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!


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.