 Solution

Analytically continue the free-particle return kernel using $M=I$ and $t=-i\hslash \beta$. This fixesSet $a=2{\pi }^{2}I/({\hslash }^2\beta )$. The Gaussian Fourier integral in Poisson resummation isApplying it to the winding sum gives ${\sum }_m{e}^{-a{m}^2}=\sqrt{\pi /a}{\sum }_n{e}^{-{\pi }^2{n}^2/a}$. The prefactors cancel exactly, $2\pi {\mathcal{Z}}_0\sqrt{\pi /a}=1$, and ${\pi }^2/a=\beta {\hslash }^2/(2I)$. Thereforein agreement with the spectral trace. This is momentum-winding duality of the quantum rotor. At low temperature the momentum ground state dominates; at high temperature the winding-zero sector gives $\mathcal{Z}\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$.