 Solution

Measure intermediate time $u$ from $t^{\prime }$. The classical path with the fixed endpoints isIt satisfies $m{\ddot{x}}_c=-\alpha$. Direct integration of $m{\dot{x}}_c^2/2-\alpha x_c$ givesFor an arbitrary path write $x=x_c+\eta$ with $\eta (0)=\eta (T)=0$. Integrating the cross term by parts gives $-\int (m{\ddot{x}}_c+\alpha )\eta du=0$. Thus $S[x]=S_c+\int m{\dot{\eta }}^{2}du/2$ exactly. The fluctuation path integral is the free endpoint-fixed one, so its determinant supplies $\sqrt{m/(2\pi i\hslash T)}$. Multiplication by $\theta (T)e^{i{S}_c/\hslash }$ reproduces the previous kernel. This exact classical-path factorization of a quadratic path integral is possible because a linear potential introduces no additional quadratic fluctuation term.