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in an excersise I have to compute a few line integrals but one of them I can't solve. It is not even written as a line integral but the others are. I am talking about:

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

Is there any possibility to write that integral in the form $\int_{\gamma}f(z)dz$ with a line $\gamma:[0,2\pi]\rightarrow\mathbb{C}$? Maybe then I can use some results of Cauchy.

Thanks and regards N.Sch

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  • $\begingroup$ this is as the problem is given? $\endgroup$ – qbert May 4 '16 at 20:52
  • $\begingroup$ The problem is solving the integral with methods from our complex analysis lecture. So i thought it makes sense to write it as a line integral. $\endgroup$ – user337060 May 4 '16 at 20:54
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Writing $z = e^{it}$, you can recognize this as

$$\int_{\gamma} e^z \, \frac{dz}{i}$$

where $\gamma$ is the unit circle oriented counterclockwise.

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  • $\begingroup$ Yes, that helps. And because $C$ is star-shaped and exp/i holomorphic, the integral is 0, right? $\endgroup$ – user337060 May 4 '16 at 21:08
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    $\begingroup$ Yes. ${}{}{}{}{}$ $\endgroup$ – user296602 May 4 '16 at 21:09
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Let $f = \exp$ be the exponential function, and consider the integral of this function over the unit circle $S^1$. This circle can be parameterized by the curve

$$ \gamma(t) = e^{it} $$

It then follows that

$$ \int_{S^1} f(z) dz = \int_0^{2 \pi} (f \circ \gamma)(t) \gamma'(t) dt = i \int_0^{2\pi} e^{e^{it}} e^{it} dt $$

So

$$ \int_0^{2\pi} e^{e^{it}} e^{it} dt = -i \int_{S^1} f(z) dz $$

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