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I'm trying to evaluate this integral but i'm having a lot of problems with the standards method.

$$\int_0^1 \log \left( \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1} \right) \frac{dx}{x} $$

I've tried integration by parts, taylor expansion of the $\log$ function and some substitution.

Wolfram says that the results is $\frac{\pi^2}{3}$. I think that it divide the integral in four parts factorizing the argument of the logarithm with complex numbers, anyway it seems to take too long as approach. Do you have any suggestions?

I've also noticed that the function is even so we could also write the integral as $$ \frac{1}{2} \int_{-1}^1 \log \left( \frac{x^2+\sqrt{3}x+1}{x^2-\sqrt{3}x+1} \right) \frac{dx}{x} $$

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    $\begingroup$ Both the numerator and denominator are invariant under the transformation $p(x) \mapsto x^2 p(\frac{1}{x})$, which suggests you might get some mileage out of applying the substitution $u = \frac{1}{x}$ and rearranging. $\endgroup$ – Travis Jan 7 at 13:13
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Let $$I(\theta)=\int_0^1\frac{\ln(x^2+2x\cos\theta+1)}xdx,$$ differentiating it gives $$I'(\theta)=\int_0^1-\frac{2\sin\theta}{1+2x\cos\theta+x^2}dx\\ =-\theta$$ (since $\theta\in(-\pi,\pi)$)
The integral need to find is $$I\left(\frac16\pi\right)-I\left(\frac56\pi\right)=\int_{5/6\pi}^{1/6\pi}-\theta d\theta=\frac13\pi^2$$

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  • $\begingroup$ Brilliant, i'll add this trick to my collection thank you $\endgroup$ – Fabio Jan 7 at 13:24
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    $\begingroup$ @Fabio one can actually generalize this method to see that $$\int_0^1\log\bigg(\frac{x^2+ax+1}{x^2-ax+1}\bigg)\frac{\mathrm{d}x}{x}=\frac{\pi^2}2-\pi\arccos\frac{a}2$$ $\endgroup$ – clathratus Jan 10 at 20:02
  • $\begingroup$ I didn't noticed, besides allow to give a more elegant answer I guess this will be quite useful in the future, thanks for the observation $\endgroup$ – Fabio Jan 10 at 20:08
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Let us consider that for any $\alpha\in S^1$

$$ \int_{0}^{1}\frac{\log(1-\alpha x)}{x}\,dx = -\sum_{n\geq 1}\int_{0}^{1}\frac{\alpha^n x^{n}}{nx}\,dx=-\sum_{n\geq 1}\frac{\alpha^n}{n^2}=-\text{Li}_2(\alpha) $$ and by setting $\zeta=\exp\left(\frac{2\pi i}{12}\right)$ we have

$$ \int_{0}^{1}\log\left(\frac{1-\sqrt{3}x+x^2}{1+\sqrt{3}x+x^2}\right)\,\frac{dx}{x}=\int_{0}^{1}\left[\log(1-\zeta x)+\log(1-\zeta^{11} x)-\log(1-\zeta^5 x)-\log(1-\zeta^7 x)\right]\frac{dx}{x} $$ hence your integral equals $$ \text{Li}_2(\zeta)+\text{Li}_2(\zeta^{11})-\text{Li}_2(\zeta^5)-\text{Li}_2(\zeta^7)=2\text{Re}\,\text{Li}_2(\zeta)-2\text{Re}\,\text{Li}_2(\zeta^5). $$ Last trick: $\text{Re}\,\text{Li}_2(e^{i\theta})$ is an elementary function, since it is the formal primitive of the sawtooth wave $\sum_{n\geq 1}\frac{\sin(nx)}{n}$ (piecewise-linear), hence a periodic and piecewise-parabolic function. Putting everything together, the outcome is simply $2\zeta(2)=\frac{\pi^2}{3}$.

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  • $\begingroup$ That's impressive! But i don't understand some things sorry for the ignorance: what's the set $S^1$? $\endgroup$ – Fabio Jan 7 at 13:31
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    $\begingroup$ @Fabio: the unit circle centered at the origin of the complex plane. $\endgroup$ – Jack D'Aurizio Jan 7 at 13:52

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