We are asked to show $$\int_0^{\infty}\frac{\log(x)}{x^3+1}dx=-\frac{2\pi^2}{27}$$, and $$\int_0^{\infty}\frac{1}{x^3+1}dx=\frac{2\pi}{3\sqrt{3}}$$ By integrating around a pie slice with angle $\frac{2\pi}{3}$. If we denote $\Gamma_R$ as the outer arc of this slice with radius $R$, and $\gamma_\epsilon$ as the inner arc of this sector with radius epsilon, we see these vanish as $R\rightarrow\infty, \epsilon \rightarrow 0$. There is only one residue in this domain, namely $e^{i\pi/3}$. The Residue here equals: $$Res[\frac{\log(x)}{x^3+1},e^{i\pi/3}]=\frac{\log(e^{i\pi/3})}{e^{i2\pi/3}+e^{i2\pi/3}+e^{i2\pi/3}}=\frac{1+{i\pi/3}}{3e^{i2\pi/3}}$$ So our total integral over this domain equals: $$2\pi i\frac{1+{i\pi/3}}{3e^{i2\pi/3}}=\int_\epsilon^{R}\frac{\log(x)}{x^3+1}dx+\int_R^{\epsilon}\frac{\log(x)}{x^3+1}dx+\int_R^{\epsilon}\frac{i\frac{2\pi}{3}}{x^3+1}dx$$ by making the change of variables $z=re^{i\theta}, 0\le\theta\le\frac{2\pi}{3}$ for the upper line segment of the pie slice... but clearly there is a problem somewhere since the two log integrals cancel. What is the issue?


Integrating $ \displaystyle f(z) = \frac{\log z}{z^{3}+1}$ around the contour you described and using the principal branch of the logarithm,

$$ \begin{align} \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx + \int_{\infty}^{0} \frac{\log x + \frac{2 \pi i}{3}}{(xe^{2 \pi i /3})^{3} + 1} \ e^{2 \pi i /3} \ dx &= 2 \pi i \ \text{Res}[f(z), e^{i \pi/3}] \\ &= 2 \pi i \lim_{z \to e^{i \pi /3}} \frac{\log z}{3z^{2}} \\ &= -\frac{2 \pi^{2}}{9} e^{-2 \pi i /3}. \end{align}$$

Rearranging, we have

$$ (1-e^{2 \pi i/3}) \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx - \frac{2 \pi i}{3} e^{2 \pi i/3} \int_{0}^{\infty} \frac{1}{x^{3}+1} \ dx = -\frac{2 \pi^{2}}{9} e^{-2 \pi i /3}. $$

And multiplying both sides of the equation by $e^{-i \pi /3}$,

$$ -2i \sin \left( \frac{\pi}{3} \right) \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx - \frac{2 \pi i}{3} e^{i \pi /3} \int_{0}^{\infty} \frac{1}{x^{3}+1} \ dx = \frac{2 \pi^{2}}{9}$$


$$ -i \sqrt{3} \int_{0}^{\infty} \frac{\log x}{x^{3}+1} \ dx +\left(\frac{\pi \sqrt{3}}{3} - \frac{i\pi}{3} \right) \int_{0}^{\infty} \frac{1}{1+x^{3}} \ dx = \frac{2 \pi^{2}}{9}.$$

Now equate the real and imaginary parts on both sides of the equation and solve the system of equations to obtain the answers.

  • $\begingroup$ Oh shoot! I made two stupid mistakes I think. First, I thought $\log(1)=1$ for some reason instead of 0, and second I forgot to add in the $e^{2i\pi/3}$ when I did the change of variables. Also, I have another question. When we do problems with keyhole domains (not applicable here), and we say the integral goes from $\epsilon$ to $R$, isn't that inaccurate since we don't start the line a direct epsilon distance to the right of the origin (although we get closer and closer to doing so as $\epsilon$ gets smaller and smaller)? $\endgroup$ – user2154420 Apr 28 '14 at 21:39
  • $\begingroup$ It should technically be written as $\lim_{\epsilon \to 0^{+}} \int_{\epsilon}^{R} f(x) \ dx$. $\endgroup$ – Random Variable Apr 28 '14 at 21:50

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