# computation of a line integral

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

• this is as the problem is given? – qbert May 4 '16 at 20:52
• 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. – user337060 May 4 '16 at 20:54

Writing $z = e^{it}$, you can recognize this as

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

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

• Yes, that helps. And because $C$ is star-shaped and exp/i holomorphic, the integral is 0, right? – user337060 May 4 '16 at 21:08
• Yes. ${}{}{}{}{}$ – user296602 May 4 '16 at 21:09

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

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