I am hoping for some clarification regarding my understanding of how the Residue Theorem applies in slightly more complicated contours (specifically regarding multiple valued functions).
I am working out of "Fundamentals of Complex Analysis - Saff", in Chapter 6.6. The following diagram is provided when trying to solve for $I = \int_0^\infty \frac{dx}{\sqrt{x}(x + 4)}$.
The argument is then as follows:
Over the entire contour, we can apply the Residue Theorem and so $\int_\Gamma = 2\pi i (Res(f; -4)) = 2\pi i\frac{(-i)}{2} = \pi$.
He then presents the argument that $\int_{\gamma_1} = \int_{\gamma_2}$. We can also bound the function on both $C_\rho$ and $C_\epsilon$, and as such show that as $\epsilon\to0$ and $\rho\to\infty$ both of these integrals must go to $0$.
As such, we conclude that $2I = \pi \implies I = \frac{\pi}{2}$.
All of these parts make sense to me, and it seems to be a reasonable approach to this integral.
Question: Why is it that we cannot apply the Residue Theorem to $C_\rho$? I think that the argument is that the Residue Theorem requires the contour to be closed. In this case, while $C_\rho$ appears to approach a closed curve, there is still the cut on the non-positive real axis [i.e. it approaches a circle, but never contains the point $\rho$].
Is that the rationale for why this result does not work, or am I missing something else here?
Thinking about it that seems to make sense, but I want to make sure I am not overlooking some subtleties (or not-so-subtle-ties)!
Thank you!
