Let $f(z)$ be analytic in the upper half plane $\operatorname{Im} \geq z$, except for a finite number of poles on the real axis.
For certain integrals $I :=\int_{-\infty}^{\infty} f(x) dx$, we can use contour integration, along the real interval $(-R,R)$ as $R\to \infty$ and a semicircle $\Gamma$ joining the ends of this interval. Our functions $f(x)$ are chosen such that the contribution along $\Gamma$ can be shown to vanish as $R\to \infty$; leaving us to use the residue theorem to compute $I$.
If we introduce a simple pole $z_0$ along the real axis, we can treat it as a separate case by excluding it from our initial contour by defining the principal value $P$
$$ P \equiv \lim_{\rho \to 0}\ \int_{-\infty}^{z_0-\rho} f(x)dx\ + \int_{z_0+\rho}^{\infty} f(x)dx \ . $$
So far, so good.
Our contour $\mathcal C$ now consists of $\Gamma$ and $P$, with an open end at the pole. We close $\mathcal C$ by introducing an indentation in the form of a semicircle $\gamma$ of radius $\rho$ around $z_0$ in the upper half plane, where $\rho$ is the same as in $P$.
I'm wondering why the text requires $\gamma$ to be in the upper half-plane.
My guess is that it's because we want to exclude $z_0$ from the interior of $\mathcal C$; so we can use the residue theorem to compute the contribution of $P$ and $\gamma$ to $\oint_{\mathcal C} f(z)dz$.
But, say, instead, we require $\gamma$ to be in the lower half-plane, thereby including $z_0$ in the interior of $\mathcal C$. Accordingly, the residue theorem now includes the contribution of $z_0$ to $I$, without the need to consider the separate case of $\gamma$.
But, since we defined our $f$ to be analytic only in the upper half-plane, the residue theorem doesn't need to apply for my previous line.
Supplement.
The contribution of $\gamma$, $\int_{\gamma}f(z)dz$, can be shown to equal $-ia_{-1}\pi$, as $\rho \to 0$.
TL;DR: I'm wondering why the text requires $\gamma$ to be in the upper half-plane.