Why do so many half-circle perimeters contribute $0$ to the answer in contour integration?

Question

Oftentimes, when one is calculating the integral over the real line for an irritating integrand, it is convenient to use contour integration in the complex plane. Then, we very often find ourselves integrating over the half-circle in the upper (or, less canonically, lower) half-plane. These almost always (colloquial, not Lebesgue language) go to zero. What is the intuitive reason behind this? I see no reason to expect it, but of course I understand that we usually end up with

$$\dfrac{R^n}{R^m+\mathcal{O}(R^l)},$$

where $l<m$ and $n<m$, and sending $R\to\infty$ is why it goes to $0$, but what intuitive way of thinking can lead us there? I'm looking for something along the lines of "of course the upper half-circle contributes zero because (short quip)", similar to how I tell my students that the dependence of the implicit derivative on both $x$ and $y$ for a circle makes sense, since circles are not functions and we need both values to determine which point we are talking about.

I am aware of this question:

why does the circle at infinity not contribute the integral?

but it dealt more with the specific problem than the general intuition, I feel.

This question also exists, and is helpful:

Intuitive reason for why many complex integrals vanish when the path is "blown-up"?

But I still think such intuition must exist somewhere, and found the answers there helpful, but not satisfying.

Answer 1

It could be that the phenomenon you ask about is an illusion; perhaps it "almost always" happens for the integrals that can be done this way, not for a random improper integral. A thoughtful calculus student thinks that most continuous functions are differentiable, while some do not have antiderivatives, when the reality is just the opposite.

Answer 2

The reason the answer by Ullrich says the phenomenon you are asking about is an illusion is because you only see that method applied in cases where the method works. It's like students learning calculus for the first time thinking derivatives are "easy" and integrals are "hard" because the actual functions they meet in a calculus course are the kind that they know from school, which all have simple derivative formulas. Analysts know that, in contrast to the impression of calculus students, integration is a better operation than differentiation because the integral of a small function is small (intuitively).

In any case, you apply the method of half-circle contours in cases where the function being integrated gets small along large semicircles as the radius gets large and where that method actually succeeds with the Residue theorem. Imagine trying to prove the Gaussian integral $\int_{-\infty}^{\infty} e^{-x^2/2}\,dx$ is $\sqrt{2\pi}$ by setting up a complex contour integral $\int_{\gamma_R} e^{-z^2/2}\,dz$ along some semicircular contour $\gamma_R$ that includes the real interval $[-R,R]$ and some other piece connecting $-R$ and $R$ through the upper or lower half-plane. You're not going to be able to make progress with this by the Residue theorem, intuitively, because the function is entire: there is no pole at which to compute a nonzero residue! And also the function $e^{-z^2/2}$ does not get small everywhere when $|z|$ is large: if $z = iy$ is purely imaginary, then $e^{-z^2/2} = e^{y^2/2}$ is huge as $|z| = |y| \to \infty$.

It is possible to compute the Gaussian integral using the Residue Theorem, as shown in the tenth proof here, but it involves a contour integral of a function that is not simply $e^{-z^2/2}$.