The potential of a measure $\mu_t$ is given by $U(t,x) = \int\vert y-x \vert \mu_t(dx)$. Now I want to compute the potential function $U(t,x)$ of $(\mu_t)_{t\geq0}$ which is the family of marginals of a Brownian motion (law of a centered Gaussian variable with variance $t$).
I know the solution but I have no idea how the get to it... Please can someone give an explanation?
It should be: $U(t,x) = 2 \sqrt{t} (\frac{e^{-x^2 / 2t}}{\sqrt{2t}} + \frac{x}{\sqrt{t}}\Phi(\frac{x}{\sqrt{t}}) - \frac{x}{2\sqrt{t}})$.