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Convergence of $int_mathbb{R^n} frac1{x^p}dx$

Mathematics Asked by user854662 on January 25, 2021

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 $|x|^{n-1}$?

One Answer

I think you mean $intlimits_{mathbb{R}^n}frac{dx_1...dx_n}{|x|^p}$. You could try to make spherical change of variables (in both cases). Other ways - Cronrod-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: $$x_1=rcos(phi_{n-1})cos(phi_{n-2})cdot...cdotcos(phi_2)cos(phi_1),$$ $$x_2=rcos(phi_{n-1})cos(phi_{n-2})cdot...cdotcos(phi_2)sin(phi_1),$$ $$x_2=rcos(phi_{n-1})cos(phi_{n-2})cdot...cdotsin(phi_2),$$ $$.............$$ $$x_{n-1}=rcos(phi_{n-1})sin(phi_{n-2}),$$ $$x_{n}=rsin(phi_{n-1}).$$

$G={(r,phi):~rin(0,+infty),~ phi_1 in (-pi,pi),~ phi_i in (-pi/2,pi/2)$ 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 $det Phi^{'}=r^{n-1}cos^{n-2}(phi_{n-1})cdot...cdotcos^2(phi_3)cos(phi_2)$. Note that $|x|=r$.

Now we can rewrite your integral as follows: $$intlimits_{mathbb{R}^n}frac{dx_1...dx_n}{|x|^p}=intlimits_{G}frac{det Phi^{'}}{r^p}drdphi_{n-1}...dphi_1=intlimits_{G}frac{r^{n-1}cos^{n-2}(phi_{n-1})cdot...cdotcos^2(phi_3)cos(phi_2)}{r^p}drdphi_{n-1}...dphi_1=$$ $$=underbrace{Bigg(prodlimits_{i=2}^{n-1}intlimits_{-pi/2}^{pi/2}cos^{i-1}(phi_i)dphi_iBigg)}_{V_n in (0,+infty)}cdotintlimits_{0}^{+infty}r^{n-1-p}dr.$$ The last integral diverges for any p.

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

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

Answered by Timur B. on January 25, 2021

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