How to calculate $\int _0^1 \int _0^1\left(\frac{1}{1-xy} \ln (1-x)\ln (1-y)\right) \,dxdy$

Question

Let us calculate the sum $$ \displaystyle{\sum_{n=1}^{+\infty}\left(\frac{H_{n}}{n}\right)^2}, $$ where $\displaystyle{H_{n}=1+\frac{1}{2}+\cdots+\frac{1}{n}}$ the $n$-th harmonic number.

My try

The main idea because the actions are many. However, we go directly to the sum without the individual decays. $$ \begin{split} {\left(\frac{H_{n}}{n}\right)^2} =\left(\frac{H_{n}}{n}\right) \left(\frac{H_{n}}{n}\right) & = \left(-\int _0^1 x^{n-1}\ln (1-x) dx\right) \left(-\int _0^1 y^{n-1} \ln (1-y)dy\right)\\ & =\int _0^1 \int _0^1 x^{n-1}y^{n-1} \ln (1-x)\ln (1-y) dxdy \end{split}$$ So the sum sought, with transfer within the integral, is $$ \begin{split} \int _0^1 \int _0^1 &\left( \sum_{n=1}^{\infty}x^{n-1}y^{n-1} \ln (1-x)\ln (1-y)\right) \,dxdy \\ &= \int _0^1 \int _0^1 \left(\frac{1}{1-xy} \ln (1-x)\ln (1-y)\right) \,dxdy \end{split} $$ My main question is how to evaluate the final integral, not the main question (sum).

Answer 1

$$ \begin{align} &\hspace{5mm}\int _0^1 \int _0^1 \frac{\ln (1-x)\ln (1-y) }{1-xy} \,dxdy \\&=\int_0^1 \ln(1-x)\frac{Li_2(\frac x{1-x})}{x}dx =\int_0^1 {Li_2(\frac x{1-x})}d[Li_2(1)-Li_2(x)]\\ & \overset{\textit{ibp}}=\int_0^1 [Li_2(x)-Li_2(1)]\frac{\ln(1-x)}xdx\\ &\qquad\qquad +\int_0^1 [Li_2(x)-Li_2(1)]\frac{\ln(1-x)}{1-x}\overset{ibp}{dx}\\ & = -\frac12 [Li_2(x)-Li_2(1)]^2\bigg|_0^1 + \frac12\int_0^1\frac{\ln^3(1-x)}x dx\\ & = \ \frac12Li_2^2(1)+3Li_4(1)= \frac12\left(\frac{\pi^2}{6}\right)^2+3\left( \frac{\pi^4}{90}\right) =\frac{17\pi^4}{360} \end{align} $$

Answer 2

\begin{align}J&=\int _0^1 \int _0^1 \frac{\ln (1-x)\ln (1-y) }{1-xy} \,dxdy\\ &\overset{u=1-xy,v=\frac{1-x}{1-xy}}=\int_0^1\int_0^1\frac{\ln\left(\frac{u(1-v)}{1-uv}\right)\ln(uv)}{1-uv}dudv\\ &=\underbrace{\int_0^1\int_0^1\frac{\ln u\ln(uv)}{1-uv}dudv}_{z(v)=uv}+\underbrace{\int_0^1\int_0^1\frac{\ln\left(1-v\right)\ln(uv)}{1-uv}dudv}_{z(v)=uv}-\\&\underbrace{\int_0^1\int_0^1\frac{\ln\left(1-uv\right)\ln(uv)}{1-uv}dudv}_{z(u)=uv}\\ &=\int_0^1 \frac{\ln u}{u}\left(\int_0^u\frac{\ln z}{1-z}dz\right)du+\int_0^1\frac{\ln(1-v)}{v}\left(\int_0^v \frac{\ln z}{1-z}\right)dv-\\&\int_0^1 \frac{1}{v}\left(\int_0^v\frac{\ln(1-z)\ln z}{1-z}\right)dv\\ &\overset{\text{IBP}}=-\frac{1}{2}\int_0^1 \frac{\ln^3 u}{1-u}du+\int_0^1 \frac{\ln(1-v)}{v}\left(-\ln(1-v)\ln v+\int_0^v\frac{\ln(1-z)}{z}dz\right)+\\&\int_0^1\frac{\ln(1-v)\ln^2 v}{1-v}dv\\ &=3\zeta(4)-\underbrace{\int_0^1\frac{\ln^2(1-v)\ln v}{v}dv}_{z=1-v}+\frac{1}{2}\zeta(2)^2+\int_0^1\frac{\ln(1-v)\ln^2 v}{1-v}dv\\ &\boxed{=3\zeta(4)+\frac{1}{2}\zeta(2)^2=\frac{17\pi^4}{360}} \end{align}

NB: i assume that, \begin{align}\int_0^1\frac{\ln(1-x)}{x}dx&=-\zeta(2)=-\frac{\pi^2}{6}\\ \int_0^1 \frac{\ln^3 x}{1-x}dx&=-6\zeta(4)\\ \zeta(4)&=\frac{\pi^4}{90} \end{align}