Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
add comment

2 Answers

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?

share|improve this answer
    
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 $r\in[0,\infty[$? –  Thomas E. 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. –  Thomas E. 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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.