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?