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1)$\int_\mathbb{R^n}$ $\frac{1}{x^p}dx$

For what values of $p$ would this converge? I mean should p be less than the dimension of $\mathbb{R^n}$ or what?

  1. My second question is, if I have $\int_\mathbb{B(0,\alpha)} \frac1{|x|^{\beta}}dx$ s.t $\alpha>0$ and $B(0,\alpha) \subset \mathbb{R^n}$ and I made a change of variable s.t $r=||x||_2:=|x|$, then can you put a slight hint for the integral structure pls?

$x=(x_1,x_2,....,x_n)$

My sol: $r=|x|$ , then $dx=\frac{|x|^n}{x} dr$ but my Dr. continued to put $dx=|x|^{n-1}dr:=r^{n-1} dr$ , but how would $\frac{|x|^n}{x}$ give $|x|^{n-1}$?

sara
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    What does $x^p$ mean if $x\in\Bbb R^n$? – anon Dec 08 '20 at 20:27
  • I think you meant the integral of $\frac{dx}{||x||^p}$ rather than $\frac{dx}{x^p}$. If that's the case, I'd suggest to take integrate on $\mathbb{R}^n \setminus {0}$ and do a spherical change of coordinates and then the answer should be easier. – Keen Dec 08 '20 at 20:38
  • @runway44 I edited my question – sara Dec 09 '20 at 04:38
  • @Keen Yes I edited my question and my Dr. wants the value of this integral with this change of variable can you help me pls? – sara Dec 09 '20 at 04:39

1 Answers1

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I think you mean $\int\limits_{\mathbb{R}^n}\frac{dx_1...dx_n}{|x|^p}$. You could try to make spherical change of variables (in both cases). Other ways - Kronrod-Federer theorem (private case - representation as integral over sphere) or distributional function $F_{|x|}(t)$ if you know what it is. After all: in the first case integral diverges for any $p$, in the second converges iff $p<n$.

Look, we have $x=(x_1,...,x_n) \in \mathbb{R}^n$. Let $r \in \mathbb{R}$ and $\phi=(\phi_1,...,\phi_{n-1}) \in \mathbb{R}^{n-1}$. We have spherecal change $x=\Phi(r,\phi)$ given by: $$x_1=r\cos(\phi_{n-1})\cos(\phi_{n-2})\cdot...\cdot\cos(\phi_2)\cos(\phi_1),$$ $$x_2=r\cos(\phi_{n-1})\cos(\phi_{n-2})\cdot...\cdot\cos(\phi_2)\sin(\phi_1),$$ $$x_2=r\cos(\phi_{n-1})\cos(\phi_{n-2})\cdot...\cdot\sin(\phi_2),$$ $$.............$$ $$x_{n-1}=r\cos(\phi_{n-1})\sin(\phi_{n-2}),$$ $$x_{n}=r\sin(\phi_{n-1}).$$

$G=\{(r,\phi):~r\in(0,+\infty),~ \phi_1 \in (-\pi,\pi),~ \phi_i \in (-\pi/2,\pi/2)$ if $ i>1\}$. Then $\Phi(G)=\mathbb{R}^n \setminus \{x:x_1 \leq 0, x_2=0\}=Y$ and $\Phi:G \longrightarrow Y$ is diffeomorfism with $\det \Phi^{'}=r^{n-1}\cos^{n-2}(\phi_{n-1})\cdot...\cdot\cos^2(\phi_3)\cos(\phi_2)$. Note that $|x|=r$.

Now we can rewrite your integral as follows: $$\int\limits_{\mathbb{R}^n}\frac{dx_1...dx_n}{|x|^p}=\int\limits_{G}\frac{\det \Phi^{'}}{r^p}drd\phi_{n-1}...d\phi_1=\int\limits_{G}\frac{r^{n-1}\cos^{n-2}(\phi_{n-1})\cdot...\cdot\cos^2(\phi_3)\cos(\phi_2)}{r^p}drd\phi_{n-1}...d\phi_1=$$ $$=\underbrace{\Bigg(\prod\limits_{i=2}^{n-1}\int\limits_{-\pi/2}^{\pi/2}\cos^{i-1}(\phi_i)d\phi_i\Bigg)}_{V_n \in (0,+\infty)}\cdot\int\limits_{0}^{+\infty}r^{n-1-p}dr.$$ The last integral diverges for any p.

In the second case we have $$\int\limits_{\mathbb{R}^n}\frac{dx_1...dx_n}{|x|^p}=V_n\int\limits_{0}^{\alpha}r^{n-1-p}dr,$$ which converges iff $p<n$.

Addition: $V_n$ is Hausdorf measure of $(n-1)$-dimentional shpere with radius = 1.

Timur B.
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  • Sorry but take a look for my edited question pls. And I want such change of variable ($r=||x||_2=|x|$ , I tried with no result ... can you help me pls. – sara Dec 09 '20 at 04:46
  • My second comment is how did you know that in the 1st case $p>n$ and the second $p<n$? Otherwise could you give me link – sara Dec 09 '20 at 04:47
  • I have edited and add solution. – Timur B. Dec 09 '20 at 14:13