Solving $\int_{-\pi}^\pi \delta[E-A-B\cos x-C\cos (2x)]$

Question

I'm trying to evaluate a delta function integral for the density of states- $$g(E)\propto \int_{-\pi}^\pi \delta[E-A-B\cos x-C\cos (2x)]$$ I don't know how to go about this. There's also one subtle difficulty. Suppose $C=0$ so that $$\int_{-\pi}^\pi \delta[E-A-B\cos x] $$ Now, If we let $B\cos x=y\rightarrow dy=-B\sin x \ dx$

$$\int \frac{dy}{-B\sin x}\delta (E-A-y)$$ The problem is with limits; since $B\cos(\pm \pi)=-B$, the upper limit and lower limits are matched, which makes integral zero, which isn't physical. What I'm missing here?

Answer 1

Assuming the integral to be over $x$, there are two issues with your original approach:

1. After your substitution, $y = B \cos (x)$, the integral should no longer include a $x$-dependency. One should write: \begin{align} dy &= -B \sin (x) dx \\ &= -B \sqrt{1 - \cos^2 (x)} dx\\ &= - \sqrt{B^2 - B^2 \cos^2 (x)} dx \\ \Leftrightarrow dx &= -\frac{dy}{\sqrt{B^2 - y^2}} \end{align}

2. As you already said, the limits of the integration match.

Instead, you could use $\cos (2x) = 2 \cos^2 (x) - 1$, s.t.: \begin{align} I :=& \int_{-\pi}^{+\pi} dx \, \delta [E - A - B \cos (x) - C \cos (2x)]\\ =& \int_{-\pi}^{+\pi} dx \, \delta [E - A - (B-C) \cos (x) - 2C \cos^2 (x)] \end{align}

Now, we only need to find all $x \in [-\pi, +\pi]$ that fulfill: \begin{equation} E - A - (B-C) \cos (x) - 2C \cos^2 (x) = 0 \\ \Leftrightarrow \cos^2 (x) + \frac{B-C}{2C} \cos (x) + \frac{A-E}{2C} = 0 \end{equation}

Use the quadratic formula: \begin{equation} \cos(x) = - \frac{B-C}{4C} \pm \sqrt{\left( \frac{B-C}{4C}\right)^2 - \frac{A-E}{2C}} \\ \Leftrightarrow x = \arccos \left( - \frac{B-C}{4C} \pm \sqrt{\left( \frac{B-C}{4C} \right)^2 - \frac{A-E}{2C}} \right) \end{equation}

So your integral $I$ gives: \begin{equation} I = \begin{cases} 1 & \text{if} &-\pi \leq \arccos \left( - \frac{B-C}{4C} \pm \sqrt{\left( \frac{B-C}{4C} \right)^2 - \frac{A-E}{2C}} \right) \leq +\pi \\ 0 & \text{else}& \end{cases} \end{equation}