Related problem: Understanding Dogbone contour example
Here, we need to compute $\int_{0}^3 \frac{x^\frac{3}{4}(3-x)^\frac{1}{4}}{5-x}dx$ using the dogbone contour. I am trying to understand https://en.wikipedia.org/wiki/Contour_integration.
First, the Wiki article (Example 6 – logarithms and the residue at infinity) chooses
$z^\frac{3}{4} = \exp(\frac{3}{4}\log z)\,(-\pi \leq \arg z < \pi) \tag{1}$
, and
$(3-z)^\frac{1}{4} = \exp(\frac{1}{4}\log(3-z))\,(0 \leq \arg z < 2\pi) \tag{2}$.
It seems that these two $\log$ are different, will they conflict?
Furthermore, when computing residues, it says:
$f(z) := z^\frac{3}{4}(3-z)^\frac{1}{4} \tag{3}$
$Res_{z=5} \frac{f(z)}{5-z} = -5^{\frac{3}{4}}\exp(\frac{1}{4}\log(-2)) \tag{4}$.
What is the $\log(-2)$, especially the argument in Eq.(4)? Common sense tells me $\log(-2) = \log(2) + i\pi$, but in such an argument system I am quite unsure what the argument of $-2$ is. Which $\log$ does Eq.(4) pick, the log in Eq.(1)/(2), or another mysterious log? It seems that branch cut and points always confuse me, and they even confuse some very advanced integral players.
This post is not about problem solving. I appreciate your help on the clarifications.