In order to prove the normalisation property of a Lorentzian function,
$L = \dfrac{1}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{b}{(z-a)^2+b^2} dz = 1$
we take a closed contour on the upper-half complex plane. This means we only consider the $z=a+ib$ pole when finding residues. I know this has to do with the winding number, but can you give a more physical explanation of why we do this?
Do we still use contours that cover the upper-half plane when $f(z)$ in
$F = \dfrac{1}{\pi}\displaystyle \int_{-\infty}^\infty \dfrac{b\,f(z)}{(z-a)^2+b^2} dz$
contains poles at say, $z=-c+id$ where $d=(2n+1)\pi i, n\in \mathbb{Z}$?
In your first example,
$$L=\frac1{\pi} \int_{-\infty}^{\infty} dx \frac{b}{(x-a)^2+b^2} = 1 $$
you could just as easily close a contour in the lower half-plane. In that case, though, the winding number is $-1$ rather than $1$. That is, assuming $b \gt 0$,
$$\int_{-\infty}^{\infty} dx \frac{b}{(x-a)^2+b^2} = - i 2 \pi \operatorname*{Res}_{z=a-i b} \frac{b}{(z-a)^2+b^2} = \pi$$
In your more general example, it depends on the behavior of $|f(z)|$ on the semicircular arc of radius $R$ as $R \to \infty$ in the upper half plane. For example, if $f$ is periodic on the real line, it tends to be exponential on the imaginary axis. If that exponential behavior is increasing rather than decreasing, then you cannot close in that half plane to determine the value of the integral.