 Solution

Use the same metric signature as above and define the Minkowski generating functional byHere $\hslash =1$. Integration by parts writes the quadratic action as $S_0=\frac{1}{2}\varphi K\varphi$ with $K=-\square -m^2+i0$; the infinitesimal imaginary part damps the oscillatory Gaussian functional integral and selects the vacuum Feynman propagator. Abbreviating repeated spacetime integrations, completing the square givesTranslation of the regulated integration variables therefore yieldsThe source-independent normalization contains the regulated functional determinant and its oscillatory phase; it cancels in $Z_0[J]/Z_0[0]$. Define ${\Delta }_F=i{K}^{-1}$. The Gaussian evaluation of a free scalar generating functional is equivalently $Z_0[J]/Z_0[0]=\exp [-J{\Delta }_{F}J/2]$.

With Fourier convention $e^{-ip\cdot x}$, $-\square$ acts as $p^2$. The inverse-kernel equation $K{K}^{-1}=I$ then givesThis is the time-ordered free two-point correlation function: differentiating twice with respect to $J$ and multiplying by $1/i^2$ recovers ${\Delta }_F$. It obeys $(\square +m^2){\Delta }_F=-i{\delta }^{(d)}$ in the distributional limit. The $i0$ is essential: the algebraic denominator alone would not distinguish Feynman, advanced and retarded Green functions.