Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution. If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$. But what contour can we use suitable for the $\log^2x $ term?
Thanks
Best Answer
Consider the integral \begin{align} I = \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx \end{align} Now consider the factorization of $x^{2} - x + 1$ which is $(x - a)(x-b)$ where $a$ and $b$ are $e^{\pi i/3}$ and $e^{-\pi i/3}$, respectively. With this in mind it is seen that \begin{align} \frac{1}{x^{2} - x + 1} = \frac{1}{a-b} \left( \frac{1}{x - a} - \frac{1}{x-b} \right). \end{align} This can also be expanded into series form and is \begin{align} \frac{1}{x^{2} - x + 1} = \frac{1}{a-b} \ \sum_{n=0}^{\infty} \left( - \frac{1}{a^{n+1}} + \frac{1}{b^{n+1}} \right) x^{n}. \end{align} Now consider the integral \begin{align} I_{n} &= \int_{0}^{1} x^{n} \ln^{2}(x) \ dx = \partial_{n}^{2} \int_{0}^{1} x^{n} \ dx \\ &= \partial_{n}^{2} \left( \frac{1}{n+1} \right) \\ &= \frac{2}{(n+1)^{3}}. \end{align}
Since the components are built the desired integral is seen as the following. \begin{align} I &= \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx \\ &= \frac{1}{a-b} \ \sum_{n=0}^{\infty} \left( - \frac{1}{a^{n+1}} + \frac{1}{b^{n+1}} \right) \ \int_{0}^{1} x^{n} \ln^{2}(x) \ dx \\ &= \frac{1}{a-b} \ \sum_{n=0}^{\infty} \left( - \frac{1}{a^{n+1}} + \frac{1}{b^{n+1}} \right) \frac{2}{(n+1)^{3}} \\ &= \frac{2}{a-b} \ \sum_{n=1}^{\infty} \left( - \frac{1}{a^{n}} + \frac{1}{b^{n}} \right) \frac{1}{n^{3}} \\ &= \frac{2}{a-b} \ \sum_{n=1}^{\infty} \left( \frac{a^{n}-b^{n}}{(ab)^{n}} \right) \frac{1}{n^{3}} \\ &= \frac{2}{a-b} \ \sum_{n=1}^{\infty} \left( \frac{a^{n}}{n^{3}} - \frac{b^{n}}{n^{3}} \right) \\ &= \frac{2}{a-b} \left[ Li_{3} (a) - Li_{3}(b) \right], \end{align} where $Li_{3}(x)$ is the trilogarithm. Utilizing the results \begin{align} Li_{3}(a) &= Li_{3}(e^{\pi i/3}) = \frac{1}{3} \zeta(3) + \frac{5 \pi^{3} i }{162} \\ Li_{3}(b) &= Li_{3}(e^{-\pi i/3}) = \frac{1}{3} \zeta(3) - \frac{5 \pi^{3} i }{162} \\ a-b &= e^{\pi i /3} - e^{- \pi i/3} = \sqrt{3} i \end{align} then \begin{align} I &= \frac{2}{\sqrt{3} i} \cdot \frac{5 \pi^{3} i}{81} = \frac{10 \pi^{3}}{81 \sqrt{3}}. \end{align} Hence \begin{align} \int_{0}^{1} \frac{\ln^{2}(x)}{1 - x + x^{2}} \ dx = \frac{10 \pi^{3}}{81 \sqrt{3}} . \end{align}