0

This answer to Why can't Wolfram Alpha calculate $\int_0^{2\pi}\sqrt{(a-\cos\theta)^2+\sin^2\theta}\ d\theta$? includes the following excerpt:

$$\Re(a (a+2))>-1\land \Re((a-2) a)>-1$$ where appear ellptic integrals of the second kind. In fact, this reduces to $$I=4(a+1)E\left(\frac{4 a}{(a+1)^2}\right)$$

Here $E$ is the Complete Elliptic Integral of the Second Kind:

$$E(a) = \int_0^{\pi/2} \sqrt{1 - a \sin^2 t} \ \ dt$$

The expression inside the square root of my expression $(a-\cos\theta)^2+\sin^2\theta$ reduces to $a^2 - 2a \cos \theta + 1$ but I don't see how it ends up in the proper form for $E$.

uhoh
  • 1,864

2 Answers2

3

Note: I get a different result, but I think the idea is correct.

We have that $\cos(2\alpha)=1-2\sin^2(\alpha)$ so \begin{align} a^2-2a\cos(\theta)+1 &=a^2-2a(1-2\sin^2(\theta/2))+1\\ &=a^2-2a+1+4a\sin^2(\theta/2)\\ &=(a-1)^2+4a\sin^2(\theta/2)\\ &=(a-1)^2 \left(1+\frac{4a}{(a-1)^2}\sin^2(\theta/2)\right) \end{align} So \begin{align} I(a) &=\int_{0}^{2\pi}\sqrt{a^2-2a\cos(\theta)+1}\mathrm{d}\theta \\ &=\int_0^{2\pi} \sqrt{(a-1)^2 \left(1+\frac{4a}{(a-1)^2}\sin^2(\theta/2)\right)}\mathrm{d}\theta \\ &=\sqrt{(a-1)^2} \int_0^{2\pi} \sqrt{ \left(1+\frac{4a}{(a-1)^2}\sin^2(\theta/2)\right)}\mathrm{d}\theta \\ &=2\sqrt{(a-1)^2} \int_0^{\pi} \sqrt{ \left(1+\frac{4a}{(a-1)^2}\sin^2(t)\right)}\mathrm{d}t \\ &=4\sqrt{(a-1)^2} \int_0^{\pi/2} \sqrt{ \left(1+\frac{4a}{(a-1)^2}\sin^2(t)\right)}\mathrm{d}t \\ &=4\sqrt{(a-1)^2}E\left(-\frac{4a}{(a-1)^2}\right) \end{align}

Botond
  • 11,938
  • I'd like to accept this but I'm hesitating because it gives the wrong sign compared to the expression in the question and by checking numerically. Any thoughts on why that might be? – uhoh Aug 20 '20 at 13:01
2

$$A=(a-\cos(\theta))^2+\sin^2(\theta)=(a^2+1)-2 a \cos (\theta )$$ $$\cos (\theta )=1-2 \sin ^2\left(\frac{\theta }{2}\right)$$ $$A=(a-1)^2+4a\sin ^2\left(\frac{\theta }{2}\right)=(a-1)^2 \left(1+\frac{4 a }{(a-1)^2}\sin ^2\left(\frac{\theta }{2}\right)\right)$$

Claude Leibovici
  • 260,315