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

Question

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

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 '19 at 13:20
Wait, is your work for your integral or for a different problem? — cmk, May 22 '19 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 '19 at 13:23
Ok. I just don't know how to calculate it so I tried something different and now I see it doesn't help. I'm looking for some other direction. — J. Doe, May 22 '19 at 13:35

score 3 · Answer 1 · answered May 22 '19 at 15:33

3

You can do it without any complex analysis, starting with $$e^{e^{it}} =\sum_{n=0}^\infty\frac{e^{int}}{n!}.$$

answered May 22 '19 at 15:33

David C. Ullrich

89,985

score 2 · Accepted Answer · answered May 22 '19 at 15:01

2

On unit circle $$\int_0^{2\pi} e^{e^{it}} dt=\int_{|z|=1} e^{z} \dfrac{dz}{iz}=\dfrac{1}{i}\int_{|z|=1} \dfrac{e^{z} dz}{z}=2\pi$$

answered May 22 '19 at 15:01

Nosrati

29,995

1

Why does $\int _{|z|=1}{e^z \over z} dz = 2\pi i$? – J. Doe May 22 '19 at 19:12
2

@J.Doe By Cauchy integral formula or residue theorem. – Nosrati May 22 '19 at 19:20

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

2 Answers2