I have:
$E[X]= \int_{2}^{\infty } x e^{-\frac{1}{2}(\frac{x-1}{1})^{2}}dx$
I reorganized this to:
$E[X]= \int_{2}^{\infty } x* e^{-\frac{1}{2}}*e^{(x-1)^2}dx$
I would use U substition:
$\\u = x-1\\ du = 1 dx\\$
then I would get:
$E[X]= \int_{u= 2-1}^{\infty } x* e^{-\frac{1}{2}}*e^{u^{2}} du$
But at this point I completely stuck. I am not sure how to integrate this, besides, since the integral should be evaluated over infinity, I am pretty sure first term would be infinity. And no matter what second term is, the end result would be infinity, what I find it really strange, because intuition says to me the expected value should be a real number.
$$I=\int_2^\infty xe^{-(x-1)^2/2}dx=\int_1^\infty(u+1)e^{-u^2/2}du=e^{-1/2}\int_1^\infty(u+1)e^{-u^2}$$ $$=e^{-1/2}\left(\int_1^\infty ue^{-u^2}du+\int_1^\infty e^{-u^2}du\right)$$ The first integral is easy with another substituion which should be obvious, then the second one can be related to the gaussian integral and error function