Calculation double Integral over Ball (optical size)

Question

I hope that someone can help me with the following problem. I have to show that $$\int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|^2}dxdy=4\pi^2~,$$ with $B_1(0)\subset\mathbb{R}^3$. I have no idea how to calculate those integrals, the common(for me) tricks won't help.

Answer 1

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.

Answer 2

I have a solution for $$ \frac{1}{2} \int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|}dxdy=\frac{4 \pi^2}{3} \frac{4}{5} $$ By Newton theorem, you have for every spherical symmetric function g (so g(y)=f(|y|): $$ \int \frac{g(y)}{|x-y|}d y = \int_0^\infty \frac{f(r)4\pi r^2}{max\{|x|,r\}} d r $$

With this in mind and the spherically transformation, you get

$$ \frac{1}{2} \int_{B_1(0)}\int_{B_1(0)}\frac{1}{|x-y|}dxdy = \frac{1}{2} \int_{B_1(0)}\int_0^1 \frac{4\pi r^2}{max\{|x|,r\}} d r \ d x \\ = \frac{1}{2} \int_0^1 4\pi t^2 \int_0^1 \frac{4\pi r^2}{max\{t,r\}} d r \ d t = \frac{1}{2} 4^2 \pi^2\int_0^1 t^2 \bigg(\int_0^t r^2 \frac{1}{t} dr + \int_t^1 r \ dr\bigg) dt=\frac{4 \pi^2}{3} \frac{4}{5} $$

Answer 3

Let us recall the Laplace's multipole expansion formula into Legendre polynomials: $$\frac{1}{|\vec{r}'-\vec{r}|}=\frac{1}{r}\sum_{n=0}^\infty \left(\frac{r'}{r}\right)^n P_n(Y)\qquad ;r\geq r',$$

where I have denoted $Y:=\hat{r}\bullet\hat{r}'$. Recall as well their orthogonality relation $$\oint P_n(Y)P_m(Y) d\Omega = \frac{4\pi}{2n+1} \delta_{nm}.$$

Denote our integral $L$, by symmetry, we can write $$I = \int\int \frac{dV'dV}{|\vec{r}'-\vec{r}|^2} = 2\int\int \theta(r-r')\frac{dV'dV}{|\vec{r}'-\vec{r}|^2}.$$

Expanding $1/|\cdot|$ into series (then squared), we get $$I=2\sum_{nm}\int_0^1\int_0^r \frac{1}{r^2} \left(\frac{r'}{r}\right)^n \left(\frac{r'}{r}\right)^m P_n(Y) P_m(Y) r'^2 r^2 d\Omega'd\Omega dr' dr.$$

By orthogonality, we get immediatelly $$I=32\pi^2\sum_{n=0}^\infty\int_0^1\int_0^r \frac{1}{r^2}\left(\frac{r'}{r}\right)^{2n} \frac{1}{2n+1} r'^2 r^2 dr' dr=8\pi^2\sum_{n=0}^\infty\frac{1}{(2n+1)(2n+3)}.$$

By partial fraction decomposition and due to telescopicity of the resulting sum, we get $$I=4\pi^2\sum_{n=0}^\infty\frac{1}{2n+1}-\frac{1}{2n+3}=4\pi^2\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots\right)=4\pi^2$$

Note: that the partial result in terms of only $r'$ and $r$ with $\sum_n$ can be summed up into the integral given by @Robert Z

Answer 4

I assume that $B_1(0)$ is the ball of radius $1$ about the origin. Since this space has nice spherical symmetry, it will probably be easier to convert to spherical coordinates. Let our change of coordinate map be $$ H(\rho,\theta,\phi) = (\rho\cos\theta\cos\phi, \rho\sin\theta\cos\phi, \rho\sin\phi), \qquad \rho \in [0,1], \theta \in [0,2\pi], \phi \in [-\pi/2,\pi/2] $$ so we wish to integrate $$ \int_{-\pi \over 2}^{\pi \over 2} \int_0^{2\pi}\!\!\!\int_0^1 {1 \over |r\cos\theta\cos\phi-r\sin\theta\cos\phi|^2} \thinspace r^2\cos\phi\thinspace drd\theta d\phi $$ Now absolute value is usually defined as $|x|=\sqrt{x^2}$, so we have $$ \begin{align} |r\cos\theta\cos\phi-r\sin\theta\cos\phi|^2 &= r^2\cos^2\theta\cos^2\phi + r^2\sin^2\theta\cos^2\phi-2r^2\cos^2\phi\sin\theta\cos\theta \\\ &= r^2\cos^2\phi(1- 2\sin\theta\cos\theta) \end{align}$$ Then our integral is $$ \int_{-\pi \over 2}^{\pi \over 2} \int_0^{2\pi}\!\!\!\int_0^1 {1 \over r^2\cos^2\phi(1- 2\sin\theta\cos\theta)} \thinspace r^2\cos\phi\thinspace drd\theta d\phi $$ easily simplifying to $$ \int_{-\pi \over 2}^{\pi \over 2} \int_0^{2\pi}\!\!\!\int_0^1 {1 \over \cos\phi}{1 \over (1- 2\sin\theta\cos\theta)} \thinspace drd\theta d\phi $$ Clearly $r$ integrates out. By the standard tricks, $$ \int {1 \over \cos x}\thinspace dx = \log(\tan x + \sec x) $$ and $$ \int {1 \over 1-2\sin x \cos x} = {\sin x \over \cos x - \sin x} $$ So the antidifferentiation results in $$ \begin{align} &\log(\tan\phi-\sec\phi)\bigg|_{-\pi \over 2}^{\pi \over 2} \times {\sin \theta \over \cos \theta - \sin\theta}\bigg|_0^{2\pi} \\ \end{align} $$ which is not defined. Hence the integral does not exist.