Consider $\Psi_{xx}+(\lambda-u(x))\Psi=0$ and $u(x)=-U_0\delta(x)$, where $U_0$ is a constant and $\delta(x)$ is Dirac's delta function.
Integrating the equation from $-\varepsilon$ to $+\varepsilon$, one gets $$ [\Psi_x]_{-\varepsilon}^\varepsilon=\int_{-\varepsilon}^\varepsilon (\lambda+U_0\delta(x))\Psi\, dx. $$
Now it is said that, if $\varepsilon\to 0$, one gets $$ [\Psi_x]=-U_0\Psi(0),\tag{*} $$ where $[\cdot]$ denotes the change in $\Psi_x$, provided $\Psi$ is continuous at $x=0$.
Can someone explain to me how we get $(*)$? I do not see it. In particular, where does the (assumed) continuity of $\Psi$ at $x=0$ come into play?
If you take the fourier transform wrt $x$: $$\mathcal{F}[\varphi(x,t)]\{\omega\}=\int_{-\infty}^\infty \varphi e^{-j\omega x}\,dx=\hat{\varphi}(\omega,t)$$ and assume that: $$\lim_{x\to\pm\infty}\varphi=\lim_{x\to\pm\infty}\varphi_x=0$$ you arrive at the equation: $$(\lambda-\omega^2)\hat{\varphi}+U_0\varphi(0,t)=0$$ then you can use the inverse fourier transform and boundary conditions for $\varphi$.