Deriving physics result for definite integral involving Dirac delta function

Question

I'm reading electrodynamics physics notes that describe a cavity of length $L$. The cavity is said to lie along the $z$-axis from $-L$ to $0$. There is a reflective mirror at $z = 0$, represented by an increase in the dielectric permittivity $\epsilon(z) = \epsilon_0[1 + \Lambda \delta(z)]$, where $\Lambda$ is a parameter that determines the reflectivity of the window.

To find how the reflective mirror affects the electromagnetic field, the notes integrate Maxwell's equations across the mirror (where there is an increase in the dielectric permittivity). The notes state that, from

$$\dfrac{\partial{E}(z, t)}{\partial{z}} = - \mu_0 \dot{H}(z, t),$$

we get that

$$E(0^+, t) = E(0^-, t),$$

and from

$$-\dfrac{\partial{H}(z, t)}{\partial{z}} = \epsilon(z) \dot{E}(z, t),$$

we get that there is a discontinuity of

$$H(0^+, t) - H(0^-, t) = - \Lambda \epsilon_0 \dot{E}(0, t)$$

Attempting to calculate this last one, I have

$$\begin{align} \int_{0^+}^{0^-} -\dfrac{\partial{H}(z, t)}{\partial{z}} \ dz = \int_{0^+}^{0^-} \epsilon(z) \dot{E}(z, t) \ dz \\ \Rightarrow -[H(0^-, t) - H(0^+, t)] = [\Lambda \epsilon_0 \dot{E}(z, t)]^{0^-}_{0^+} - \int_{0^+}^{0^-} \Lambda \epsilon_0 \dfrac{\partial{\dot{E}}(z, t)}{\partial{z}} \ dz \end{align}$$

I got $-[H(0^-, t) - H(0^+, t)]$ by the fundamental theorem of calculus, and I got $[\Lambda \epsilon_0 \dot{E}(z, t)]^{0^-}_{0^+} - \int_{0^+}^{0^-} \Lambda \epsilon_0 \dfrac{\partial{\dot{E}}(z, t)}{\partial{z}} \ dz$ by integration by parts, where $\int_{0^+}^{0^-} \epsilon_0[1 + \Lambda \delta(z)] = \Lambda \epsilon_0$, since, as I understand the Dirac delta function, $\int_{0^+}^{0^-} \delta(z) \ dz = 1$.

Continuing, I got that

$$\begin{align} [\Lambda \epsilon_0 \dot{E}(z, t)]^{0^-}_{0^+} - \int_{0^+}^{0^-} \Lambda \epsilon_0 \dfrac{\partial{\dot{E}}(z, t)}{\partial{z}} \ dz = \Lambda \epsilon_0 [\dot{E}(0^-, t) - \dot{E}(0^+, t)] - \Lambda \epsilon_0 \int_{0^+}^{0^-} \dfrac{\partial{\dot{E}}(z, t)}{\partial{z}} \ dz \end{align}$$

Using the result from above that $E(0^+, t) = E(0^-, t)$, we finally get

$$\Lambda \epsilon_0 [\dot{E}(0^-, t) - \dot{E}(0^+, t)] - \Lambda \epsilon_0 [\dot{E}(0^-,t) - \dot{E}(0^+, t)] = 0$$

But we wanted $- \Lambda \epsilon_0 \dot{E}(0, t)$, as shown in the notes above. What's going on here? Did I do something wrong?

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EDIT

Using the advice in the chat from the user Eric, my new work is as follows:

$$\begin{align} \int_{0^-}^{0^+} \epsilon(z) \dot{E}(z, t) \ dz = \int_{0^-}^{0^+} \left\{ \epsilon_0 [1 + \Lambda \delta(z)] \right\} \dot{E}(z, t) \ dz \\ = \int_{0^-}^{0^+} \epsilon_0 \dot{E}(z, t) \ dz + \int_{0^-}^{0^+} \epsilon_0 \Lambda \delta(z) \dot{E}(z, t) \ dz \end{align}$$

Using the property $\int f(x) \delta(x) \ dx = f(0)$, we get that

$$\begin{align} \int_{0^-}^{0^+} \epsilon_0 \dot{E}(z, t) \ dz + \int_{0^-}^{0^+} \epsilon_0 \Lambda \delta(z) \dot{E}(z, t) \ dz = 0 + \epsilon_0 \Lambda \dot{E}(0, t) \end{align}$$

Now, we also have that

$$\int_{0^-}^{0^+} - \dfrac{\partial{H}(z, t)}{\partial{z}} \ dz = -[H(0^+, t) - H(0^-, t)]$$

So we would then have that

$$-[H(0^+, t) - H(0^-, t)] = \epsilon_0 \Lambda \dot{E}(0, t) \\ \Rightarrow H(0^+, t) - H(0^-, t) = -\epsilon_0 \Lambda \dot{E}(0, t),$$

as required.

But this leaves me with three questions:

1. Why do we have $\int_{0^-}^{0^+}$ instead of $\int_{0^+}^{0^-}$?
2. Why is $\int_{0^-}^{0^+} \dot{E}(z, t) \ dz = 0$ (or just $\int_{0^-}^{0^+} E(z, t) \ dz$)?
3. What exactly is this property $\int f(x) \delta(x) \ dx = f(0)$? Is this the Heaviside function?

Question 2 is a physics question (so I'll leave it for physics.stackexchange), but 1 and 3 are mathematics questions.

Answer 1

I can answer your first and third questions3

$\boldsymbol{1:}$

This is purely a notational matter. $0^-$ is a notation that means $0$, from the left, so to speak. In other words we would say $$f(0^-):=\lim_{x\to 0 \\ \text{from the left}}f(x)$$ When integrating, we like to integrate from left to right, so it makes much more sense to write $\int_{0^-}^{0^+}$ than the other way around. A more rigorous way of dealing with these kinds of integrals would be to let $a>0$ be a small positive number and consider the integrals $$\int_{-a}^{a}\epsilon(z)\dot{E}(z,t)\mathrm dz$$ And analyze the limiting case $a\to 0$.

$\boldsymbol{3:}$

I'm not sure in what capacity you've encountered the Dirac delta before, but it is defined as satisfying the property $$\int_U f(x)\delta(x)\mathrm dx=f(0)$$ Where $U$ is some open set containing $0$. Formally speaking, $\delta$ is a distribution, which is a map from functions on $\mathbb{R}$ to $\mathbb{R}$ itself.

In fact, let $V$ be the real vector space of all piecewise continuous functions from $\mathbb{R}\to\mathbb{R}$. (Checking that this is actually a vector space is very easy.) Then we can write very succinctly, $$\delta\in V^*$$ Where $V^*$ is the dual space.