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Prove that $$\int_{0}^{1} \frac{\ln \left ( x^2+x+1 \right )}{x}\mathrm dx=\frac{\pi^2}{9}.$$

As I understand, I can do this: $$\large 1- x^3 = (1-x)(1+x+x^2) \Rightarrow x^2 +x+1 = \frac{1-x^3}{1-x}, $$ this gives $$f(x)= \frac{1}{x} \ln \left(\frac {1-x^3}{1-x}\right) =\frac{1}{x} (\ln(1-x^3)-\ln(1-x)).$$
But what shall I do next? Thank you for your help very much!

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You recognized a crucial fact, i.e. that $x^2+x+1$ is a cyclotomic polynomial.
For any $m\geq 1$ we have $$ \int_{0}^{1}\frac{-\log(1-x^m)}{x}\,dx = \sum_{n\geq 1}\int_{0}^{1}\frac{x^{mn-1}}{n}\,dx = \frac{1}{m}\sum_{n\geq 1}\frac{1}{n^2} = \frac{\zeta(2)}{m}\tag{1}$$ hence $$ \int_{0}^{1}\frac{\log\Phi_3(x)}{x}\,dx =\frac{2}{3}\zeta(2)=\color{red}{\frac{\pi^2}{9}}.\tag{2}$$


In general, given $$ \Phi_n(x) = \prod_{d\mid n}(1-x^d)^{\,\mu\left(\frac{n}{d}\right)} \tag{3}$$ we have $$\begin{eqnarray*} \int_{0}^{1}\frac{\log\Phi_n(x)}{x}\,dx&=&-\zeta(2)\sum_{d\mid n}\frac{1}{d}\cdot\mu\left(\frac{n}{d}\right)\\&=&-\frac{\zeta(2)}{n}\sum_{d\mid n}d\cdot\mu(d)\\&=&-\frac{\zeta(2)}{n}\prod_{p\mid n}(1-p)\\&=&\frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)}{n^2}\prod_{p\mid n}p\\&=&\color{red}{\frac{\zeta(2)(-1)^{\omega(n)+1}\varphi(n)\,\text{rad}(n)}{n^2}}. \tag{4}\end{eqnarray*}$$

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  • $\begingroup$ can we also use Frullani's integral to calculate this inequality ? $\endgroup$ – user448747 Jan 5 '18 at 19:37
  • $\begingroup$ @FatsWallers: it is not needed, one may derive $(1)$ by just enforcing the substitution $x\mapsto z^{1/m}$. $\endgroup$ – Jack D'Aurizio Jan 5 '18 at 19:41

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