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If $K$ is the zero-Dirichlet heat kernel on an interval for $w_t=w_{xx}-cw$, integration by parts givesThe opposite endpoint signs are essential. Nonzero boundary values are limits of the full formula; evaluating each homogeneous sine mode at the boundary prematurely loses the forcing.

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 $e^{-S/\hslash }$.

Put $a=\alpha /2$ and apply the Dirichlet gauge transform for constant drift $w(x,t)=e^{ax}u(x,t)$. Taking derivatives directly removes the advection term and gives the damped heat equation $w_t=w_{xx}-a^{2}w$. Thus $w_0(y)=e^{ay}{u}_0(y)$, while the two Dirichlet boundary conditions become $w(0,t)=g_0(t)$ and $w(L,t)=e^{aL}h(t)$.

The Dirichlet heat kernel on an interval and its damped version areThe method of images gives an alternative, often better at short times:To determine the boundary signs, multiply the equation for $w$ by the backward heat kernel $K_a(x,y,t-s)$ and use integration by parts in $y$. Since the kernel vanishes at $y=0,L$, the surviving boundary expression is $K_{a,y}(x,0,t-s)g_0(s)-K_{a,y}(x,L,t-s)e^{aL}h(s)$. Consequently an integral representation containing only the given data isThis is a Dirichlet boundary-forcing heat-kernel formula. For $0<x<L$ the short-time Gaussian decay makes the forcing integral well defined. The endpoint values are interior limits of the complete formula: evaluating a termwise sine series at an endpoint before taking the time integral loses the nonzero boundary values. At the initial corners a continuous classical solution requires $u_0(0)=g_0(0)$ and $u_0(L)=h(0)$; otherwise the same formula describes the solution away from those corners.

Here is a second integral representation, useful when the integral transforms are explicit. Extend $g_0,h$ by zero after $T$; the values of the extension cannot affect a solution at $t<T$. Write their Laplace transforms as $G(p),H(p)$ and set $q=\sqrt{p+a^2}$ using the principal square root. The transformed Dirichlet Green function for $-{\partial }_x^2+q^2$ isIt vanishes at both endpoints and its first derivative in $x$ jumps by $-1$, which verifies the sign of its unit source. The resolvent kernel for Dirichlet advection-diffusion on an interval gives the transformed solutionThe Bromwich inversion formula therefore yieldsThe square-root notation creates no genuine branch singularity here: all three kernels are even in $q$. Their only spatial resolvent operator poles are $p=-a^2-q_n^2$. This transformed formula and the causal heat kernel formula represent the same solution in the usual smooth-data class.

Reflecting an interval repeatedly across its endpoints maps a reflected path to a free path in an image interval. A return trajectory from $q\in (0,L)$ ends at $q+2rL$ after an even number of wall reflections, or $-q+2rL$ after an odd number. An infinite square well has reflection amplitude $-1$, so the corresponding Dirichlet heat kernel on an interval is the difference of direct and mirrored free kernels. Individual straight paths illustrate the geometry; the path integral sums all time-sliced paths with the same endpoints.