= Solution
Write $L_0\psi=-U\psi$ and use the outgoing inverse of $L_0$. Adding the homogeneous incoming wave yields
$$
\psi=e^{ikx}-\frac{im}{k\hbar^2}\int e^{ik|x-x'|}U(x')\psi(x')\,dx'.
$$
For the attractive delta potential, let $\eta=m\alpha/(k\hbar^2)$. The equation reduces to $\psi(x)=e^{ikx}+i\eta e^{ik|x|}\psi(0)$ and $\psi(0)=1/(1-i\eta)$. Separating the two half-lines gives
$$
\boxed{t=\frac1{1-i\eta},\qquad r=\frac{i\eta}{1-i\eta},\qquad
\psi=\begin{cases}e^{ikx}+re^{-ikx},&x<0,\\te^{ikx},&x>0.\end{cases}}
$$
Both sides have the same <group velocity>, so
$$
\boxed{T=\frac1{1+\eta^2},\qquad R=\frac{\eta^2}{1+\eta^2},\qquad T+R=1.}
$$
The same amplitudes satisfy continuity and the exact delta-potential <derivative> jump $\psi'(0^+)-\psi'(0^-)=-2m\alpha\psi(0)/\hbar^2$, providing an independent local check on the sign.
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