積分

$$\begin{array}{rcl}{\int }_0^1{\displaystyle \frac{\ln (1-x)\ln x}{x}}dx& =& {\int }_0^1{\displaystyle \frac{\ln (1-x)\ln x}{1-x}}(1-x\mapsto x)\\ & =& -{\int }_0^1\sum _{n=0}^{\mathrm{\infty }}{H}_n{x}^n\cdot {\displaystyle \frac{\partial }{\partial t}}{x}^{t}dx{|}_{t=0}\\ & =& -\sum _{n=0}^{\mathrm{\infty }}{H}_n{\displaystyle \frac{\partial }{\partial t}}{\int }_0^1{x}^{n+t}dx{|}_{t=0}\\ & =& -\sum _{n=0}^{\mathrm{\infty }}{H}_n{\displaystyle \frac{\partial }{\partial t}}{\displaystyle \frac{1}{n+t+1}}{|}_{t=0}\\ & =& \sum _{n=0}^{\mathrm{\infty }}{\displaystyle \frac{H_n}{(n+1{)}^2}}\\ & =& \zeta (1,2)\\ & =& \zeta (3)\end{array}$$
よってこの積分の答えは$\zeta (3)$となります.