Prove In $\mathbb R^n$, $\int_{\{|y| \geq 1 \}} \frac{1}{|y|^n}dy$ = $\int_{1}^{\infty} \frac{1}{r^n} r^{n-2}dr$.

Question

In $\mathbb R^n$, $\int_{\{|y| \geq 1 \}} \frac{1}{|y|^n}dy$ = $\int_{1}^{\infty} \frac{1}{r^n} r^{n-2}dr$. My question is: why is $r^{n-2}$. I think by Jacobi, it should be $r^{n-1}$.

Answer 1

Yes, you are correct. For $n\geq 1$ and $a>0$ it should be $$\int_{\{|y| \geq 1 \}} \frac{1}{|y|^a}dy=\int_{r=1}^{\infty}\frac{1}{r^a}\cdot |S_{n-1}(r)| dr =|S_{n-1}(1)|\int_{r=1}^{\infty}\frac{1}{r^a}\cdot r^{n-1} dr\\=|S_{n-1}(1)|\int_{r=1}^{\infty}\frac{dr}{r^{a-n+1} }dr$$ where $S_{n-1}(r)=\{y\in\mathbb{R}^n: |y|= r\}$ the $(n-1)$-sphere of radius $r$ centred at the origin. Note that by homogeneity $|S_n(r)|=|S_n(1)|r^{n-1}$. Moreover the above integral is convergent iff $a-n>0$ (so in your case for $a=n$ it gives $+\infty$).