ラマヌジャンの公式集：級数の関係式

\begin{align} \Li_2(1-z)+\Li_2\l(1-\frac1z\r)&=-\frac12(\log z)^2\\ \Li_2(1-z)+\Li_2\l(-\frac1z\r)&=-\frac{\pi^2}2-\frac12(\log z)^2\\ \Li_2(z)+\Li_2(1-z)&=\frac{\pi^2}6-\log(z)\log(1-z)\\ \Li_2(z)+\Li_2(-z)&=\frac12\Li_2(z^2)\\ \x_2(z)+\x_2\l(\frac{1-z}{1+z}\r)&=\frac{\pi^2}8-\frac12\log(z)\log\l(\frac{1-z}{1+z}\r)\\ \Li_2\l(\frac z{1-w}\r)+\Li_2\l(\frac w{1-z}\r) &=\Li_2(z)+\Li_2(w)+\Li_2\l(\frac{zw}{(1-z)(1-w)}\r)+\log(1-z)\log(1-w)\\ \Li_2(e^{-z})&=\frac{\pi^2}6+z\log z-z+\sum^\infty_{n=1}\frac{B_n}{(n+1)!n}z^{n+1}\\ \Li_2(1-e^{-z})&=\sum^\infty_{n=0}\frac{B_n}{(n+1)!}z^{n+1}\\ \end{align}

\begin{align} \Li_2\l(\frac12\r)&=\frac{\pi^2}{12}-\frac12(\log2)^2\\ \Li_2\l(\frac{\sqrt5-1}2\r)&=\frac{\pi^2}{10}-\log^2\l(\frac{\sqrt5-1}2\r)\\ \Li_2\l(\frac{3-\sqrt5}2\r)&=\frac{\pi^2}{15}-\log^2\l(\frac{\sqrt5-1}2\r)\\ \x_2(\sqrt2-1)&=\frac{\pi^2}{16}-\frac14\log^2(\sqrt2-1)\\ \x_2\l(\frac{\sqrt5-1}2\r)&=\frac{\pi^2}{12}-\frac34\log^2\l(\frac{\sqrt5-1}2\r)\\ \x_2(\sqrt5-2)&=\frac{\pi^2}{24}-\frac34\log^2\l(\frac{\sqrt5-1}2\r) \end{align}

\begin{align} \Li_3(z)+\Li_3(1-z)+\Li_3\l(1-\frac1z\r) &=\z(3)+\frac{\pi^2}6\log z+\frac16(\log z)^3-\frac12(\log z)^2\log(1-z)\\ \Li_3(-z)-\Li_3\l(-\frac1z\r)&=-\frac16(\log z)^3-\frac{\pi^2}6\log z\\ \Li_3(z)+\Li_3(-z)&=\frac14\Li_3(z^2) \end{align}

\begin{align} \Li_3\l(\frac12\r)&=\x_3(1)+\frac16(\log2)^3-\frac{\pi^2}{12}\log2\\ \Li_3\l(\frac{3-\sqrt5}2\r)&=\frac45\z(3)-\frac23\log^3\l(\frac{\sqrt5-1}2\r)+\frac{2\pi^2}{15}\log\l(\frac{\sqrt5-1}2\r) \end{align}

$$f(x)=\sum^\infty_{n=1}\frac{h_n}{2n-1}x^{2n-1}$$
とおくと
$$f\l(\frac x{2-x}\r)=\frac18\log^2(1-x)+\frac12\Li_2(x)$$

\begin{align} f\l(\frac13\r)&=\frac{\pi^2}{24}-\frac18(\log2)^2\\ f\l(\frac1{\sqrt5}\r)&=\frac{\pi^2}{20}\\ f(\sqrt5-2)&=\frac{\pi^2}{30}-\frac38\log^2\l(\frac{\sqrt5-1}2\r) \end{align}

