Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$

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

Solution 1:

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}

Solution 2:

Substituting $u = 1/x$ and averaging with the original integral gives that $$ \int_0^{1} \frac{\ln^2 x}{x^2+2x\cos\varphi+1} dx = \frac{1}{2}\int_0^{\infty} \frac{\ln^2 x}{x^2+2x\cos\varphi+1} dx $$

For real $\varphi$ and suitable complex $a$, we can prove using contours that $$ I(a) := \int_0^{\infty} \frac{x^{a}}{x^2+2x\cos\varphi+1} dx = \frac{\pi}{\sin(\pi a)} \frac{\sin(a\varphi)}{\sin(\varphi)}$$

Differentiating twice gives (and this is the most tedious part of the calculation) $$ I''(a) := \int_0^{\infty} \frac{x^{a}\ln^2x}{x^2+2x\cos\varphi+1} dx = \frac{2\pi}{\sin \varphi} \left[ -\phi ^2 \csc (\pi a) \sin (a \phi )+\pi ^2 \csc ^3(\pi a) \sin (a \phi )-2 \pi \phi \cot (\pi a) \csc (\pi a) \cos (a \phi )+\pi ^2 \cot ^2(\pi a) \csc (\pi a) \sin (a \phi ) \right] $$ Letting $a$ tend to $0$ gives $I''(0) = \dfrac{2\varphi \left(\pi ^2-\varphi ^2\right)}{3 \sin \varphi}$. For the integral under discussion, we have $\cos \varphi = -1/2$ so we may choose $\varphi = 2\pi/3$. This gives $I''(0) =\dfrac{20 \pi ^3}{81\sqrt 3}$. Remembering the factor $1/2$ gives the desired answer.

Solution 3:

Real Part

Substituting $x\mapsto1/x$ says $$ \int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x =\int_1^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{1} $$ Therefore, $$ \int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x =\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\tag{2} $$

Contour Integration Part

Putting the branch cut for $\log(z)$ along the positive real axis, and using the contour $$ \gamma=[0,R]e^{i\epsilon}\cup Re^{i[\epsilon,2\pi-\epsilon]}\cup[R,0]e^{-i\epsilon}\tag{3} $$ as $R\to\infty$ and $\epsilon\to0$, $\log(z)=\log(x)$ on the outbound segment and $\log(z)=\log(x)+2\pi i$ on the inbound segment and he integral around huge circular arc vanishes. Therefore, $$ \begin{align} \int_\gamma\frac{\log(z)^3}{z^2-z+1}\mathrm{d}z &=\int_0^\infty\frac{-6\pi i\log(x)^2+12\pi^2\log(x)+8\pi^3i}{x^2-x+1}\mathrm{d}x\\ &=\frac{124\pi^3}{27\sqrt3}\cdot2\pi i\tag{4} \end{align} $$ where $\dfrac{124\pi^3}{27\sqrt3}$ is the sum of the residues of $\dfrac{\log(z)^3}{z^2-z+1}$ at $e^{i\pi/3}$ and $e^{i5\pi/3}$.

Combining Real and Complex Analysis

Therefore, using $(2)$ and the imaginary part of $(4)$, we get $$ \begin{align} \int_0^1\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x &=\frac12\int_0^\infty\frac{\log(x)^2}{x^2-x+1}\mathrm{d}x\\ &=\frac12\left[-\frac{124\pi^3}{81\sqrt3} +\frac{4\pi^2}{3}\int_0^\infty\frac1{x^2-x+1}\mathrm{d}x\right]\\ &=\frac12\left[-\frac{124\pi^3}{81\sqrt3} +\frac{4\pi^2}{3}\frac{4\pi}{3\sqrt3}\right]\\ &=\frac{10\pi^3}{81\sqrt3}\tag{5} \end{align} $$