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 ?
the function $1/|x|^α$ in $R^n$
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?
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0It 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 $\tilde{f}$. – 2012-08-24
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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0@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. – 2012-08-24