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

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – G. Smith, GiorgioP, Jon Custer, tpg2114
If this question can be reworded to fit the rules in the help center, please edit the question.

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.