I have some schematic notes on computing the effective action and I would like someone to help me fill the gaps.
We start with \begin{equation*} \int{}\mathcal{D}\phi\,e^{-iS[\phi]} \end{equation*} employing the background field method we write \begin{equation*} \phi=\phi_0+\Delta\phi \end{equation*} so we have \begin{equation*} \int{}\mathcal{D}(\Delta\phi)\,e^{-iS[\phi_0+\Delta\phi]} \end{equation*} Taylor expanding around $\phi_0$
$$S[\phi_0+\Delta\phi]=S[\phi_0]+\int{}d^4x_1\,\frac{\delta{}S}{\delta\phi(x_1)}\Delta\phi(x_1)$$ $$+\frac{1}{2}\int{}d^4x_1d^4x_2\frac{\delta^2S}{\delta\phi(x_1)\delta\phi(x_2)}\Delta\phi(x_1)\Delta\phi(x_2)+$$ $$\frac{1}{3!}\int{}d^4x_1d^4x_2d^4x_3\frac{\delta^3S}{\delta\phi(x_1)\delta\phi(x_2)\delta\phi(x_3)}\Delta\phi(x_1)\Delta\phi(x_2)\Delta\phi(x_3)+\ldots$$ since $\phi_0$ satisfies the equations of motion the linear term in $\Delta\phi$ vanishes. Then we have
$$e^{-iS[\phi_0]}\int{}\mathcal{D}(\Delta\phi)e^{-i\frac{1}{2}\int{}d^4x_1d^4x_2\frac{\delta^2S}{\delta\phi(x_1)\delta\phi(x_2)}\Delta\phi(x_1)\Delta\phi(x_2)+\ldots}$$
from here on my notes neglect terms cubic,quartic... in $\Delta\phi$. Can anybody tell me why?.
Also, after this it is written $$e^{-iS[\phi_0]}det(\ldots)$$ where the dots represent (I think) a functional determinant of something. Can anybody tell me what goes inside the determinant, and where this comes from?