Path integral on $\mathbb C\setminus(-\infty,-1]$

Question

Let $\gamma$ a path such that $\gamma(0) = -i$ and $\gamma(1) = i$. Moreover, $\gamma(t) \in \mathbb{C}\setminus(-\infty,-1] \ \forall t \in [0,1]$.

How can I compute $\displaystyle\int_\gamma \! \frac{1}{z+1} \, \mathrm{d}z$?

I rewrite $$\int_\gamma \! \frac{1}{z+1} \, \mathrm{d}z = \int_0^1 \! \frac{\gamma'(t)}{\gamma(t)+1} \, \mathrm{d}t$$ How can I proceed from here?

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{\gamma}{\dd z \over z + 1} & \,\,\,\stackrel{z\ =\ \ic t}{=}\,\,\, \int_{-1}^{1}{\ic\,\dd t \over \ic t + 1} = \ic\int_{-1}^{1}{1 - \ic t \over 1 + t^{2}}\,\dd t = 2\ic\int_{0}^{1}{\dd t \over 1 + t^{2}} = 2\ic\,{\pi \over 4} = \bbx{{\pi \over 2}\,\ic} \end{align}