= Solution
Apply the <Poisson summation formula> in the convention supplied, taking
$$
f(x)=\exp\left(-\frac{2\pi^2i\hbar T}{\Lambda}x^2\right),\qquad\Theta=\Delta.
$$
Completing the square in its <Fourier transform> and evaluating the analytically continued <Gaussian integral> gives
$$
\widetilde f(k)=\sqrt{\frac{\Lambda}{2\pi i\hbar T}}\exp\left(\frac{i\Lambda k^2}{2\hbar T}\right).
$$
Thus the spectral sum becomes
$$
\boxed{K(\theta_f,\theta_i;T)=\sqrt{\frac{\Lambda}{2\pi i\hbar T}}\sum_{m\in\mathbb Z}e^{iS_m/\hbar}.}
$$
The square-root branch is fixed by continuation from positive imaginary-time damping. Each term is the classical contribution of part (b), multiplied by the same fluctuation prefactor. Since the <action> is quadratic, writing $\theta=\theta_m+\eta$ with $\eta(0)=\eta(T)=0$ removes the cross term and leaves a free Gaussian fluctuation integral independent of $m$. This <real-time winding representation of a quantum rotor kernel> is exact, rather than only a leading semiclassical approximation.
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