Computing the Lebesgue integral over a ball

Question

Let $\alpha \in \mathbb{R}$ and $\lambda_n$ the Lebesgue measure on $\mathbb{R^n}$.

Define $f:\mathbb{R^n}\backslash(0)\to\mathbb{R}, \ f(x)=\left\lVert x\right\rVert^\alpha$

For which $\alpha$ is $f$ Lebesgue integrable on $B_1(0):=\lbrace x \in \mathbb{R^n}:\left\lVert x\right\rVert \leq 1 \rbrace$

and for which $\alpha$ on $\mathbb{R^n}\backslash B_1(0)$?

Also, how to compute $\int_{B_1(0)}f \ d\lambda_n$ and $\int_{\mathbb{R^n}\backslash B_1(0)}f \ d\lambda_n$ if $f$ is integrable?

I tried:

$f$ is Lebesgue integrable on $B_1(0) \Leftrightarrow\int_{B_1(0)}|f| \ d\lambda_n<\infty$.

So if $\alpha<\infty$, then $f(x)$ is Lebesgue integrable.

For computing the integrals, I thought about using the transformation formula:

$\int_{B_1(0)}\left\lVert x\right\rVert^\alpha \ d\lambda_n=\alpha \cdot \int_{0}^{1}r^n \ d \lambda_r$

Here I don't know how to continue. Is this method correct or is there another way to do this?

Answer 1

Consider spherical shells $S_{a,b}:=\{x\in{\mathbb R}^n\,|\,a\leq\|x\|\leq b\}$. One then has $$\int_{S_{a,b}}\|x\|^\alpha {\rm d}(x)=\omega_{n-1}\int_a^b r^\alpha\>r^{n+1}\>dr\ ,$$ whereby $\omega_{n-1}$ denotes the $(n-1)$-dimensional surface area of $S^{n-1}\subset{\mathbb R}^n$. Now see what happens when $a\to0+$ or $b\to\infty$ for various values of $n$ and $\alpha$.