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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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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