I'm working through Brown-Churchill exercises and one of the problems is eluding me. The task is to evaluate the $\int_0^\infty \frac{1}{x^3+1}$ using residues and a contour they gave which goes from the origin of the complex plane to $+R$ on the real axis, then counterclockwise along a circular path to $Re^{2\pi i/{3}}$ and then back to the origin along a straight line. My argument follows:
Using a hypothetical parameterization of the contour shown, the application of the residue theorem $\int_C f(z)dz=2\pi i\sum_{z_k}Res_{z=z_k}{f(z)}$ may be written $\int_0^R f(x)dx+\int_{C_R}f(z)dz=2\pi i\sum_{z_k}Res_{z=z_k}{f(z)}$ where $C_R$ is the counterclockwise path from $R$ to $O$ along the contour and $x$ is a real variable. It is necessary now to examine the behaviour as $R\to\infty$. Observing $|z^3+1|\geq||z^3|-1|=R^3-1$ for $R>1$, $|f(z)|=\frac{1}{|z^3+1|}\leq\frac{1}{R^3-1}$ implies via the maximum modulus theorem (with $M=\frac{1}{R^3-1}$ and path length $L=(\frac{2\pi}{3}+1) R$ ) that $|\int_{C_R}f(z)dz|\leq \frac{(\frac{2\pi}{3}+1)R}{R^3-1}$ which clearly limits to $0$ as $R\to\infty$, and therefore $\int_{C_R}f(z)dz$ does as well. Applying this to the statement of the residue theorem above and taking all other occurrences of $R$ to $\infty$, $\int_0^\infty f(x)dx=2\pi i\sum_{z_k}Res_{z=z_k}{f(z)}$. Now it is necessary to calculate the residues. There is only one singularity within the contour: the third root of unity $z=e^{\frac{i\pi}{3}}$. The residue may be calculated by the fourth row of Marsden's table 4.1, where it is $g(z_0)/h'(z_0)=\frac{1}{3e^{2i/3}}=-\frac{1}{6}(1+i\sqrt{3})$. This gives $\int_0^\infty f(x)dx=\frac{-2\pi i}{6}(1+i\sqrt{3})=-\pi i/3 +\pi\sqrt{3}/3$
This answer is not real, which is my problem. I've seen this question, which is literally the exact problem and is a method I could apply, but I still do not understand why my argumentation about the contour integral over the straight off-axis segment going to zero is incorrect. Thanks in advance.
