On page 81 of Stein and Shakarchi's Complex Analysis, there is an inequality, \begin{equation} \left\lvert \int_{A_{R}} f \right\rvert \leq \int_{0}^{2\pi}\left\lvert\frac{e^{a(R+it)}}{1+e^{R+it}} \right\rvert {\rm{d}}t \leq Ce^{(a-1)R}, \end{equation} where $f(z)=e^{az}/(1+e^z)$ and $A_R$ is the vertical path from $z=R\in\mathbb{R}$ to $z=R+2\pi i\in \mathbb{C}$.
Can any one demonstrate how the second step is obtained? I appreciate the help.
First note that in the book we will let $R\to \infty$, so we assume that $R$ is large. In particular, we let $2 < e^R$.
Note that by triangle inequality
$$e^R = \left\lvert e^{R+it} \right\rvert = \left\lvert 1+ e^{R+it} - 1\right\rvert \le \left\lvert 1+ e^{R+it}\right\rvert + 1 \le \left\lvert 1+ e^{R+it}\right\rvert + \frac 12 e^R$$
This implies $$ e^R \le 2 \left\lvert 1+ e^{R+it}\right\rvert$$ and so $$\left\lvert\frac{e^{a(R+it)}}{1+e^{R+it}} \right\rvert = \frac{e^{aR}}{\left\lvert 1+ e^{R+it} \right\rvert}\le 2\frac{e^{aR}}{e^R} = 2 e^{(a-1)R} . $$ So after the integration, you have the desired inequality with $C = 4\pi$.