二重対数関数の公式
実数$x$に対して二重対数関数${\mathrm{Li}_2{\left( x \right)}}$を以下のように定義する.
$$\begin{align}
{\mathrm{Li}_2{\left( x \right)}}\coloneqq\sum_{k=1}^\infty\dfrac{x^k}{k^2}\end{align}$$
次の式が成立する. (但し, 収束条件はある.) $$\begin{align} &{\mathrm{Li}_2{\left( x \right)}}=-\int_0^x\dfrac{\ln{\left( 1-t \right)}}{t}dt \\ &{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( -x \right)}}=\dfrac{1}{2}{\mathrm{Li}_2{\left( x^2 \right)}} \\ &{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( 1-x \right)}}=\dfrac{\pi^2}{6}-{\left( \ln x \right)}\ln{\left( 1-x \right)} \\ &{\mathrm{Li}_2{\left( 1-x \right)}}+{\mathrm{Li}_2{\left( 1-\dfrac{1}{x} \right)}}=-\ln^2{\left( x \right)} \\ &{\mathrm{Li}_2{\left( -x \right)}}-{\mathrm{Li}_2{\left( 1-x \right)}}+\dfrac{1}{2}{\mathrm{Li}_2{\left( 1-x^2 \right)}}=-\dfrac{\pi^2}{12}-{\left( \ln x \right)}\ln{\left( x+1 \right)} \\ &{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( \dfrac{1}{x} \right)}}=-\dfrac{\pi^2}{6}-\dfrac{1}{2}\ln^2{\left( -x \right)}\end{align}$$
- 1つ目
$$\begin{align} &-\int_0^x\dfrac{\ln{\left( 1-t \right)}}{t}dt \\ &=-\int_0^x\dfrac{1}{t}{\left( -\sum_{k=1}^\infty\dfrac{t^k}{k} \right)}dt \\ &=\sum_{k=1}^\infty{\left[ \dfrac{t^k}{k^2} \right]}_0^x \\ &=\sum_{k=1}^\infty\dfrac{x^k}{k^2} \\ &={\mathrm{Li}_2{\left( x \right)}}\end{align}$$
- 2つ目
$$\begin{align} &{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( -x \right)}} \\ &=\sum_{n=1}^\infty{\left( \dfrac{x^n}{n^2}+\dfrac{{\left( -x \right)}^n}{n^2} \right)} \\ &=\sum_{m=1}^\infty{\left( \dfrac{x^{2m-1}}{{\left( 2m-1 \right)}^2}-\dfrac{x^{2m-1}}{{\left( 2m-1 \right)}^2} \right)} +\sum_{k=1}^\infty{\left( \dfrac{x^{2k}}{{\left( 2k \right)}^2}+\dfrac{x^{2k}}{{\left( 2k \right)}^2} \right)} \\ &=\dfrac{1}{2}{\mathrm{Li}_2{\left( x^2 \right)}}\end{align}$$
- 3つ目
$$\begin{align} &{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( 1-x \right)}} \\ &=-\int_0^x\dfrac{\ln{\left( 1-t \right)}}{t}dt-\int_0^{1-x}\dfrac{\ln{\left( 1-t \right)}}{t}dt \\ &=\int_0^x-\dfrac{\ln{\left( 1-t \right)}}{t}dt+\int_1^x\dfrac{\ln t}{1-t}dt \\ &=\int_0^1-\dfrac{\ln{\left( 1-t \right)}}{t}dt+\int_1^x{\left( -\dfrac{\ln{\left( 1-t \right)}}{t}+\dfrac{\ln t}{1-t} \right)}dt \\ &={\mathrm{Li}_2{\left( 1 \right)}}+{\Bigl[ -{\left( \ln t \right)}\ln{\left( 1-x \right)} \Bigr]}_1^x \\ &=\zeta\pare{2}-{\left( \ln {\left( x \right)} \right)}{\left( 1-x \right)}+\lim_{a\to1+0}{\Bigl[ {\left( \ln a \right)}\ln{\left( 1-a \right)} \Bigr]} \\ &=\dfrac{\pi^2}{6}-{\left( \ln x \right)}\ln{\left( 1-x \right)}\end{align}$$
