contour integration - Example of Improper integral in complex analysis

I'm doing this example of Cauchy principle value

$$ \int_0^\infty \frac{dx}{x^3+1}=\frac{2\pi}{3\sqrt{3}} $$

After some steps i got,

$$ \int_{[0,R]+C_R} \frac{dz}{z^3+1}=2\pi i(B_1)\text{
where }B_1= \operatorname{Res}_{z=z_0}\frac{1}{z^3+1} $$

also I got that $\displaystyle \bigg|\int_{C_R} \frac{dz}{z^3+1}\bigg|\to 0 \text{ as } R \to \infty$

There is problem to to finding residue at $z_0=\displaystyle \frac{1+\sqrt{3}i}{2}$

Here i am considering the following contour:

please help me.thanks in advance.

Answer

The contour is good. Two things though:

1) You have to consider the integral along the angled line of the wedge contour. The angle of the contour was chosen to preserve the integrand. 2) Write $z=e^{i 2 \pi/3} x$ and get that the contour integral is

$$\left(1-e^{i 2 \pi/3}\right) \int_0^{\infty} \frac{dx}{x^3+1} = i 2 \pi \frac{1}{3 e^{i 2 \pi/3}}$$

The term on the right is the residue at the pole $z=e^{i\pi/3}$ times $i 2\pi$. I used the fact that, if $f(z)=a(z)/b(z)$, then the residue of a simple pole $z_k$ of $f$ is $a(z_k)/b'(z_k)$.

Note that $e^{i 2 \pi/3}-e^{i 4 \pi/3}=i \sqrt{3}$. The result follows.