= Solution
Analytically continue the supplied free propagator using $M=I$ and $t=-i\hbar\beta$. This gives $Z_0=\sqrt{I/(2\pi\hbar^2\beta)}$ without an arbitrary measure constant.
Apply the <Poisson summation formula> to $h(x)=e^{-Ax^2}$, with $A=2\pi^2I/(\hbar^2\beta)$. Its <Gaussian integral> is
$$
\int_{-\infty}^\infty dx\,e^{-Ax^2+2\pi inx}=\sqrt{\frac\pi A}\,e^{-\pi^2n^2/A}.
$$
The prefactors cancel exactly because $2\pi Z_0\sqrt{\pi/A}=1$. Therefore the <momentum-winding duality of the quantum rotor> gives
$$
\boxed{2\pi Z_0\sum_me^{-Am^2}=\sum_ne^{-\beta\hbar^2n^2/(2I)},}
$$
in agreement with the spectrum. The <thermal winding dominance of a free rotor> interchanges between the two representations. At low temperature the momentum sum is dominated by n=0 and $Z\to1$. At high temperature the winding sum is dominated by m=0 and $Z\sim\sqrt{2\pi I/(\beta\hbar^2)}$, the classical angular phase-space integral. Winding and momentum labels are dual descriptions; keeping only m=0 at low temperature would be incorrect.
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