I don't really see how that would work.
I tried setting u=ax, v=bx^2. Then i would use
$$ \Gamma(1) = \int_{0}^{\infty} e^{-u}du $$
$${\rm erf}(x) = \int_{-\infty}^{x} e^{-v^2}dv $$
I'm not exactly sure how to use that in
$$ \int_{-\infty}^{+\infty} e^{-u} e^{-v^2} du$$
without another...
How would one evaluate $$\Phi = \int_{-\infty}^{+\infty} e^{-(ax+bx^2)} dx$$.
I was trying to change it into a product of an error function and a gamma function, but I needed an extra dx. Any other ideas?