Delta function and potential step [closed]

I have a potential consisting of an attractive delta funtion well located at the origin and a superimposed with a potential step at the origin, just like: With $$V(x)=-\lambda \delta (x) +V_0 H(x)$$

They ask me to find the analytical expression for the eigenfunctions in different cases.

My problem is one doubt about one integral:

Trying to resolve (for example $$0), I know that, applying the conditions, I have to solve:

$$\frac{-\hbar^2}{2m}\int_{-\epsilon}^{\epsilon}\frac{d^2 \psi(x)}{dx^2}dx+\int_{-\epsilon}^{\epsilon}V(x)\psi(x)dx=E\int_{-\epsilon}^{\epsilon}\psi(x)dx$$

Taking $$\lim_{\epsilon \rightarrow 0}$$ We finnaly have:

$$\Delta \left( \frac{d\psi(x)}{dx}\right)=\frac{2m}{\hbar}\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}V(x)\psi(x)dx$$

My doubt appears when I try to solve the integral at the right side of the expression with $$V(x)=-\lambda \delta (x) +V_0 H(x)$$

$$\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}V(x)\psi(x)dx=\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}\left[-\lambda \delta (x) +V_0 H(x)\right]\psi(x)dx$$

Having finally: $$\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}\left[-\lambda \delta (x) +V_0 H(x)\right]\psi(x)dx=\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}-\lambda \delta(x)\psi(x)dx + \lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}V_0 H(x) \psi(x)dx=-\lambda\psi(0) +\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}V_0 H(x) \psi(x)dx$$

How do I solve $$\lim_{\epsilon\rightarrow 0}\int_{-\epsilon}^{\epsilon}V_0 H(x) \psi(x)dx$$ It's just equal to $$0$$?

closed as off-topic by G. Smith, GiorgioP, Jon Custer, tpg2114♦Jun 14 at 10:49

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Yes the integral is zero. Since the value of the step function is zero for $$x<0$$, and equal to 1 for $$x\geq0$$, you can write the integral as,
$$\int_{-\epsilon}^{\epsilon} V_0 H(x) \psi(x) dx = \int_{0}^{\epsilon} V_0 \psi(x) dx$$
Since $$\psi(x)$$ is continuous at $$x=0$$, integral of $$\psi(x)$$ is also continuous at zero. Hence taking the limit $$\epsilon$$ going to zero yields zero as the answer.