Context: I want to compute the Dirichlet integral with the help of $f(z)=\frac{\exp(iz)}{z}$ but without residues, Jordan's lemma etc, so everything by hand. While I understood how to show that something converges to $0$, I still don't understand how to show that something converges to a specific value.
I am confused about the rules of switching limits and integrals. In particular:
I want to show that the integral of $f(z)=\frac{\exp(iz)}{z}$ along the curve $\gamma:[0,\pi] \to \mathbb C$, $\gamma(t)=\epsilon \exp(it)$ - thus the small semicircle around $0$ - goes to $i\pi$.
So it boils down to showing $lim_{\epsilon \to0} \int_{0}^{\pi} \exp(\epsilon i e^{it} ) dt=\pi$
I know that one method would be showing that $\exp(\epsilon i e^{it} )$ converges uniformly to $1$ but I think formally i might get problems as
$|\exp(\epsilon i e^{it} )|=|\frac{1}{\exp(\epsilon \sin(t))}|$ and the convergence won't be uniformly as $\sin(t)$ can be small on $[0,\pi]$
Is it possible to use dominated convergence by showing that $|\frac{1}{\exp(\epsilon \sin(t))}|$ is integrable?
$\int_{0}^{\pi} |\frac{1}{\exp(\epsilon \sin(t))}| dt$
$\le \int_{0}^{\pi} |\frac{1}{1+\epsilon \sin(t)}| dt$
$= \int_{0}^{\pi/2} |\frac{1}{1+\epsilon \sin(t)}|dt+\int_{\pi/2}^{\pi} |\frac{1}{1+\epsilon \sin(t)}|dt$ and by convexity/concave of $\sin$
$\le \int_{0}^{\pi/2} |\frac{1}{1+\epsilon \frac{2}{\pi} t}|dt+\int_{\pi/2}^{\pi} |\frac{1}{1+\epsilon (-\frac{2}{\pi} t+2)}|dt$
$=\int_{0}^{\epsilon} \frac{1}{1+u}\frac{2}{\epsilon \pi}du+\int_{1}^{0} \frac{1}{1+\epsilon u} \frac{-\pi}{2} du$
$=\pi (\frac{\ln \epsilon+1}{\epsilon})$ $<\infty$
So by DCT: $lim_{\epsilon \to0} \int_{0}^{\pi} \exp(\epsilon i e^{it} ) dt=\int_{0}^{\pi} lim_{\epsilon \to0} \exp(\epsilon i e^{it} ) dt = \int_{0}^{\pi} 1dt=\pi$
So, my question is: Is this a formally correct way or were all these computation useless? Or is there a standard method for doing this?
You have a mistake in the evaluation of the last integrals. And for the dominated convergence theorem, you need a dominating function that is independent of $\epsilon$.
But you are thinking too complicated. On $[0,\pi]$, we have $\sin t \geqslant 0$, and we have $\epsilon > 0$, so
$$\biggl\lvert\frac{1}{\exp (\epsilon \sin t)}\biggr\rvert \leqslant 1.$$
Since the domain of integration has finite measure, that shows that the dominated convergence theorem is applicable.
Also, the convergence of $\exp (i\epsilon e^{it})$ to $1$ is uniform. For every $z \in \mathbb{C}$ we have
$$\lvert e^z - 1\rvert = \Biggl\lvert \sum_{n = 1}^{\infty} \frac{z^n}{n!}\Biggr\rvert \leqslant \sum_{n = 1}^{\infty} \frac{\lvert z\rvert^n}{n!} = e^{\lvert z\rvert} - 1,$$
so we see
$$\lvert \exp(\epsilon ie^{it}) - 1\rvert \leqslant e^{\epsilon} - 1.$$
When $\epsilon \leqslant 1$, we can bound the right hand side by $(e-1)\cdot \epsilon$.