 Solution

Use unfolding a reflected interval trajectory: reflect the interval across each wall so that a bouncing trajectory becomes a free trajectory in an image interval. A return path starting at $q$ ends at $q+2rL$ after an even number of reflections, or at $-q+2rL$ after an odd number. Each hard-wall reflection contributes phase $-1$, giving the positive direct images and negative reflected images of the Dirichlet heat kernel on an interval.

The method of images therefore givesHere $\tau$ has inverse-energy units, since $\beta =1/(k_{B}T)$. In physical imaginary time $u=\hslash \tau$, the same dimensionless action is $\mathcal{A}=S_E/\hslash$, with $S_E={\int }_0^{\hslash \beta }(m/2)(dq/du{)}^{2}du$. Thus the question's convention $S=(m/(2\hslash )){\int }_0^{\beta }({\partial }_{\tau }q{)}^{2}d\tau$ has exactly the weight $e^{-S/\hslash }$.