EXERCISE: Let $C_p$ is a semi-circle $\{z\in \mathbb{C}:\left| {z - 1} \right| = p,\operatorname{Im}(z)>0\}$, described in the counterclockwise direction. Prove that: $$I=\mathop {\lim }\limits_{p \to {0^ + }} \int\limits_{{C_p}} {\left( {\frac{1}{{z - 1}} + \frac{{{e^z}}}{{z + 1}}} \right)d} z = \pi i$$
if $C_p$ is a circle $z\in \mathbb{C}:\left| {z - 1} \right| = p$, I can prove that $I=\operatorname{Res}(f,1)=2\pi i$. But $C_p$ is a semi-circle, can some one help me?
On
For any $p > 0$, $$ \int_{C_p} \frac{1}{z - 1} \, dz = \pi i $$ can be evaluated directly, with a parametrization $z = 1 + pe^{it}$, $0 \le t \le pi$, of the semi-circle. And $$ \lim_{p \to 0^+} \int_{C_p} \frac{e^z}{z + 1} \, dz = 0 $$ because the integrand is continuous (and therefore bounded) in a neighborhood of $z=1$, and the length of $C_p$ approaches zero for $p \to 0$.
Let $f$ be a primitive of $\dfrac{e^z}{z+1}$ near $1$. Then\begin{align}\lim_{p\to0^+}\int_{C_p}\left(\frac1{z-1}+\frac{e^z}{z+1}\right)\,\mathrm dz&=\lim_{p\to0^+}\left(\int_{C_p}\frac{\mathrm dz}{z-1}+\int_{C_p}f'(z)\,\mathrm dz\right)\\&=\lim_{p\to0^+}\left(\int_0^\pi\frac{(1+pe^{it})'}{1+pe^{it}-1}\,\mathrm dt+f(1-p)-f(1+p)\right)\\&=\lim_{p\to0^+}\left(\int_0^\pi i\,\mathrm dt+f(1-p)-f(1+p)\right)\\&=\lim_{p\to0^+}\bigl(\pi i+f(1-p)-f(1+p)\bigr)\\&=\pi i+f(0)-f(0)\\&=\pi i.\end{align}