Ok. I gave it another shot but I don't seem to be able to come up with a reasonable solution.
\begin{equation}
\frac{1}{\sqrt{4\pi Dt}}\int_{-\infty}^\infty x e^{-\frac{(x-t)^2}{4Dt}} dx=\frac{1}{\sqrt{4\pi Dt}} \left( \int_{-\infty}^\infty (x-dt) e^{-\frac{(x-dt)^2}{4Dt}} dx - dt...
Homework Statement
I'm trying to find the expected value of a probability distribution.
Homework Equations
\int_{-\infty}^\infty xP(x,t) = \int_{-\infty}^\infty x \frac{1}{\sqrt{4\pi Dt}}e^{-\frac{(x-dt)^2}{4Dt}}dx
The Attempt at a Solution
I expect the value to be something like...