Deriving the change in the Helmholtz free energy in the context of the free energy perturbation method I am reading Free Energy Calculations: Theory and Applications in Chemistry and Biology by Chipot and Pohorille.  At the beginning of the text (page 19, for example), the authors define the Helmholtz free energy $A$ and changes $\Delta A$ in it for two states (states 0 and 1):
$$\begin{align}
A&=-\beta^{-1} \ln Q(N, V, T) \;\;\;\textbf{(1.14)} \\
\Delta A&=-\beta^{-1} \ln \frac{Q_1}{Q_0} \;\;\;\textbf{(1.15)} \\
\Delta A&=-\beta^{-1} \ln \frac{Z_1}{Z_0} \;\;\;\textbf{(1.16)} 
\end{align}$$
where $Q_i$ is a partition function:
$$Q_i = \frac{1}{N! h^{3N}} \int_{\Gamma_i} \exp[-\beta \mathcal{H}(d\vec{x}, d\vec{p})] d\vec{x} d\vec{p}_x$$
and $Z_i$ is a configurational integral:
$$Z_i = \int_{\Gamma_i} \exp[-\beta U(\vec{x})] d\vec{x}$$
where $U$ is the potential energy.
Next (page 20), the authors write a derivation for $\Delta A$, in the framework of free energy perturbation (FEP):

Equation (1.15) indicates that our ultimate focus in calculating
  $\Delta A$ is on determining the ratio $Q_1/Q_0$, or equivilently
  $Z_1/Z_0$, rather than on individual partition functions.  On the
  basis of computer simulations, this can be done in several ways.  One
  approach consists of transforming (1.16) as follows:
$$\begin{align} \Delta A&=-\beta^{-1} \ln \frac{\int \exp[-\beta U_1(\vec{x})] d\vec{x}}{\int \exp[-\beta U_0(\vec{x})] d\vec{x}} \\ &=-\beta^{-1} \ln \exp\left\{-\beta [U_1(\vec{x}) - U_0(\vec{x})]\right\} P_0(\vec{x}) \\ &=-\beta^{-1} \ln \langle \exp\left\{-\beta [U_1(\vec{x}) - U_0(\vec{x})]\right\} \rangle_{0} \end{align}$$
Here, the systems 0 and 1 are described by the potential energy
  functions, $U_0(\vec{x})$ and $U_1(\vec{x})$, respectively. 
  Generalization to conditions in which systems 0 and 1 are at two
  different temperatures is straight forward.  $\beta_0$ and $\beta_1$
  are equal to $(k_B T_0)^{-1}$ and $(k_B T_1)^{-1}$, respectively. 
  $P_0(\vec{x})$ is the probability density function of finding system 0
  in the microstate defined by positions $\vec{x}$ of the particles:
$$P_0(\vec{x}) = \frac{\exp[-\beta_0 U_0(\vec{x})]}{Z_0} \;\;\;\textbf{(1.19)}$$

My question is, what happened to the integrals in going from (1.18) to the next step?  $P_0(\vec{x})$ only contains an integral in its denominator (i.e., in $Z_0$), and not also in the numerator.  Could you please perhaps provide one more step between equation (1.18) and the equation that immediately follows it?  Thanks for your time.
 A: There has to be an integral missing in your book citation, line 2 of $\Delta A =\ldots$ should read 
$\Delta A = \beta^{-1} \ln \left[\int \exp{-\beta[U_1(\vec x) - U_0(\vec x)]} P_0(\vec x) d \vec x \right]$. Other than that, the calculation is as follows:
$$\begin{align*} 
\Delta A &=-\beta^{-1} \ln \frac{\int \exp[-\beta U_1(\vec{x})] d\vec{x}}{\int exp[-\beta U_0(\vec{x})] d\vec{x}} \\ \\ 
&= -\beta^{-1} \ln \frac{\int \exp[-\beta U_1(\vec{x})] \frac{\exp(\beta U_0(\vec x))}{\exp(\beta U_0(\vec x))} d\vec{x}}{\int \exp[-\beta U_0(\vec{x})] d\vec{x}} \\ \\
&= -\beta^{-1} \ln \frac{\int \exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] \exp[-\beta U_0(\vec x)] d\vec{x}}{\int \exp[-\beta U_0(\vec{x})] d\vec{x}} \\ \\
&= -\beta^{-1} \ln \frac{\int \exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] \exp[-\beta U_0(\vec x)] d\vec{x}}{Z_0} \\ \\
&= -\beta^{-1} \ln \int \exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] p_0(\vec x) d\vec{x} \\ \\
&= -\beta^{-1} \ln \langle\exp[-\beta [U_1(\vec{x}) - U_0(\vec x)]] \rangle
\end{align*}$$
Long, faulty latex story short: multiply the integrand in the nominator by $1=\frac{\exp[\beta U_0(\vec x)]}{\exp[\beta U_0(\vec x)]}$, unite the two "top" exponentials, convert $\frac{1}{\exp[\beta U_0(\vec x)]} = \exp[-\beta U_0(\vec x)]$ which , with $Z_0$ from the big denominator, gives $P_0(\vec x)$ and write the integral as the ensemble average.
