Integration of $\exp(-z^2)/(z-z_0)$ I am interested into the following integral
$I = \int_{-\infty}^{+\infty} \frac{f(z)}{z-z_0} dz$
where $f$ is typically a Gaussian function $e^{-z^2}$.
I am tempted to simply use Cauchy formula to get
$I = 2\pi i f(z_0) = 2\pi i \ e^{-z_0^2}$
But I am not sure that the contribution of the half circle containing $z_0$ goes to $0$ as the radius of the circle $R$ goes to $+\infty$ and I could not find a description of the analytic properties of the complex Gaussian to justify this integration on the internet. On top of that, Mathematica gives me $0$ as a result too..
Can the integral be carried out like written or is not so simple? Thank you.


One can deform the real line contour in the complex plane, but it is rather dubious that such a defomation facilitates evaluation of the integral of interest via Cauchy's Integral Theorem.  However, we can use Feynman's trick to evaluate the integral in terms of the Imaginary Error Function.  To that end we now proceed.


We first note that 
$$\begin{align}
\int_{-\infty}^\infty \frac{e^{-z^2}}{z-z_0}\,dz&=2z_0 \int_0^\infty \frac{e^{-x^2}}{x^2-z_0^2}\,dx\tag1
\end{align}$$

Second, let $I(a)$ be defined as
$$\begin{align}
I(a)&=2z_0 \int_0^\infty \frac{e^{-ax^2}}{x^2-z_0^2}\,dx\tag2
\end{align}$$
Observe that $I(1)$ is simply the integral of interest in $(1)$.

Third, differentiate $(2)$ to find 
$$\begin{align}
I'(a)&=-2z_0 \int_0^\infty \frac{x^2\,e^{-ax^2}}{x^2-z_0^2}\,dx\\\\
&=-2z_0 \int_0^\infty \frac{(x^2-z_0^2+z_0^2)\,e^{-ax^2}}{x^2-z_0^2}\,dx\\\\
&=-2z_0\int_0^\infty e^{-ax^2}\,dx-z_0^2I(a)\\\\
I'(a)+z_0^2I(a)&=-z_0\sqrt{\pi/a}\tag3
\end{align}$$

Now, solving the ODE in $(3)$ for $I(a)$ we find that
$$I(a)=I(0)e^{-az_0^2}-2\sqrt{\pi}e^{-az_0^2}\int_0^{\sqrt{a}z_0}e^{x^2}\,dx\tag4$$
Using the initial condition $I(0)=i\pi\,\text{sgn}\left(\text{Im}\left(z_0\right)\right)$ and setting $a=1$ in $(4)$ yields the coveted result
$$\begin{align}
\int_{-\infty}^\infty \frac{e^{-z^2}}{z-z_0}\,dz&=i\pi\,\text{sgn}\left(\text{Im}\left(z_0\right)\right)e^{-z_0^2}-2\sqrt{\pi}e^{-z_0^2}\int_0^{z_0}e^{x^2}\,dx\\\\
&=\bbox[5px,border:2px solid #C0A000]{i\pi\,\text{sgn}\left(\text{Im}\left(z_0\right)\right)e^{-z_0^2}-\pi e^{-z_0^2}\text{erfi}(z_0)}\tag5
\end{align}$$
where $\text{sgn}(x)$ is the Sign Function and $\text{erfi}(z)$ is the Imaginary Error Function.

EXAMPLES:
As an example, if $z_0=2+i5$, the result is $i\pi e^{21-20i}-\pi e^{21-20i} \text{erfi}(2+i5)$, which is verified using Wolfram Alpha (See Here).
And if $z_0=2-i5$ the result is $-i\pi e^{21+20i}-\pi e^{21+20i} \text{erfi}(2-i5)$, which was also verified using Wolfram Alpha.
Thus, we see that the sign of the imaginary part of the answer depends on the sign of $\text{Im}(z_0)$.

Note that if $z_0$ is purely real, we can evaluate $(1)$ as a principal value.  Setting $z_0$ to $x_0\in \mathbb{R}$ in $(5)$ shows that 
$$\text{PV}\int_{-\infty}^\infty \frac{e^{-z^2}}{z-x_0}\,dz=-\pi e^{-x_0^2}\text{erfi}(x_0)$$
