Li}}_3(1-z)+{\mathrm{Li}}_3(1-\frac{1}{z})& =\zeta (3)+\frac{{\pi }^2}{6}\log z+\frac{1}{6}(\log z{)}^3-\frac{1}{2}(\log z{)}^2\log (1-z)\\ {\mathrm{Li}}_3(-z)-{\mathrm{Li}}_3(-\frac{1}{z})& =-\frac{1}{6}(\log z{)}^3-\frac{{\pi }^2}{6}\log z\\ {\mathrm{Li}}_3(z)+{\mathrm{Li}}_3(-z)& =\frac{1}{4}{\mathrm{Li}}_3(z^2)\end{array}$$

$$\begin{array}{rl}{\mathrm{Li}}_3\left(\frac{1}{2}\right)& ={\chi }_3(1)+\frac{1}{6}(\log 2{)}^3-\frac{{\pi }^2}{12}\log 2\\ {\mathrm{Li}}_3\left(\frac{3-\sqrt{5}}{2}\right)& =\frac{4}{5}\zeta (3)-\frac{2}{3}{\log }^3\left(\frac{\sqrt{5}-1}{2}\right)+\frac{2{\pi }^2}{15}\log \left(\frac{\sqrt{5}-1}{2}\right)\end{array}$$

$$f(x)=\sum _{n=1}^{\mathrm{\infty }}\frac{h_n}{2n-1}{x}^{2n-1}$$
とおくと
$$f\left(\frac{x}{2-x}\right)=\frac{1}{8}{\log }^2(1-x)+\frac{1}{2}{\mathrm{Li}}_2(x)$$

$$\begin{array}{rl}f\left(\frac{1}{3}\right)& =\frac{{\pi }^2}{24}-\frac{1}{8}(\log 2{)}^2\\ f\left(\frac{1}{\sqrt{5}}\right)& =\frac{{\pi }^2}{20}\\ f(\sqrt{5}-2)& =\frac{{\pi }^2}{30}-\frac{3}{8}{\log }^2\left(\frac{\sqrt{5}-1}{2}\right)\end{array}$$

$$g(z)=\sum _{n=1}^{\mathrm{\infty }}\frac{H_n}{(n+1{)}^2}{z}^{n+1}$$
とおくと
$$\begin{array}{rl}g(1-z)& =\frac{1}{2}(\log z{)}^2\log (1-z)+{\mathrm{Li}}_2(z)\log z-{\mathrm{Li}}_3(z)+\zeta (3)\\ g(1-z)-g(1-\frac{1}{z})& =\frac{1}{6}(\log z{)}^3\\ g(1-z)& =\frac{1}{2}(\log z{)}^2\log (1-z)-\frac{1}{3}(\log z{)}^3-{\mathrm{Li}}_2\left(\frac{1}{z}\right)\log z-{\mathrm{Li}}_3\left(\frac{1}{z}\right)+\zeta (3)\\ g(-z)+g(-\frac{1}{z})& =-\frac{1}{6}(\log z{)}^3-{\mathrm{Li}}_2(-z)\log z+{\mathrm{Li}}_3(-z)+\zeta (3)\end{array}$$

$$h(z)=\sum _{n=1}^{\mathrm{\infty }}\frac{H_n}{(n+1{)}^3}{z}^{n+1}$$
とおくと
$$\begin{array}{rl}h(1-z)-h(1-\frac{1}{z})& =-\frac{1}{24}(\log z{)}^4+\frac{1}{6}(\log z{)}^3\log (1-z)+\zeta (3)\log z-2{\mathrm{Li}}_4(z)+{\mathrm{Li}}_3(z)\log z+\frac{{\pi }^2}{45}\\ h(-z)-h(-\frac{1}{z})& =-\frac{1}{24}(\log z{)}^4-{\mathrm{Li}}_3(-z)\log z+2{\mathrm{Li}}_4(-z)+\zeta (3)\log z+\frac{7{\pi }^4}{360}\end{array}$$

