 Solution

For first-order excitation in a linearly expanding infinite well, retain only the ground-state amplitude on the right. For the second level, the coupling coefficient is $4\dot{L}/(3L)$. With $L=vt$,The relative dynamical phase is $-\alpha /t$ plus a constant. Up to an irrelevant phase,The lower endpoint is an oscillatory improper limit: substituting $u=\alpha /s$ gives ${\int }_{\alpha /t}^{\infty }{e}^{-iu}du/u$. A physical initial state cannot occupy a zero-width well. Prepare the ground state at $t_0>0$, integrate from $t_0$, and then take the formal $t_0\downarrow 0$ limit of the probability. For $t/\alpha \ll 1$, endpoint asymptotics give $P\sim 16t^2/(9{\alpha }^2)$; the first-order expression is reliable only while total excitation remains small.