The following is what I need help with:
How do you go from the above: ([(x/y)^2]e^(-x^2/2y^2) dx)
to: -xe^(-x^2/2y^2) + integral e^(-x^2/2y^2) dx
It seems like the first step is to do it by parts:
Let u = x, dv = (x/y^2) e^(-x^2/(2y^2)) dx
du = dx, v = -e^(-x^2/(2y^2)).
So, ∫ (x/y)^2 e^(-x^2/(2y^2)) dx
= -x e^(-x^2/(2y^2)) - ∫ -e^(-x^2/(2y^2)) dx
= -x e^(-x^2/(2y^2)) + ∫ e^(-x^2/(2y^2)) dx
However, what confuses me is how you get from dv to v. Doesn't that require us to do integration by parts again? Is there some kind of Calculus trick going on?
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.