$$\begin{array}{rl}F(x)& =\sum _{n=1}^{\mathrm{\infty }}\frac{h_n}{(2n{)}^2}{x}^{2n}\\ G(x)& =\sum _{n=1}^{\mathrm{\infty }}\frac{h_n}{(2n{)}^3}{x}^{2n}\\ H(x)& =\sum _{n=1}^{\mathrm{\infty }}\frac{H_n}{2n-1}{x}^{2n-1}\end{array}$$
とおくと
$$\begin{array}{rl}F\left(\frac{1-x}{1+z}\right)& =\frac{1}{8}(\log x{)}^2\log \left(\frac{1-x}{1+x}\right)+\frac{1}{2}{\chi }_2(x)\log x+\frac{1}{2}({\chi }_3(1)-{\chi }_3(x))\\ G(x)+G\left(\frac{1-x}{1+x}\right)& =F(x)\log x+F\left(\frac{1-x}{1+x}\right)\log \left(\frac{1-x}{1+x}\right)-\frac{1}{16}(\log x{)}^2{\log }^2\left(\frac{1-x}{1+x}\right)+G(1)\\ H\left(\frac{1-x}{1+x}\right)& =(\log 2-1)\log x+\frac{1+x}{1-x}\log \frac{4x}{(1+x{)}^2}+\frac{1}{4}(\log x{)}^2+\frac{{\pi }^2}{12}+{\mathrm{Li}}_2(-x)\end{array}$$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{H_n}{n^2{2}^n}& =\zeta (3)-\frac{{\pi }^2}{12}\log 2\\ \sum _{n=1}^{\mathrm{\infty }}\frac{H_n}{(2n-1{)}^2}& =\frac{3}{2}{\chi }_3(1)\\ \sum _{n=1}^{\mathrm{\infty }}\frac{h_n}{n^2}& =2{\chi }_3(1)\\ \sum _{n=1}^{\mathrm{\infty }}\frac{h_n}{2n-1}(\sqrt{5}-2{)}^{2n-1}& =\frac{{\pi }^2}{60}+\frac{3}{4}{\log }^2\left(\frac{\sqrt{5}-1}{2}\right)+(\sqrt{5}+2)\log 4\\ & +(3\sqrt{5}+5+\log 2)\log \left(\frac{\sqrt{5}-1}{2}\right)\end{array}$$

$${\mathrm{Li}}_2(-\frac{1}{z})+{\mathrm{Li}}_2\left(\frac{1}{z+1}\right)=-\frac{1}{2}{\log }^2(1+\frac{1}{z})$$

$$\begin{array}{rl}{\mathrm{Li}}_2\left(\frac{1}{3}\right)-\frac{1}{6}{\mathrm{Li}}_2\left(\frac{1}{9}\right)& =\frac{{\pi }^2}{18}-\frac{1}{6}(\log 3{)}^2\\ {\mathrm{Li}}_2(-\frac{1}{2})+\frac{1}{6}{\mathrm{Li}}_2\left(\frac{1}{9}\right)& =-\frac{{\pi }^2}{18}+(\log 2)(\log 3)-\frac{1}{2}(\log 2{)}^2-\frac{1}{3}(\log 3{)}^2\\ {\mathrm{Li}}_2\left(\frac{1}{4}\right)+\frac{1}{3}{\mathrm{Li}}_2\left(\frac{1}{9}\right)& =\frac{{\pi }^2}{18}+2(\log 2)(\log 3)-2(\log 2{)}^2-\frac{2}{3}(\log 3{)}^2\\ {\mathrm{Li}}_2(-\frac{1}{3})-\frac{1}{3}{\mathrm{Li}}_2\left(\frac{1}{9}\right)& =-\frac{{\pi }^2}{18}+\frac{1}{6}(\log 3{)}^2\\ {\mathrm{Li}}_2(-\frac{1}{8})+{\mathrm{Li}}_2\left(\frac{1}{9}\right)& =-\frac{1}{2}{(\log \frac{9}{8})}^2\end{array}$$

