# Contour integration help: $\int_{\gamma}e^zdz$ with $\gamma(\theta)$=$e^{i\theta}$

I am stuck integrating $$\int_{\gamma}e^zdz$$ with $\gamma(\theta)$=$e^{i\theta}$ and $0\le\theta\le2\pi$. I got up to $$\int_{\gamma}e^zdz = \int_{0}^{2\pi}e^{e^{i\theta}}\cdot ie^{i\theta}d\theta = \int_{0}^{2\pi}e^{\cos\theta+i\sin\theta}\cdot i(\cos\theta+i\sin\theta)d\theta.$$

Thanks.

-
Have you seen the Cauchy Integral Theorem?? – ncmathsadist Dec 1 '11 at 20:04
Ray: could you explain the meaning of your Edit and why @Sasha's post does not answer your needs? – Did Dec 2 '11 at 12:43

The integrand $\mathrm{e}^z$ is holomorphic for $\vert z \vert \le 1$ (check that), therefore the integral vanishes by the Cauchy integral theorem (wiki).
Now let's look at it like you do: $$\begin{eqnarray} \int_0^{2 \pi} \mathrm{e}^{\cos(\theta)} \cos(\theta + \sin(\theta)) \mathrm{d} \theta &=& \int_0^{2 \pi} \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right) \mathrm{d} \theta = \left. \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right|_0^{2\pi} = 0 \end{eqnarray}$$ Indeed: $$\begin{eqnarray} \mathrm{e}^{\cos(\theta)} \cos(\theta + \sin(\theta)) &=& \mathrm{e}^{\cos(\theta)} \cos(\theta) \cos(\sin(\theta)) - \mathrm{e}^{\cos(\theta)} \sin(\theta) \sin(\sin(\theta) \\ &=& \mathrm{e}^{\cos(\theta)} \cdot \frac{\mathrm{d} \sin(\sin(\theta))}{\mathrm{d} \theta} + \frac{\mathrm{d} \mathrm{e}^{\cos(\theta)}}{\mathrm{d} \theta} \cdot \sin(\sin(\theta)) \\ &=& \frac{\mathrm{d}}{\mathrm{d} \theta} \left( \mathrm{e}^{\cos(\theta)} \sin(\sin(\theta)) \right) \end{eqnarray}$$