juanberrio8a

Answered

2022-06-27

I'm looking to show that $\frac{1}{|x{|}^{n}}$ is not integrable on the subset of ${\mathbb{R}}^{n}$ where $|x|\ge 1$. It's easy to do on $\mathbb{R}$, and I think I need to apply Fubini's theorem for the general case, but not sure how to do so in n-dimensional space. Thanks

Answer & Explanation

knolsaadme

Expert

2022-06-28Added 16 answers

Via some computation and Tonelli's theorem, we have for measurable $f:{\mathbb{R}}^{n}\to [0,\mathrm{\infty}]$ that

${\int}_{{\mathbb{R}}^{n}}f(x)\phantom{\rule{thinmathspace}{0ex}}dx={\int}_{{S}^{n-1}}{\int}_{0}^{\mathrm{\infty}}f(r\omega ){r}^{n-1}\phantom{\rule{thinmathspace}{0ex}}dr\phantom{\rule{thinmathspace}{0ex}}dS(\omega ),$

where $\phantom{\rule{thinmathspace}{0ex}}dS$ is "surface measure" on ${S}^{n-1}$. Now you can plug in $f(x)=\frac{1}{|x{|}^{n}}{\chi}_{\{x:|x|\ge 1\}}$ and compute it's integral.

Most Popular Questions