$${\int }_0^1\log \left(\frac{1+\sqrt{1+4x}}{2}\right)\frac{dx}{x}=\frac{{\pi }^2}{15}$$

$$\begin{array}{rl}\frac{1}{2}(\arctan x{)}^2& =\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}\frac{h_{2n}}{2n}{x}^{2n}\\ \frac{1}{2}(\arcsin x{)}^2& =\sum _{n=1}^{\mathrm{\infty }}\frac{1}{4n^2{\beta }_n}{x}^{2n}\\ \frac{1}{3!}(\arcsin x{)}^3& =\sum _{n=1}^{\mathrm{\infty }}\left(\sum _{k=1}^n\frac{1}{(2k-1{)}^2}\right)\frac{{\beta }_n}{2n+1}{x}^{2n+1}\\ \frac{1}{4!}(\arcsin x{)}^4& =\sum _{n=2}^{\mathrm{\infty }}\left(\sum _{k=1}^{n-1}\frac{1}{(2k{)}^2}\right)\frac{1}{4n^2{\beta }_n}{x}^{2n}\end{array}$$

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}{\sin }^{2n+1}x& =x\log |2\sin x|+\frac{1}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin 2nx}{n^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}{\tan }^{2n+1}x& =x\log |\tan x|+\sum _{n=0}^{\mathrm{\infty }}\frac{\sin (4n+2)x}{(2n+1{)}^2}\end{array}$$

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}& =\frac{\pi }{2}\log 2\\ \sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{2^n(2n+1{)}^2}& =\frac{\pi }{4\sqrt{2}}\log 2+\frac{1}{\sqrt{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{2^{2n+1}(2n+1{)}^2}& =-\frac{{\pi }^2}{6\sqrt{3}}+\frac{3\sqrt{3}}{4}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+1{)}^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{2^{2n}(2n+1{)}^2}& =\frac{\pi }{3\sqrt{3}}\log 3-\frac{2{\pi }^2}{27}+\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+1{)}^2}\end{array}$$

$$\begin{array}{rl}{\int }_0^1\frac{\arctan x}{x}dx& =\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\\ {\int }_0^{1/\sqrt{3}}\frac{\arctan x}{x}dx& =-\frac{\pi }{12}\log 3-\frac{5{\pi }^2}{18\sqrt{3}}+\frac{5\sqrt{3}}{4}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+1{)}^2}\\ {\int }_0^{\sqrt{2}-1}\frac{\arctan x}{x}dx& =\frac{\pi }{8}\log (\sqrt{2}-1)-\frac{{\pi }^2}{16}+\sqrt{2}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(4n+1{)}^2}\\ {\int }_0^{2-\sqrt{3}}\frac{\arctan x}{x}dx& =\frac{\pi }{12}\log (2-\sqrt{3})+\frac{2}{3}{\int }_0^1\frac{\arctan x}{x}dx\end{array}$$

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}({\cos }^{2n+1}x-{\sin }^{2n+1}x)& =\frac{\pi }{2}\log (2\cos x)-\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}{\sin }^{2n+1}2x\\ \sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}({\cos }^{2n+1}x+{\sin }^{2n+1}x)& =\frac{\pi }{2}\log (2\cos x)+\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}{\tan }^{2n+1}x\end{array}$$

$$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{8n^3{\beta }_n}{\sin }^{2n}x=\frac{x^2}{2}\log |2\sin x|+\frac{x}{2}\sum _{n=1}^{\mathrm{\infty }}\frac{\sin 2nx}{n^2}+\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{\cos 2nx}{n^3}-\frac{1}{4}\zeta (3)$$

$$\psi (x)={\int }_0^x\frac{\arcsin t}{t}dt$$
とおくと
$$\psi \left(\frac{3}{5}\right)-\frac{1}{2}\psi \left(\frac{24}{25}\right)=\frac{\pi }{2}\log 2+2\psi \left(\frac{1}{\sqrt{5}}\right)-2\psi \left(\frac{2}{\sqrt{5}}\right)$$

