Let's concentrate on $$\int_0^\pi e^{iRe^{i\theta}} i d\theta$$
If $R \to \infty$, this integrand converges pointwise to $0$; plus, the modulus of the function is $= e^{-R\sin\theta} \le e^{-\sin\theta}$ $\in L^1([0, \pi])$ so it is dominated and applying lebesgue I find $$\lim_{R \to \infty} \int_0^\pi e^{iRe^{i\theta}} i d\theta = 0$$
If $R \to 0$, the integrand converges pointwise to $i$, and it is still dominated by $e^{-\sin\theta}$ so applying lebesgue I find $$\lim_{R \to 0} \int_0^\pi e^{iRe^{i\theta}} i d\theta = \int_0^\pi i d\theta = \pi i$$
It this all correct? I am unsure wether I used the dominated convergence theorem correctly.
It is worthwhile computing the upper bound explicitly.
You have $| e^{iRe^{i\theta}} | = e^{-R \sin \theta}$, and if $\theta \in [0,\pi]$, we see that $e^{-R \sin \theta} \le 1$ for all $R \ge 0$. Since $\theta \mapsto 1$ is integrable on $[0,\pi]$, you can apply the dominated convergence theorem.