Delta function integral

I encountered an issue when doing some problems in solid state physics, and I spent a whole day trying to clear this up, unsuccessfully. I'm posting it here because my issue is purely mathematical.

Namely, there appears an integral of the Dirac $\delta$-function of this form: $$n(E)=\frac{Na}{\pi}\int_\frac{-\pi}{a}^\frac{\pi}{a} \delta(E-E_s+2J\cos ka)dk$$

where $N,a,E,E_s,J,k$ are real.

This is supposed to be:

$$n(E)=\frac{N}{\pi J\sqrt{1-(\frac{E-E_s}{2J})^2}}(\theta(E-E_b)-\theta(E-E_t))$$

where $E_b=E_s-2J, E_t=E_s+2J$, and $\theta$ is the Heaviside step function.

Now, the integral of $\delta$ is either $1$ or $0$, depending whether the argument vanishes inside the integral limits or not. How exactly is this fact transformed into those terms involving $\theta$? I have tried manipulating using definitions of $\theta$, but to no avail.

I suppose this may be very elementary, but I'd be grateful for any help.

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Who is $a$ that doesn't appear in the result? –  checkmath Feb 20 '12 at 9:40
sos440's answer has all the computations; brief answers to your conceptual questions: The integral of $\delta$ is either $1$ or $0$ if you integrate it over its argument, but you're integrating over something else, so you first have to transform the integral as in sos440's computation before you can apply the $0$/$1$ rule. And regarding the step functions: their difference is indeed either $0$ or $1$, and is $1$ iff the zeros of the argument of the delta function lie in the interval of integration. If so, you get sos440's answer, and if not, $0$. –  joriki Feb 20 '12 at 9:46
Am I missing something, or should it be $E-E_S$ in the integrand? –  Ansgar Esztermann Feb 20 '12 at 9:47
@AnsgarEsztermann Yes, thank you. Corrected. –  vedran Feb 20 '12 at 9:56
@joriki Now, things are becoming clearer. Thank you! :) –  vedran Feb 20 '12 at 9:58

We first derive a simple formula for delta function integral. Let assume $\Omega$ is an open set and $f : \Omega \to \mathbb{R}$ is a $C^1$-function with finite zeros $x_1, \cdots, x_n$. We also assume that $f'(x_i) \neq 0$. Then there are sufficiently small neighborhoods $\mathcal{U}_i$ of $x_i$ such that

1. $\mathcal{U}_i \cap \mathcal{U}_j = \varnothing$ if $i \neq j$ and
2. $f$ has local $C^1$-inverse $g_i$ on each $\mathcal{U}_i$.

Then \begin{align*} \int_{\Omega} \delta(f(x)) \; dx & = \sum_{i} \int_{\mathcal{U}_i} \delta(f(x)) \; dx \\ & = \sum_{i} \int_{f(\mathcal{U}_i)} \delta(y) |g_i'(y)| \; dy \\ & = \sum_{i} |g_i'(0)| \\ & = \sum_{i} \frac{1}{|f'(x_i)|}. \end{align*}

Now let's return to the original problem. It is sufficient to assume that $|E - E_s| < 2J$, since we have $n(E) = 0$ for $|E - E_s| > 2J$. Then we may let $$\frac{E - E_s}{2J} = -\cos\alpha$$ with $0 < \alpha < \pi$. Then \begin{align*} n(E) &= \frac{N}{\pi} \int_{-\pi}^{\pi} \delta(E-E_s + 2J \cos x) \; dx \qquad (x = ka) \\ &= \frac{N}{\pi} \left[ \left.\frac{1}{2J|\sin x|}\right|_{x = -\alpha} + \left.\frac{1}{2J|\sin x|}\right|_{x = \alpha} \right] \\ &= \frac{N}{\pi J \sin \alpha} \\ &= \frac{N}{\pi J\sqrt{1-(\frac{E-E_s}{2J})^2}}, \end{align*} as desired.

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The formula for for delta function integral is most helpful and illuminating! Still, your end result is missing the $\theta$ terms, or are you claiming that $\theta(E-E_b)-\theta(E-E_t)=1$ for all $E$? ($E_t,E_B$ as defined above) –  vedran Feb 20 '12 at 9:53
@vedran, Note that $\theta(E−E_b)−\theta(E−E_t) = 1$ if $|E-E_s|<2J$ and $\theta(E−E_b)−\theta(E−E_t) = 0$ if $|E-E_s| > 2J$. So the step function term is just a rephrasal of the case expression. –  Sangchul Lee Feb 20 '12 at 10:04
So the rephrasal was what was bothering me. :/ As I said, this was rather elementary. Thank you so much for clearing this up for me! –  vedran Feb 20 '12 at 10:10