 Solution

After integration by parts, the quadratic action isWith the pole prescription appropriate to the conventions in the question, its inverse kernel isFor $t>0$ close the contour in the lower half-plane and for $t<0$ close it in the upper half-plane. The enclosed pole in each case giveswhich equivalently satisfies $({\partial }_t^2+{\omega }^2)D(t)=i\delta (t)$.

The source-dependent Gaussian functional integral is evaluated by translating the integration variable by the classical sourced solution. Completing the square givesChanges in the sign of the source term or of the path-integral phase move factors of $i$ between $D$ and the exponent but leave the contraction rules equivalent.