Consider the integral: $$\int_0^\infty\frac{dx}{1+x^n}$$ where $n \gt 1 $ for the integral to exist. I've been trying to use contour integration to solve this and with a nice pizza slice shaped contour the integral is easily evaluated and there are many videos on youtube that use the same technique. However, there is one thing that I just seem to not understand: The function $z^n$, for non-integer $n$ obviously has a branch cut which connects $z=0$ to the $z=\infty$, $the \ point \ at \ infinity$. But the pizza slice shaped contour doesn't really respect this fact. So I don't know we're allowed to use the Residue Theorem in such a case (I don't even understand "why" at even an intuitive level). Here are the two explanations that I've got, which are all quite plausable:
- A branch cut is not needed in such a scenario because our contour does not consist of a full circle, so the multi-valuedness is not a problem to begin with.
- Say that the branch cut is chosen to the left so that it only coincides with our contour at the origin. The integrand, at the origin just evaluates to be $1$, i.e. it doesn't diverge, Therefore it only contributes to the integral infinitesimally.
Are there any other explanations? Can we explain the already existing ones in a clearer way?
To be honest, I really don't know what you mean by "a branch cut connecting $z=0$ to $z=\infty$". I don't think that makes sense.
If you want to define $z\mapsto z^n$ as an analytic function for some fixed $n>1$ you will be forced to restrict your domain and have a "branch cut", yes. Standard choices would be $\Bbb C\setminus[0,\infty)$ or $\Bbb C\setminus(-\infty,0]$. You could use the latter one here.
Observe that your domain of integration is always in the first quadrant and $z\mapsto z^n$ can be defined in an analytic way here, both on the path of integration itself and on the interior of the contour. There are no obstructions to using the residue theorem in this examples, none at all. $z=0$ and $z=\infty$ are not contained in your integration at any point (for fixed $\epsilon,r$, which is how you should rigorously do things) so, even if you do have a valid complaint about this cut from $0\to\infty$ (again, your meaning was quite unclear to me) it wouldn't matter here!
To repeat, based on my reading of your point $(2)$: $0$ is never contained in the domain of integration. But even if it were, this is sometimes acceptable; the Cauchy theorems and the residue theorems are valid (iirc) if the function is merely continuous on the boundary of the contour (and, of course, holomorphic on the interior) and $z\mapsto z^n$ is a continuous function on, say, the closed first quadrant (which does contain zero!). There is no issue. You do need to be rigorous about taking $\epsilon\to0^+$ i.e. checking the small semicircular integrals vanish, but this is true and not too hard to check; technically you should also check that $\lim_{\epsilon\to0^+\\r\to\infty}\int_\epsilon^r\frac{1}{1+x^n}\,\mathrm{d}x=\int_0^\infty\frac{1}{1+x^n}\,\mathrm{d}x$ - this is either easy or true by definition, depending on how you think about improper integrals.