= 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 $L=vt$,
$$
\frac{E_2-E_1}{\hbar}=\frac{\alpha}{t^2},\qquad
\alpha=\frac{3\pi^2\hbar}{2mv^2}.
$$
The relative <dynamical phase> is $-\alpha/t$ plus a constant. Up to an irrelevant phase,
$$
a_2(t)\simeq\frac43\int_0^t\frac{e^{-i\alpha/s}}{s}\,ds,
\qquad
\boxed{P_{1\to2}\simeq\frac{16}{9}\left|\int_0^t\frac{e^{-i\alpha/s}}s\,ds\right|^2}.
$$
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 $t_0\downarrow0$ limit of the <probability>. For $t/\alpha\ll1$, endpoint asymptotics give $P\sim16t^2/(9\alpha^2)$; the first-order expression is reliable only while total excitation remains small.
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