Derivation of Crooks's Fluctuation Theorem

[TeethWhitener]

I'm reading through Crooks's paper:
https://journals.aps.org/pre/abstract/10.1103/PhysRevE.60.2721
as well as a review paper by Mansour et al.:
https://aip.scitation.org/doi/10.1063/1.4986600
trying to figure out their derivation of the fluctuation theorem (section II of both papers). I had a question about the math behind one specific part of the derivation. The derivation is essentially identical in both papers, but I'll use Mansour's notation.

We have a stochastic process with a forward-time path through phase space denoted $\mathbf{X}$, and a time-reversed path $\mathbf{\overline{X}}$. We define a quantity
$$Z(\mathbf{X}) = \ln{\frac{P(\mathbf{X})}{P(\mathbf{\overline{X}})}}$$
where $P(\mathbf{X})$ is the probability density of $\mathbf{X}$. Then $Z$ is just the logarithm of the ratio of the forward-time path probability to the reverse-time path probability (in thermodynamic terms, this is related to the entropy change of a process). To find the probability that we will observe a particular entropy production, $\zeta <Z<\zeta +d\zeta$, denoted $P(\zeta)d\zeta$, we take the ensemble average $\langle\delta(\zeta-Z(\mathbf{X}))\rangle$ over all paths:
$$P(\zeta)=\int{d[\mathbf{X}]\delta(\zeta-Z(\mathbf{X}))P(\mathbf{X})}$$
where $d[\mathbf{X}]=d[\mathbf{\overline{X}}]$ is a suitable path integral measure and $\delta(\zeta-Z(\mathbf{X}))$ is the Dirac delta function. Then we multiply by $\frac{P(\mathbf{\overline{X}})}{P(\mathbf{\overline{X}})}$ and use the definition of $Z$ to get:
$$P(\zeta)=\int{d[\mathbf{X}]\delta(\zeta-Z(\mathbf{X}))P(\mathbf{\overline{X}})}\exp[Z(\mathbf{X})]$$
This part is fine. The next part is where I'm stumped. The authors use the delta function to pull the exponential outside the integral:
$$\int{d[\mathbf{X}]\delta(\zeta-Z(\mathbf{X}))P(\mathbf{\overline{X}})}\exp[Z(\mathbf{X})]=\exp(\zeta)\int{d[\mathbf{X}]\delta(\zeta-Z(\mathbf{X}))P(\mathbf{\overline{X}})} $$
I don't understand how they can pull the exponential out without getting rid of the integral altogether. I'm familiar with the delta function's use as:
$$\int f(x)\delta(x-\tau) dx =f(\tau)$$
but I've never seen a case where you can use the delta function on only one piece of an integrand, like they're doing above. Is this some quirk of delta functions with path integrals? Or am I missing something obvious?

Any help would be great, thanks!

 

[TeethWhitener]

Oh wait a second, is this just because $\int {\delta(x)dx} = 1$? So that $\int {f(x)\delta(x-\tau)dx} = f(\tau)\int {\delta(x-\tau)dx}$? Man I'd feel really dumb if that were the case.

Edit:
So is it true in general that $\int {f(x)g(x)\delta(x-\tau)dx} = f(\tau)\int {g(x)\delta(x-\tau)dx}$?

 

[Hiero]

Oh wait a second, is this just because $\int {\delta(x)dx} = 1$? So that $\int {f(x)\delta(x-\tau)dx} = f(\tau)\int {\delta(x-\tau)dx}$? Man I'd feel really dumb if that were the case.

Edit:
So is it true in general that $\int {f(x)g(x)\delta(x-\tau)dx} = f(\tau)\int {g(x)\delta(x-\tau)dx}$?

I should preface: that paper and your OP are beyond me.
But I can say this much: you can go a step farther;
$$\int F(x)\delta (x-\tau)dx = F(\tau)$$
Where $\delta$ is the Dirac delta function.
This of course includes the case where F(x) = f(x)g(x)
(Also, this is assuming $\tau$ is in the interval of integration.)
This is partly because (like you said) $\int \delta (x)dx=1$ , but of course it wouldn’t be true for any function $\delta$ which satisfies that equation; the key detail is that $\delta(x-\tau)$ is zero everywhere except at $x=\tau$

Edit:
Oops, I just noticed you said you’re familiar with this in the OP, sorry! (I glossed over it once I got lost.)

What you said is also true:
$$\int f(x)g(x)\delta (x-\tau)dx = f(\tau)\int g(x)\delta (x-\tau)dx $$
Because both sides simplify to $f(\tau)g(\tau)$

Not sure why that would be useful... but it’s indeed true.

 

[TeethWhitener]

Not sure why that would be useful... but it’s indeed true.

Thanks for your help. It’s useful because after the variable transformation $\mathbf{X}\rightarrow\mathbf{\overline{X}}$, you get:
$$\exp (\zeta) \int {d[\mathbf{X}] \delta(\zeta -Z(\mathbf{X}))P(\mathbf{\overline{X}}) }= \exp (\zeta) \int {d[\mathbf{\overline{X}}] \delta(\zeta +Z(\mathbf{\overline{X}}))P(\mathbf{\overline{X}}) }= \exp(\zeta)P(- \zeta)$$
from the previous definition of $P(\zeta)$. This gives the fluctuation theorem immediately:
$$\frac{P(\zeta)}{P(-\zeta)}=\exp(\zeta)$$