= Solution
After integration by parts, the quadratic action is
$$
S_0[x]=-\frac12\int dt\,x(t)(\partial_t^2+\omega^2)x(t).
$$
With the pole prescription appropriate to the conventions in the question, its inverse kernel is
$$
D(t)=\int\frac{dE}{2\pi}\frac{i\,e^{-iEt}}{\omega^2-E^2-i\epsilon}.
$$
For $t>0$ close the <contour integration>[contour] in the lower half-plane and for $t<0$ close it in the upper half-plane. The enclosed pole in each case gives
$$
D(t-t')=\frac1{2\omega}e^{i\omega|t-t'|},
$$
which 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 gives
$$
Z_0[J]=Z_0[0]\exp\left[
-\frac12\int dt\,dt'\,J(t)D(t-t')J(t')
\right].
$$
Changes 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.
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