= Solution
Use the same <metric signature> as above and define the Minkowski <generating functional> by
$$
Z_0[J]=\int\mathcal D\phi\,
\exp\left(iS_0[\phi]+i\int d^dx\,J\phi\right).
$$
Here $\hbar=1$. <Integration by parts> writes the quadratic action as $S_0=\tfrac12\phi K\phi$ with $K=-\Box-m^2+i0$; the infinitesimal imaginary part damps the oscillatory <Gaussian functional integral> and selects the vacuum <Feynman propagator>. Abbreviating repeated spacetime integrations, <completing the square> gives
$$
\frac12\phi K\phi+J\phi
=\frac12(\phi+K^{-1}J)K(\phi+K^{-1}J)-\frac12JK^{-1}J.
$$
Translation of the regulated integration variables therefore yields
$$
\boxed{Z_0[J]=Z_0[0]\exp\left[-\frac i2\int dx\,dy\,J(x)K^{-1}(x,y)J(y)\right].}
$$
The source-independent normalization contains the regulated <functional determinant> and its oscillatory phase; it cancels in $Z_0[J]/Z_0[0]$. Define $\Delta_F=iK^{-1}$. The <Gaussian evaluation of a free scalar generating functional> is equivalently $Z_0[J]/Z_0[0]=\exp[-J\Delta_FJ/2]$.
With Fourier convention $e^{-ip\cdot x}$, $-\Box$ acts as $p^2$. The inverse-kernel equation $KK^{-1}=I$ then gives
$$
\boxed{\widetilde\Delta_F(p)=\frac{i}{p^2-m^2+i0},\qquad
\Delta_F(x-y)=\int\frac{d^dp}{(2\pi)^d}\,
\frac{i\,e^{-ip\cdot(x-y)}}{p^2-m^2+i0}.}
$$
This 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 $(\Box+m^2)\Delta_F=-i\delta^{(d)}$ in the distributional limit. The $i0$ is essential: the algebraic denominator alone would not distinguish Feynman, advanced and retarded <Green functions>.
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