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
On
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
$$ \int_0^{2\pi} e^{e^{it}} e^{it} dt = -i \int_{S^1} f(z) dz $$
Writing $z = e^{it}$, you can recognize this as
$$\int_{\gamma} e^z \, \frac{dz}{i}$$
where $\gamma$ is the unit circle oriented counterclockwise.