# Integration by substitution in n dimension

I just read the proof of Hardy-Littlewood-Sobolev inequality abaout fractional integral operator. Then I found the following identity (but don't understand it)

$$\int_{% %TCIMACRO{\U{211d} }% %BeginExpansion \mathbb{R} %EndExpansion ^{n}\backslash B\left( x,R\right) }\frac{1}{\left\vert x-y\right\vert ^{c(n-1)}}dy=\int_{S^{n-1}}\int_{R}^{\infty }\frac{1}{r^{^{c(n-1)}}}% r^{n-1}drd\sigma$$ where B(x,R) denote n dimensional ball, centered on $x$ and radius $R$. I guess that identity is generalization of substitution formula. Could you help me to understand that identity?

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The left integral is over all of space except the ball. $y$ is a point outside the ball and ranges over the volume of integration. $|x-y|$ is the distance between the points. The right integral is changing to spherical coordinates. $r$ is the distance between $x$ and $y$, so $r=|x-y|$. The inner integral is over the radius, and ranging from $R$ to $\infty$ avoids the ball nicely. $d\sigma$ represents all the angular variables ($n-$1 of them). The $r^{n-1}$ factor represents the radial variation of the surface area of an $n-1$ sphere. Presumably the next thing to happen is to do the $d\sigma$ integral getting the surface of the $n-1$ sphere ($4\pi$, for example, if $n=3$).
@beginner: I'm not sure what singular variable you mean, but in 2 dimensions the area element is $r\;dr\; d\theta$. The $r$ is $r^{2-1}$, the $dr$ you see, and $d\theta$ is the $d\sigma$. In 3 dimensions it is $r^2\; dr\; \sin \theta\; d\theta\; d\phi$ with $\sin \theta\; d\theta\; d\phi$ being $d\sigma$ –  Ross Millikan Jul 24 '12 at 3:29
@beginner: In my example above, whether the factor is $\sin \theta$ or $\cos \theta$ depends whether you measure up from the $xy$ plane (sin) or down from $z$ (cos). The idea is the same, but I think down from $z$ is more standard. –  Ross Millikan Jul 24 '12 at 13:00