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}\frac{1+2^{2n+1}}{5^{n+1}}& =\frac{\pi }{2\sqrt{5}}\log \frac{4}{\sqrt{5}}+\frac{1}{\sqrt{5}}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{2^{2n+1}(2n+1{)}^2}\\ \sum _{n=1}^{\mathrm{\infty }}\frac{1}{8n^3{\beta }_n}& =\frac{{\pi }^2}{8}\log 2-\frac{1}{2}{\chi }_3(1)\\ \sum _{n=1}^{\mathrm{\infty }}\frac{1}{2^{n+4}{n}^3{\beta }_n}& =\frac{{\pi }^2}{64}\log 2-\frac{5}{16}{\chi }_3(1)+\frac{\pi }{8}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\end{array}$$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}\frac{h_n}{(2n{)}^2}{\tan }^{2n}x& =\frac{x^2}{2}\log |\tan x|+x\sum _{n=0}^{\mathrm{\infty }}\frac{\sin (4n+2)x}{(2n+1{)}^2}+\frac{1}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{\cos (4n+2)x}{(2n+1{)}^3}-\frac{1}{2}{\chi }_3(1)\\ & =2\sum _{n=1}^{\mathrm{\infty }}\frac{1}{8n^3{\beta }_n}{\sin }^{2n}x-\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{8n^3{\beta }_n}{\sin }^{2n}2x\end{array}$$

$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{8n^3{\beta }_n}({\cos }^{2n}x+{\sin }^{2n}x)=-\frac{{\pi }^2}{8}\log (2\cos x)& +\frac{\pi }{2}\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}{\cos }^{2n+1}x\\ & +\frac{1}{4}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{8n^3{\beta }_n}{\sin }^{2n}2x-\frac{1}{2}{\chi }_3(1)\end{array}$$

　$x{e}^{i\theta }+y{e}^{i\phi }=1$において
$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n\theta }{n^2}{x}^n+\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n\phi }{n^2}{y}^n& =\frac{{\pi }^2}{6}-(\log x)(\log y)+\theta \phi \\ \sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\theta }{n^2}{x}^n+\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\phi }{n^2}{y}^n& =-\phi \log x-\theta \log y\end{array}$$

　$1/x{e}^{i\theta }+1/y{e}^{i\phi }=1$において
$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n\theta }{n^2}{x}^n+\sum _{n=1}^{\mathrm{\infty }}\frac{\cos n\phi }{n^2}{y}^n& =\frac{1}{8}\log (1-2x\cos \theta +x^2)\log (1-2y\cos \phi +y^2)\\ & -\frac{1}{2}\arctan \left(\frac{x\sin \theta }{1-x\cos \theta }\right)\arctan \left(\frac{y\sin \phi }{1-y\cos \phi }\right)\\ \sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\theta }{n^2}{x}^n+\sum _{n=1}^{\mathrm{\infty }}\frac{\sin n\phi }{n^2}{y}^n& =-\frac{1}{4}\log (1-2x\cos \theta +x^2)\arctan \left(\frac{y\sin \phi }{1-y\cos \phi }\right)\\ & -\frac{1}{4}\log (1-2y\cos \phi +y^2)\arctan \left(\frac{x\sin \theta }{1-x\cos \theta }\right)\end{array}$$

　$x{e}^{i\theta }+y{e}^{i\phi }+xy{e}^{i(\theta +\phi )}=1$において
$$\begin{array}{rl}\sum _{n=1}^{\mathrm{\infty }}\frac{\cos (2n+1)\theta }{(2n+1{)}^2}{x}^{2n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{\cos (2n+1)\phi }{(2n+1{)}^2}{y}^{2n+1}& =\frac{{\pi }^2}{8}-\frac{1}{2}(\log x)(\log y)+\frac{1}{2}\theta \phi \\ \sum _{n=1}^{\mathrm{\infty }}\frac{\sin (2n+1)\theta }{(2n+1{)}^2}{x}^{2n+1}+\sum _{n=1}^{\mathrm{\infty }}\frac{\sin (2n+1)\phi }{(2n+1{)}^2}{y}^{2n+1}& =-\frac{1}{2}\phi \log x-\frac{1}{2}\theta \log y\end{array}$$

