Integration by parts in statistical physics In the book Equilibrium Statistical Physics: Phases of Matter and Phase Transitions by Marc Baus and Carlos F. Tejero at page 49 (English version) says
3.2 Liouville's Equation
[...]
and hence
$$ a(\textbf{r},t)\equiv \int \text{d}q\int\text{d}p a(q,p;\textbf{r},t)\rho(q,p),$$
may be written as
$$ a(\textbf{r},t) = \int \text{d}q \int \text{d}p \left[ \text{e}^{-L_N t} a(q,p; \textbf{r},0)\right] \rho (q,p) \overset{?}{=} \int \text{d}q\int\text{d}p a(q,p;\textbf{r},0) \left[ \text{e}^{L_N t} \rho (q,p)\right], $$
where to go from the first to the second expression an integration by parts has been performed (since $L_N$ is a linear differential operator), and it has been admitted that $\rho(q,p)$ and all of its derivatives vanish at the limits of integration.
How can I go from the first to the second expression? I don't understand how the book has made the integration by parts. 
*In particular, if
$$ \int x y' =xy-\int x' y.$$
Who is who?
Just like this:

from Statistical Mechanics of Nonequilibrium Liquids by Evans and Morriss. (Page 48)
 A: I think that the answer is:
$$ \int \text{d}\Gamma \text{ }a(\Gamma)\text{ } L \text{ } \rho (\Gamma) = \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }\dot{\Gamma } \dfrac{\partial}{\partial \Gamma} \rho (\Gamma)\tag{1}$$
because of the Chain rule and the Hamilton's equations. The measure of the phase space $\Gamma$ is d$\Gamma =\text{d}q\text{ d}p.$ We know that
$$ \int \text{d}\Gamma \text{ }a(\Gamma)\text{ } L \text{ } \rho (\Gamma) =  \int \text{d}\Gamma \text{ }\rho(\Gamma)\text{ }\left[- L \right]\text{ } a(\Gamma) \tag{2}$$
The equation (2) is obtained by integrating by part once 
$$ \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }\dot{\Gamma } \dfrac{\partial}{\partial \Gamma} \rho (\Gamma) = \left. a(\Gamma) \text{ }\dot{\Gamma} \text{ } \rho(\Gamma) \right|_S-\int \text{d}\Gamma \rho(\Gamma) \dfrac{\partial}{\partial \Gamma}\left[ a(\Gamma)\dot{\Gamma}\right] = -\int \text{d}\Gamma \rho (\Gamma) \dot{\Gamma}\dfrac{\partial}{\partial \Gamma} a(\Gamma). \tag{3}$$
So, by induction, we have 
$$ \int \text{d}\Gamma \text{ }a(\Gamma)\text{ } L^n \text{ } \rho (\Gamma) = \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }(-1)^n L^n \rho (\Gamma),\quad \forall n \in \mathbb{N}\tag{4}.$$
Note that, in order to get the equation (4), we've been made $n$ integrations by parts:
$$ \int \text{ }\text{d}\Gamma \text{ } a(\Gamma) \text{ }\dot{\Gamma }^n \dfrac{\partial^n}{\partial \Gamma^n} \rho (\Gamma) = \left. a(\Gamma) \text{ }\dot{\Gamma}^n \dfrac{\partial^{n-1}}{\partial \Gamma^{n-1}} \text{ } \rho(\Gamma) \right|_S-\int \text{d}\Gamma \dfrac{\partial^{n-1}}{\partial \Gamma^{n-1}}\rho(\Gamma) \dfrac{\partial}{\partial \Gamma}\left[ a(\Gamma)\dot{\Gamma}^n\right] $$  $$ =...= (-1)^n\int \text{d}\Gamma \rho (\Gamma) \dot{\Gamma}^n\dfrac{\partial^n}{\partial \Gamma^n} a(\Gamma). \tag{3}$$
Therefore, we can express the exponential function $\text{e}^{\pm Lt}$ as
$$ e^{\pm Lt} = 1 \pm tL + \dfrac{t^2}{2}L^2 \pm \dfrac{t^3}{6} L^3 + O(L^4)$$
and apply the previous result.
