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This morning I decided to look up some complex analysis, and I came across this Wikipedia section, where the following integral is evaluated:

$$\int_0^3 \frac{x^{3/4}(3-x)^{1/4}}{5-x}dx$$

There are a couple things I do not quite understand and I was wondering whether someone could clear them up for me. For convenience I shall copy and paste the sections that are troubling me.

Here is the drawing alongside:

enter image description here

We will construct $f(z)$ so that it has a branch cut on $[0, 3]$, shown in red in the diagram. To do this, we choose two branches of the logarithm, setting $z^{\frac{3}{4}} = \exp \left (\frac{3}{4}\log(z) \right )$ where $-\pi \le \arg(z) < \pi $ and $(3-z)^{\frac{1}{4}} = \exp \left (\frac{1}{4} \log(3-z) \right )$ where $0 \le \arg(3-z) < 2\pi$

  • I don't understand the branch choice. Looking at the contour, I would have written the opposite: $0\leq \arg z<2\pi$ and $-\pi\leq \arg (3-z)<\pi$. Why is this flawed thinking? Sorry if that's a very silly question.

Let $z=r$ (in the limit, i.e. as the two green circles shrink to radius zero), where $0 ≤ r ≤ 3$. Along the upper segment, we find that $f(z)$ has the value: $$r^{\frac{3}{4}} \exp(\tfrac{3}{4}0 \pi i) (3-r)^{\frac{1}{4}} \exp(\tfrac{1}{4}2 \pi i) = i \, r^{\frac{3}{4}} (3-r)^{\frac{1}{4}}$$ and along the lower segment, $$r^{\frac{3}{4}} \exp(\tfrac{3}{4}0 \pi i) (3-r)^{\frac{1}{4}} \exp(\tfrac{1}{4}0 \pi i) = r^{\frac{3}{4}} (3-r)^{\frac{1}{4}}$$ It follows that the integral of $\dfrac{f(z)}{5-z}$ along the upper segment is $-iI$ in the limit, and along the lower segment, $I$.

  • I do not understand why it is $-iI$ along the upper segment rather than $iI$.

$$(1-i) I = -2\pi i \left( \mathrm{Res}\left( \frac{f(z)}{5-z},5\right) + \mathrm{Res}\left( \frac{f(z)}{5-z} ,\infty\right)\right)$$

  • This is where I realised that I must be completely misunderstanding the idea. From what I understood, neither the singularity at $2$ nor that at $\infty$ are in the contour, so I do not see how they come into play and how the residue theorem is being used. I realise this is the whole point of all the previous working, but I don't understand what is going on.

  • Also, if instead the poles were not on the real line - for instance, if we were considering:

$$\int_0^3 \frac{f(x)}{1+x^4}\,dx$$

How would one adapt the residue calculations? Would we consider all the poles? A big thank you to anyone who answers.

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1 Answer 1

The branch $0 \le \arg(3-z) < 2 \pi$ means that there is a cut where $3-z$ is real and positive. Therefore, the cut is on $(-\infty,3]$.

Just above the cut we're integrating from right to left. Thus the reason for the negative sign.

And in this problem, the usual residue theorem for interior domains is not being used. This is because there are non-isolated singularities inside of the contour. What is being used is the residue theorem for exterior domains which includes the idea of a residue at infinity.

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