4つ目
$$\begin{align} &{\mathrm{Li}_2{\left( 1-x \right)}}+{\mathrm{Li}_2{\left( 1-\dfrac{1}{x} \right)}} \\ &={\mathrm{Li}_2{\left( 1-x \right)}}-\int_0^{1-1/x}\dfrac{\ln{\left( 1-t \right)}}{t}dt \\ &={\mathrm{Li}_2{\left( 1-x \right)}}-\int_1^x\dfrac{\ln{\left( \frac{1}{t} \right)}}{1-\frac{1}{t}}\cdot\dfrac{dt}{t^2} \\ &={\mathrm{Li}_2{\left( 1-x \right)}}-\int_1^x\dfrac{\ln t}{t}dt-\int_1^x\dfrac{\ln t}{t-1}dt \\ &={\mathrm{Li}_2{\left( 1-x \right)}}-{\left[ \dfrac{1}{2}\ln^2 t \right]}_1^x+\int_0^{1-x}\frac{\ln{\left( 1-t \right)}}{t}dt \\ &={\mathrm{Li}_2{\left( 1-x \right)}}-\dfrac{1}{2}\ln^2 x-{\mathrm{Li}_2{\left( 1-x \right)}} \\ &=-\dfrac{1}{2}\ln^2 x\end{align}$$5つ目
$$\begin{align} {\mathrm{Li}_2{\left( x \right)}}&=-{\mathrm{Li}_2{\left( 1-x \right)}}+\dfrac{\pi^2}{6}-{\left( \ln x \right)}\ln{\left( x+1 \right)} \\ {\mathrm{Li}_2{\left( x^2 \right)}}&=2{\mathrm{Li}_2{\left( x \right)}}+2{\mathrm{Li}_2{\left( -x \right)}} \\ {\mathrm{Li}_2{\left( 1-x^2 \right)}}+{\mathrm{Li}_2{\left( x^2 \right)}}&=\dfrac{\pi^2}{6}-2{\left( \ln x \right)}{\left( \ln{\left( 1-x \right)}+\ln{\left( 1+x \right)} \right)} \end{align}$$
$$\begin{align} &\Longrightarrow\ \dfrac{1}{2}{\mathrm{Li}_2{\left( 1-x^2 \right)}}+{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( -x \right)}}=\dfrac{\pi^2}{12}-{\left( \ln x \right)}\ln{\left( 1-x \right)}-{\left( \ln x \right)}\ln{\left( 1+x \right)} \\ &\Longrightarrow\ \dfrac{1}{2}{\mathrm{Li}_2{\left( 1-x^2 \right)}}-{\mathrm{Li}_2{\left( 1-x \right)}}+\dfrac{\pi^2}{6}-{\left( \ln x \right)}\ln{\left( 1+x \right)}+{\mathrm{Li}_2{\left( -x \right)}}=\dfrac{\pi^2}{12}-{\left( \ln x \right)}\ln{\left( 1-x \right)}-{\left( \ln x \right)}\ln{\left( 1+x \right)} \\ &\Longrightarrow\ {\mathrm{Li}_2{\left( -x \right)}}-{\mathrm{Li}_2{\left( 1-x \right)}}+\dfrac{1}{2}{\mathrm{Li}_2{\left( 1-x^2 \right)}}=-\dfrac{\pi^2}{12}-{\left( \ln x \right)}\ln{\left( x+1 \right)}\end{align}$$
- 6つ目
