I'm not sure on the proper descriptions for what I'm trying to do here. Essentially, if at all possible, I want to simplify this as much as possible to obtain something along the lines of $\Lambda \propto M^N$ or $\Lambda \approx M^N$.
Not to say I'm expecting it to just be a power like that, but those are the variables I'm interested in.
I've already gone some way on the simplification but the integration is a complicated nightmare that I just don't know where to begin with.
Firstly, I've gotten $x = n t M^{-1/2} $ where $n$ is just a bundle of other constants from an earlier step and $t$ is time, but the time is unimportant so may be taken as another constant, and $y = n \frac{M^{1/2}}{2\tau}$ where again $\tau$ is just some other constant.
So the integral then:
$$ \Lambda(x, y) = \int_0^x 2z \exp\left(-2zy + z^2\right) \ dz $$
I have frankly no idea where to go from here, I've thrown it at various integral engines and the result is, ugly. I assume there must be a few tricks for removing more of the stuff I'm disinterested in but I lack the mathematical knowledge to know so.
Thanks for any and all help! To be clear I'm not trying to actually evaluate this integral, I can already do that with quads, I wish to find the relation within it. I'm also going to be doing a similar thing again with another variable instead of $M$, such that $x = ntv^{1/2}$ and $y=n\frac{v^{-1/2}}{2\tau}$. i.e. the variable of interest is then $v$ which has an inverse relationship with $x$ and $y$ to $M$.
Would a viable alternative approach be to evaluate the integral at a set value of $t$ for a range of values of $M$ and just see if I can find rough $\approx \propto$ relation "manually"?
For the origin of these equations see Arnett 1982
Note that $$ z^2-2zy = (z-y)^2-y^2, $$ Apply that and change variables using $u = z-y, du = dz$ to get $$ \begin{split} \Lambda(x,y) &= \int_0^x 2z \exp\left(-2zy + z^2\right) dz \\ &= e^{-y^2} \int_0^x 2z \exp\left((z-y)^2\right) dz \\ &= e^{-y^2} \int_{-y}^{x-y} 2(u+y) \exp\left(u^2\right) dz, \end{split} $$ which now splits into 2 integrals, one you can take by substitution and the other is somewhat similar to the error function or cdf of the Normal distribution, I am sure you can find a special function that relates to it.