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 $I = \int_0^\infty \frac{dx}{\sqrt{x}(x + 4)}$.

Contour Integral Diagram

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!


1 Answer 1


The residue theorem requires the function to be analytic on the contour but as we know the logarithm fails to obey this constraint around the origin. Moreover the sketch you have for the contour is a bit misleading. It should be like this

enter image description here

Now since we are avoiding the branch cut this will make the residue theorem work.

  • 1
    $\begingroup$ Nice Picture! (+1) $\endgroup$
    – Mark Viola
    Apr 16, 2017 at 5:54
  • $\begingroup$ @Dr.MV Thanks Mark... $\endgroup$ Apr 16, 2017 at 6:00

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