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

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    $\begingroup$ There is something wrong. The result shall be $2\pi$ $\endgroup$ – Von Neumann May 22 at 13:19
  • $\begingroup$ In my attampt I tried to solve something similar with the hope it will help somehow. This is a different integral. @VonNeumann $\endgroup$ – J. Doe May 22 at 13:20
  • $\begingroup$ Wait, is your work for your integral or for a different problem? $\endgroup$ – cmk May 22 at 13:22
  • $\begingroup$ For the integral @cmk $\endgroup$ – J. Doe May 22 at 13:22
  • $\begingroup$ The work you showed is not the same as the problem you were given. They're clearly different integrals. $\endgroup$ – cmk May 22 at 13:23
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On unit circle $$\int_0^{2\pi} e^{e^{it}} dt=\int_{|z|=1} e^{z} \dfrac{dz}{iz}=\dfrac{1}{i}\int_{|z|=1} \dfrac{e^{z} dz}{z}=2\pi$$

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    $\begingroup$ Why does $\int _{|z|=1}{e^z \over z} dz = 2\pi i$? $\endgroup$ – J. Doe May 22 at 19:12
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    $\begingroup$ @J.Doe By Cauchy integral formula or residue theorem. $\endgroup$ – Nosrati May 22 at 19:20
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You can do it without any complex analysis, starting with $$e^{e^{it}} =\sum_{n=0}^\infty\frac{e^{int}}{n!}.$$

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See this post for a similar question.

If $f(z) = e^{z}$ (a function whose Maclaurin series has an infinite radius of of convergence), we get $$ \text{Re} \int_{0}^{2 \pi} e^{e^{i \theta}} \, d \theta = \int_{0}^{2 \pi} e^{\cos \theta} \cos (\sin \theta) \, d \theta = 2 \pi (1) = 2 \pi.$$

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