Let $f(z)=\frac{e^{e^{iz}}}{z}$ and let $\Gamma$ be the semicircle of radius $R$ in the upper half plane. Evaluate $\lim_{R\to\infty}\int_\Gamma f(z)dz$.
I managed to show that
$$\lim_{R\to\infty}\left|\int_\Gamma f(z)dz\right|\leq\lim_{R\to\infty}\int_0^\pi e^{e^{-R\sin(\theta)}\cos(R\cos(\theta))}d\theta=\pi.$$
Is it possible that the integral goes to 0 in the limit and if so, how would you show that?
Best Answer
No, but it is possible to show that the integral converges to $i\pi$:
$$ \int_{\Gamma_R} \frac{e^{e^{iz}}}{z} \, \mathrm{d}z = i \int_{0}^{\pi} \exp( e^{-R\sin\theta + iR\cos\theta}) \, \mathrm{d}\theta \quad \xrightarrow[R\to\infty]{\text{DCT}} \quad i \int_{0}^{\pi} \exp(0) \, \mathrm{d}\theta = i\pi. $$
As a corollary, we can verify that
$$ \mathrm{PV}\! \int_{-\infty}^{\infty} \frac{e^{e^{ix}}}{x} \, \mathrm{d}x = (e - 1)\pi i. $$
(This can also be proved by a purely real-analytic method, though.)