Limit as $r$ tends to zero of integral $\int_C \frac{e^{iz}-1}z \mathrm dz$

Question

Let $\mathcal C$ be a semi-circle of center $O(0,0)$ and radius $R$, such that $y \ge 0$. Find the limit as $R$ tends to zero of:

$$\int_{\mathcal C} \frac{e^{iz}-1}z \mathrm dz$$

How can I find this?

Answer 1

Note that $$\bigl|e^{iz}-1\bigr|=2\left|\sin{z\over2}\right|\ .$$It follows that $$\left|\int_{\cal C}{e^{iz}-1\over z}\>dz\right|\leq\int_{\cal C}\left|{\sin(z/2)\over z/2}\right|\>|dz|\ .$$ As $\lim_{z\to0}{\sin z\over z}=1$ and ${\rm length}({\cal C})\to0$ when $R\to 0+$ the right hand side tends to $0$ when $R\to 0+$.