$$g(z)=\sum^\infty_{n=1}\frac{H_n}{(n+1)^2}z^{n+1}$$
とおくと
\begin{align} g(1-z)&=\frac12(\log z)^2\log(1-z)+\Li_2(z)\log z-\Li_3(z)+\z(3)\\ g(1-z)-g\l(1-\frac1z\r)&=\frac16(\log z)^3\\ g(1-z)&=\frac12(\log z)^2\log(1-z)-\frac13(\log z)^3-\Li_2\l(\frac1z\r)\log z-\Li_3\l(\frac1z\r)+\z(3)\\ g(-z)+g\l(-\frac1z\r)&=-\frac16(\log z)^3-\Li_2(-z)\log z+\Li_3(-z)+\z(3) \end{align}

$$h(z)=\sum^\infty_{n=1}\frac{H_n}{(n+1)^3}z^{n+1}$$
とおくと
\begin{align} h(1-z)-h\l(1-\frac1z\r) &=-\frac1{24}(\log z)^4+\frac16(\log z)^3\log(1-z) +\z(3)\log z-2\Li_4(z)+\Li_3(z)\log z+\frac{\pi^2}{45}\\ h(-z)-h\l(-\frac1z\r)&=-\frac1{24}(\log z)^4-\Li_3(-z)\log z +2\Li_4(-z)+\z(3)\log z+\frac{7\pi^4}{360} \end{align}

\begin{align} F(x)&=\sum^\infty_{n=1}\frac{h_n}{(2n)^2}x^{2n}\\ G(x)&=\sum^\infty_{n=1}\frac{h_n}{(2n)^3}x^{2n}\\ H(x)&=\sum^\infty_{n=1}\frac{H_n}{2n-1}x^{2n-1} \end{align}
とおくと
\begin{align} F\l(\frac{1-x}{1+z}\r) &=\frac18(\log x)^2\log\l(\frac{1-x}{1+x}\r)+\frac12\x_2(x)\log x+\frac12(\x_3(1)-\x_3(x))\\ G(x)+G\l(\frac{1-x}{1+x}\r) &=F(x)\log x+F\l(\frac{1-x}{1+x}\r)\log\l(\frac{1-x}{1+x}\r)-\frac1{16}(\log x)^2\log^2\l(\frac{1-x}{1+x}\r)+G(1)\\ H\l(\frac{1-x}{1+x}\r)&=(\log2-1)\log x+\frac{1+x}{1-x}\log\frac{4x}{(1+x)^2}+\frac14(\log x)^2+\frac{\pi^2}{12}+\Li_2(-x) \end{align}

\begin{align} \sum^\infty_{n=1}\frac{H_n}{n^22^n}&=\z(3)-\frac{\pi^2}{12}\log2\\ \sum^\infty_{n=1}\frac{H_n}{(2n-1)^2}&=\frac32\x_3(1)\\ \sum^\infty_{n=1}\frac{h_n}{n^2}&=2\x_3(1)\\ \sum^\infty_{n=1}\frac{h_n}{2n-1}(\sqrt5-2)^{2n-1} &=\frac{\pi^2}{60}+\frac34\log^2\l(\frac{\sqrt5-1}2\r)+(\sqrt5+2)\log4\\ &\qquad+(3\sqrt5+5+\log2)\log\l(\frac{\sqrt5-1}2\r) \end{align}

$$\Li_2\l(-\frac1z\r)+\Li_2\l(\frac1{z+1}\r)=-\frac12\log^2\l(1+\frac1z\r)$$

\begin{align} \Li_2\l(\frac13\r)-\frac16\Li_2\l(\frac19\r) &=\frac{\pi^2}{18}-\frac16(\log3)^2\\ \Li_2\l(-\frac12\r)+\frac16\Li_2\l(\frac19\r) &=-\frac{\pi^2}{18}+(\log2)(\log3)-\frac12(\log2)^2-\frac13(\log3)^2\\ \Li_2\l(\frac14\r)+\frac13\Li_2\l(\frac19\r) &=\frac{\pi^2}{18}+2(\log2)(\log3)-2(\log2)^2-\frac23(\log3)^2\\ \Li_2\l(-\frac13\r)-\frac13\Li_2\l(\frac19\r) &=-\frac{\pi^2}{18}+\frac16(\log3)^2\\ \Li_2\l(-\frac18\r)+\Li_2\l(\frac19\r) &=-\frac12\l(\log\frac98\r)^2 \end{align}

