二重対数関数の公式

• 1つ目
  $$\begin{array}{rl}& -{\int }_0^x{\displaystyle \frac{\ln (1-t)}{t}}dt\\ & =-{\int }_0^x{\displaystyle \frac{1}{t}}(-\sum _{k=1}^{\mathrm{\infty }}{\displaystyle \frac{t^k}{k}})dt\\ & =\sum _{k=1}^{\mathrm{\infty }}{\left[{\displaystyle \frac{t^k}{k^2}}\right]}_0^x\\ & =\sum _{k=1}^{\mathrm{\infty }}{\displaystyle \frac{x^k}{k^2}}\\ & ={\mathrm{Li}}_2\left(x\right)\end{array}$$

• 2つ目
  $$\begin{array}{rl}& {\mathrm{Li}}_2\left(x\right)+{\mathrm{Li}}_2(-x)\\ & =\sum _{n=1}^{\mathrm{\infty }}({\displaystyle \frac{x^n}{n^2}}+{\displaystyle \frac{{(-x)}^n}{n^2}})\\ & =\sum _{m=1}^{\mathrm{\infty }}({\displaystyle \frac{x^{2m-1}}{{(2m-1)}^2}}-{\displaystyle \frac{x^{2m-1}}{{(2m-1)}^2}})+\sum _{k=1}^{\mathrm{\infty }}({\displaystyle \frac{x^{2k}}{{\left(2k\right)}^2}}+{\displaystyle \frac{x^{2k}}{{\left(2k\right)}^2}})\\ & ={\displaystyle \frac{1}{2}}{\mathrm{Li}}_2\left(x^2\right)\end{array}$$

• 3つ目
  $$\begin{array}{rl}& {\mathrm{Li}}_2\left(x\right)+{\mathrm{Li}}_2(1-x)\\ & =-{\int }_0^x{\displaystyle \frac{\ln (1-t)}{t}}dt-{\int }_0^{1-x}{\displaystyle \frac{\ln (1-t)}{t}}dt\\ & ={\int }_0^x-{\displaystyle \frac{\ln (1-t)}{t}}dt+{\int }_1^x{\displaystyle \frac{\ln t}{1-t}}dt\\ & ={\int }_0^1-{\displaystyle \frac{\ln (1-t)}{t}}dt+{\int }_1^x(-{\displaystyle \frac{\ln (1-t)}{t}}+{\displaystyle \frac{\ln t}{1-t}})dt\\ & ={\mathrm{Li}}_2\left(1\right)+{[-(\ln t)\ln (1-x)]}_1^x\\ & =\zeta \left(2\right)-(\ln \left(x\right))(1-x)+\underset{a\to 1+0}{lim}[(\ln a)\ln (1-a)]\\ & ={\displaystyle \frac{{\pi }^2}{6}}-(\ln x)\ln (1-x)\end{array}$$

• 4つ目
  $$\begin{array}{rl}& {\mathrm{Li}}_2(1-x)+{\mathrm{Li}}_2(1-{\displaystyle \frac{1}{x}})\\ & ={\mathrm{Li}}_2(1-x)-{\int }_0^{1-1/x}{\displaystyle \frac{\ln (1-t)}{t}}dt\\ & ={\mathrm{Li}}_2(1-x)-{\int }_1^x{\displaystyle \frac{\ln \left(\frac{1}{t}\right)}{1-\frac{1}{t}}}\cdot {\displaystyle \frac{dt}{t^2}}\\ & ={\mathrm{Li}}_2(1-x)-{\int }_1^x{\displaystyle \frac{\ln t}{t}}dt-{\int }_1^x{\displaystyle \frac{\ln t}{t-1}}dt\\ & ={\mathrm{Li}}_2(1-x)-{[{\displaystyle \frac{1}{2}}{\ln }^{2}t]}_1^x+{\int }_0^{1-x}\frac{\ln (1-t)}{t}dt\\ & ={\mathrm{Li}}_2(1-x)-{\displaystyle \frac{1}{2}}{\ln }^{2}x-{\mathrm{Li}}_2(1-x)\\ & =-{\displaystyle \frac{1}{2}}{\ln }^{2}x\end{array}$$

