= Solution
Work in units $\hbar=1$ and with the particle <mass> absorbed into the stated kinetic term. Divide the interval into $N$ pieces of length $\Delta=T/N$. For a suitable self-adjoint <Hamiltonian operator>, the <Trotter product formula> gives the regulated limit
$$
e^{-i\hat HT}=\lim_{N\to\infty}\left(e^{-i\Delta\hat p^2/2}e^{-i\Delta V(\hat q)}\right)^N.
$$
Insert $1=\int dq_j\,|q_j\rangle\langle q_j|$ between neighboring factors. Because $V(\hat q)$ is diagonal in the <position operator> basis, the given <free-particle propagator> yields
$$
\begin{aligned}
K(q',q;T)
&=\lim_{N\to\infty}(2\pi i\Delta)^{-N/2}\int\prod_{j=1}^{N-1}dq_j\,
\exp\left[i\sum_{j=0}^{N-1}\left\{\frac{(q_{j+1}-q_j)^2}{2\Delta}-\Delta V(q_j)\right\}\right],\\
q_0&=q,\qquad q_N=q'.
\end{aligned}
$$
The exponent is the time-sliced <action>. Its continuum notation is
$$
\boxed{K(q',q;T)=\int_{q(0)=q}^{q(T)=q'}\mathcal Dq\,e^{iS[q]},\qquad
S[q]=\int_0^T\left(\frac12\dot q^2-V(q)\right)dt.}
$$
The measure is defined by the displayed finite-dimensional limits and their normalization, not by a translation-invariant flat measure on an infinite-dimensional space. The oscillatory limit needs a damping/analytic-continuation prescription. This is the <time-sliced configuration-space path integral> derived from the operator kernel.
For a quadratic <potential energy>, write $q=q_c+\eta$, where the <classical path> satisfies $\ddot q_c+V'(q_c)=0$ and the endpoint conditions, and $\eta(0)=\eta(T)=0$. Expansion is exact:
$$
S[q_c+\eta]=S[q_c]+\int_0^T(\dot q_c\dot\eta-V'(q_c)\eta)dt
+\frac12\int_0^T(\dot\eta^2-V''\eta^2)dt.
$$
<Integration by parts> makes the linear term equal to $-[\int_0^T(\ddot q_c+V'(q_c))\eta\,dt]$, since the endpoint term vanishes. It is zero by the <Euler-Lagrange equation>. Also $V''$ is constant, independent of the endpoints. Thus <classical-path factorization of a quadratic path integral> is exact:
$$
\boxed{K(q',q;T)=N(T)e^{iS[q_c]},\qquad
N(T)=\int_{\eta(0)=\eta(T)=0}\mathcal D\eta\,
\exp\left[-\frac i2\int_0^T\eta\left(\frac{d^2}{dt^2}+V''\right)\eta\,dt\right].}
$$
Only the classical <action> carries the endpoint dependence.
Let $D=\partial_t^2+V''$ on homogeneous <Dirichlet boundary conditions>. Expand $\eta$ in its normalized eigenfunctions, with <eigenvalues> $\lambda_n$. The fluctuation integral becomes a product of regulated integrals $\int da_n\exp(-i\lambda_na_n^2/2)$. Each is proportional to $\lambda_n^{-1/2}$, with a prescribed complex phase. Relative to $D_0=\partial_t^2$, the normalization is therefore
$$
\boxed{N(T)=\frac1{\sqrt{2\pi iT}}\left(\frac{\det_D D}{\det_D D_0}\right)^{-1/2}.}
$$
The subscript denotes the Dirichlet operator domain, and the square-root branch is continued from the free kernel with its convergence prescription. Equivalently one can use <determinants> of $-D$ and $-D_0$. For $V''=\omega^2$, the <Dirichlet oscillator determinant ratio> is
$$
\prod_{n=1}^{\infty}\left(1-\frac{\omega^2T^2}{\pi^2n^2}\right)
=\frac{\sin\omega T}{\omega T},\qquad
N(T)=\left(\frac{\omega}{2\pi i\sin\omega T}\right)^{1/2}.
$$
This also checks the free limit. At conjugate times, the fluctuation operator has a zero <eigenvalue> and the ordinary finite-prefactor expression fails; the kernel is obtained by distributional continuation. Away from these times the classical boundary problem is unique and the stated factorization applies directly.
For the infinite-time oscillator with an <external source>, use vacuum boundary conditions and the <Feynman i-epsilon prescription>. For a smooth <source> of compact support, define
$$
A_\epsilon=-\partial_t^2-m^2+i\epsilon,\qquad
G_\epsilon=A_\epsilon^{-1},\qquad\epsilon>0.
$$
The imaginary term damps the real-field integral. Integration by parts gives $S_\epsilon[q]=\tfrac12qA_\epsilon q+Jq$, with the pairings understood as time integrals. <Completing the square> gives
$$
\frac12qA_\epsilon q+Jq
=\frac12(q+G_\epsilon J)A_\epsilon(q+G_\epsilon J)-\frac12JG_\epsilon J.
$$
Translate the integration variable. The source-independent <Gaussian functional integral> is $N$, so
$$
\boxed{\int\mathcal Dq\,e^{iS[q]}=N\exp\left[-\frac i2\int dt\,dt'\,J(t)G(t-t')J(t')\right].}
$$
On <Fourier modes> $e^{i\omega t}$, $A_\epsilon$ has <eigenvalue> $\omega^2-m^2+i\epsilon$, hence
$$
\boxed{G(t)=\lim_{\epsilon\downarrow0}\frac1{2\pi}\int_{\mathbb R}\frac{e^{i\omega t}}{\omega^2-m^2+i\epsilon}\,d\omega.}
$$
For $m>0$, closing the contour above for $t>0$ and below for $t<0$ gives $G(t)=-ie^{-im|t|}/(2m)$. The physical time-ordered oscillator <quantum field theory propagator> is $iG$, not $G$ itself. The shifted stationary path is $q_c=-GJ$ and obeys $\ddot q_c+m^2q_c=J$ with the same vacuum prescription. A retarded inverse would define a different boundary problem. Dividing by the zero-source integral removes $N$ and gives the <vacuum source functional of a quantum oscillator>. At zero <mass> an additional infrared prescription is needed for the inverse on constant modes.
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