A multiple integral problem in $6$D: $I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$

Question

Let $B_r$ to be a ball centered at origin with radius $r$ in dimension $3$. My problem is the following multiple integral: $$I=\iint_{B_r\times B_r}\frac{1}{|x-y|}dxdy$$ The result says that $I$ is exactly equivalent to $r^5$. The lower bound for $I$ is trivial but I have trouble in the upper bound. Moreover, can this result be extended to dimension $n$?

Answer 1

Solution for 3-dimension space

$I(A,B)=\int\limits_{V_{x}}\int\limits_{V_y} d^3\textbf x\,d^3\textbf y \frac{1}{|\textbf x-\textbf y|}=\int\limits_{V_{y}}(\int\limits_{V_x} d^3\textbf x\, \frac{1}{|\textbf x-\textbf y|})d^3\textbf y=$$=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1\left(\int\limits_{0}^B\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2d\phi_2\frac{1}{\sqrt{R^2+r^2-2rR\cos\theta_2}}\right)$ $$I(A,B)=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}r^2dr\sin\theta_1d\theta_1d\phi_1F(B,r)$$ $A$ and $B$ are the radii of balls; for the convenience $B\geqslant{A}$ and $z$ axis of the polar system of coordinates ($X$ space) is oriented along the vector $\textbf y$.

Integrating over $\theta_2$ and $\phi_2$

$$F(B,r)=2\pi\int\limits_{0}^B\int\limits_{0}^{\pi}R^2dR\sin\theta_2d\theta_2\frac{1}{\sqrt{R^2+r^2-2rR\cos\theta_2}}=$$$$\frac{2\pi}{\sqrt{2r}}\int\limits_{0}^B\int\limits_{-1}^1\frac{R^2}{\sqrt{R}\sqrt{\frac{r^2+R^2}{2Rr}-x}}dxdR=$$ $$=\frac{2\pi}{r}\int\limits_{0}^BR\Bigl(R+r-|R-r|\Bigr)dR=\frac{2\pi}{r}\Bigl(rB^2-\frac{r^3}{3}\Bigr)$$

$$I(A,B)=\int\limits_{0}^A\int\limits_{0}^{2\pi}\int\limits_{0}^{\pi}\frac{2\pi}{r}\Bigl(rB^2-\frac{r^3}{3}\Bigr)r^2dr\sin\theta_1d\theta_1d\phi_1=\frac{8\pi^2}{3}A^3\Bigl(B^2-\frac{1}{5}A^2)\quad (B>A)$$ For $B=A$ $$I=\frac{32\pi^2}{15}A^5$$

Answer 2

By setting $(ra,rb )=(x ,y)$ we see that $|x-y|^{-1}dxdy = r^{-1}|a-b| (r^3da)(r^3db) = r^5 |a-b|^{-1}dadb$ and the integration domain becomes the product of unit balls. So if the integral is finite when $r=1$, then for general $r$ it scales like $r^5$. In higher dimensions the answer is of course $r^{2n-1}. $

As for this unit radius case we can change variables to $u=x-y$ and $v=x+y$. Being a linear orthogonal change of coordinates it introduces a constant scale factor which I ignore. Then the integral is like

$$\int_{B-B}\int_{B+B}\frac{ dudv }{|u|}$$ Which is finite because $1/|u|$ is locally integrable in dimension 3. Written in this form it could be computed explicitly but I’ll leave that to you.

Answer 3

Partial solution

We can also try to find a general solution for $\textbf x$ and $\textbf y$ in $\textbf R^n$

The element of volume in spherical coordinates in $\textbf R^n$ is $dV=r^{n-1}dr \,d\Omega$, where $d\Omega=\sin^{n-2}\theta_1\sin^{n-3}\theta_2...\sin\theta_{n-2}\,d\theta_1...d\theta_{n-2}\,d\phi$, where integration over $\theta_1...\theta_{n-2}$ goes from $0$ to $\pi$, and over $\phi$ - from $0$ to $2\pi$

If we have a full spherical symmetry in $\textbf R^n$ $\int d\Omega$ can be expressed in closed form:

$$\int_{-\infty}^{\infty}..\int_{-\infty}^{\infty}e^{-x_1^2-x_2^2-...-x_n^2}dx_1...dx_n=\pi^{\frac{n}{2}}=\int d\Omega\int_{0}^{\infty}e^{-r^2}r^{n-1}dr=\frac{1}{2}\Gamma(\frac{n}{2})\int d\Omega$$ $$ \int d\Omega=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}$$

On the other hand

$$\int d\Omega=\int_0^\pi..\int_0^\pi\int_0^{2\pi}\sin^{n-2}\theta_1\sin^{n-3}\theta_2...\sin\theta_{n-2}\,d\theta_1...d\theta_{n-2}\,d\phi=\int_0^\pi\sin^{n-2}\theta_1\,d\theta_1\int d\Omega\,'$$

Where $\theta_1$ is the polar angle measured from axis $Z$ and $\int d\Omega\,'$ means integration over all other angles in $\textbf R^n$

$$\int_0^\pi\sin^{n-2}\theta_1\,d\theta_1=\int_0^1(1-t)^{\frac{n-3}{2}}t^{-\frac{1}{2}}dt = B(\frac{n-1}{2},\frac{1}{2})=\frac{\sqrt{\pi}\Gamma(\frac{n-1}{2})}{\Gamma(\frac{n}{2})}$$

