Does it possible to show that this integral $$\int_0^\pi \mathrm{e}^{-\mathrm ir\mathrm{e}^{-\mathrm i t}} \,\mathbf dt$$ tends to $\pi$ as $r\to 0$ without the dominated convergence theorem ?
Thank for answers.
(edit $\int_0^\pi \mathrm{e}^{-\mathrm ir\mathrm{e}^{-\mathrm i t}} \,\mathbf dt$ sorry)
Considering Euler identity $e^{it} = \cos(t) + i \sin(t)$, you can express the integral as: $\int\limits_{0}^{\pi} e^{r\sin t} \cos(r \cos t) \, dt + i\int\limits_{0}^{\pi} e^{r\sin t} \sin(r \cos t) \, dt$ Using Taylor series expansion of $\exp$, $\sin$ and $\cos$ functions at $r = 0$, you can prove that the imaginary part vanishes and the real part of the integral converges to $\pi$. I hope this answers your question
The function $(e^z-1)/z$ has a removable singularity at $z=0$. It follows that there is a constant $C>0$ such that $$ |e^z-1|\le C\,|z|,\quad|z|\le1. $$ Then if $|r|\le1$ we have $$ |e^{ire^{it}}-1|\le C\,r. $$
The exponential function is an entire function, hence
$$g(r)=\int_{0}^{\pi}\exp\left(-ir e^{-it}\right)\,dt = \sum_{n\geq 0}\frac{(-ir)^n}{n!}\int_{0}^{\pi}e^{-nit}\,dt =\pi+\sum_{m\geq 0}\frac{-2i(-ir)^{2m+1}}{(2m+1)(2m+1)!}$$
is an entire function too, with $\lim_{r\to 0}g(r) = g(0) = \pi.$
Actually, $g(r) = \pi-2\,\text{Si}(r)$: this also proves
$$ \lim_{r\to -\infty}g(r)=2\pi,\qquad \lim_{r\to +\infty}g(r) = 0.$$
