A question on the Zeta function of Riemann I am reading a paper "A note on the Riemann zeta function" by F. T. Wang, Bull. Amer. Math. Soc $(1946)$.
Let $K$ be the unit semi circle with center $z=0$ lying in the right half plane $\Re(z)>0$ and let $C$ be the broken line consisting of three segments $L_1$ $(0\leq x\leq T, y=-T)$, $L_2$ $(0\leq x\leq T, y=-T)$ and $L_3$ $(x=T, -T\leq y\leq T)$.
Let $\Gamma$ be a contour describing $C$, $K$ and the corresponding part of the imaginary axis, and let $\rho_v=\beta_v+i\gamma_v$ (the zeros of the zeta function $\zeta(1/2+z)$ whose real part $0\leq \beta_v<1/2$) be a point interior to $\Gamma$, and $\log(z-\rho_v)$ be taken as its principal value. We write $C_1$ as a contour describing $\Gamma$ in positive direction to the point $i\gamma_v$, then along a segment $y=\gamma_v$, $0<x<\beta_v-r$, and describing a small circle with center $z=\rho_v$, radius $r$ , then going back along the negative side   of the segment to $i\gamma_v$, and then along $\Gamma$ to the starting point.
By Cauchy's theorem we get $$\int_{C_1}\frac{\log(z-\rho_v)}{z^2}dz=0 \tag{1}$$
Hence $$\frac{1}{2\pi i} \int_{\Gamma}\frac{\log(z-\rho_v)}{z^2}dz=-\int_{0}^{\beta_v}\frac{dx}{(x+i\gamma_v)^2}\tag{2}$$
where the integral around the small circle with center $z=\rho_v$, radius $r$, tends to zero as $r\to 0$.

Question: How did we get equation $(2)$? Also I am struggling to understand the above highlighted portion which describes the contour $C_1$

If possible please draw the contour. Any insights will be welcome. Thank you!

For $\Gamma$ and $C_1$ see Fig.1, 2 below.

Let $L^+$, $L^-$ be the upper edge and the lower edge of the segment $y=\gamma_v$,$0<x<\beta_v$ respectively.
Since $$
0=\int_{C_1}\frac{\log(z-\rho_v)}{z^2}dz=\int_\Gamma \frac{\log(z-\rho_v)}{z^2}dz+\int_{L^+} \frac{\log(z-\rho_v)}{z^2}dz+\int_{L^-} \frac{\log(z-\rho_v)}{z^2}dz,
$$
we have $$
\int_\Gamma \frac{\log(z-\rho_v)}{z^2}dz=-\int_{L^+} \frac{\log(z-\rho_v)}{z^2}dz-\int_{L^-} \frac{\log(z-\rho_v)}{z^2}dz.
$$
The value of $\log(z-\rho_v)$ on $L^+$ is $\log|x+i\gamma_v|+i\pi$ and the value on $L^-$ is $\log|x+i\gamma_v|-i\pi$. So we get
$$
\int_{L^+} \frac{\log(z-\rho_v)}{z^2}dz=\int_0^{\beta_v} \frac{\log|x+i\gamma_v|+i\pi}{(x+i\gamma_v)^2}dx$$
and$$ 
\int_{L^-} \frac{\log(z-\rho_v)}{z^2}dz=\int_{\beta_v}^0 \frac{\log|x+i\gamma_v|-i\pi}{(x+i\gamma_v)^2}dx.$$
Therefore we have $$
\int_{L^+} \frac{\log(z-\rho_v)}{z^2}dz+\int_{L^-} \frac{\log(z-\rho_v)}{z^2}dz=2\pi i \int_0^{\beta_v} \frac{dx}{(x+i\gamma_v)^2},$$
which leads $(2)$.
