A step of change of variables I don't understand when show derivative of a function in $L^n$?

Question

How (16) can goes to (17) and how to show $loglog(1+\frac{1}{|x|})\in L^n$ for $n>1$?

My attempt:

Let $|x|=r$, then $(x^2)^{\frac{1}{2}}=r$. Differentiate on both sides, I get $dx=\frac{|x|}{x}dr$.

Then I integrate (16), $$\int_{B(0,1)}|Du|^ndx=\int_{B(0,1)}C^n\frac{1}{log(1+\frac{1}{|x|})^n}\frac{1}{|x|^n}dx=\int_0^1C^n\frac{1}{log(1+\frac{1}{r})^n}\frac{1}{r^n}\frac{r}{x}dx$$ I don't know how to proceed to get (17)?

I think I also need to show $loglog(1+\frac{1}{|x|})\in L^n$ for $n>1$ which is not in the proof. But I don't know how to start.

Can anyone help about this? Thanks!

Answer 1

You are doing a change to spherical coordinates in $n$ dimensions. The Jacobian of the transformation is bounded by $r^{n-1}$. $$ \int_{B(0,R)}\phi(r)\,dx\le C_n\int_0^R|\phi(r)|\,r^{n-1}\,dr. $$

Answer 2

To go from (16) to (17):

Substitution $|x|=r$ and introducing spherical coordinates, because you integrate over the unit sphere $B(0,1)$ with radius 1. Now applying Integration over the sphere yields: $\int_{B(0,1)}|Du|^ndx < C \int_{Angles} \int_0^1 \frac{1}{r^nlog(1+\frac{1}{r})^n} d \phi r^{n-1}dr$. The volume elemnt of an $n$-dimensional sphere is proportional to $r^{n-1}dr$. The Integration over the angles is greater than 1 so the inequality-signs are preserved.