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I am interested in finding the average Euclidean distance from a point $(x,y)\in\mathbb{D}_2$, the unit disk $\{(u,v):u^2+v^2\leq 1\}\subseteq\mathbb{R}^2$, to the disk $\mathbb{D}_2$. This amounts to essentially calculating the integral $$\iint_{\mathbb{D}_2} \|(x,y)-\omega\|_2d\omega.$$ By rotational symmetry this gives an integral of the form $$\iint_{\mathbb{D}_2}\|(a,0)-\omega\|_2d\omega$$ where $0\leq a \leq 1$. I believe this can be converted to an elliptic integral of the second kind where the integral over the angle $\theta$ appears in the form $$\int_{0}^{2\pi}\sqrt{1-\frac{4ar}{(r+a)^2}\cos^2(\theta/2)}d\theta.$$ I am not sure how to proceed from here.

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Let me set

$$k^2=\frac{4ar}{(r+a)^2} <1. $$

If we replace $\theta$ with $2\theta$ we reduce this to an integral;

$$ 2 \underbrace{\int_0^\pi \sqrt{1-k^2\cos^2\theta} d\theta}_{=I}. $$

Now set

$$ x=\cos\theta $$

so that $$ dx=-\sqrt{1-x^2} d\theta $$


$$ I=\int_{-1}^1\frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx= 2\underbrace{\int_0^1 \frac{\sqrt{1-k^2x^2}}{\sqrt{1-x^2}} dx}_{=:E_2(k)}. $$

The integral $E_2(k)$ is called Jacobi's complete elliptic integral of the second kind. There is no simple formula for it but you can have a look at the beautiful book by H. McKean and V. Moll, Elliptic Curves. Function Theory, Geometry Arithmetic Cambridge University Press,1997.

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