<s>[INST] Solving the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$ $$\int_0^{\infty} \frac{dx}{1+x^3}$$
So far I have found the indefinite integral, which is:
$$-\frac{1}{6} \ln |x^2-x+1|+\frac{1}{\sqrt{3}} \arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{1}{3}\ln|x+1|$$
Now what do I need to do in order to calculate the improper integral?
 [/INST] As this one has been solved using a keyhole contour I thought I would show that it can also be done using a slice contour. Let $$f(z) = \frac{1}{1+z^3}.$$
The slice consists of three parts parameterized by $R$, which is real and goes to infinity. The first part $\Gamma_1$ is a line going from zero to $R$ along the real axis. The next one is a counterclockwise arc $\Gamma_2$ going from $R$ to $Re^{2i\pi/3}.$ The last one namely $\Gamma_3$ is a line back to the origin from  $Re^{2i\pi/3}.$ along the ray at angle $2i\pi/3.$
Now we have the following bound on the integral along the arc:
$$\left|\int_{\Gamma_2} f(z) dz\right| \le 2\pi R/3 \frac{1}{R^3-1}
\in O(1/R^2) \rightarrow 0 \quad \text{as} \quad R \rightarrow\infty.$$
Moreover along $\Gamma_3$ setting $z = te^{2i\pi/3}$ we have
$$\int_{\Gamma_3} f(z) dz = \int_R^0 \frac{1}{1+t^3 e^{2i\pi}} e^{2i\pi/3} dt =
- e^{2i\pi/3} \int_0^R \frac{1}{1+t^3} dt.$$
Now with $Q$ being the integral we are looking for, we thus have in the limit
$$\int_{\Gamma_1} f(z) dz = Q \quad \text{and}\quad
\int_{\Gamma_3} f(z) dz = -  e^{2i\pi/3} Q.$$
Applying the Cauchy residue theorem to the slice contour, we obtain
$$ Q (1 -  e^{2i\pi/3} ) = 2\pi i 
\operatorname{Res}(f(z); z = e^{i\pi/3}) $$
where $$\operatorname{Res}(f(z); z = e^{i\pi/3}) =
\lim_{z\to e^{i\pi/3}} \frac{z-e^{i\pi/3}}{1+z^3} =
\lim_{z\to e^{i\pi/3}} \frac{1}{3z^2} =\frac{1}{3} e^{- 2i\pi/3}.$$ 
It follows that
$$ Q= \frac{1}{3} 2\pi i \frac{e^{- 2i\pi/3}}{1 -  e^{2i\pi/3}} =
\frac{1}{3}2\pi i \frac{e^{- 3i\pi/3}}{e^{-i\pi/3} -  e^{i\pi/3}} =
\frac{1}{3}\pi \frac{1}{\sin(\pi/3)} =
\frac{1}{3}\pi \frac{2}{\sqrt 3} = \frac{2\pi}{3\sqrt{3}}.$$
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