$\lim \int_0^{\infty} \frac1n e^{-\frac{t}{n}} dt \ne \int_0^{\infty} \lim \frac1n e^{-\frac{t}{n}} dt$ in complex as in real?

A First Course in Complex Analysis by Matthias Beck, Gerald Marchesi, Dennis Pixton, and Lucas Sabalka

This is probably related to a real analysis or even elementary analysis classic counterexample that I forgot. Anyhoo, why doesn't the switch hold?

1. I was about to say that Prop 7.27 doesn't apply because $\gamma = G$. Unfortunately, the inclusion is loose.

2. Um, does $G$ have to be a region? If so, then I guess $\mathbb R_{\ge 0}$ is not a region? If not, then perhaps

3. $f_n(t)$ is not continuous because the left hand limit at $t=0$ does not exist because $f_n(t)$ is not defined for $t<0$? In that case, what changes if we redefine $f_n: \mathbb R \to \mathbb R$ ?

4. There's also that $f_n$'s codomain isn't C. As with #3, what changes if we redefine $f_n: \mathbb R_{\ge 0} \to \mathbb C$ ?

5. In re #3 and #4, what changes if we redefine $f_n: \mathbb R \to \mathbb C$ ?

We can extend the $\{f_n\}$ to functions $\tilde f_n:\mathbb C\to\mathbb R$ which converge uniformly on $\mathbb C$, by defining $\tilde f_n(z)=f_n(\text{Re}(z))$ for $Re(z)\geq0$ and $\tilde f_n(z)=\frac{1}{n}$ otherwise, so the issue isn't with domain of the functions.

Furthermore, changing the codomain to be $\mathbb C$ doesn't affect the behavior of the function, as it converges uniformly whenever we consider the codomain to be $\mathbb R_{>0}$, $\mathbb R$, $\mathbb C$, or any other reasonable subset of $\mathbb C$.

The issue is with the "contour" over which we are integrating. The theorem applies to paths, whose images are compact subsets of $\mathbb C$, and $\mathbb R_{\geq0}$ is not a compact subset of $\mathbb C$.

• Thanks Aweygan! Prop 7.27 that path is piecewise smooth. Where does it say in Prop 7.27 anything (implicitly I guess) about compactness? – BCLC Jul 23 '18 at 13:12
• I am unsure of what you are asking in your second comment. – Aweygan Jul 23 '18 at 13:21
• Check the definition of a piecewise smooth curve in the text you link to. They are defined as images of smooth functions over a finite number of disjoint closed (hence compact) intervals, so the images of the curves are compact subsets of $\mathbb C$, hence the image of the path is compact in $\mathbb C$. – Aweygan Jul 23 '18 at 13:26
• If you check your real analysis book, any analogous theorem would probably require that you be integrating over a bounded interval $[a,b]$, not over $[0,\infty)$. – Aweygan Jul 23 '18 at 13:27
• @BCLC You have it now! The integral extends over the entire positive reals and so the referenced theorem does not apply. However, there is a theorem that would permit the interchange of the limit and the integral IF the improper integral converged uniformly. Here, the improper integral does not converge uniformly despite the fact that sequence $f_n(x)$ converges uniformly. – Mark Viola Jul 23 '18 at 16:23

Note that in the proof of given, a key quantity is $\text{length}(\gamma)$. Uniform convergence of $f_n$ is good enough to switch the order when the measure of $G$ is finite: $$\lim_{n\to\infty}\int_Gf_n(x)\,\mathrm{d}x=\int_G\lim_{n\to\infty}f_n(x)\,\mathrm{d}x$$ For the example in the title, the length of the path, $[0,\infty)$ is not finite.

• Good catch and explanation. Thanks! – BCLC Jul 23 '18 at 15:17