In The Quantum Theory of Fields, Volume II, Weinberg defines the BRST variation $\delta_\theta$ as follows: \begin{align} \delta_\theta \psi & = i t_\alpha \theta \omega_\alpha \psi \tag{15.7.7} \\ \delta_\theta \boldsymbol{A}_{\alpha \mu} & = \theta D_\mu \omega_\alpha = \theta\left[\partial_\mu \omega_\alpha + C_{\alpha \beta \gamma} \boldsymbol{A}_{\beta \mu} \omega_\gamma\right] \tag{15.7.8} \\ \delta_\theta \omega_\alpha^* & = - h_\alpha \tag{15.7.9} \\ \delta_\theta \omega_\alpha & = -\frac{1}{2} C_{\alpha \beta \gamma} \omega_\beta \omega_\gamma \tag{15.7.10} \\ \delta_\theta h_\alpha & = 0 \tag{15.7.11} \end{align}
The auxiliary fields $h_\alpha$ are introduced to express the gauge-fixing term in a Gaussian form: $$ \exp \left(-\frac{i}{2 \xi} \int d^4 x\, f_\alpha f_\alpha\right) \propto \int\left[\prod_{\alpha, x} d h_\alpha(x)\right] \exp \left[\frac{i \xi}{2} \int h_\alpha h_\alpha\right] \exp \left[i \int d^4 x\, f_\alpha h_\alpha\right]. \tag{pg. 28} $$
This allows the action to be written as: $$ I_{\mathrm{NEW}} = \int d^4 x \left( \mathscr{L} + \omega_\alpha^* \Delta_\alpha + h_\alpha f_\alpha + \frac{1}{2} \xi h_\alpha h_\alpha \right). \tag{15.7.6} $$
Weinberg then discusses the case of pure electrodynamics, taking the gauge-fixing function to be $f = \partial_\mu A^\mu$, and writes: "To see how this works in practice, let us consider the simple example of pure electrodynamics. Taking the gauge-fixing function as $f = \partial_\mu A^\mu$ and integrating over the auxiliary field $h$, the BRST transformation (15.7.8)-(15.7.10) is here" $$ s A_\mu = \partial_\mu \omega, \quad s \omega^* = \frac{\partial_\mu A^\mu}{\xi}, \quad s \omega = 0. \tag{15.7.32} $$ My question is:
What does Weinberg mean by "integrating out the auxiliary field $h$", and how does this lead to the expression $s \omega^* = \partial_\mu A^\mu/\xi$?
