# Integrating $\int_\Gamma exp(z) \,dz$ along the upper half unit circle directly

Let $\Gamma$ be the upper half of the unit circle $|z| = 1$, directed from $z = 1$ to $z = -1$. One way to evaluate $\int_\Gamma exp(z) \,dz$ is to apply the the Fundamental Theorem of Calculus. Indeed, noting that $exp(z)$ is entire and $exp '(z) = exp(z)$ for all $z \in \mathbb{C}$, we have:

$\int_\Gamma exp(z) \,dz = exp(-1) - exp(1) = \frac{1 - e^2}{e}$.

Alternatively if we parameterise $\Gamma$ as $z(\theta) = e^{i \theta}$, $0 \leq \theta \leq \pi$ and attempt to address the integral directly we have:

$\int_\Gamma exp(z) \,dz = i \int_0^\pi exp(e^{i\theta})e^{i\theta} = i \int_0^\pi e^{cos(\theta)+i\theta}(cos(sin(\theta)) + i sin(sin(\theta)) \,d\theta$.

Is there a way in which we can evaluate this integral simply? If not, is there a better parameterisation that will allow us to evaluate $\int_\Gamma exp(z) \,dz$ without appealing to FToC?

(Note: $exp(x + i y) = e^x(cos(y) + i sin(y))$ denotes the complex exponential function.)

Starting with $i\int_0^\pi e^{e^{i\theta}}e^{i\theta}d\theta$, let $u=e^{i\theta}$. Then we have

$$i\int_0^\pi e^{e^{i\theta}}e^{i\theta}d\theta=\int_1^{-1}e^udu=e^{-1}-e.$$

Since $e^z$ is entire, we can deform the contour of integration from the upper half circle of radius 1 in counterclockwise direction to the real axis from 1 to -1.

You are better off just leaving things alone:

$$\int_{\Gamma} dz \, e^z = i \int_0^{\pi} dt \, e^{i t} \, e^{e^{i t}}$$

Note that, no matter what the parametrization, we will be using the FToC anyhow. By expanding the exponentials, you are simply overcomplicating the problem.

Because $e^z$ is analytic, it doesn't matter what parametrization we use. The answer only depends on the endpoints, thus:

$$e^{-1}-e^1$$