 Solution

Analytically continue the supplied free propagator using $M=I$ and $t=-i\hslash \beta$. This gives $Z_0=\sqrt{I/(2\pi {\hslash }^2\beta )}$ without an arbitrary measure constant.

Apply the Poisson summation formula to $h(x)=e^{-A{x}^2}$, with $A=2{\pi }^{2}I/({\hslash }^2\beta )$. Its Gaussian integral isThe prefactors cancel exactly because $2\pi Z_0\sqrt{\pi /A}=1$. Therefore the momentum-winding duality of the quantum rotor givesin 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 $Z\to 1$. At high temperature the winding sum is dominated by m=0 and $Z\sim \sqrt{2\pi I/(\beta {\hslash }^2)}$, the classical angular phase-space integral. Winding and momentum labels are dual descriptions; keeping only m=0 at low temperature would be incorrect.