limit of $\Lambda$ introduced in Eq. (\[Lambda\]). The right hand of Eq. (\[Z\_0S\]) forms the continuum limit of the lattice expression in the last line of Eq. (\[lnZ0\]) with the number of steps $b$ taken by the particle assuming the role of the Schwinger proper time parameter $s$: $b a^2/2d \equiv s$ as in Eq. (\[identification\]).

On account of the path integral (\[pathintegral\]) with $x=x'$, the last two factors in the integrand at the right hand of Eq. (\[Z\_0S\]) can be written as a path integral over *closed* worldlines $$\label{pathintegralo}
\mathrm{e}^{-s m^2} \left( \frac{1}{4 \pi s} \right)^{d/2} = \oint
\mathrm{D}x(s') \, \exp\left\{-\int_0^{s}\mathrm{d} s'
\left[\tfrac{1}{4} {\dot {x}}^2(s') + m^2 \right] \right\},$$ where the notation $\oint \mathrm{D} x(s')$ refers to closed paths. In this way, $\ln Z_0$ expressed as a path integral reads[@StoneThomas] $$\begin{aligned}
\label{lnZ0pi}
\ln Z_0 &=& \ln \mathrm{Det}^{-N/2}\left( - \partial_\mu^2 + m^2 \right)
   \nonumber \\ &=& \frac{N}{2}  \Omega \int_0^\infty \frac{\mathrm{d}s}{s}
   \oint \mathrm{D}x(s') \, \exp\left\{-\int_0^{s} \mathrm{d} s'
   \left[\tfrac{1}{4} \dot{x}^2(s') + m^2 \right] \right\},\end{aligned}$$ where the argument of the exponential function is given by (minus) the classical action (\[W0\]) of the point particle tracing out its trajectories in spacetime. An easy mnemonic for remembering the main ingredients of this formula is to 