Counter exchanging limit and integral Background
I came across this answer on Math SE which claimed it made a lot of sense to switch limit and integral. In response I came up with the following counter-examples:


*

*$\lim_{w \to 0} \int_0^\infty we^{-wt}\ dt = 1$ but $\int_0^\infty \lim_{w \to 0} we^{-wt}\ dt = 0$.

*$\lim_{n \to \infty} \int_0^\infty \frac{t^{n+1}}{n!} e^{-t}\ dt = \infty$ but $\int_0^\infty \lim_{n \to \infty} \frac{t^{n+1}}{n!} e^{-t}\ dt = 0$.

*$\lim_{n \to \infty} \int_0^1 n^3 t^n (1-t)\ dt = \infty$ but $\int_0^1 \lim_{n \to \infty} n^3 t^n (1-t)\ dt = 0$.

*$\lim_{r \to \infty} \int_{-\infty}^\infty \frac{r(rx)^3}{(rx)^4-(rx)^3+1}\ dx = \lim_{r \to \infty} \int_{-\infty}^\infty \frac{y^3}{y^4-y^3+1}\ dy \approx 2$ and $\lim_{r \to \infty} \frac{r(rx)^3}{(rx)^4-(rx)^3+1} = \frac{1}{x}$ for any $x \ne 0$. [Newly added]
Unlike the claimed method in that post, the first three counter-examples have continuous limit functions.
The third has the advantage of having the integral over a finite interval, to show that interchanging is still not valid without additional assumptions on the function sequence.
The fourth uses exactly the same technique as the claimed method (with a discontinuous limit function) and also gives an integral that is constant and independent of the limit!
Of course, these examples do not satisfy the dominated convergence theorem in various ways.
Question
Do you have any good counter-examples that actually show up in real-world problems and have an experimentally measurable implication for the real world? I somehow get the impression that the functions in the real-world might be too nice (wave functions are infinitely differentiable when you really factor everything in; in particular there are no such things as square potential wells because it is impossible to have a discontinuous change in potential!) I would be satisfied with examples in classical mechanics if the effect is measurable.

Consider the retarted propagator of a  free right moving quantum mechanical particle with linear dispersion
$$
G(\omega,k,\delta)=\frac{1}{\omega-vk+i \delta}
$$ 
(the $+i \delta$ is there for causality reasons and have set to  be $0$ in the end (in a limiting procedure))
Calculating the corresponding density of states requires the calculation of $\rho(\omega)=\int_{\mathbb{R}} dk G(k,\omega)$ which only makes sense if we calculate it like
$$
\lim_{\delta\rightarrow 0}\int_{\mathbb{R}} dk G(k,\omega,\delta)
$$
instead of
$$
\int_{\mathbb{R}} dk \lim_{\delta\rightarrow 0} G(k,\omega,\delta)
$$
(the second quantity clearly diverges). The reason is that we are handeling quantities which are strictly speaking only defined in a distributional sense.
