1

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?

ThymeTravel
  • 115
  • The best way? First find a rigorous definition of the delta "function". Then a definition of a "composition" of the form $\cos(u(\theta))$. Without such a definition, you can prove nothing. – GEdgar May 30 '20 at 20:03
  • 1
    There is a formula: $$\delta(f(x)) = \sum_{f(x_i)=0} \frac{\delta(x-x_i)}{|f'(x_i)|}.$$ The summation notation here means that a sum is taken over every zero $x_i$ of $f$. – md2perpe May 30 '20 at 20:09
  • As GEdgar says, essentially you should study distributions. However you can get away with a change of variable ($\cos(\theta)=x$). Watch out that the function is not invertible on the whole interval, so you have to be a bit careful and split the integral in regions where $\cos$ can be inverted. The result is a general formula that you can find in Wikipedia (I'm pretty sure) – lcv May 30 '20 at 20:12
  • ..or up here (from md2perpe) – lcv May 30 '20 at 20:13
  • This is probably handled in the answers to Dirac Delta Function of a Function. – robjohn May 30 '20 at 21:18

1 Answers1

4

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 $$ \int_{0}^{2\pi} \delta(\cos(\theta)-k) = \int_{0}^{2\pi} \left( \frac{\delta(\theta-\arccos(k))}{\sqrt{1-k^2}} + \frac{\delta(\theta-(2\pi-\arccos(k)))}{\sqrt{1-k^2}} \right) d\theta = \frac{2}{\sqrt{1-k^2}} . $$

md2perpe
  • 26,770
  • Thanks, this was really helpful. How were the limits used in the integration? What would happen if we chose another interval such as to $0$ to $\pi$ – would this halve the answer? – ThymeTravel May 30 '20 at 20:53
  • The only thing that matters regarding the limits is the number of zeros between them. So integrating from $0$ to $\pi$ would halve the result. – md2perpe May 30 '20 at 21:00