# Questions regarding the complex logarithm and complex integration

I'm currently preparing for an exam in complex analysis and I don't quite feel comfortable with some exercises, mostly those including a complex logarithm and some "unusual" paths to integrate along. I have specific questions for the following exercise:

Compute $\int_0^\infty \frac{x^\alpha}{1+x^2} \, \mathrm dx$ for $-1 < \alpha < 1$.

The standard solution suggests to evaluate this integral by integrating along the boundary of the following set: $\{z \in \mathbb{C}|\, \varepsilon < |z| < R \} \setminus \{z \in \mathbb{C} | \, \Re(z) \geq 0, - \varepsilon \leq \Im (z) \leq \varepsilon\}$. This seems like a very odd path to integrate along.

Question 1: How come we integrate along this path? I mean, is there some easily comprehensible motivation to use this path or did just someone try his luck with a lot of paths and eventually, this path proved to be the most useful (as the integral along the small and big "partial" circle vanish)?

Now the standard solution defines $z^\alpha := e^{\log(z) \alpha}$ where they choose a branch of the logarithm on $\mathbb{C} \setminus [0, \infty)$ with $\log(z) = \log|z| + i \theta, \theta \in (0, 2 \pi)$.

Question 2: I understand a complex logarithm can only be holomorphic function on a simply connected, open subset of $\mathbb{C}\setminus\{0\}$. Why do we leave out the positive real axis in this case? After all, our original integral is along the positive real axis whereas here we use a logarithm which is not even defined there.

Next, the standard solution states that the integrals along the partial circles vanishes and this is just a simple calculation which I have managed to do myself. However, then, it is stated that the remaining horizontal path in the upper half plane converges to our original integral whereas the remaining horizontal path in the lower half plane converges to the original integral times $(-e^{2 \pi i \alpha})$.

Question 3: The minus sign makes perfect sense. However, why does the integral in the lower half plane not just converge to the original integral times $-1$? Also, why do both integrals even converge to the original integral up to some constant? After all, our complex logarithm is not even defined for positive real $z$, so both integrals will always be taken along some non-real path. I really don't see why they should become identical with the original integral up to a constant.

Thank you very much in advance for any help.

-
Do you mean $\{z \in \mathbb{C}|\, \varepsilon < |z| < R \} \setminus \{z \in \mathbb{C} | \, \Re(z) \geq 0, - \varepsilon \leq \Im (z) \leq \varepsilon\}$? –  Ron Gordon Jan 28 '13 at 13:41
@rlgordonma: Yes, I do. I fixed the typo. –  Huy Jan 28 '13 at 13:42

1) The Residue Theorem covers singularities from poles; it does not address branch point singularities resulting from multivalued "functions". The near-circular hole of radius $\epsilon$ in the center of the contour to meant to avoid the branch point at the origin. Recall that you are also integrating about a near-circle of radius $R$, on which the integral will vanish as $R$ gets big. This is then a closed contour defining a set in which the integrand is holomorphic except for isolated poles, or in this case, pole.
2) A branch cut is an ugly solution to the problem of defining a single-valued function from an inherently multivalued one like $z^{\alpha}$. But it can be defined whereever it is convenient for the solution of a problem. Here, it is defined on the positive real axis because it is there that $\theta$ is defined to be $0$.
3) You get the factor of $(-e^{2 \pi i \alpha})$ because you are parametrizing $z=(e^{i 2 \pi}) t$; that is, the effect of the multivaluedness of $z^{\alpha}$ comes into play here. Why? Because you have gone through a rotation of $2 \pi$ (almost) to get to the piece of the contour in the lower-half plane.