二重対数関数の特殊値

${\varphi }^2-\varphi -1=0$を用いる.
$$\begin{array}{r}{\mathrm{Li}}_2(-x)-{\mathrm{Li}}_2(1-x)+{\displaystyle \frac{1}{2}}{\mathrm{Li}}_2(1-x^2)=-{\displaystyle \frac{{\pi }^2}{12}}-(\ln x)\ln (1+x)\end{array}$$
に$x=\varphi$を代入して,
$$\begin{array}{rl}& {\mathrm{Li}}_2(-\varphi )-{\mathrm{Li}}_2(1-\varphi )+{\displaystyle \frac{1}{2}}{\mathrm{Li}}_2(-\varphi )=-{\displaystyle \frac{{\pi }^2}{12}}-(\ln \varphi )\ln (1+\varphi )\\ ⟺& {\displaystyle \frac{3}{2}}{\mathrm{Li}}_2(-\varphi )-{\mathrm{Li}}_2(1-\varphi )=-{\displaystyle \frac{{\pi }^2}{12}}-2{\ln }^2\varphi ,\end{array}$$
また,
$$\begin{array}{r}{\mathrm{Li}}_2\left(x\right)+{\mathrm{Li}}_2\left({\displaystyle \frac{1}{x}}\right)=-{\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{1}{2}}{\ln }^2(-x)\end{array}$$
に, $x=-\varphi$を代入すると,
$$\begin{array}{rl}& {\mathrm{Li}}_2(-\varphi )+{\mathrm{Li}}_2(-{\displaystyle \frac{1}{\varphi }})=-{\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{1}{2}}{\ln }^2\varphi \\ ⟺& {\mathrm{Li}}_2(-\varphi )+{\mathrm{Li}}_2(1-\varphi )=-{\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{1}{2}}{\ln }^2\varphi \end{array}$$
これらを, ${\mathrm{Li}}_2(-\varphi ),{\mathrm{Li}}_2(1-\varphi )$について解き直すと,
$$\begin{array}{r}{\mathrm{Li}}_2(-\varphi )=-{\displaystyle \frac{{\pi }^2}{10}}-{\ln }^2\varphi ,{\mathrm{Li}}_2(1-\varphi )={\mathrm{Li}}_2(-{\displaystyle \frac{1}{\varphi }})=-{\displaystyle \frac{{\pi }^2}{15}}+{\displaystyle \frac{1}{2}}{\ln }^2\varphi \end{array}$$
が分かる.

定義通り計算する.
$$\begin{array}{rl}& {\mathrm{Li}}_2\left(i\right)\\ & =\sum _{k=1}^{\mathrm{\infty }}{\displaystyle \frac{i^k}{k^2}}\\ & =\sum _{n=1}^{\mathrm{\infty }}({\displaystyle \frac{i^{4n}}{{\left(4n\right)}^2}}+{\displaystyle \frac{i^{4n-3}}{{(4n-3)}^2}}+{\displaystyle \frac{i^{4n-2}}{{(4n-2)}^2}}+{\displaystyle \frac{i^{4n-1}}{{(4n-1)}^2}})\\ & ={\displaystyle \frac{1}{16}}\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}-{\displaystyle \frac{1}{4}}\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{{(2n-1)}^2}}+i\sum _{n=1}^{\mathrm{\infty }}({\displaystyle \frac{1}{{(4n-3)}^2}}-{\displaystyle \frac{1}{{(4n-1)}^2}})\\ & ={\displaystyle \frac{1}{16}}\cdot {\displaystyle \frac{{\pi }^2}{6}}-{\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{{\pi }^2}{8}}+i\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{{(-1)}^{n-1}}{{(2n-1)}^2}}\\ & =-{\displaystyle \frac{{\pi }^2}{48}}+iC\end{array}$$