# the function $1/|x|^α$ in $R^n$

How to show that in $\mathbb{R}^n$ the function $x \mapsto 1/|x|^α$ is in $L^1(B)$ iff $\alpha＜n$ and in $L^1(\mathbb{R}^n\setminus B)$ iff $\alpha>n$ by using polar coordinates ?

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Remember that, for any radially symmetric function $f$, we can compute its integral as $$\int_{\mathbb{R}^n} f \, d\mathcal{L}^n = \mathcal{H}^{n-1} (S^{n-1})\int_0^{+\infty} f(r) r^{n-1}\, dr,$$ where $\mathcal{L}^n$ is the standard Lebesgue measure in $\mathbb{R}^n$ and $\mathcal{H}^{n-1} (S^{n-1})$ is the surface measure of the unit sphere. If you choose $f(x)=\frac{1}{|x|^\alpha}\chi_B(x)$, where $\chi_B$ is the characteristic function of $B$, can you conclude without further help?

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Hups. This time you beat me by 1 minute! How did you manage to do that ;-) – user20266 Aug 24 '12 at 10:45
If $f:\mathbb{R}^{n}\to\mathbb{R}$, then what does $f(r)$ mean for $r\in[0,\infty[$? – T. Eskin Aug 24 '12 at 11:42
@ThomasE., Since $f$ is radially symmetric, $f(x)$ depends only on $|x|$. Here, $f(r)$ technically means $f(r\cdot u)$, where $u$ is a unit vector in $\mathbb R^n$. – B R Aug 24 '12 at 12:14
@BR. Thanks, that makes it a lot more clear. – T. Eskin Aug 24 '12 at 12:16
It is customary to slightly abuse notation. When $f$ is radially symmetric, there exists a function $\tilde{f}$ of one real variable such that $f(x)=\tilde{f}(|x|)$ for every $x$. Usually people "confuse" $f$ with $\tilde{f}$. – Siminore Aug 24 '12 at 13:20

The n-dimensional volume element in polar coordinates is $r^{n-1} d\xi dr$ with $d\xi$ the volume element of the $n-1$ sphere. Hence $$\int_{\mathbb{R}^n} \frac{1}{|x|^\alpha} dx= \int_{S^{n-1}}\int_0^\infty r^{n-1-\alpha}dr d\xi= C(n) \int_0^\infty r^{n-1-\alpha}dr$$ Now just calculate.

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For the purpose of this question, you may want do provide a formula for any ball $\Bbb B_n(0,R)$ rather than for its special case $\Bbb R^n$. – Ilya Aug 24 '12 at 11:20
@Ilya I think it is quite obvious how to modify the integral for the case of a ball (or the complement of a ball, which was asked for, too. – user20266 Aug 24 '12 at 12:09