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I found a double integral involving a Dirac distribution of a sine function,

$\int_{-1}^{1} \Big( \int_{0}^{2\pi} g(\theta,\epsilon)\delta(\epsilon-\frac{1}{3}\sin\theta)d\theta\Big)f(\epsilon)d\epsilon$

(both $g$ and $f$ are continuous and differentiable in the integration domain)

I have doubts about:

  • the order in which the variables can be integrated
  • the Dirac delta having more than one root

Integrating $\epsilon$ first: If one 'moves' $f(\epsilon)d\epsilon$ inside the $\theta$ integral and then considers $\delta$ as a function of $\epsilon$ and solves the $\epsilon$ integral just by setting $\epsilon= \frac{1}{3}\sin\theta$ then this results in the $\theta$ integral of both $g$ and $f$ as functions of theta only. Notice $f$ was outside the $\theta$ integral at the bigenning (as a function of $\epsilon$ only), but now we have $f(\frac{1}{3}\sin\theta)$ and it has to be integrated respect to $\theta$ too.

Integrating $\theta$ first: What would be the correct way to do this given that the argument of $\delta$ (seen as a function of $\theta$) has more than one root? (when $-\frac{1}{3}<\epsilon<\frac{1}{3}$)

For a Dirac delta of the form $\delta(g(x))$ with $g$ having simple roots $x_i$ in the integration domain:

$\delta(g(x))= \sum_i \frac{\delta(x-x_i)}{|g'(x_i)|}$

But it is not very clear to me how to use this and get to the same answer one gets by doing the $\epsilon$ integral first

Aside: The specific forms of $f$ and $g$ are $g(\theta,\epsilon)=\cos^2\theta \frac{1-\frac{1}{3}\epsilon\sin \theta}{1-(\frac{1}{3}\sin \theta)^2}$ and $f(\epsilon)=\frac{1}{a+\cosh\epsilon}$

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  • $\begingroup$ You should be able to swap the order. What is precisely your question? $\endgroup$ Commented Jan 8, 2019 at 1:15

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Essentially both ways you suggest are valid, the results from both iterated integrals will be the same. If we fix $\theta \in [0, 2 \pi]$ and integrate wrt $\epsilon$, there is always one root $\epsilon_1 = \sin(\theta)/3 \in (-1, 1)$. If we integrate wrt $\theta$ first, there are no roots if $|\epsilon| > 1/3$ and two simple roots in $(0, 2 \pi)$ if $0 < |\epsilon| < 1/3$. In the latter case, the roots are $\theta_1 = \pi - \arcsin 3 \epsilon$ and $\theta_2 = 2 \pi H(-\epsilon) + \arcsin 3 \epsilon$. We obtain $$I = \int_0^{2 \pi} f {\left( \frac {\sin \theta} 3 \right)} \,g {\left( \theta, \frac {\sin \theta} 3 \right)} d\theta = \\ \int_{-1/3}^{1/3} \frac {3 f(\epsilon)} {\sqrt {1 - 9 \epsilon^2}} \,(g(\pi - \arcsin 3 \epsilon, \epsilon) + g(2 \pi H(-\epsilon) + \arcsin 3 \epsilon, \epsilon)) \,d\epsilon.$$ For your $f$ and $g$, this becomes $$I = \int_0^{2 \pi} \frac {\cos^2 \theta} {a + \cosh \frac {\sin \theta} 3} d\theta = 12 \int_0^{1/3} \frac {\sqrt {1 - 9 \epsilon^2}} {a + \cosh \epsilon} d\epsilon.$$

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