Convert elliptic: $F\left(\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)\biggr\rvert 1-\frac{4}{R^2}\right)$ to complete: $K\left(1-\frac{4}{R^2}\right)$

Question

I explored the convolution of $\arcsin$ distributions SE. Here I found the following identity empirical (allot of trial, error and luck):

typo: $1-\frac{1}{R^2}$ should be $1-\frac{\color{red}{4}}{R^2}$. As pointed out by first answer.

$$F\left(\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right) \biggr\rvert 1-\frac{\color{red}{4}}{R^2}\right) \stackrel{?}{=}\frac{1}{2}\cdot K\left(1-\frac{\color{red}{4}}{R^2}\right) \tag{1}$$

With $F$ the incomplete and $K$ the complete elliptic integral and: $-2 \leq R \leq 2$. I only have numerical confirmation with WA online and Python (see: SE).

Looking at definitions: $$F(\varphi \rvert k^2)=\int_{0}^{\varphi} \frac{ \ d \theta}{\sqrt{1-k^2 \sin^2(\theta)}}$$ $$K(k)=\int_{0}^{\frac{\pi}{2}}\frac{ \ d \theta}{\sqrt{1-k^2 \sin^2(\theta)}} $$

If the identity $(1)$ is correct how to interpret: $$\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right) \neq \frac{\pi}{2}$$

Seems maybe related to: SE, this article is currently outside my math level.

Seeking a proof (comments: hints/tips to continue studying). Do not know where to start. Note that I am novice with elliptic integrals.

Answer 1

This is not true.

What is true is that $$F\left(\sin ^{-1}\left(\sqrt{\frac{|R]}{|R]+2}}\right)|1-\frac{\color{red}{4}}{R^2}\right)=\frac 12 K\left(1-\frac{\color{red}{4}}{R^2}\right)$$ There are many interrelations between elliptic functions that, playing with the derivatives, you could show for example (I consider $R>0$) $$R \left(2 R E\left(\tan ^{-1}\left(\sqrt{\frac R2}\right)|1-\frac{4}{R^2}\right)-R+2\right)-8 F\left(\tan ^{-1}\left(\sqrt{\frac R2}\right)|1-\frac{4}{R^2}\right)=$$ $$R^2 E\left(1-\frac{4}{R^2}\right)-4 K\left(1-\frac{4}{R^2}\right)$$

Answer 2

The following is not a full answer but gives me more understanding exploring the function.

Incomplete elliptic integral: $$F\left(\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right) \biggr\rvert 1-\frac{4}{R^2}\right) = \int_{0}^{\varphi}\frac{d \theta}{\sqrt{1-\left( 1- \frac{4}{R^2} \right) \sin^{2} \left( \theta\right)}}$$

First I evaluated the range of: $\varphi=\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)$. Maximum value $\varphi$:

$$\lim_{R \rightarrow \infty} \sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)=\lim_{R \rightarrow \infty} \sin^{-1}\left(\sqrt{\frac{1}{1+2/R}}\right)=\frac{\pi}{2} $$

Plot: $$Z=\frac{1}{\sqrt{1-\left( 1- \frac{4}{R^2} \right) \sin^{2} \left( \theta\right)}}$$

Observations (from plot):

• $F(a,b)$ from question is periodic with: $\pi$.
• Integrating $\int Z \ d\theta$ gives $F(a,b)$ from $\varphi=0$ till $\varphi=\sin^{-1}\left(\sqrt{\frac{R}{R+2}}\right)$. $K(b)$ for $varphi=0$ till $\varphi=\pi/2$ giving: $F(a|b)=\frac{1}{2}K(b)$.