# Double integral of Dirac delta distribution with more than one root

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)

• 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 $$f(\epsilon)=\frac{1}{a+\cosh\epsilon}$$

• You should be able to swap the order. What is precisely your question? – InertialObserver Jan 8 '19 at 1:15

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 $$I = \int_0^{2 \pi} \frac {\cos^2 \theta} {a + \cosh \frac {\sin \theta} 3} d\theta = 12 \int_0^{1/3} \frac {\sqrt {1 - 9 \epsilon^2}} {a + \cosh \epsilon} d\epsilon.$$