If we let the semi-cricle blow up to infinity and the radius of the tiny circle encircling the branch point at origin go to zero, by residue theorem we have:
$$\int_\gamma + \int_{AB} + \int_{BC} + \int_{CD} + \int_{EF} = 2\pi i \times Residue $$
Since the function has no residue, RHS = 0.
We know that $\int_{CD} \rightarrow 0$, but how do I show that $\int_{AB} = \int_{EF} \rightarrow 0$ ?
Just as in Jordan's lemma http://en.wikipedia.org/wiki/Jordan's_lemma , you prove that $$ \int_{AB}+\int_{EF}\le\pi\frac1{\sqrt{R}}, $$ where $R$ is the radius of the large semicircle.