Solve an Integral one open and one complex limit If we pose an integral
$2ie^{-\zeta^2}\int\limits_{-\infty}^{i\zeta} e^{-x^2} dx$
where $\zeta$ is a complex number. For imaginary $\zeta$ this comes down to the error function.
I would appreciate help on how to treat it for complex $\zeta$ and is there any suitable approximation if $Re(\zeta)\to0$
Thank you in advance...
 A: The integral can be written in terms of the error function as
$$\int_{-\infty}^{i\zeta} e^{-t^2}dt=\frac{\sqrt{\pi}}{2}(1+erf(i \zeta))$$
If use of the error function with complex arguments is unacceptable, then we will proceed along a different line.
Write $i\zeta=x+iy$, where $x$ and $y$ are real numbers.  Then, we have that
$\int_{-\infty}^{i\zeta} e^{-t^2}dt=\int_{-\infty}^{x+iy} e^{-t^2}dt$
Here, we can take the integration path to be from $-\infty$ to $x$ along the real axis and from $x$ to $x+iy$ parallel to the imaginary axis.  Then, we can write
$$\begin{align}
\int_{-\infty}^{i\zeta} e^{-t^2}dt&=\int_{-\infty}^{x} e^{-t^2}dt+\int_{x}^{x+iy} e^{-t^2}dt=\\\\
&\frac{\sqrt{\pi}}{2}(erf(x)+1)+\int_{x}^{x+iy} e^{-t^2}dt
\end{align}$$
The integral on the right-hand side can be re-written in a number of ways.  Let's make the substitution $t=x+is$ so that $dt=ids$ and the limits of integration go from $s=0$ to $s=y$.  Then
$$\begin{align}
\int_{x}^{x+iy} e^{-t^2}dt&=i\int_{0}^{y} e^{-(x+is)^2}ds\\\\
&=e^{-x^2}\int_0^y e^{s^2}(\sin(2xs)+i\cos(2xs))ds\\\\
&=e^{-x^2}\int_0^y e^{s^2}\,\sin(2xs)ds+ie^{-x^2}\int_0^y e^{s^2}\,\cos(2xs)ds
\end{align}$$
Thus, we have 
$$\begin{align}
\int_{-\infty}^{i\zeta} e^{-t^2}dt&=\frac{\sqrt{\pi}}{2}(erf(-\zeta_i)+1)\\\\
&-e^{-\zeta_i^2}\int_0^{\zeta_r} e^{s^2}\,\sin(2\zeta_i s)ds\\\\
&+ie^{-\zeta_i^2}\int_0^{\zeta_r} e^{s^2}\,\cos(2\zeta_i s)ds
\end{align}$$
where $y=\zeta_r$ and $-x=\zeta_i$ are the real and imaginary parts of $\zeta$, respectively.
This now casts the answer in terms of the error function of a real argument $x$, the integral of $e^{s^2}\sin(2xs)$ and $e^{s^2}\cos(2xs)$.  These latter two integrals can, of course, be expressed in terms of the error function with complex arguments.
Note, if $\zeta_r =0$, then
$$\begin{align}
\int_{-\infty}^{i\zeta} e^{-t^2}dt&=\frac{\sqrt{\pi}}{2}(erf(-\zeta_i)+1)
\end{align}$$
