Evaluate \(\displaystyle \int_0^{2\pi} e^{e^{i\theta}}\,d\theta\), carefully explaining your solution.
Solution
Use the uniformly convergent exponential series on the unit circle: \[ e^{e^{i\theta}} =\sum_{n=0}^\infty\frac{e^{in\theta}}{n!}. \] Termwise integration is therefore valid, and \[ \int_0^{2\pi}e^{e^{i\theta}}\,d\theta =\sum_{n=0}^\infty\frac1{n!} \int_0^{2\pi}e^{in\theta}\,d\theta. \] Every term with \(n\ge1\) integrates to \(0\), while the \(n=0\) term contributes \(2\pi\). Hence \[ \boxed{\displaystyle \int_0^{2\pi}e^{e^{i\theta}}\,d\theta=2\pi.} \]