integration - Arc contribution in $int_{-infty}^infty mathrm{d}z frac{e^{-z^2}}{z-1}$

Consider an improper integral with a pole on the integration contour at say $z=1$,

$$

\tag{1} I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{e^{-z^2}}{z-1+i\epsilon},~~~~~\epsilon>0.
$$
Let $$f(z) = \frac{e^{-z^2}}{z-1+i\epsilon}$$
then
$$
\sum_{residues~inside~\Gamma} = 0 = \oint_\Gamma f(z) = I+\left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z),
$$
where the total contour is $\Gamma\equiv (-R,R)+\Gamma_\epsilon+\Gamma_\infty$ with $R\rightarrow \infty$.

Thus

$$
I = - \left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z).
$$
The contour $\Gamma_\epsilon$ is a semicircle centered about $z = 1$ of radius $\epsilon$. Its contribution is given by
$$
\int_{\Gamma_\epsilon} \mathrm{d}z ~f(z) = i (\theta_2-\theta_1)~ \mathrm{Res}(f;z=1) = \frac{-i\pi}{e}.
$$

Evaluating $(1)$ in Mathematica and taking the $\epsilon\rightarrow 0 $ limit gives
$$

I = e^{(\epsilon +i)^2} \left(-\pi \text{erfi}(1-i \epsilon )+\log (-1+i \epsilon )+\log
\left(\frac{i}{\epsilon +i}\right)-2 i \pi \right)
\\
\longrightarrow
-\frac{\pi (\text{erfi}(1)+i)}{e}~~~(\text{as}~ \epsilon \rightarrow 0).
$$

Thus apparently,
$$ \tag{2}
\int_{\Gamma_\infty} \mathrm{d}z \, f(z) = \frac{2\pi i}{e}+\frac{\mathrm{erfi}(1)}{e}.

$$

Can anyone derive this contribution from the semicircle at infinity? I.e. is $(2)$ correct and how about generalizations of $(1)$ to integrals of the form

$$\tag{3}
I = \int_{-\infty}^\infty \mathrm{d}z\ \frac{z^n e^{-z^2}}{(z-a+i\epsilon)(z-b-i\epsilon)},~~~~~\epsilon>0,a,b\in\mathbb{R},n\in \mathbb{N}.
$$

---

Note, erfi is defined as
$\mathrm{erfi}(z) \equiv \mathrm{erf}(iz)/i$ with the familiar error function.