Thus $\int d\Omega\,'=\frac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})}$. We had to separate the integration over $\theta_1$, because we will integrate over this angle further. $$I=\int\limits_{V_{x}}\int\limits_{V_y} d^n\textbf x\,d^n\textbf y \frac{1}{|\textbf x-\textbf y|}=\int d\Omega_x\int_0^Ar^{n-1}dr\int d\Omega_y\,'\int_0^BR^{n-1}dR\int_0^\pi\frac{\sin^{n-2}\theta \,d\theta}{\sqrt{R^2+r^2-2Rr\cos\theta}}$$

This expression can be simplified if $A=B$ $$I(A,n)=A^{2n-1}\int d\Omega_x\int_0^1x^{n-1}dx\int d\Omega_y\,'\int_0^1y^{n-1}dy\int_0^\pi\frac{\sin^{n-2}\theta \,d\theta}{\sqrt{x^2+y^2-2xy\cos\theta}}$$ $$I(A,n)=\frac{2\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{2\pi^{\frac{n-1}{2}}}{\Gamma(\frac{n-1}{2})}A^{2n-1}J(n)$$ where $$J(n)=\int_0^1\int_0^1x^{n-1}y^{n-1}dxdy\int_0^\pi\frac{\sin^{n-2}\theta \,d\theta}{\sqrt{x^2+y^2-2xy\cos\theta}}$$

Using the Gamma function property $\,\Gamma(z)\Gamma(z+\frac{1}{2}) =2^{1-2z}\sqrt{\pi}\,\Gamma(2z)$ $$I(A,n)=\frac{2^n\pi^{n-1}}{(n-2)!}A^{2n-1}J(n)$$

To evaluate the integral $J(n)$, we will use the decomposition of the square root in a series of Legendre polynomials $$J(n)=\int_0^\pi\sin^{n-2}\theta \,d\theta\int_0^1x^{n-1}dx\biggl(\int_0^x\frac{y^{n-1}dy}{x\sqrt{1+(\frac{y}{x})^2-2\frac{y}{x}\cos\theta}}+\int_x^1\frac{y^{n-1}dy}{y\sqrt{1+(\frac{x}{y})^2-2\frac{x}{y}\cos\theta}}\biggr)$$ $$=2\int_0^\pi\sin^{n-2}\theta \,d\theta\int_0^1x^{n-2}dx\int_0^x\frac{y^{n-1}dy}{\sqrt{1+(\frac{y}{x})^2-2\frac{y}{x}\cos\theta}}$$

But $$\frac{1}{\sqrt{1+t^2-2t\cos\theta}}=\sum_{k=0}^{\infty} \,t^k \,P_k(\cos\theta)$$ where $t<1$, and $P_k\,(z)=\frac{1}{k!2^k}\frac{d^k}{dz^k}(z^2-1)^k$ - Legendre polynomial of degree $k$

$$J(n)=2\sum_{k=0}^{\infty}\int_0^\pi\sin^{n-2}\theta \,P_k\,(\cos\theta)\,d\theta\int_0^1x^{n-2-k}dx\int_0^xy^{n-1+k}dy$$ $$=\frac{2}{2n-1}\sum_{k=0}^{\infty}\frac{1}{n+k}\int_{-1}^1(1-z^2)^{\frac{n-3}{2}}P_k(z)dz$$

I'm afraid there is no a closed form for this integral, though it can be evaluated for every given $n$. For instance, for odd $n\,$ there is the most simple case, because $\,k\leqslant n-3$ (integral with higher $k$ vanishes).

For $n=3$ the only surviving term is with $k=0$; $\,J(n=3)=\frac{2}{5}\frac{2}{3}=\frac{4}{15}$ and $I(A, n=3)=\frac{32\pi^2}{15}A^5$ - as was evaluated above for $R^3$

For $n=5$ there are two surviving terms ($k=2$ and $k=0$); $\,J(n=5)=\frac{2}{9}(\frac{4}{15}-\frac{4}{15}\frac{1}{7})=\frac{2}{9}\frac{8}{35}$ and $I(a,n=5)=\frac{2^8}{35*3^3}\pi^4A^9$

For arbitrary $n$:

$$I(A,n)=\frac{\pi^{n-1}2^{n+1}}{(n-2)!(2n-1)}A^{2n-1}\sum_{k=0}^{\infty}\frac{1}{n+k}\int_{-1}^1(1-z^2)^{\frac{n-3}{2}}P_k(z)dz$$ As the last point we can evaluate asymptotics for big $n>>1$ $$\int_{-1}^1(1-z^2)^{\frac{n-3}{2}}P_k(z)dz=\int_{-1}^1e^{\frac{n-3}{2}\log(1-z^2)}P_k(z)dz\sim\int_{-\infty}^\infty e^{-\frac{n-3}{2}z^2}P_k(z)dz\sim\sqrt{\frac{2\pi}{n-3}}P_k(0)$$ $$J(n)\sim\frac{2}{2n-1}\sqrt{\frac{2\pi}{n-3}}\sum_{k=0}^{\infty}\frac{1}{n+k}P_k(0)\sim\frac{\sqrt{2\pi}}{n^{\frac{5}{2}}}\sum_{k=0}^{\infty}P_k(0)=\frac{\sqrt{\pi}}{n^{\frac{5}{2}}}$$ $$I(A,n)\sim\frac{\pi^{n-\frac{1}{2}}2^n}{(n-2)!\,n^{\frac{5}{2}}}A^{2n-1}\,, \,n>>1$$