Integrating $\frac{1}{1+z^3}$ over a wedge to compute $\int_0^\infty \frac{dx}{1+x^3}$. 
Compute $\displaystyle\int_0^\infty \frac{dx}{1+x^3}$ by integrating $\dfrac{1}{1+z^3}$ over the contour $\gamma$ (defined below) and letting $R\rightarrow \infty$.
The contour is $\gamma=\gamma_1+\gamma_2+\gamma_3$ where $\gamma_1(t)=t$ for $0\leq t \leq R$, $\gamma_2(t)=Re^{i\frac{2\pi}{3}t}$ for $0\leq t \leq 1$, and $\gamma_3(t)=(1-t)Re^{i\frac{2\pi}{3}}$ for $0\leq t \leq 1$.

So, the contour is a wedge, and by letting $R\rightarrow \infty$ we're integrating over one third of the complex plane.  I believe this means we are integrating over the entire complex plane under the substitution $u=x^3$.  There are poles at $-\zeta$ for each third root of unity $\zeta$, so there's only one pole in this wedge.  I'll just refer to that pole as $-\zeta$.
I guess this means that we can use the residue theorem to say $$\int_{\gamma}\frac{1}{1+z^3}dz=2\pi i\eta(\gamma,-\zeta)\operatorname{Res}\left(\frac{1}{1+z^3},-\zeta\right)=2\pi i \lim_{z\rightarrow -\zeta}\left[(z+\zeta)\frac{1}{1+z^3}\right]$$
I can't evaluate this limit.  Also I don't see how it involves $R$, which I'm supposed to be taking a limit of. I suspect I've done something wrong.

What's the problem? How do I proceed?

Also, after I do properly evaluate this integral, I am assuming that its value is supposed to be $\displaystyle\int_0^\infty\frac{dx}{1+x^3}$.  Why? (I think I know why conceptually but I need to see how one rigorously writes that out.)
(Note: This is exam review, not homework.)
 A: Hints: Firstly, $1+z^3 = (z+\zeta)(z+\zeta^2)(z+1)$ by factorizing the polynomial.
Secondly, $$\int _0^1\frac{1}{1+z^3} z'(t) \mathrm d t $$ on the contour $z = R(1-t)\exp(2\pi i/3)$ can be related to the real integral you were originally looking for just by substituting this expression for $z$ in.
A: The easier way out to compute $\displaystyle \int_0^{\infty} \dfrac{dx}{1+x^3}$ is as follows. We have
$$I = \int_0^{\infty} \dfrac{dx}{1+x^3} = \int_{\infty}^0 -\dfrac1{x^2}\dfrac{dx}{1+1/x^3} = \int_0^{\infty} \dfrac{xdx}{1+x^3}$$
We hence have
$$2I = \int_0^{\infty} \dfrac{1+x}{1+x^3} dx = \int_0^{\infty}\dfrac{dx}{1-x+x^2} = \int_0^{\infty} \dfrac{dx}{\left(\dfrac{\sqrt3}2 \right)^2+ \left(x - \dfrac12\right)^2}$$
Hence, we get that
$$2I = \left. \dfrac2{\sqrt3}\arctan\left(\dfrac{2x-1}{\sqrt3}\right) \right \vert_0^{\infty} = \dfrac2{\sqrt3}\left(\dfrac{\pi}2 + \dfrac{\pi}6\right) \implies I = \dfrac{2 \pi}{3\sqrt3}$$
A: $$\frac{1}{x^3+1}=\frac{1}{3(x+1)}-\frac{x-2}{3(x^2-x+1)}$$
But
$$\frac{x-2}{x^2-x+1}=\frac{1}{2}\frac{2x-1}{x^2-x+1}-\frac{\frac{3}{2}}{\frac{3}{4}+\left(x-\frac{1}{2}\right)^2}=\frac{(x^2-x+1)'}{x^2-x+1}-\frac{4}{3}\frac{\frac{3}{2}}{1+\left(\frac{2}{\sqrt3}\left(x-\frac{1}{2}\right)\right)^2}=$$
$$=\frac{(x^2-x+1)'}{x^2-x+1}-\sqrt3\,\frac{\frac{2}{\sqrt3}dx}{1+\left(\frac{2}{\sqrt3}\left(x-\frac{1}{2}\right)\right)^2}$$
Finally:
$$\int\limits_0^\infty\frac{dx}{x^3+1}=\left.\left[\frac{1}{3}\log\frac{\sqrt{x^2-x+1}}{x+1}+\sqrt 3\arctan\frac{2}{\sqrt 3}\left(x-\frac{1}{2}\right)\right]\right|_0^\infty=$$
$$0+\sqrt3\,\left(\frac{\pi}{2}-\arctan\left(-\frac{1}{\sqrt3}\right)\right)=\sqrt3\left(\frac{\pi}{2}+\frac{\pi}{6}\right)=\frac{2\pi}{\sqrt3}$$
