Random Variable wrote:

Let \(\displaystyle f(z) = \frac{\sqrt[3]{z^{2}(z-1)}}{(1+z)^{3}}\) where \(\displaystyle 0 \le \arg(z), \arg(z-1) \le 2 \pi\).

Then integrating around a dumbbell contour,

\(\displaystyle 2 \pi i a_{-1} + \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x) e^{i \pi}}}{(1+z)^{3}} \ dx + \int_{1}^{0} \frac{\sqrt[3]{(xe^{2 \pi i})^{2} (1-x) e^{i \pi}}}{(1+z)^{3}} \ dx = 2 \pi i \text{Res} [f(z),-1]\)

where $a_{-1}$ is the coefficient of the $\frac{1}{z}$ term in the Laurent expansion at infinity, that is, \(\displaystyle -\text{Res}[f(z), \infty]\).

\(\displaystyle 0 + 2 i \sin \left( \frac{\pi}{3} \right) \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+z)^{3}} \ dx = - \frac{2 \pi i }{9} \lim_{z \to -1} x^{2} \Big( x^{2}(x-1) \Big)^{-5/3} = - \frac{2 \pi i}{9} \Big( (e^{\pi i})^{2} 2 e^{\pi i} \Big)^{-5/3}\)

\(\displaystyle = - \frac{2 \pi i}{9} 2^{-5/3} e^{-5 \pi i} = \frac{\pi i}{9} 2^{-2/3}\)

Which implies

\(\displaystyle \int_{0}^{1} \frac{\sqrt[3]{x^{2} (1-x)}}{(1+x)^{3}} \ dx = \frac{\pi \ 2^{-2/3}}{9 \sqrt{3}} = \frac{ \pi \sqrt[3]{2}}{18 \sqrt{3}}\)

RV, buddy, may I ask you something?. Or, whomever wants to add to the dialogue. It's driving me nuts. I am a little fuzzy about how to do these types of integrals when one uses contours. In other words, determining the arguments with a dogbone. How in the world do you determine the arguments?. I have noticed they are different depending on the problem.

Sometimes you use \(\displaystyle -\pi<arg(1-z)\leq \pi\) and other times \(\displaystyle 0<arg(1-z)\leq 2\pi\) or something like that. My inequalities may not be exact.

Take the last one you just done. You used \(\displaystyle e^{\pi i}\) inside the radical and said the argument for z and 1-z are between 0 and 2pi. Why not make the argument for z between -pi and pi... as it is many times I have seen?.

Take this one:

\(\displaystyle \int_{0}^{1}\frac{\sqrt{x(1-x)}}{(1+x^{2})^{2}}=\frac{\pi}{8}\sqrt{\sqrt{2}-1}\)

I found the sum of the residues at -i and i to be \(\displaystyle \frac{\pi i}{4}\sqrt{\sqrt{2}+1}\)

which would mean we would have to find something like \(\displaystyle 2(\sqrt{2}+1)i\) in the two integrals above and below the branch so it could be divided into the residue and get the correct result.

But, what arguments would you use for this one?. I have looked at that diagram they have on Wikipedia for 'methods of contour integration', but I do not understand why the circle around 3 is 0 below the branch and 2pi above, and the circle around 0 is pi above and -pi below the branch.

There is something about the argument business with these that has be a little befuddled.

I can make them work, but not sure why we use what we do and when.

For instance, I saw one of your old posts on SE where achillehui said that the argument for \(\displaystyle \sqrt{1-x^{2}}\) was \(\displaystyle \pi/2\). From where is he getting this?. How do you tell what the argument for something like this is?. I am missing something probably obvious with this stuff, but it is driving me crazy.