Integration of Multiplied Logarithms

$\begin{aligned}\int_0^1\ln(x)\ln(1-x)\ dx&=-\int_0^1\ln(x)\sum_{n=1}^\infty\frac{x^n}{n}\ dx&&\text{using series expansion for }\ln(1-x)\\ &=-\int_\infty^0(-e^{-t})(-t)\sum_{n=1}^\infty\frac{e^{-tn}}{n}\ dt&&\text{letting }x=e^{-t}\\ &=\int_0^\infty\sum_{n=1}^\infty\frac{te^{-t(n+1)}}{n}\ dt\\ &=\left[-\sum_{n=1}^\infty\frac{te^{-t(n+1)}}{n(n+1)}\right]_0^\infty+\int_0^\infty\sum_{n=1}^\infty\frac{e^{-t(n+1)}}{n(n+1)}\ dt&&\text{using integration by parts}\\ &=\left[-\sum_{n=1}^\infty\frac{t}{n(n+1)e^{t(n+1)}}\right]_0^\infty+\left[-\sum_{n=1}^\infty\frac{e^{-t(n+1)}}{n(n+1)^2}\right]_0^\infty\\ &=-0+0-0+\sum_{n=1}^\infty\frac{1}{n(n+1)^2}&&\text{note that }e^{t(n+1)}>>t,\text{ so }\lim_{t \to \infty}\frac{t}{e^{t(n+1)}}=0\\ &=\sum_{n=1}^\infty\frac{1}{n(n+1)}-\sum_{n=1}^\infty\frac{1}{(n+1)^2}\\ &=\sum_{n=1}^\infty\left(\frac{1}{n}-\frac{1}{n+1}\right)-\sum_{n=2}^\infty\frac{1}{n^2}\\ &=1-\left(\frac{\pi^2}{6}-1\right)&&\text{using telescoping series and the result of the Basel problem}\\ &=\boxed{2-\frac{\pi^2}{6}\approx0.355}\end{aligned}$

$\begin{aligned} I & = \int_0^1 \ln x \ln(1-x) \ dx & \small \color{#3D99F6} \text{Using integration by parts} \\ & = (x\ln x - x)\ln (1-x) \bigg|_0^1 + \int_0^1 \frac {x\ln x - x}{1-x} dx \\ & = \lim_{x \to 0}(-\ln x) + \int_0^1 \frac {x\ln x}{1-x} dx - \color{#3D99F6} \int_0^1 \frac x{1-x} dx & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - \lim_{x \to 0}\ln x - \int_0^1\left(\ln x - \frac {\ln x}{1-x}\right) dx - \color{#3D99F6} \int_0^1 \frac {1-x}x dx \\ & = - \lim_{x \to 0}\ln x - (x\ln x - x) \bigg|_0^1 +{\color{#3D99F6}\int_0^1 \frac {\ln x}{1-x} dx} - \int_0^1\left(\frac 1x-1\right) dx & \small \color{#3D99F6} \text{By }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = - \lim_{x \to 0}\ln x +1 +{\color{#3D99F6}\int_0^1 \frac {\ln (1-x)}x dx} - (\ln x -x)\bigg|_0^1 & \small \color{#3D99F6} \text{Dilogarithm function Li}_2 (z) = - \int_0^z \frac {\ln (1-t)} t dt \\ & = - \lim_{x \to 0}\ln x +1 {\color{#3D99F6}-\text{Li}_2 (1)} + 1 + \lim_{x \to 0}\ln x & \small \color{#3D99F6} \text{also Li}_s (z) = \sum_{k=1}^\infty \frac {z^k}{k^s} \implies \text{Li}_2(1) = \sum_{k=1}^\infty \frac 1{k^2} = \zeta (2), \\ & = 2 {\color{#3D99F6}-\zeta (2)} = 2 {\color{#3D99F6}-\frac {\pi^2}6} \approx \boxed{0.355} & \small \color{#3D99F6} \text{where }\zeta (\cdot) \text{ is Riemann zeta function.} \end{aligned}$

We have $\begin{aligned} F(b) \; = \; \int_0^1 \ln x (1-x)^{b-1}\,dx & = \; \frac{d}{da} \int_0^1 x^{a-1}(1-x)^{b-1}\,dx \Big|_{a=1} \; = \; \frac{\partial}{\partial a}B(a,b)\Big|_{a=1} \; = \; B(a,b)\big[\psi(a) - \psi(a+b)\big]\Big|_{a=1} \\ & = \; \frac{\Gamma(b)}{\Gamma(b+1)}\big[\psi(1) - \psi(b+1)\big] \; = \; \frac{\psi(1) - \psi(b+1)}{b} \end{aligned}$ and hence $\begin{aligned} \int_0^1 \ln x \ln(1-x)\,dx & = \; F'(1) \; = \; \Big(- \frac{\psi(1) - \psi(b+1)}{b^2} - \frac{\psi'(b+1)}{b}\Big)\Big|_{b=1} \; = \; -\psi(1) + \psi(2) - \psi'(2) \\ & = \; -(-\gamma) + (1 - \gamma) - (-1 + \zeta(2)) \; = \; \boxed{2 - \zeta(2)} \end{aligned}$