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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.

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  • $\begingroup$ You are doing fine. You used it to change the order of limit and integration! $\endgroup$ – Mhenni Benghorbal Jan 28 '15 at 15:27
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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.

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  • $\begingroup$ Good :) eventually I cleared it up but it's a nice answer to the question :) $\endgroup$ – Ant Oct 23 '15 at 8:03

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