Interested in a closed form solution for
$$ \int_0^{\infty} \exp\left(- \left(\frac{x-z_3}{z_4}\right)^2 \right) \text{Erf}(z_1 x + z_2) dx $$
I started by changing the integral to a more nicely looking integral with the substation $u = \frac{x-z_3}{z_4}$.
$$ \int_c^{\infty} \exp(-u^2) \text{Erf}(au + b) du $$
I tried solving the last integral using the derivative under integral sign for the following parameters:
Parameter c:
$$ \frac{\partial I(c)}{\partial c} = - \exp(-c^2) \text{Erf}(ac + b) $$
Too bad the resulting indefinite integral w.r.t c doesn't seem to have a closed form solution. [See this]
Parameter a:
$$ \frac{\partial I(a)}{\partial a} = \frac{2}{\sqrt{\pi} }\int_c^{\infty} e^{-u^2} u e^{-(au+b)^2} du $$ Proceeding with integration by parts:
$$ \begin{aligned} \frac{\partial I(a)}{\partial a} &= -2a\frac{e^{-u^2}}{2} e^{-(au+b)^2} \Bigg\rvert_c^\infty + a \int_c^{\infty}e^{-u^2} e^{-(au+b)^2}dy \\ & = -a e^{-c^2} e^{-(ac +b)^2} + a \frac{\sqrt{\pi}e^{-\frac{b^2}{1+a^2}}}{2 \sqrt{1+a^2}} \left[1 - \text{Erf}\left(\frac{ab+c+a^2c}{\sqrt{1+a^2}} \right)\right] \end{aligned} $$
Again, this results into indefinite integrals with no closed form solutions.
A similar issue arises when when the parameter under differetiation is b. If the Integral limits were from $-\infty$ to $\infty$, a similar approach to this can be used to find a closed form.
Is this a trivial integral of some sort under some substitution that I'm not able to see?
