= Solution
Analytically continue the free-particle return kernel using $M=I$ and $t=-i\hbar\beta$. This fixes
$$
\boxed{\mathcal Z_0=\sqrt{\frac{I}{2\pi\hbar^2\beta}}.}
$$
Set $a=2\pi^2I/(\hbar^2\beta)$. The Gaussian Fourier integral in <Poisson resummation> is
$$
\int_{-\infty}^\infty dx\,e^{-ax^2+2\pi inx}=\sqrt{\frac\pi a}\,e^{-\pi^2n^2/a}.
$$
Applying it to the winding sum gives $\sum_me^{-am^2}=\sqrt{\pi/a}\sum_ne^{-\pi^2n^2/a}$. The prefactors cancel exactly, $2\pi\mathcal Z_0\sqrt{\pi/a}=1$, and $\pi^2/a=\beta\hbar^2/(2I)$. Therefore
$$
\boxed{\mathcal Z=\sum_{n\in\mathbb Z}e^{-\beta\hbar^2n^2/(2I)},}
$$
in 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 $\mathcal Z\simeq\sqrt{2\pi I/(\beta\hbar^2)}$.
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