$$\begin{align} &{\mathrm{Li}_2{\left( x \right)}}+{\mathrm{Li}_2{\left( \dfrac{1}{x} \right)}} \\ &=\int_x^0\dfrac{\ln{\left( 1-t \right)}}{t}dt+\int_{1/x}^0\dfrac{\ln{\left( 1-t \right)}}{t} \\ &=\int_x^{-1}\dfrac{\ln{\left( 1-t \right)}}{t}dt+\int_{1/x}^{-1}\dfrac{\ln{\left( 1-t \right)}}{t}dt+2\int_{-1}^0\dfrac{\ln{\left( 1-t \right)}}{t}dt \\ &=2{\mathrm{Li}_2{\left( -1 \right)}}\int_x^{-1}\dfrac{\ln{\left( 1-t \right)}}{t}dt+\int_x^{-1}t\ln{\left( \dfrac{t-1}{t} \right)}\dfrac{-dt}{t^2} \\ &=-2\sum_{k=1}^\infty\dfrac{{\left( -1 \right)}^{k-1}}{k^2}+\int_x^{-1}{\left( \dfrac{\ln{\left( 1-t \right)}}{t}-\dfrac{\ln{\left( 1-t \right)}}{t}+\dfrac{\ln{\left( -t \right)}}{t} \right)}dt \\ &=-\dfrac{\pi^2}{6}+{\left[ \dfrac{1}{2}\ln^2{\left( -t \right)}^2 \right]}_x^1=-\dfrac{\pi^2}{6}-\dfrac{1}{2}\ln^2{\left( -x \right)}\end{align}$$
$$\begin{align} {\mathrm{Li}_2{\left( \dfrac{1}{2} \right)}}&=\dfrac{\pi^2}{12}-\dfrac{\ln^2 2}{2} \\ {\mathrm{Li}_2{\left( \dfrac{1}{3} \right)}}-\dfrac{1}{6}{\mathrm{Li}_2{\left( \dfrac{1}{9} \right)}}&=\dfrac{\pi^2}{18}-\dfrac{1}{6}\ln^2 3 \\ {\mathrm{Li}_2{\left( -\dfrac{1}{2} \right)}}+\dfrac{1}{6}{\mathrm{Li}_2{\left( \dfrac{1}{9} \right)}}&=-\dfrac{\pi^2}{18}+{\left( \ln 2 \right)}\ln 3-\dfrac{1}{2}\ln^2 2-\dfrac{1}{3}\ln^2 3 \\ {\mathrm{Li}_2{\left( \dfrac{1}{4} \right)}}+\dfrac{1}{3}{\mathrm{Li}_2{\left( \dfrac{1}{9} \right)}}&=\dfrac{\pi^2}{18}-\dfrac{1}{6}\ln^2 3 \\ {\mathrm{Li}_2{\left( -\dfrac{1}{8} \right)}}+{\mathrm{Li}_2{\left( \dfrac{1}{9} \right)}}&=\dfrac{1}{2}\ln^2{\left( \dfrac{9}{8} \right)} \\ {\mathrm{Li}_2{\left( -1 \right)}}&=-\dfrac{\pi^2}{12} \\ {\mathrm{Li}_2{\left( 0 \right)}}&=0 \\ {\mathrm{Li}_2{\left( -\phi \right)}}&=-\dfrac{\pi^2}{10}-\ln^2\phi \\ {\mathrm{Li}_2{\left( -\dfrac{1}{\phi} \right)}}&=-\dfrac{\pi^2}{15}+\dfrac{1}{2}\ln^2\phi \\ {\mathrm{Li}_2{\left( \dfrac{1}{\phi^2} \right)}}&=\dfrac{\pi^2}{15}-\ln^2\phi \\ {\mathrm{Li}_2{\left( \dfrac{1}{\phi} \right)}}&=\dfrac{\pi^2}{10}-\ln^2\phi \\ \mathrm{Li}_2\pare{i}&=-\dfrac{\pi^2}{48}+iC \\ \mathrm{Li}_2\pare{1+i}&=\dfrac{\pi^2}{16}+i\pare{C+\dfrac{\pi\ln 2}{4}}\end{align}$$
$$\begin{align}
\mathrm{Li}_2\pare{\dfrac{1}{2}} +{\mathrm{Li}_2{\left( 1-\dfrac{1}{2} \right)}}=\dfrac{\pi^2}{6}-{\left( \ln{\left( \dfrac{1}{2} \right)} \right)}\ln{\left( 1-\dfrac{1}{2} \right)}
\\
\therefore \mathrm{Li}_2\left(\dfrac{1}{2}\right)=\dfrac{\pi^2}{12}-\dfrac{\ln^2 2}{2}\end{align}$$
(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
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