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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?

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  • $\begingroup$ For some special $a,b$ it is possible to write the integral in terms of special functions, for example if $a=1,\,b=0$ is trivial and if $b=0$ the integral is linked to the Owen's T function, but honestly I would not bet on a closed form for generic $a,b$. $\endgroup$ – Marco Cantarini Jul 6 '17 at 9:45
  • $\begingroup$ Just checking: How does differ from your earlier question? $\endgroup$ – Andrew D. Hwang Jul 6 '17 at 11:16
  • $\begingroup$ @MarcoCantarini I was afraid of that too. $\endgroup$ – Adel Bibi Jul 6 '17 at 12:24
  • $\begingroup$ @AndrewD.Hwang The previous question is indefinite. This limits the number of tools one can use. For instance, the problem in hand can be solved for the integral limits $(-\infty,\infty)$ but apparently not for ($0,\infty$). + The previous question has been edited now to be on an approximation. $\endgroup$ – Adel Bibi Jul 6 '17 at 12:26
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    $\begingroup$ I gave you a series for math.stackexchange.com/questions/2345874/… which is maybe useful for an approximation. $\endgroup$ – user90369 Jul 7 '17 at 22:46

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