= Solution
Make the <Wick rotation> $t'=-i\tau$, and write $T_E=\beta\hbar$ for the final <imaginary time>, where $\beta$ is <inverse temperature> and $\hbar$ is the <reduced Planck constant>. Then $dt'=-i\,d\tau$ and $dx/dt'=i\,dx/d\tau$. Thus the continued <action> is
$$
S=i\int_0^{T_E}\left[\frac m2\left(\frac{dx}{d\tau}\right)^2+V(x)\right]d\tau
=iS_{\mathrm{Euc}},
\qquad
\boxed{S_{\mathrm{Euc}}[x]=\int_0^{\beta\hbar}\left[\frac m2\dot x^2+V(x)\right]d\tau.}
$$
Both the <kinetic energy> and the <potential energy> enter this <Euclidean action> with positive signs relative to their real-time definitions. Also $e^{-iH(-i\beta\hbar)/\hbar}=e^{-\beta H}$, so the <Euclidean path integral> is
$$
\boxed{\langle x_f|e^{-\beta H}|x_i\rangle
=\int_{x(0)=x_i}^{x(T_E)=x_f}\mathcal D x\,
e^{-S_{\mathrm{Euc}}[x]/\hbar}.}
$$
Its <boundary conditions> fix the two endpoint <positions>. They need not coincide: identifying and integrating over them would instead give the <thermal trace>.
The normalization of $\mathcal D x$ can be fixed without continuing a formal measure. Apply the <Trotter product formula> to $H=\hat p^2/(2m)+V(\hat x)$, divide $T_E$ into $M$ intervals of size $\epsilon=T_E/M$, and integrate the intermediate <positions>. The resulting <time-sliced configuration-space path integral> is
$$
\lim_{M\to\infty}
\left(\frac{m}{2\pi\hbar\epsilon}\right)^{M/2}
\int_{\mathbb R^{M-1}}\prod_{j=1}^{M-1}dx_j\,
\exp\!\left[-\frac1\hbar\sum_{j=1}^M
\left\{\frac{m(x_j-x_{j-1})^2}{2\epsilon}+\epsilon V(x_{j-1})\right\}\right],
\quad x_0=x_i,\quad x_M=x_f.
$$
This also explains the upper limit $\beta\hbar$ rather than $\beta$ when the <reduced Planck constant> has not been set to one.
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