Calculation double Integral over Ball (optical size)

The integral is connected with the mean inverse squared distance between two points within a unit ball. Following the method used by Christian Blatter in this post we have that, $$\begin{align*} \int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|^2}dxdy &=|B_1(0)|^2 \int_0^1\int_0^1\int_0^\pi \frac{f_R(r) f_S(s)f_\Theta(\theta) \ d\theta\ ds\ dr}{r^2+s^2-2rs\cos\theta}\\ &=\left(\frac{4\pi}{3}\right)^2 \int_0^1\int_0^1\int_0^\pi \frac{3r^2\cdot 3s^2\cdot1/2\sin(\theta) \ d\theta\ ds\ dr}{r^2+s^2-2rs\cos\theta}\\ &=8\pi^2 \int_0^1\int_0^1r^2 s^2\left(\int_0^\pi \frac{\sin(\theta) \ d\theta}{r^2+s^2-2rs\cos\theta}\right)\ ds\ dr\\ &=4\pi^2 \int_0^1r\left(\int_0^1 s\ln\left(\frac{(r+s)^2}{(r-s)^2}\right)\ ds\right)\ dr\\ &=4\pi^2 \int_0^1r\left(\ln\left(\frac{1+r}{1-r}\right)(1-r^2)+2r\right)\ dr\\ &=4\pi^2. \end{align*}$$ and we are done.

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Integration

Absolute Value

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