How to perform a Gaussian functional integral? I'm completely beginner to the quantum field theory and try to learn the basics of functional integrals. However, I could not understand clearly. Could someone please explain the idea with the help of the following example:
$$\begin{equation}
\mathcal{Z} = \int[dE][dx] \exp \left[ - \int_{0}^{\beta} d\tau \left( \frac{1}{2}m\dot{x}^{2} + \frac{1}{2}m\omega_{0}^{2}x^{2} - eEx + \frac{1}{2g}(\dot{E}^{2} + \omega_{LC}^{2}E^{2}) \right) \right]
\end{equation}$$
Let's say I want to perform the integral over $[dx]$ of partition function $\mathcal{Z}$. Could someone explain with some initial steps how to do that?
 A: Let's start with
$$ \mathcal{Z}=\int \exp(-x\cdot D\cdot x-2x\cdot y) \mathcal{D}[x] , $$
where $\cdot$ represents an integral contraction (see below), and $D$ is some operator/kernel. For the example above, it is something like
$D\propto \partial_t^2-\omega_0^2$, after partial integration of the $\dot{x}^2$-term.
There are basically three steps: 1) complete the square:
$$ \mathcal{Z}=\int \exp(-(x+y\cdot D^{-1})\cdot D\cdot (x+D^{-1}\cdot y)+y\cdot D^{-1}\cdot y) \mathcal{D}[x] . $$
2) shift $y\cdot D^{-1}$ into the field variable:
$$ \mathcal{Z}=\int \exp(-x'\cdot D\cdot x'+y\cdot D^{-1}\cdot y) \mathcal{D}[x'] . $$
3) evaluate the Gaussian integral:
$$ \mathcal{Z}=\exp(y\cdot D^{-1}\cdot y)\int \exp(-x'\cdot D\cdot x') \mathcal{D}[x']=\exp(y\cdot D^{-1}\cdot y)\frac{1}{\sqrt{\det\{D\}}} . $$
Definition of integral contraction: $$ x\cdot D\cdot x = \int x(t) D(t,t') x(t') dtdt' . $$
A: In general for a real symmetric N x N matrix A and vector x we have
$$\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} \cdots \int_{-\infty}^{+\infty} d x_{1} d x_{2} \cdots d x_{N} e^{-\frac{1}{2} x \cdot A \cdot x+J \cdot x}=\left(\frac{(2 \pi)^{N}}{\operatorname{det}[A]}\right)^{\frac{1}{2}} e^{\frac{1}{2} J \cdot A^{-1} \cdot J}$$
Which you can derive by diagonalizing A with an orthogonal transformation i.e $A=O^{-1} \cdot D \cdot O$
Set $y_{i}=O_{i j} x_{j}$ and the expression in the exponential becomes $-\frac{1}{2} y \cdot D \cdot y+(O J) \cdot y$
The integrals over $x_i$ just become integrals over $y_i$ and so we just have integrals of the form $$\int_{-\infty}^{+\infty} d y_{i} e^{-\frac{1}{2} D_{i i} y_{i}^{2}+(O J)_{i} y_{i}}$$
and plugging into the general result for these types of integrals, namely $$\int_{-\infty}^{+\infty} d x e^{-\frac{1}{2} a x^{2}+J x}=\left(\frac{2 \pi}{a}\right)^{\frac{1}{2}} e^{J^{2} / 2 a}$$
and using the fact that $(O J) \cdot D^{-1} \cdot(O J)=J \cdot O^{-1} D^{-1} O \cdot J=J \cdot A^{-1} \cdot J$
we get the right hand side of the first equation $$\left(\frac{(2 \pi)^{N}}{\operatorname{det}[A]}\right)^{\frac{1}{2}} e^{\frac{1}{2} J \cdot A^{-1} \cdot J}$$
Source: Zee QFT