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

\begin{align} \frac12(\arctan x)^2&=\sum^\infty_{n=1}(-1)^{n-1}\frac{h_{2n}}{2n}x^{2n}\\ \frac12(\arcsin x)^2&=\sum^\infty_{n=1}\frac1{4n^2\b_n}x^{2n}\\ \frac1{3!}(\arcsin x)^3&=\sum^\infty_{n=1}\l(\sum^n_{k=1}\frac1{(2k-1)^2}\r)\frac{\b_n}{2n+1}x^{2n+1}\\ \frac1{4!}(\arcsin x)^4&=\sum^\infty_{n=2}\l(\sum^{n-1}_{k=1}\frac1{(2k)^2}\r)\frac1{4n^2\b_n}x^{2n} \end{align}

\begin{align} \sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}\sin^{2n+1}x &=x\log|2\sin x|+\frac12\sum^\infty_{n=1}\frac{\sin2nx}{n^2}\\ \sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\tan^{2n+1}x &=x\log|\tan x|+\sum^\infty_{n=0}\frac{\sin(4n+2)x}{(2n+1)^2} \end{align}

\begin{align} \sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}&=\frac\pi2\log2\\ \sum^\infty_{n=0}\frac{\b_n}{2^n(2n+1)^2} &=\frac\pi{4\sqrt2}\log2+\frac 1{\sqrt2}\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\\ \sum^\infty_{n=0}\frac{\b_n}{2^{2n+1}(2n+1)^2} &=-\frac{\pi^2}{6\sqrt3}+\frac{3\sqrt3}4\sum^\infty_{n=0}\frac1{(3n+1)^2}\\ \sum^\infty_{n=0}\frac{\b_n}{2^{2n}(2n+1)^2} &=\frac\pi{3\sqrt3}\log3-\frac{2\pi^2}{27}+\sum^\infty_{n=0}\frac1{(3n+1)^2} \end{align}

\begin{align} \int^1_0\frac{\arctan x}xdx&=\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\\ \int^{1/\sqrt3}_0\frac{\arctan x}xdx &=-\frac\pi{12}\log3-\frac{5\pi^2}{18\sqrt3}+\frac{5\sqrt3}4\sum^\infty_{n=0}\frac1{(3n+1)^2}\\ \int^{\sqrt2-1}_0\frac{\arctan x}xdx &=\frac\pi8\log(\sqrt2-1)-\frac{\pi^2}{16}+\sqrt2\sum^\infty_{n=0}\frac{(-1)^n}{(4n+1)^2}\\ \int^{2-\sqrt3}_0\frac{\arctan x}xdx &=\frac\pi{12}\log(2-\sqrt3)+\frac23\int^1_0\frac{\arctan x}xdx \end{align}

\begin{align} \sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}(\cos^{2n+1}x-\sin^{2n+1}x) &=\frac\pi2\log(2\cos x)-\frac12\sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}\sin^{2n+1}2x\\ \sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}(\cos^{2n+1}x+\sin^{2n+1}x) &=\frac\pi2\log(2\cos x)+\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\tan^{2n+1}x\\ \end{align}

$$\sum^\infty_{n=1}\frac1{8n^3\b_n}\sin^{2n} x =\frac{x^2}2\log|2\sin x|+\frac x2\sum^\infty_{n=1}\frac{\sin2nx}{n^2} +\frac14\sum^\infty_{n=1}\frac{\cos2nx}{n^3}-\frac14\z(3)$$

