Let $x(t)$ be a stationary Gaussian process with derivative $v(t)$. Let the stationary distribution be $p(x,v) = e^{-A x^2 - B v^2}$. Let the normalized autocorrelation function of $x(t)$ be $\rho(t)$. Then
$$ \int_{-\infty}^{\infty} |v| p(0,v)\, dv = \frac{1}{\pi} \sqrt{- \rho''(0)}.$$
For context, this is a step in the proof of Rice's Formula. I'd appreciate help deriving it.
From the definition $\rho(t) = \langle x(t') x(t'+t)\rangle$ (up to normalization) we find $\rho''(0) = \langle x(t') x''(t')\rangle$. But I have no way of computing this or relating it to the left-hand side. If I square the left-hand side I get something that looks like $\langle x'(t') x'(t')\rangle$, but I don't see the relation with the right-hand side.
This is an example of the dot-shifting formula, or in your case dash-shifting. For any stationary time correlation function we can differentiate with respect to $t'$, the time origin, and the result must be zero: $$ \frac{d}{d t'} \langle A(t') B(t'+t) \rangle = \langle A'(t') B(t'+t) \rangle + \langle A(t') B'(t'+t) \rangle = 0 $$ So, setting $t=0$ we get $$ \langle A'(t')B(t')\rangle = -\langle A(t')B'(t')\rangle $$ For your case, let $A=x'$ and $B=x$. $$ \langle x''(t')x(t')\rangle = -\langle x'(t')x'(t')\rangle $$ This should be the missing link between the left hand side and the right hand side.