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In the solution of a complex analysis problem I'm working through the following comment was made:

$\displaystyle-\lim_{\epsilon \rightarrow 0} \int_{\pi}^{0} i e^{i\epsilon \displaystyle e^{i \theta}} d\theta = \pi i, \;\;\;\;\;$ since the limit can be taken under the integral sign...

I'm not sure how to justify this.

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2 – user17762 Jan 16 '13 at 20:49
up vote 2 down vote accepted

Since $$|\exp(i\varepsilon e^{i\theta})| = e^{-\varepsilon \sin\theta},$$ the integrands are bounded by an integrable function, independently of $\varepsilon$.

Lebesgue's dominated convergence theorem shows that the limit is $-i\pi$, since the integrand converges pointwise to $i$ as $\varepsilon \to 0$.

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