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If someone could please explain how to do these kinds of integrals, where the boundaries of integration are not $_{-}^{+}\infty$ and the function is not even, that is how to choose the complex function and the contour?

$$\int_{-\infty}^{0} \frac{x}{x^3-1} \, \mathrm{d}x$$

Jackob
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    With $z:=-x^3$ we want $\tfrac13\int_0^\infty\frac{z^{-1/3}dz}{1+z}$. What we need next is a keyhole contour. The argument is similar to this one. You should get $\frac{2\pi}{3\sqrt{3}}$. – J.G. Dec 08 '21 at 18:19
  • For this problem, make a contour around $e^{2 i \pi / 3}$ by adding in a ray extending from $0$ to the upper right at an angle of $60^\circ$. The integral along the ray will equal the desired integral, so the answer will be half the residue at $e^{2 i \pi / 3}$. – Joseph Camacho Dec 08 '21 at 18:19

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