= Solution
Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta\hbar\omega$ and $q=\sqrt{m\omega/\hbar}\,x$. The exact answer is
$$
\boxed{R=\exp[-q^2\tanh(z/2)].}
$$
At high <temperature>, $z\ll1$. Expanding the <hyperbolic tangent> gives
$$
\log R=-q^2\left(\frac z2-\frac{z^3}{24}+O(z^5)\right),
\qquad
\boxed{R\simeq\exp\!\left(-\frac{\beta m\omega^2x^2}{2}\right).}
$$
The displayed leading expression is the ratio of classical <Boltzmann factors> for <potential energy> $V(x)=m\omega^2x^2/2$ and $V(0)=0$. Indeed the classical <canonical ensemble> gives
$$
p_{\rm cl}(x)\ \propto\ \int_{-\infty}^{\infty}dp\,
e^{-\beta[p^2/(2m)+V(x)]}
\ \propto\ e^{-\beta V(x)}.
$$
The <momentum> <Gaussian integral> is independent of $x$ and cancels in the ratio. This is the <classical phase-space limit of the thermal kernel>. Here $\beta=1/(k_BT)$, with $k_B$ the <Boltzmann constant>; the associated classical position <variance> is $1/(\beta m\omega^2)$. At fixed $x$, $R$ ultimately tends to one as <temperature> tends to infinity, while its width grows.
At low <temperature>, $z\gg1$. Now $\tanh(z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$
$$
R=e^{-q^2}\left[1+2q^2e^{-z}+O(e^{-2z})\right],
\qquad
\boxed{R\simeq\exp\!\left(-\frac{m\omega x^2}{\hbar}\right).}
$$
Unlike the classical high-<temperature> expression, this limiting width remains nonzero at zero <temperature>, reflecting the <ground state> of the <quantum harmonic oscillator>.
\Image[/past-exam-of-the-advanced-quantum-theory-path-integrals-and-many-particle-physics-course-of-the-university-of-oxford/2024-oscillator-thermal-paths.png]
{title=Imaginary-time stationary paths and the crossover of the diagonal thermal kernel}
{description=Original plots of the <equal-endpoint Euclidean oscillator path> and the <diagonal thermal kernel of a quantum harmonic oscillator>, divided by its value at zero. Increasing $z=\beta\hbar\omega$ lowers the <temperature>: the path spends more of its interior near zero, and the ratio approaches the finite-width <ground state> profile.}
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