= Solution
As $x\to-\infty$, the scattered kernel is $e^{-ikx}e^{ikx'}$; as $x\to+\infty$, it is $e^{ikx}e^{-ikx'}$. Therefore $\psi_s\sim e^{ikx}+r e^{-ikx}$ on the left and $t e^{ikx}$ on the right, with
$$
r=-\frac{im}{k\hbar^2}\int e^{ikx'}U(x')\psi_s(x')dx',\quad
t=1-\frac{im}{k\hbar^2}\int e^{-ikx'}U(x')\psi_s(x')dx'.
$$
Equal asymptotic wavenumbers give $R=|r|^2$, $T=|t|^2$. Squaring the first <amplitude> proves the requested expression. These are current ratios, which would require a <velocity> factor if the two asymptotic potentials differed.
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