I don't understand the following example.
The second term on the right-hand side is $\pi$, since $$\lim \limits _{r \to 0} \int \limits _{\gamma _r} \frac {\Bbb e ^{\Bbb i z} - 1} {z^2} \Bbb d z = - \frac 1 2 2 \pi \Bbb i \ \text{Res}(f, 0) = -\pi \Bbb i ^2 = \pi. $$
Here, $\gamma_r$ denotes the upper semicircle from $-r$ to $r$.
I think that by the residue theorem, $\displaystyle\int \limits _{\gamma_r} \dfrac {\Bbb e ^{\Bbb i z} - 1} {z^2} \Bbb dz = 2 \pi \Bbb i \ \text{Res} (f,0)$, where $f(z)=\dfrac {\Bbb e ^{\Bbb i z} - 1} {z^2}$, and $\text{Res} (f,0) = \Bbb i$. Would anyone explain me where does that $-\dfrac 1 2$ come from?
Thank you.
The minus comes from the orientation of the half-circle: it is from $-r$ to $r$, which means counter-clockwise. The $\dfrac 1 2$ comes from the fact that the integral is computed only on a half-circle; to use the residue theorem, you have to add the lower half-circle in order to get a closed contour.