So in applying the residue theorem to solve improper real integrals, we agree to take our semicircles to be as large or as small as necessary such that all the poles we wish to work with lie inside the closed path.
Somehow, this has an air of hand-waviness to it. Is there any formal justification for why we can just assume off the bat that our limits of integration are sufficiently large or small? I understand the reasons why we want to do this, but not really why we're allowed to.
Thank you in advance.
It's not ``that we assume off the bat that our limits of integration are sufficiently large or small", it's that we extend the integral so the residue theorem applies. Once we have that, we then play with the components of our closed curve to simplify our expression and solve more complicated integrals.
Consider the integral
$ \int_\mathbb{R} f(x) dx $
where $f(x)$ is well defined on $\mathbb{R}$ but isn't nice for integrating. We can extend this integral into the complex plane as a line integral of some closed curve, i.e we look at
$\oint_\gamma f(z) dz$
For the sake of having an example, lets take our path $\gamma$ as a semi circle about $z=0$ with radius $R$. By linearity of the integral, we can break up the the path into two integrals, the bit on the real axis, and the arc in the complex plane.
$\oint_\gamma f(z) dz = \int_{-R}^R f(x) dx + \int_{arc} f(z) dz$
What's nice about this? We have the residue theorem so we know
$\int_{-R}^R f(x) dx + \int_{arc} f(z) dz = 2 \pi i \sum Residues(f)$
This means that we need to take a limit to play around with this, i.e
$\int_\mathbb{R} f(x) dx = \lim_{R \to \infty} 2 \pi i \sum Residues(f) - \int_{arc} f(z) dz$
Residues are something we can calculate quite easily and arcs generally go to zero in calculations. This is why we care about this method.