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. 
 A: To check your work: if you care about a fast solution, start with considering $I(a)= \int_{-\infty}^{\infty}\frac{e^{-a(x^2-1)}}{x-1}\textrm{dx}$, where by differentiating with respect to $a$ and then integrating back, we're lead to $I(1)=-\sqrt{\pi}\int_0^1 \frac{e^a}{\sqrt{a}}\textrm{d}a=-\pi\operatorname{erfi}(1).$ Hence, we have 
$$\int_{-\infty}^{\infty}\frac{e^{-x^2}}{x-1}\textrm{dx}=-\frac{\pi}{e}\operatorname{erfi}(1).$$
Note: the integrals above have a meaning only in the CPV sense as seen described at https://en.wikipedia.org/wiki/Cauchy_principal_value. 
Literature: related to Faddeeva function (it's sometimes referred to as the plasma dispersion function) - https://en.wikipedia.org/wiki/Faddeeva_function.
A: Near the top of the arc, the integrand blows up exponentially. I would avoid using that arc.

Real Method
By substituting $z\mapsto-z$, we get
$$
\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2}}{z-1}\mathrm{d}z
=-\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2}}{z+1}\mathrm{d}z\tag1
$$
Therefore,
$$
\begin{align}
\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2-1}e^{-2z}}{z}\mathrm{d}z
&=-\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2-1}e^{2z}}{z}\mathrm{d}z\tag2\\
&=-\frac1e\int_{-\infty}^\infty e^{-z^2}\frac{\sinh(2z)}z\,\mathrm{d}z\tag3
\end{align}
$$
Explanation:
$(2)$: substitute $z\mapsto z+1$ on the left and $z\mapsto z-1$ on the right of $(1)$
$(3)$: average the right and left of $(2)$ 
Setting
$$
f(a)=\int_{-\infty}^\infty e^{-z^2}\frac{\sinh(az)}z\,\mathrm{d}z\tag4
$$
we have $f(0)=0$ and
$$
\begin{align}
f'(a)
&=\int_{-\infty}^\infty e^{-z^2}\cosh(az)\,\mathrm{d}z\tag5\\
&=\int_{-\infty}^\infty e^{-z^2}e^{az}\,\mathrm{d}z\tag6\\
&=e^{a^2/4}\int_{-\infty}^\infty e^{-z^2}\,\mathrm{d}z\tag7\\[3pt]
&=\sqrt\pi\,e^{a^2/4}\tag8
\end{align}
$$
Explanation:
$(5)$: take the derivative under the integral
$(6)$: $\cosh(ax)$ is the even part of $e^{ax}$
$(7)$: substitute $z\mapsto z+a/2$
$(8)$: evaluate the integral
Thus,
$$
\begin{align}
\operatorname{PV}\int_{-\infty}^\infty\frac{e^{-z^2}}{z-1}\mathrm{d}z
&=-\frac1e\int_{-\infty}^\infty e^{-z^2}\frac{\sinh(2z)}z\,\mathrm{d}z\tag9\\
&=-\frac{\sqrt\pi}e\int_0^2e^{a^2/4}\,\mathrm{d}a\tag{10}\\
&=-\frac{2\sqrt\pi}e\int_0^1e^{a^2}\,\mathrm{d}a\tag{11}\\[3pt]
&=-\frac\pi{e}\,\operatorname{erfi}(1)\tag{12}
\end{align}
$$
Explanation:
$\phantom{0}(9)$: apply $(3)$
$(10)$: apply $(8)$
$(11)$: substitute $a\mapsto2a$
$(12)$: evaluate the integral

