Evaluate \[ \int_0^{2\pi} e^{e^{i\theta}}\,d\theta. \]
Solution
1. Under the substitution\[ z=e^{i\theta}, \]the integral becomes\[ \frac1i\oint_{\abs{z}=1}\frac{e^z}{z}\,dz. \]
As \(\theta\) runs from \(0\) to \(2\pi\), the point \(z=e^{i\theta}\) traverses the unit circle once counterclockwise. Moreover, \[ dz=iz\,d\theta, \qquad d\theta=\frac{dz}{iz}. \] Substituting this into the integral gives the displayed contour integral.
2. One has\[ \frac1i\oint_{\abs{z}=1}\frac{e^z}{z}\,dz = 2\pi. \]
By Cauchy’s integral formula, \[ \oint_{\abs{z}=1}\frac{e^z}{z}\,dz = 2\pi i\,e^0 = 2\pi i. \] Multiplying by \(1/i\) gives \(2\pi\).
3. Therefore\[ \boxed{ \int_0^{2\pi} e^{e^{i\theta}}\,d\theta=2\pi }. \]
Combine steps 1 and 2.
4. Q.E.D.
Step 3 is the required evaluation.