• 5つ目
  $$\begin{array}{rl}{\mathrm{Li}}_2\left(x\right)& =-{\mathrm{Li}}_2(1-x)+{\displaystyle \frac{{\pi }^2}{6}}-(\ln x)\ln (x+1)\\ {\mathrm{Li}}_2\left(x^2\right)& =2{\mathrm{Li}}_2\left(x\right)+2{\mathrm{Li}}_2(-x)\\ {\mathrm{Li}}_2(1-x^2)+{\mathrm{Li}}_2\left(x^2\right)& ={\displaystyle \frac{{\pi }^2}{6}}-2(\ln x)(\ln (1-x)+\ln (1+x))\end{array}$$
  $$\begin{array}{rl}& ⟹{\displaystyle \frac{1}{2}}{\mathrm{Li}}_2(1-x^2)+{\mathrm{Li}}_2\left(x\right)+{\mathrm{Li}}_2(-x)={\displaystyle \frac{{\pi }^2}{12}}-(\ln x)\ln (1-x)-(\ln x)\ln (1+x)\\ & ⟹{\displaystyle \frac{1}{2}}{\mathrm{Li}}_2(1-x^2)-{\mathrm{Li}}_2(1-x)+{\displaystyle \frac{{\pi }^2}{6}}-(\ln x)\ln (1+x)+{\mathrm{Li}}_2(-x)={\displaystyle \frac{{\pi }^2}{12}}-(\ln x)\ln (1-x)-(\ln x)\ln (1+x)\\ & ⟹{\mathrm{Li}}_2(-x)-{\mathrm{Li}}_2(1-x)+{\displaystyle \frac{1}{2}}{\mathrm{Li}}_2(1-x^2)=-{\displaystyle \frac{{\pi }^2}{12}}-(\ln x)\ln (x+1)\end{array}$$

• 6つ目
  $$\begin{array}{rl}& {\mathrm{Li}}_2\left(x\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{1}{x}}\right)\\ & ={\int }_x^0{\displaystyle \frac{\ln (1-t)}{t}}dt+{\int }_{1/x}^0{\displaystyle \frac{\ln (1-t)}{t}}\\ & ={\int }_x^{-1}{\displaystyle \frac{\ln (1-t)}{t}}dt+{\int }_{1/x}^{-1}{\displaystyle \frac{\ln (1-t)}{t}}dt+2{\int }_{-1}^0{\displaystyle \frac{\ln (1-t)}{t}}dt\\ & =2{\mathrm{Li}}_2(-1){\int }_x^{-1}{\displaystyle \frac{\ln (1-t)}{t}}dt+{\int }_x^{-1}t\ln \left({\displaystyle \frac{t-1}{t}}\right){\displaystyle \frac{-dt}{t^2}}\\ & =-2\sum _{k=1}^{\mathrm{\infty }}{\displaystyle \frac{{(-1)}^{k-1}}{k^2}}+{\int }_x^{-1}({\displaystyle \frac{\ln (1-t)}{t}}-{\displaystyle \frac{\ln (1-t)}{t}}+{\displaystyle \frac{\ln (-t)}{t}})dt\\ & =-{\displaystyle \frac{{\pi }^2}{6}}+{[{\displaystyle \frac{1}{2}}{\ln }^2{(-t)}^2]}_x^1=-{\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{1}{2}}{\ln }^2(-x)\end{array}$$

$$\begin{array}{r}{\mathrm{Li}}_2\left({\displaystyle \frac{1}{2}}\right)+{\mathrm{Li}}_2(1-{\displaystyle \frac{1}{2}})={\displaystyle \frac{{\pi }^2}{6}}-(\ln \left({\displaystyle \frac{1}{2}}\right))\ln (1-{\displaystyle \frac{1}{2}})\\ \therefore {\mathrm{Li}}_2\left({\displaystyle \frac{1}{2}}\right)={\displaystyle \frac{{\pi }^2}{12}}-{\displaystyle \frac{{\ln }^{2}2}{2}}\end{array}$$
(残りは省略.)

(残りの特殊値も別記事に一部証明を書きました.) こちら