I found a nice discussion of Brownian motion in the Feynman lectures, reproduced online here: https://www.feynmanlectures.caltech.edu/I_41.html
Feynman considers a particle undergoing a Brownian motion, treated in one dimension in the first instance, such that the equation of motion (Newton's second law) is $$ m \ddot{x} = F - \alpha \dot{x}, $$ where $F$ is a rapidly fluctuating force and $\alpha$ is a constant viscous drag coefficient. (He uses the letter $\mu$ but I prefer to use $\alpha$ because $\mu$ is widely used for mobility.) (This is also called the Langevin equation). At one point in the argument we want to know the value of $$ \left\langle \frac{d}{dt}(x \dot{x}) \right\rangle, \tag{1} $$ where I believe the average is over all paths setting out from a given start point and continuing for a given time $t$. Feynman states that this quantity will be zero, but the reason he gives is not quite convincing to me. He asserts
"Now $x$ times the velocity has a mean that does not change with time, because when it gets to some position it has no remembrance of where it was before, so things are no longer changing with time. So this quantity, on the average, is zero."
The trouble is that the phrase "$x$ times the velocity has a mean that does not change with time" is the statement $$ \frac{d}{dt} \langle x \dot{x} \rangle = 0. \tag{2} $$ This quantity is indeed zero, but it is not self-evident that it is equal to (1).
Given (2), one way to obtain (1) would be to show that $$ \frac{d}{dt} \langle x \dot{x} \rangle = \left\langle \frac{d}{dt} (x \dot{x}) \right\rangle . \tag{3} $$
My question is: is (3) easy to prove (without assuming (1)!)? (If so please provide proof). Or is there some better way to prove that the quantity in (1) is equal to zero?
To forestall answers merely claiming "you can reverse order of integration and differentiation" here is why that alone is not the answer: $$ \frac{d}{dt} \langle x v \rangle = \frac{d}{dt} \iint x v f(x,v,t) dx dv $$ where $f(x,v,t)$ is the probability density function. This gives $$ \frac{d}{dt} \langle x v \rangle = \iint \frac{d}{dt} \left( x v f \right ) dx dv \\ = \iint f \frac{d}{dt}(xv) + (xv) \frac{df}{dt} dx dv = \langle \frac{d}{dt}(xv) \rangle + \iint x v \frac{df}{dt} dx dv $$ so either I have muddled something about the distribution function, or one has to show that the extra term is zero (keeping in mind that $\langle x v \rangle$ is not zero).