$$\psi(x)=\int^x_0\frac{\arcsin t}tdt$$
とおくと
$$\psi\l(\frac35\r)-\frac12\psi\l(\frac{24}{25}\r) =\frac\pi2\log2+2\psi\l(\frac1{\sqrt5}\r)-2\psi\l(\frac2{\sqrt5}\r)$$

\begin{align} \sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}\frac{1+2^{2n+1}}{5^{n+1}} &=\frac\pi{2\sqrt5}\log\frac4{\sqrt5}+\frac1{\sqrt5}\sum^\infty_{n=0}\frac{(-1)^n}{2^{2n+1}(2n+1)^2}\\ \sum^\infty_{n=1}\frac1{8n^3\b_n}&=\frac{\pi^2}8\log2-\frac12\x_3(1)\\ \sum^\infty_{n=1}\frac1{2^{n+4}n^3\b_n}&=\frac{\pi^2}{64}\log2-\frac5{16}\x_3(1)+\frac\pi8\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2} \end{align}

\begin{align} \sum^\infty_{n=1}(-1)^{n-1}\frac{h_n}{(2n)^2}\tan^{2n}x &=\frac{x^2}2\log|\tan x|+x\sum^\infty_{n=0}\frac{\sin(4n+2)x}{(2n+1)^2} +\frac12\sum^\infty_{n=0}\frac{\cos(4n+2)x}{(2n+1)^3}-\frac12\x_3(1)\\ &=2\sum^\infty_{n=1}\frac1{8n^3\b_n}\sin^{2n}x -\frac14\sum^\infty_{n=1}\frac1{8n^3\b_n}\sin^{2n}2x\\ \end{align}

\begin{align} \sum^\infty_{n=1}\frac1{8n^3\b_n}(\cos^{2n}x+\sin^{2n}x) =-\frac{\pi^2}8\log(2\cos x)&+\frac\pi2\sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}\cos^{2n+1}x\\ &+\frac14\sum^\infty_{n=1}\frac1{8n^3\b_n}\sin^{2n}2x-\frac12\x_3(1) \end{align}

　$xe^{i\t}+ye^{i\vp}=1$において
\begin{align} \sum^\infty_{n=1}\frac{\cos n\t}{n^2}x^n+\sum^\infty_{n=1}\frac{\cos n\vp}{n^2}y^n &=\frac{\pi^2}6-(\log x)(\log y)+\t\vp\\ \sum^\infty_{n=1}\frac{\sin n\t}{n^2}x^n+\sum^\infty_{n=1}\frac{\sin n\vp}{n^2}y^n &=-\vp\log x-\t\log y \end{align}

　$1/xe^{i\t}+1/ye^{i\vp}=1$において
\begin{align} \sum^\infty_{n=1}\frac{\cos n\t}{n^2}x^n+\sum^\infty_{n=1}\frac{\cos n\vp}{n^2}y^n &=\frac18\log(1-2x\cos\t+x^2)\log(1-2y\cos\vp+y^2)\\ &\qquad-\frac12\arctan\l(\frac{x\sin\t}{1-x\cos\t}\r)\arctan\l(\frac{y\sin\vp}{1-y\cos\vp}\r)\\ \sum^\infty_{n=1}\frac{\sin n\t}{n^2}x^n+\sum^\infty_{n=1}\frac{\sin n\vp}{n^2}y^n &=-\frac14\log(1-2x\cos\t+x^2)\arctan\l(\frac{y\sin\vp}{1-y\cos\vp}\r)\\ &\qquad-\frac14\log(1-2y\cos\vp+y^2)\arctan\l(\frac{x\sin\t}{1-x\cos\t}\r) \end{align}

