= Solution
Use <position eigenstates> normalized by $\langle q|q'\rangle=\delta(q-q')$, and the free <Hamiltonian operator> $\widehat H=\widehat p^2/(2m)$. The time-sliced <path integral> is the matrix element of the <time-evolution operator>:
$$
\boxed{\mathcal I[q^{(i)},q^{(f)}]=\langle q^{(f)}|e^{-iT\widehat H/\hbar}|q^{(i)}\rangle.}
$$
Insert <momentum eigenstates>, with $\langle q|p\rangle=(2\pi\hbar)^{-1/2}e^{ipq/\hbar}$. The <free-particle propagator> becomes
$$
\mathcal I=\int_{-\infty}^{\infty}\frac{dp}{2\pi\hbar}\exp\!\left(\frac{ip\Delta q}{\hbar}-\frac{iTp^2}{2m\hbar}\right),\qquad\Delta q=q^{(f)}-q^{(i)}.
$$
Complete the square in the exponent:
$$
\frac{ip\Delta q}{\hbar}-\frac{iTp^2}{2m\hbar}=-\frac{iT}{2m\hbar}\left(p-\frac{m\Delta q}{T}\right)^2+\frac{im(\Delta q)^2}{2\hbar T}.
$$
The <Fresnel integral>, defined by the damping prescription $T\to T-i0$, gives, for $T>0$,
$$
\boxed{\mathcal I=\left(\frac{m}{2\pi i\hbar T}\right)^{1/2}\exp\!\left(\frac{im(\Delta q)^2}{2\hbar T}\right).}
$$
The square-root branch is $i^{-1/2}=e^{-i\pi/4}$; it is fixed by the regulated <Gaussian integral> and the short-time <Dirac delta distribution> limit, rather than an arbitrary phase.
The <Euler-Lagrange equation> for the <free particle> is $m\ddot q=0$. Its unique classical path satisfying the endpoint <boundary conditions> is $q_{\rm cl}(t)=q^{(i)}+\Delta q\,t/T$. Therefore its classical <action> is
$$
\boxed{S_{\rm cl}=\int_0^T\frac m2\left(\frac{\Delta q}{T}\right)^2dt=\frac{m(\Delta q)^2}{2T}.}
$$
Substituting into the operator result yields $\mathcal I=C(T)e^{iS_{\rm cl}/\hbar}$, with $C(T)=(m/(2\pi i\hbar T))^{1/2}$ independent of the endpoint positions.
The same independence is transparent directly in the <path integral>. Write $q=q_{\rm cl}+\eta$, with $\eta(0)=\eta(T)=0$. The cross term in the <action> is $m\dot q_{\rm cl}\int_0^T\dot\eta\,dt=0$, so $S[q]=S_{\rm cl}+(m/2)\int_0^T\dot\eta^2dt$. The fluctuation integral depends only on $m,T,\hbar$. Because the <action> is quadratic, the factorization is exact; the <semiclassical propagator> already equals the full answer.
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