$$\psi (x)=\sum _{n=0}^{\mathrm{\infty }}\frac{{\beta }_n}{(2n+1{)}^2}{\sin }^{2n+1}\pi x$$
とおくと
$$\begin{array}{rl}\psi (x)& =\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{h_{n+1}}{2n+1}{\tan }^{2n+1}\pi x\\ \psi (x)+\psi (\frac{1}{2}-x)& =\frac{\pi }{2}\log (2\cos \pi x)+\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}{\tan }^{2n+1}\pi x\\ \psi (\frac{1}{2}-x)+\frac{1}{2}\psi (2x)-\psi (x)& =\frac{\pi }{2}\log (2\cos \pi x)\\ \psi (\frac{1}{2}-x)+\psi (\frac{1}{2}+x)& =\pi (1-2x)\log |2\cos \pi x|+\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}\frac{\sin 2\pi nx}{n^2}\end{array}$$

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}{\left(\frac{4x}{(1+x{)}^2}\right)}^n& =(1+x)\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}{x}^n\\ \sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}{\sin }^{2n+1}2x& =2\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}{\tan }^{2n+1}x\\ \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2{\beta }_n}{\tan }^{2n+1}2x& =2\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}{\tan }^{2n+1}x\end{array}$$

$$\begin{array}{rl}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}& =2\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}{\left(\frac{3}{4}\right)}^n& =-\frac{\pi }{3\sqrt{3}}\log 3-\frac{10{\pi }^2}{27}+5\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(3n+1{)}^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}\frac{1}{4^n}& =-\frac{\pi }{3}\log (2+\sqrt{3})+\frac{8}{3}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}\frac{1}{2^n}& =-\frac{\pi }{2\sqrt{2}}\log (2+\sqrt{3})-\frac{{\pi }^2}{4\sqrt{2}}+4\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(4n+1{)}^2}\\ \sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2{\beta }_n}(1-\frac{3}{4^{n+1}})& =\frac{\pi }{4}\log (2+\sqrt{3})\\ \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2{\beta }_n}& =\frac{{\pi }^2}{8}-\frac{1}{2}{\log }^2(1+\sqrt{2})\\ \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n}{(2n+1{)}^2{\beta }_n}\frac{1}{4^n}& =\frac{{\pi }^2}{6}-3{\log }^2\left(\frac{\sqrt{5}+1}{2}\right)\end{array}$$

$$\begin{array}{rl}{\int }_0^{\frac{\pi }{2}}x{\cos }^{n}x\sin nxdx& =\frac{\pi }{2^{n+2}}\sum _{k=1}^n\frac{1}{k}\\ {\int }_0^{\frac{\pi }{2}}{\cos }^{n}x\sin nxdx& =\frac{1}{2^{n+1}}\sum _{k=1}^n\frac{2^k}{k}\end{array}$$

$$\sum _{n=0}^{\mathrm{\infty }}\frac{1}{(2n+1{)}^2}{\left(\frac{x}{1+x}\right)}^{n+1}=\sum _{n=0}^{\mathrm{\infty }}(-1{)}^n\frac{h_{n+1}}{(2n+1){\beta }_n}{x}^{n+1}$$

$$K_n=\sum _{k=1}^{\mathrm{\infty }}\frac{1}{k^n(k+1{)}^n},A_s=(1+\cos \pi s)\zeta (s)$$
とおくと
$$K_n=\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n+k-1}{k})A_{n-k}$$

$$\begin{array}{rl}{K}_2& =\frac{{\pi }^2}{3}-3\\ K_3& =10-{\pi }^2\\ K_4& =\frac{{\pi }^4}{45}+\frac{10{\pi }^2}{3}-35\\ K_5& =126-\frac{35{\pi }^2}{3}-\frac{{\pi }^4}{9}\end{array}$$