Can the integral,
$$I = \int^{-x}_{-\infty} e^{-at^2}\ \mathrm dt,\ a \in \mathbb{C},\ Re(a) > 0,$$
be written as an error function?
I tried, by substitution,
$$\int^{-x}_{-\infty} e^{-at^2}\ \mathrm dt = \frac{1}{\sqrt{a}}\int^{-x\sqrt{a}}_{-\infty\sqrt{a}} e^{-u^2}\ \mathrm du.$$ Reducing this to a complementary error function, $erfc(z) = \frac{2}{\sqrt{\pi}} \int_{z}^{\infty}e^{-u^2}du$, seems to require that the lower bound of the integral obtained after the substitution, $-\sqrt{a}\cdot \infty$, is real.
Despite this, Mathematica computes the integral as defined above to,
$\int^{-x}_{-\infty} e^{-at^2}\ dt = \frac{\sqrt{\pi}}{2\sqrt{a}}erfc(\sqrt{a}x)$,
subject to the above constraints on the parameter $a$. The limit of the error function for $x \rightarrow \infty$ depends on the real and imaginary parts of $\sqrt{a}$. [see http://functions.wolfram.com/GammaBetaErf/Erfc/ for limiting values at infinity]. For the initial integral, $I(x \rightarrow \infty) \rightarrow 0$, independent of $a$. I would like to understand where this apparent discrepancy arises.
Any thoughts would be very helpful!