When discussing analytic continuation, my lecturer used the following example,

$$ I = \int_0^{\infty} \frac{dx}{1+x^3} $$ I have in my notes that the contour was taken as below. I must admit I was pretty tired that day so my notes are significantly lacking.

Q: I am unsure why the contour was taken as in the image. Can anyone explain? (specifically why the contour stops at at $e^{2i\pi/3}$ rather than the full semi-circle)

confusing contour


The reason this is a good choice of contour is that $(re^{2\pi i/3})^3 = r^3$.

Hence, the part of the integral along the ray with argument $2\pi/3$ can be easily related to the part along the $x$-axis (which is the integral we really are looking for).

(Just to be clear, your sector should have radius $R$ and you then want to let $R \to \infty$. In your sketch it looks like the radius is fixed to $1$, which is not a good idea.)


By the residue theorem $$-2i\pi \int \frac{1}{1+x^{3}}= -2i \pi\sum_{Res z=z_{k}}\frac{1}{1+z^{3}} $$

The right hand side contains the sums of the poles of the integrand and they are the solutions to $1+z^{3}=-$ from which we obtain $z=e^{i \pi/3}$, $z=e^{i \pi}$ and $z=e^{5i \pi/3}$.

A good example of integrating $(1+r^{n})^{-1}$ is given at the following Stack Exchange page

Show that $\int_0^ \infty \frac{1}{1+x^n} dx= \frac{ \pi /n}{\sin(\pi /n)}$ , where $n$ is a positive integer.

  • $\begingroup$ The contour in the question is not a keyhole contour and works at least as good (in fact a little better) than a keyhole contour for this particular integral. Just one residue is needed. $\endgroup$ – mrf Oct 26 '15 at 12:31
  • $\begingroup$ Ah @mrf you are right it is not a keyhole - I will remove this statement from my answer. Many thanks, Bacon. $\endgroup$ – Kevin Oct 27 '15 at 9:41

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