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

Klangen
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J. Doe
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    There is something wrong. The result shall be $2\pi$ –  May 22 '19 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 '19 at 13:20
  • Wait, is your work for your integral or for a different problem? – cmk May 22 '19 at 13:22
  • For the integral @cmk – J. Doe 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

2 Answers2

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

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

Nosrati
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