= Solution
Use units $\hbar=1$. The fluctuations about a <classical path> satisfy $f(0)=f(T)=0$. Expanding the <action functional> and integrating the term $m\dot q_c\dot f$ by parts gives
$$
S[q_c+f]=S[q_c]+\int_0^T(-m\ddot q_c-V^{\prime}(q_c))f\,dt+\frac12\int_0^T\{m\dot f^2-V^{\prime\prime}(q_c)f^2\}\,dt+O(f^3).
$$
The linear term vanishes by the <Euler-Lagrange equation> and the endpoint conditions. Integrating the quadratic <kinetic term> by parts yields the <Dirichlet fluctuation determinant of a semiclassical propagator>:
$$
\mathcal O_c=-m\frac{d^2}{dt^2}-V^{\prime\prime}(q_c(t)),\qquad K(q_1,q_0;T)\simeq e^{iS[q_c]}\int\mathcal Df\,\exp\left(\frac i2\int f\mathcal O_cf\,dt\right).
$$
A finite time-slicing first turns this into a product of ordinary <Gaussian integrals>, whose limit is proportional to $(\det\mathcal O_c)^{-1/2}$. A useful normalization relative to the free operator $\mathcal O_0=-m\,d^2/dt^2$ is
$$
D_c=\left(\frac{m}{2\pi iT}\right)^{1/2}\left(\frac{\det\mathcal O_c}{\det\mathcal O_0}\right)^{-1/2}.
$$
The oscillatory square root is fixed by the short-time limit and the <Feynman i-epsilon prescription>. Away from <zero modes> this gives the <semiclassical propagator>; if several <classical paths> contribute, their saddle contributions are summed with their corresponding determinant phases.
\b[For a general potential the prefactor depends on the endpoints through $q_c$; writing only $D(T)$ suppresses that dependence.] If $V(q)=aq^2/2+bq+c$, the Taylor expansion has no terms beyond quadratic and $V^{\prime\prime}=a$ is independent of $q_c$. The <Gaussian functional integral> is then exact, and its prefactor genuinely depends only on $T$ and the potential parameters. Singular focusing times are understood by continuation of the kernel, rather than by assigning a finite value to a zero-mode determinant.
For the <free-particle propagator>, $q_c(t)=q_0+(q_1-q_0)t/T$ and $S_c=m(q_1-q_0)^2/(2T)$. The free determinant supplies the factor $\sqrt{m/(2\pi iT)}$. Its normalization can also be checked by the momentum-space representation
$$
K_0=\int\frac{dp}{2\pi}\exp\left\{ip(q_1-q_0)-\frac{iTp^2}{2m}\right\}=\sqrt{\frac{m}{2\pi iT}}\exp\left\{\frac{im(q_1-q_0)^2}{2T}\right\},
$$
where completing the square evaluates the <Gaussian integral> and the short-time limit gives the position-space delta function.
For the gravitational potential, the original PDF contains $-mgq$; the TeX aid drops the final $q$. The <Euler-Lagrange equation> is $\ddot q_c=g$, and its endpoint solution is
$$
q_c(t)=q_0+\left(\frac{q_1-q_0}{T}-\frac{gT}{2}\right)t+\frac12gt^2.
$$
Substitution into the kinetic and potential terms gives
$$
S_c=\frac{m(q_1-q_0)^2}{2T}+\frac{mgT}{2}(q_1+q_0)-\frac{mg^2T^3}{24}.
$$
Since $V^{\prime\prime}=0$, the fluctuation operator is the free one. Thus the <quantum-mechanical propagator in a constant force> is exactly
$$
\boxed{K_g(q_1,q_0;T)=\sqrt{\frac{m}{2\pi iT}}\exp\left[i\left\{\frac{m(q_1-q_0)^2}{2T}+\frac{mgT}{2}(q_1+q_0)-\frac{mg^2T^3}{24}\right\}\right].}
$$
The <quantum-mechanical propagator> is $\langle q_1|e^{-iHT}|q_0\rangle$, with <Hamiltonian operator> $H=-\partial_q^2/(2m)+V(q)$. For imaginary time $-i\tau$, insert an orthonormal complete set of <energy eigenstates> to obtain
$$
\boxed{K(q_1,q_0;-i\tau)=\sum_n e^{-\tau E_n}\psi_n(q_1)\psi_n(q_0)^*,\qquad\tau>0.}
$$
Here $\psi_n(q)=\langle q|n\rangle$. This discrete sum assumes a discrete complete spectrum; continuous spectral components require the corresponding spectral integral. The heat-semigroup expression is well defined for an appropriate <self-adjoint> Hamiltonian bounded below. It is this spectral condition, not an unrestricted formal rotation for every conceivable potential, that justifies the imaginary-time kernel.
A path on the circle lifts to a path on the real line ending at $q_1+2\pi w$, where the integer $w$ is its <winding number>. The periodic quantum theory sums all these winding sectors with equal weight, giving the <real-time winding representation of a quantum rotor kernel>. For imaginary time, put $x=q_1-q_0$ and $y=\tau/m$ in the supplied Gaussian summation identity:
$$
K_{S^1}(q_1,q_0;-i\tau)=\sqrt{\frac{m}{2\pi\tau}}\sum_{w\in\mathbb Z}e^{-m(x+2\pi w)^2/(2\tau)}=\frac1{2\pi}\sum_{n\in\mathbb Z}e^{-\tau n^2/(2m)+inx}.
$$
Comparison with the spectral expansion determines
$$
\boxed{E_n=\frac{n^2}{2m},\qquad\psi_n(q)=\frac{e^{inq}}{\sqrt{2\pi}},\qquad n\in\mathbb Z.}
$$
The normalization uses $0\le q<2\pi$. The ground state is nondegenerate; levels with $n\ne0$ have the degeneracy $n\leftrightarrow-n$. Returning to real time replaces $e^{-\tau E_n}$ by $e^{-iTE_n}$, with the usual limiting prescription. No twisted boundary condition or flux phase is present in the stated image sum.
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