= Solution
For the \i[<amplitude>] continue to $k=i\kappa$ on the decaying physical sheet. Then $\zeta=q$ and $D=\cos(qa)+(\kappa^2-q^2)\sin(qa)/(2\kappa q)$. The double-angle identity factors its zero condition into the even and odd <Matching condition>[matching conditions] $\kappa=q\tan(qb)$ and $\kappa=-q\cot(qb)$. Thus physical upper-half-plane <amplitude> <Pole>[poles] are exactly the bound energies.
\b[The literal statement for the <probability> requires correction.] The displayed real-axis expression continued algebraically is $1/[D(k)D(-k)]$, and includes <Pole>[poles] of the opposite-boundary-condition <amplitude> too. A concrete counterexample uses $\hbar^2/(2m)=1$, $a=2$, $q=1$, $\kappa=\cot1$ and $U_0=1+\cot^2 1$. Its continued <probability> denominator vanishes at $k=i\cot1$, because $D(-k)=0$. But neither $\kappa=\tan1$ nor $\kappa=-\cot1$ holds, so $E=-\cot^2 1$ is not a bound <energy>. Keeping the <amplitude> and the decaying <boundary condition> removes this spurious identification.
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