# Calculating $\int_0^{2\pi} e^{e^{i \theta}} d\theta$

The question:

Calculate $$\int_0^{2\pi} e^{e^{it}} dt$$.

My attempt:

Notice that $$e^{it}$$ is the unit circle and we seek the integral of the image of that circle of the exponent function. I found something similar but it didn't helped. Maybe it can be improved:

Let $$\gamma$$ be the path $$z(t)=e^{it}$$, $$t\in[0,2\pi]$$ and $$f(z)=e^z$$. $$\int _\gamma f(z)dz=\int _0^{2\pi}f(z(t))\cdot\dot{z}(t)= \int_0^{2\pi}e^{e^{it}}ie^{it}dt \\ [u=e^{it}, du=ie^{it}dt] \\=\int_1^1e^udu=0$$

• There is something wrong. The result shall be $2\pi$ – Von Neumann May 22 at 13:19
• In my attampt I tried to solve something similar with the hope it will help somehow. This is a different integral. @VonNeumann – J. Doe May 22 at 13:20
• Wait, is your work for your integral or for a different problem? – cmk May 22 at 13:22
• For the integral @cmk – J. Doe May 22 at 13:22
• The work you showed is not the same as the problem you were given. They're clearly different integrals. – cmk May 22 at 13:23

On unit circle $$\int_0^{2\pi} e^{e^{it}} dt=\int_{|z|=1} e^{z} \dfrac{dz}{iz}=\dfrac{1}{i}\int_{|z|=1} \dfrac{e^{z} dz}{z}=2\pi$$
• Why does $\int _{|z|=1}{e^z \over z} dz = 2\pi i$? – J. Doe May 22 at 19:12
You can do it without any complex analysis, starting with $$e^{e^{it}} =\sum_{n=0}^\infty\frac{e^{int}}{n!}.$$
If $$f(z) = e^{z}$$ (a function whose Maclaurin series has an infinite radius of of convergence), we get $$\text{Re} \int_{0}^{2 \pi} e^{e^{i \theta}} \, d \theta = \int_{0}^{2 \pi} e^{\cos \theta} \cos (\sin \theta) \, d \theta = 2 \pi (1) = 2 \pi.$$