Nope.However, what confuses me is how you get from dv to v. Doesn't that require us to do integration by parts again?

Yep, that could be a substitution, because the factor (x/y^2) in dv is so like the derivative of the inner function of the exponential.Is there some kind of Calculus trick going on?

You need the error function to integrate the remaining integrand, where you have the same composite exponential but with no inner derivative conveniently hanging around.I also gather that this has to do with the error function (if you plug it into wolframalpha that's what you get). However, I would really like to NOT go down that route.