A Cleaner, But Still Real, Approach
$$
\begin{align}
\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-x^2}}{x-1}\,\mathrm{d}x
&=-\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-x^2}}{x+1}\,\mathrm{d}x\tag{13}\\
&=\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-x^2}}{x^2-1}\,\mathrm{d}x\tag{14}\\
&=\left.\frac1e\,\mathrm{PV}\int_{-\infty}^\infty\frac{e^{-a(x^2-1)}}{x^2-1}\,\mathrm{d}x\,\right]_{a=1}\tag{15}\\
&=\frac1e\,\mathrm{PV}\int_{-\infty}^\infty\frac1{x^2-1}\,\mathrm{d}x-\frac1e\int_0^1\int_{-\infty}^\infty e^{-a(x^2-1)}\,\mathrm{d}x\,\mathrm{d}a\tag{16}\\
&=0-\frac1e\int_0^1\sqrt{\frac\pi a}\,e^a\,\mathrm{d}a\tag{17}\\
&=-\frac{2\sqrt\pi}e\int_0^1e^{a^2}\,\mathrm{d}a\tag{18}\\
&=-\frac{2\sqrt\pi}e\frac{\sqrt\pi}2\,\operatorname{erfi}(1)\tag{19}\\[6pt]
&=-\frac\pi e\,\operatorname{erfi}(1)\tag{20}
\end{align}
$$
Explanation:
$(13)$: substitute $x\mapsto-x$
$(14)$: average the left and right sides of $(13)$
$(15)$: set up for differentiation under the integral
$(16)$: write as the integral of the derivative under the integral
$(17)$: the PV can be evaluated using a simple contour integration
$\phantom{(17)\text{:}}$ evaluate the inner integral on the right
$(18)$: substitute $a\mapsto a^2$
$(19)$: evaluate the integral
$(20)$: simplify
which agrees with $(12)$.
A: The contribution from the (superfluous) deformation of the segment on $[0,2]$ to a semicircle is cancelled by other parts of the integral along the real axis.  This deformation crosses no poles, so the path integral along $\Gamma_\epsilon + [0,2]$ is zero.
So what's going on?  In your expression
$$ I+\left(\int_{\Gamma_\epsilon}+\int_{\Gamma_\infty} \right) f(z) \text{,} $$
you have two paths from $0$ to $2$.  One runs along the real axis and is contributed by $I$.  The other runs along $\Gamma_\epsilon$.  It should come as no surprise that this contribution is overcounted in the result you have.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{-z^{2}} \over z - 1}\,\dd z & =
\mrm{P.V.}\int_{-\infty}^{\infty}{\expo{-\pars{z + 1}^{2}} \over z}\,\dd z =
\int_{0}^{\infty}
{\expo{-\pars{z + 1}^{2}} - \expo{-\pars{z - 1}^{2}} \over z}\,\dd z
\\[5mm] & =
\sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty}
\expo{-\pars{z + \sigma}^{2}}\pars{\int_{0}^{\infty}\expo{-zx}\dd x}\,\dd z
\\[5mm] & =
\sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty}\int_{0}^{\infty}
\exp\pars{-\bracks{z^{2} + 2\sigma z + 1 + xz}}\,\dd z\,\dd x
\\[5mm] & =
\sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty}\int_{0}^{\infty}
\exp\pars{-\bracks{z + \sigma + {x \over 2}}^{2} - 1 + \sigma^{2} + \sigma x + {x^{2} \over 4}}\,\dd z\,\dd x
\\[5mm] & =
\sum_{\sigma = \pm 1}\sigma\int_{0}^{\infty}
\exp\pars{\sigma x + {x^{2} \over 4}}\int_{\sigma + x/2}^{\infty}
\exp\pars{-z^{2}}\,\dd z\,\dd x
\\[5mm] & =
\int_{0}^{\infty}\exp\pars{-z^{2}}
\sum_{\sigma = \pm 1}\sigma\int_{0}^{2z - 2\sigma}
\exp\pars{{1 \over 4}\braces{\bracks{x + 2\sigma}^{\, 2} - 4\sigma^{2}}}
\,\dd x\,\dd z
\\[5mm] & =
\expo{-1}\int_{0}^{\infty}\exp\pars{-z^{2}}
\sum_{\sigma = \pm 1}\sigma\int_{2\sigma}^{2z}\exp\pars{x^{2} \over 4}
\,\dd x\,\dd z
\\[5mm] & =
2\expo{-1}\int_{0}^{\infty}\exp\pars{-z^{2}}
\sum_{\sigma = \pm 1}\sigma\int_{\sigma}^{z}\exp\pars{x^{2}}
\,\dd x\,\dd z
\\[5mm] & =
-2\expo{-1}\int_{0}^{\infty}\exp\pars{-z^{2}}
\bracks{\int_{-1}^{z}\exp\pars{x^{2}}\,\dd x -
\int_{1}^{z}\exp\pars{x^{2}}\,\dd x}\,\dd z
\\[5mm] & =
-2\expo{-1}\ \overbrace{\bracks{\int_{0}^{\infty}\exp\pars{-z^{2}}\dd z}}
^{\ds{\root{\pi} \over 2}}\
\overbrace{\int_{-1}^{1}\exp\pars{x^{2}}\,\dd x}
^{\ds{\mbox{Set}\ x = -\ic t}}
\\[5mm] & =
-2\expo{-1}\root{\pi}\int_{0}^{\ic}\exp\pars{-t^{2}}\pars{-\ic}\,\dd t
\\[5mm] & =
\ic\pi\expo{-1}
\bracks{{2 \over \root{\pi}}\int_{0}^{\ic}\exp\pars{-t^{2}}\dd t} =
\bbx{\ic\pi\expo{-1}\mrm{erf}\pars{\ic}}
\end{align}

Note that
  $\ds{\,\mrm{erf}\pars{\ic} = \ic\,\mrm{erfi}\pars{1}}$.

A: $$I=-pv.\int_{-\infty }^{\infty }\frac{e^{-x^2}}{x-1}dx=-\sum_{n=1}^{\infty }\Gamma (\frac{1-2n}{2})\\
\\
\\
=-\sum_{n=1}^{\infty }(-1)^n\frac{2^n.\sqrt{\pi }}{\prod_{n=1}^{\infty }(2m-1)}=-\sum_{n=1}^{\infty }(-1)^n\frac{2^n\sqrt{\pi }}{\frac{2^n.\Gamma (n+\frac{1}{2})}{\sqrt{\pi }}}\\
\\
\\
\therefore I=-\sum_{n=1}^{\infty }(-1)^n\frac{\pi }{\Gamma (n+\frac{1}{2})}=-\sqrt{\pi }\sum_{n=1}^{\infty }(-1)^n\frac{\Gamma (n).\Gamma (\frac{1}{2})}{\Gamma (n)\Gamma (n+\frac{1}{2})}\\
\\
\\
=-\sqrt{\pi }\sum_{n=1}^{\infty }(-1)^n\int_{0}^{1}\frac{(1-x)^{n-1}}{\sqrt{x}(n-1)!}dx\\
\\
\\
\therefore I=-\frac{\pi }{e}.ierf(i)=\frac{\pi }{e}erfi(1$$
