 Solution

Keep the same measure normalization as in the free generating functional and defineHere $W$ is the source-dependent partition function, despite the common alternative convention in which that letter denotes its logarithm. A functional derivative of $e^{\int Jx}$ brings down one factor $x(\tau )$ at the differentiated point. Expanding the interaction exponential and replacing each such factor by $\delta /\delta J(\tau )$ therefore givesCompleting the square in the free Gaussian integral gives $W_0[J]=W_0[0]\exp (\frac{1}{2}\int J{G}^{(0)}J)$ because $G^{(0)}=D^{-1}$ with periodic boundary conditions. Restoring $\hslash$ replaces the upper limit by $\hslash \beta$ and the interaction coefficients by $\lambda /(4!\hslash )$ and $\mu /(6!\hslash )$, exactly as in the source.