積分

$$ \begin{eqnarray*} \int_0^1\dfrac{\ln(1-x)\ln{x}}{x}dx&=&\int_0^1\dfrac{\ln(1-x)\ln{x}}{1-x}\ \ \ \ \ \ \ (1-x\mapsto x)\\ &=&-\int_0^1\sum_{n=0}^{\infty}H_nx^n\cdot\dfrac{\partial}{\partial t}x^tdx{\big|}_{t=0}\\ &=&-\sum_{n=0}^{\infty}H_n\dfrac{\partial}{\partial t}\int_0^1x^{n+t}dx\big|_{t=0}\\ &=&-\sum_{n=0}^{\infty}H_n\dfrac{\partial}{\partial t}\dfrac{1}{n+t+1}\big|_{t=0}\\ &=&\sum_{n=0}^{\infty}\dfrac{H_n}{(n+1)^2}\\ &=&\zeta(1,2)\\ &=&\zeta(3) \end{eqnarray*} $$
よってこの積分の答えは$\zeta(3)$となります.