　$xe^{i\t}+ye^{i\vp}+xye^{i(\t+\vp)}=1$において
\begin{align} \sum^\infty_{n=1}\frac{\cos(2n+1)\t}{(2n+1)^2}x^{2n+1} +\sum^\infty_{n=1}\frac{\cos(2n+1)\vp}{(2n+1)^2}y^{2n+1} &=\frac{\pi^2}8-\frac12(\log x)(\log y)+\frac12\t\vp\\ \sum^\infty_{n=1}\frac{\sin(2n+1)\t}{(2n+1)^2}x^{2n+1} +\sum^\infty_{n=1}\frac{\sin(2n+1)\vp}{(2n+1)^2}y^{2n+1} &=-\frac12\vp\log x-\frac12\t\log y \end{align}

$$\psi(x)=\sum^\infty_{n=0}\frac{\b_n}{(2n+1)^2}\sin^{2n+1}\pi x$$
とおくと
\begin{align} \psi(x)&=\sum^\infty_{n=0}(-1)^n\frac{h_{n+1}}{2n+1}\tan^{2n+1}\pi x\\ \psi(x)+\psi\l(\frac12-x\r)&=\frac\pi2\log(2\cos\pi x)+\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\tan^{2n+1}\pi x\\ \p\l(\frac12-x\r)+\frac12\p(2x)-\p(x)&=\frac\pi2\log(2\cos\pi x)\\ \p\l(\frac12-x\r)+\p\l(\frac12+x\r) &=\pi(1-2x)\log|2\cos\pi x|+\sum^\infty_{n=1}(-1)^{n+1}\frac{\sin2\pi nx}{n^2} \end{align}

\begin{align} \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}\l(\frac{4x}{(1+x)^2}\r)^n &=(1+x)\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}x^n\\ \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}\sin^{2n+1}2x &=2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\tan^{2n+1}x\\ \sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2\b_n}\tan^{2n+1}2x &=2\sum^\infty_{n=0}\frac1{(2n+1)^2}\tan^{2n+1}x\\ \end{align}

\begin{align} \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}&=2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\\ \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}\l(\frac34\r)^n &=-\frac\pi{3\sqrt3}\log3-\frac{10\pi^2}{27}+5\sum^\infty_{n=0}\frac1{(3n+1)^2}\\ \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}\frac1{4^n} &=-\frac\pi3\log(2+\sqrt3)+\frac83\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}\\ \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}\frac1{2^n} &=-\frac\pi{2\sqrt2}\log(2+\sqrt3)-\frac{\pi^2}{4\sqrt2}+4\sum^\infty_{n=0}\frac{(-1)^n}{(4n+1)^2}\\ \sum^\infty_{n=0}\frac1{(2n+1)^2\b_n}\l(1-\frac3{4^{n+1}}\r) &=\frac\pi4\log(2+\sqrt3)\\ \sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2\b_n} &=\frac{\pi^2}8-\frac12\log^2(1+\sqrt2)\\ \sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2\b_n}\frac1{4^n} &=\frac{\pi^2}6-3\log^2\l(\frac{\sqrt5+1}2\r) \end{align}

\begin{align} \int^{\frac\pi2}_0x\cos^nx\sin nx\ dx&=\frac\pi{2^{n+2}}\sum^n_{k=1}\frac1k\\ \int^{\frac\pi2}_0\cos^nx\sin nx\ dx&=\frac1{2^{n+1}}\sum^n_{k=1}\frac{2^k}k\\ \end{align}

$$\sum^\infty_{n=0}\frac1{(2n+1)^2}\l(\frac x{1+x}\r)^{n+1} =\sum^\infty_{n=0}(-1)^n\frac{h_{n+1}}{(2n+1)\b_n}x^{n+1}$$

$$K_n=\sum^\infty_{k=1}\frac1{k^n(k+1)^n},\quad A_s=(1+\cos\pi s)\z(s)$$
とおくと
$$K_n=\sum^n_{k=0}(-1)^k\binom{n+k-1}kA_{n-k}$$

\begin{align} 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{align}