# Evaluating the integral $\int_0^{1}\int_{0}^{2\pi}x\delta (v-x\cos\theta)dxd\theta$

I want to evaluate the following integral

$$\int_0^{1}\int_{0}^{2\pi}x\delta (v-x\cos\theta)dxd\theta$$ According to the book where I found the exercise, the answer is $$C\Theta(1-v^2)\sqrt{1-v^2}$$ where $$C$$ is some constant and $$\Theta$$ is the Heaviside function. However, I cannot obtain this result.

My own work is the following: We use the theorem $$\delta(f(\theta))=\frac{\delta(\theta-\theta_0)}{|f(\theta_0)|}$$ where $$f(\theta)=v-x\cos\theta$$ so $$\theta_0=\arccos\left(\frac{v}{x}\right)$$ and $$f'(\theta_0)=x\sin\theta_0=x\sqrt{1-\frac{v^2}{x^2}}$$. Then, $$\frac{1}{|f'(\theta_0)|}=\frac{1}{\sqrt{x^2-v^2}}$$. Thus, we are integrating $$\int_0^{1}\int_{0}^{2\pi}\frac{x}{\sqrt{x^2-v^2}}\delta(\theta-\theta_0)d\theta dx$$ which can be written as $$\int_0^1\frac{x}{\sqrt{x^2-v^2}}\Theta(\theta_0)\Theta(2\pi-\theta_0)dx$$ Now comes the part where I'm unsure what to do, as the Heaviside step function includes and $$x$$ dependence. It looks like I'm on the right track as if we disregard the Heaviside function, then the antiderivative is $$\sqrt{x^2-v^2}$$ which looks alright.

(Note, the step functions were introduced to take care of the case if the delta function is outside the domain of integration, inspired by this post: Double Integral with a Delta Function)

Any ideas how to proceed?

We can rewrite the integral by changing x to r, and it become a bit clearer how to proceed:

$$\int_{0}^{1}\int_{0}^{2\pi}\delta(v-r\cos\theta)rdrd\theta$$

Changing from polar coordinates to cartesian:

$$\int_{-1}^{1}\int_{-\sqrt{1-x^{2}}}^{\sqrt{1-x^{2}}}\delta(v-x)dydx$$

We then solve the inside integral:

$$2\int_{-1}^{1}\delta(v-x)\sqrt{1-x^{2}}dx$$ Let u = v-x: $$2\int_{v-1}^{v+1}\delta(u)\sqrt{1-(v-u)^{2}}dx$$

And finally, using properties of the Dirac Delta function: when a<0<b $$\int_{a}^{b}\delta(x)f(x)=f(0),$$ and when 0<a<b or a<b<0:$$\int_{a}^{b}\delta(x)f(x)=0,$$

So this implies that the integral is non-zero when |v|<=1, as so: $$2\Theta(1-v^{2})\sqrt{1-v^{2}}$$

• This is great! Thank you! However, I have two questions, did you not do a typo on the lower integration bound , i.e we should have $2\int_{-1}^{1}\delta(v-x)\sqrt{1-x^2}dx$? Furthermore, could you make it slightly more explicit how we obtain the Heaviside function from properties of the delta function? Thanks! Commented Feb 22 at 14:43