Integrating Dirac delta functions with trigonometric arguments

Question

I am not sure about how to integrate Dirac delta functions which have trigonometric arguments. I am currently trying to work out

$\int_{0}^{2\pi} \delta(\cos(\theta)-k)d\theta$, $\lvert k \rvert$ < $1$.

What is the best way to approach this?

Answer 1

There is a formula for $\delta(f(x))$: $$\delta(f(x)) = \sum_{f(x_i)=0} \frac{\delta(x-x_i)}{|f'(x_i)|}.$$

In our case we have $f(\theta) = \cos(\theta)-k,$ the zeros of which are $\theta_1=\arccos(k)$ and $\theta_2=2\pi-\arccos(k).$ This gives us $$\begin{align} \delta(\cos(\theta)-k) &= \frac{\delta(\theta-\arccos(k))}{|-\sin(\arccos(k))|} + \frac{\delta(\theta-(2\pi-\arccos(k)))}{|-\sin(2\pi-\arccos(k))|} \\ &= \frac{\delta(\theta-\arccos(k))}{\sqrt{1-k^2}} + \frac{\delta(\theta-(2\pi-\arccos(k)))}{\sqrt{1-k^2}} \end{align}$$ so $$ \int_{0}^{2\pi} \delta(\cos(\theta)-k) = \int_{0}^{2\pi} \left( \frac{\delta(\theta-\arccos(k))}{\sqrt{1-k^2}} + \frac{\delta(\theta-(2\pi-\arccos(k)))}{\sqrt{1-k^2}} \right) d\theta = \frac{2}{\sqrt{1-k^2}} . $$