integration over unit ball Can anyone give me some explanation about the following fact? Many thanks!
Let $B$ be the unit ball in $\mathbb{R}^d$, $d\ge 1$. Then 
$$\int_{\mathbf{z}\in B}\frac{1}{|\mathbf{z}|^c}\,\mathrm{d} \mathbf{z}<\infty$$
if and only if $c<d$.
 A: You can employ a scaling argument. First, assume that $c\ge d$, and find (with $r<1$)
$$
  \int_B \frac{1}{|z|^c}\,dz\ge\int_{rB}\frac{1}{|z|^c}\,dz=\int_B\frac{r^d}{r^c|w|^c}\,dw=r^{d-c}\int_B\frac{1}{|w|^c}\,dw=r^{d-c}\int_B\frac{1}{|z|^c}\,dz
$$
using the change of variables $z=rw$, $dz=r^d\,dw$. If the integral is finite, the first inequality is strict, but $r^{d-c}\ge1$, so you get a contradiction.
If $c<d$, let instead $A_n$ be the shell $2^{-n}B\setminus2^{-n-1}B$ and employ the change of variables $z=2w$:
$$
  \int_{A_n}\frac{1}{|z|^c}\,dz=\int_{A_{n+1}}\frac{2^d}{2^c|w|^c}\,dw=2^{d-c}\int_{A_{n+1}}\frac{1}{|z|^c}\,dz
$$
so by induction
$$
  \int_{A_n}\frac{1}{|z|^c}\,dz=2^{n(c-d)}\int_{A_0}\frac{1}{|z|^c}\,dz.
$$
Now $B=\{0\}\cup\bigcup_{n=0}^\infty A_n$ (this is a disjoint union), so
$$
  \int_B\frac1{|z|^d}\,dz
  =\sum_{n=0}^\infty\int_{A_n}\frac1{|z|^d}\,dz
  =\sum_{n=0}^\infty2^{n(c-d)}\int_{A_0}\frac{1}{|z|^c}\,dz<\infty,
$$
since $\int_{A_0}\frac{1}{|z|^c}\,dz$ is clearly finite and $\sum_{n=0}^\infty2^{n(c-d)}$ converges.
(You may think this argument ridiculously involved for something that is easily solved by direct calculation, but this sort of scaling argument is frequently useful in situations that are not amenable to direct calculation. And note that there is very little actual computation going on in this argument.)
