Let us define a curve $\gamma$ (orientated counterclockwise) as the boundary of the region $\{s \in \mathbb C : |s| ≤ R, \sigma > -\delta\}$ where $s=\sigma+it$.
$\gamma_- = \gamma \cap \{σ < 0\}$.
$\gamma'_{-} = \{|s| = R, σ < 0\}$ (a semicircle)
What is the difference between $\gamma_- $ and $\gamma'_{-}$? Is $\gamma_- = \gamma'_{-}$? If so, why is $\gamma_- = \gamma'_{-}$?
The source of confusion is this M.Sc. thesis (go to page 113, 115). I am trying to understand the equation obtained by integration of closed path of $\gamma_- $ and $\gamma'_{-}$ respectively, I could not understand why the equation is true.
The curves $\gamma_-$ and $\gamma'_-$ are not the same. See the picture on p.113: $\gamma_-$ is the part of $\gamma$ in the closed half-plane $\text{Im}(z) \le 0$, and $\gamma'_-$ is the half circle with center $0$ and radius $R$ in this half-plane. But note that the curves agree on two small parts at their beginning at $-R$ and at their end at $R$ (both points are on the real axis):
However, the author only claims that the integrals over these two curves agree. This is true because the difference of the intergrals is an integral over a closed curve completetely contained in the open half-plane $\text{Im}(z) < 0$. We integrate a function which is holomorphic on the open half-plane, thus the integral is $0$.
Edited:
The claim is $\int_{\gamma_-} = \int_{\gamma'_-}$. This is the same as $\int_{\gamma_- - \gamma'_-} = \int_{\gamma_-} - \int_{\gamma'_-} = 0$. Here $\varphi = \gamma_- - \gamma'_-$ is the closed curve obtained by going with $\gamma_-$ from $-R$ to $R$ and with $-\gamma'_-$ (which is the path in opposite direction) from $R$ to $-R$. By Theorem 2.5.3 you see that $\int_\varphi = 0$ because $g_T(s)e^{sT}(1 +s^2/R^2)$ is an entire function.