A nicer closed form? $\int_0^1 \frac{\log (x) \log \left(x^2-x+1\right)}{x^2-x+2} \, dx$

Solution 1:

After lots of simplifications I got this simpler form: $$\int_0^1\frac{\ln(x)\ln\left(x^2-x+1\right)}{x^2-x+2}\, dx\\=\frac1{648\sqrt7}\Big[71\!\;\pi^3-522\!\;\alpha\!\;\pi^2-54\!\;\big(4\!\;\alpha^3-3\!\;\alpha\ln^2\xi-24\!\;\beta\big)\\+27\!\;\pi\!\;\big(28\!\;\alpha^2-2\!\;(16\!\;\gamma+\ln^22)+(4\ln2-9\ln\xi)\cdot\ln\xi\big)\Big]$$ where

$$\alpha=\arctan\big(\sqrt7\big)$$

$$\beta=\Im\left[2\operatorname{Li}_3\left(\frac{1+i\sqrt7}4\right)-\operatorname{Li}_3\left(\frac{1-2\!\;i\sqrt3+i\sqrt7}2\right)\\+2 \operatorname{Li}_3\left(\frac{i+\sqrt3}\eta\right)-2 \operatorname{Li}_3\left(\frac{i-\sqrt3}\eta\right)-\operatorname{Li}_3\left(\frac{\xi-3\!\;i\sqrt3+i\sqrt7}8\right)\right]$$

$$\gamma=\Re\left[\operatorname{Li}_2\left(\frac{i+\sqrt3}\eta\right)\right]$$

$$\xi=5-\sqrt{21}$$

$$\eta=\sqrt3+\sqrt7$$

Solution 2:

Let \begin{align}\tag{1} a &= \frac{1+i\sqrt{3}}{2} \hspace{10mm} \alpha = \frac{1 + i \sqrt{7}}{2} \\ b &= \frac{1-i\sqrt{3}}{2} \hspace{10mm} \beta = \frac{1 - i \sqrt{7}}{2} \end{align} to take the integral \begin{align}\tag{2} I = \int_0^1 \frac{\log (x) \log \left(x^2-x+1\right)}{x^2-x+2} \, dx \end{align} into the form \begin{align}\tag{3} I = \int_0^1 \frac{\log (x) \, [ \ln(x-a) + \ln(x-b)]}{(x-\alpha)(x-\beta)} \, dx \end{align} and break up the integral into 4 integrals and see what goes from there.

One may also consider the form \begin{align}\tag{4} J(x,y; a) = \int_{0}^{1} \frac{t^{x} \, (t-a)^{y}}{(t-\alpha)(t-\beta)} \, dt \end{align} and take the first derivative with respect to $x$ and $y$. Indeed \begin{align}\tag{5} I = \partial_{x,y} \left[J(x,y; a) + J(x,y;b) \right]_{x,y=0}. \end{align} A fundamental form may be \begin{align}\tag{6} J_{1}(x,y;a;\alpha) = \int_{0}^{1} \frac{t^{x} \, (t-a)^{y}}{t- \alpha} \, dt \end{align} for which \begin{align} J(x,y;a) = \frac{1}{\alpha - \beta} \left( J_{1}(x,y;a;\alpha) - J_{1}(x,y;a;\beta) \right) \end{align}