A Proof of Classical Fluctuation-Dissipation Theorem

[Twigg]

TL;DR Summary

Found a proof of the classical version of the fluctuation-dissipation theorem, couldn't figure out one step. Involves classical transition probabilities for a stochastic, dissipative system. How does one evaluate the integral $\int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0)$.

Sorry if there's latex errors. My internet connection is so bad I can't preview.

Here's the wikipedia proof I'm referring to. I'm fine with the steps up to $$W(x,0) = W_0 (x) [1 + \beta f_0 (x(0) - \langle x \rangle_0) ]$$ where $W(x,t)$ is the probability density of finding the system at state $x$ at time $t$, $f_0$ is the magnitude of a perturbation $f_0 << kT$ on the Hamiltonian H of the form $H = H_0 - x f_0 \Theta (-t)$ where $\Theta (t)$ is the heaviside step function, $\langle \cdot \rangle_0$ is expectation value in the unperturbed equilibrium distribution (called $W_0 (x)$), and where $\beta = \frac{1}{kT}$.

I was able to reproduce the approximate, first order perturbation expression for $W(x,0)$ just fine using taylor/binomial series. It's the next step, using this distribution to prove $$\langle x(t) \rangle = \langle x \rangle_0 + \beta f_0 A(t)$$ where $A(t) = \langle [x(t) - \langle x \rangle_0][x(0) - \langle x \rangle_0] \rangle_0$ is the autocorrelation function.

I'm confident that the general procedure is to write $$\langle x(t) \rangle = \int dx \text{ } x \int dx' P(x,t|x',0) W(x',0) $$ where $P(x,t|x',0)$ is the conditional probability for state x at time t given the system is in state x' at time 0. I can see why $\int dx' P(x,t|x',0) W_0 (x') = W_0(x)$ since $W_0$ is an equilibrium distribution. I also have a hunch that $\lim_{t \rightarrow \infty} \int dx' P(x,t|x',0) W(x',0) = W_0(x)$ for any initial distribution since it's a dissipative system and will reach equilibrium eventually.

However, given all this, what I get is $$\langle x(t) \rangle = \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') [1 + \beta f_0 (x'(0) - \langle x \rangle_0) ] = \langle x \rangle_0 + \beta f_0 \left[ \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0) - \int dx \text{ } x W_0(x) \langle x \rangle_0 \right] = \langle x \rangle_0 + \beta f_0 \left[ \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0) - \langle x \rangle_0^2 \right]$$ It's the integral $$I = \int dx \text{ } x \int dx' P(x,t|x',0) W_0 (x') x'(0)$$ that gets me. I have no idea how to evaluate this, because the thing that $P(x,t|x',0)$ is multiplied by isn't a probability distribution. Can anyone point me in the right direction?

Alternatively, if you have a different concise proof, I'm all ears. I was interested in this proof because it's (relatively) short and might be easier to remember.

 

[Twigg]

Update: I'm having serious doubts with the form of the FDT stated in the wikipedia article. They claim $$S_x (\omega) = \frac{2kT}{\omega} \mathrm{Im}[\chi(\omega)]$$ However, for the vanilla Langevin equation $$m\ddot{x} = -\gamma \dot{x} + \eta$$ I am fairly confident that $$\lim_{t \rightarrow \infty} \langle \dot{x}(t+\tau) \dot{x}(t) \rangle \propto e^{-\gamma \tau}$$ In other words $$\lim_{t \rightarrow \infty} \langle \dot{x}(t+\tau) \dot{x}(t) \rangle \propto \chi(\tau)$$ and that's inconsistent with the wikipedia article's claim since the Fourier transform of $e^{-\gamma \tau}$ has non-zero real part.