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The integral is $$ \int_{-\infty}^0 \frac{x^{1/3}}{x^5-1}\mathrm dx $$

I have tried taking the typical half circle contour and finding the enclosed residues: $$ \text{Res}(f,1)=\frac{1}{5}\\ \text{Res}(f,e^{\frac{2\pi i}{5}})=\frac{e^{\frac{8\pi i}{15}}}{5}\\ \text{Res}(f,e^{\frac{4\pi i}{5}})=\frac{e^{\frac{-2\pi i}{15}}}{5} $$ By Jordan's lemma, the arc contributes nothing. However, I have a problem with the non integrable singularity at 1 on $\mathbb{R}^+$.

Can I deviate to avoid it while also keeping track of the contribution from the positive real axis? There doesn't seem to be great symmetry at $1$ thanks to the numerator. Is my contour not a good choice?

edit: Also tried changing variables to evaluate on the positive real axis, but this again introduces the bad singularity.

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  • $\begingroup$ What's your $x^{1/3}$-definition ?. Without that we don't have any clue about the whole integration meaning. $\endgroup$ – Felix Marin Sep 27 '17 at 17:50
  • $\begingroup$ @FelixMarin I don't know, it is from a qualification exam I was using to practice. Presumably, if you do valid things given a certain branch cut it is taken to be correct $\endgroup$ – qbert Sep 27 '17 at 17:51
  • $\begingroup$ I guess the Jack answer assumption $\left(x^{1/3} = \mathrm{sgn}\left(x\right)\left\vert x\right\vert^{1/3}\right)$ is a reasonable def... when we don't have enough information. $\endgroup$ – Felix Marin Sep 27 '17 at 20:08
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Assuming that $x^{1/3}$ is defined as $-(-x)^{1/3}$ on $\mathbb{R}^-$, the change of variable $x=-z^3$ bring the given integral in the form $$ \int_{0}^{+\infty}\frac{3z^3}{z^{15}+1}\,dz $$ and by setting $\frac{1}{1+z^{15}}=u$ the previous integral can be computed through Euler's Beta function. By exploiting the reflection formula for the $\Gamma$ function, $\Gamma(z)\,\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$, such integral simplifies to $$ \frac{\pi}{5\sin\frac{4\pi}{15}}=\frac{\pi}{5\cos\frac{7\pi}{30}}=\frac{4\pi}{5\sqrt{7-\sqrt{5}+\sqrt{6 \left(5-\sqrt{5}\right)}}}.$$

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  • $\begingroup$ Thank you for your answer Jack! Out of curiosity, do you see a good way to evaluate this without appealing to "special" functions? $\endgroup$ – qbert Aug 17 '17 at 16:29
  • $\begingroup$ @qbert: contour integration works, you just have to "dodge" the singularity at $z=1$ by introducing a suitable deformation of the contour, but the approach above is way more efficient in my opinion. $\endgroup$ – Jack D'Aurizio Aug 17 '17 at 17:03

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