気に入っている積分1

$$\begin{array}{rl}{\mathrm{Li}}_2(z)+{\mathrm{Li}}_2(1-z)& =-{\int }_0^z\frac{\ln (1-t)}{t}dt-{\int }_0^{1-z}\frac{\ln (1-t)}{t}dt\\ & =-{\int }_0^1\frac{\ln (1-t)}{t}dt-{\int }_1^z\frac{\ln (1-t)}{t}dt-{\int }_0^{1-z}\frac{\ln (1-t)}{t}dt\\ & =\zeta (2)-{\int }_1^z\frac{\ln (1-t)}{t}dt+{\int }_1^z\frac{\ln \left(u\right)}{1-u}du(t=1-u)\\ & =\zeta (2)-{\int }_1^z\frac{\ln (1-t)}{t}dt-[\ln \left(u\right)\ln (1-u){]}_1^z+{\int }_1^z\frac{\ln (1-u)}{u}du\\ & =\zeta (2)-\ln \left(z\right)\ln (1-z)◼\end{array}$$

命題1で$z=\frac{1}{2}$を代入すればよい$◼$

$$\begin{array}{rl}{\int }_0^1\frac{{\mathrm{L}\mathrm{i}}_2(\frac{1+x}{2})-\frac{{\pi }^2}{12}+\frac{1}{2}(\ln 2{)}^2}{x}dx& ={\int }_0^1\frac{{\mathrm{L}\mathrm{i}}_2(\frac{1+x}{2})-{\mathrm{Li}}_2(\frac{1}{2})}{x}dx(\because \text{補題2})\\ & =-{\int }_0^1\frac{dx}{x}{\int }_{\frac{1}{2}}^{\frac{1+x}{2}}\frac{\ln (1-t)}{t}dt\\ & =-{\int }_0^1\frac{dx}{x}{\int }_0^x\frac{\ln \left(\frac{1-u}{2}\right)}{1+u}du(t=\frac{1+u}{2})\\ & =-{\int }_0^1\frac{dx}{x}{\int }_0^x\frac{\ln \left(1-u\right)}{1+u}du+\ln 2{\int }_0^1\frac{dx}{x}{\int }_0^x\frac{du}{1+u}\\ & ={\int }_0^1\frac{dx}{x}{\int }_0^x\frac{du}{1+u}{\int }_0^u\frac{ds}{1-s}+\ln 2{\int }_0^1\frac{dx}{x}{\int }_0^x\frac{du}{1+u}\\ & ={\int }_{0<s<u<x<1}\frac{dx}{x}\frac{du}{1+u}\frac{ds}{1-s}+\ln 2{\int }_{0<u<x<1}\frac{dx}{x}\frac{du}{1+u}\\ & =-\zeta (\overline{1},\overline{2})-\zeta (\overline{2})\ln 2\\ & =\frac{13}{8}\zeta (3)-\frac{{\pi }^2}{6}\ln 2(\because \text{補題3})◼\end{array}$$