= Solution
The free <Euclidean action> has inverse <propagator> $D=-m\partial_\tau^2+\kappa$. Therefore the thermal <Green's function> obeys
$$
\boxed{(-m\partial_\tau^2+\kappa)G^{(0)}(\tau)=\delta_\beta(\tau),\qquad
G^{(0)}(\tau+\beta)=G^{(0)}(\tau),}
$$
where $\delta_\beta$ is the periodic <Dirac delta function>. It is continuous, has continuous first derivative away from its source, and satisfies $G'(0^+)-G'(0^-)=-1/m$. These conditions follow by integrating the differential equation across the source and using the bosonic thermal trace.
For $\Omega=\sqrt{\kappa/m}$, solving the homogeneous equation on $0<\tau<\beta$ and imposing these conditions gives
$$
\boxed{G^{(0)}(\tau)=\frac{\cosh[\Omega(\beta/2-\tau)]}{2m\Omega\sinh(\beta\Omega/2)}\quad(0\le\tau\le\beta).}
$$
In particular $g_0\equiv G^{(0)}(0)=\coth(\beta\Omega/2)/(2m\Omega)$. Equivalently $G^{(0)}=\beta^{-1}\sum_ne^{-i\omega_n\tau}/(m\omega_n^2+\kappa)$ with bosonic <Matsubara frequencies> $\omega_n=2\pi n/\beta$.
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