Find $p$ such that the integral is finite

Let $X \subset \mathbb{R}^{n}$. For which $p \in[1, \infty)$ it holds that $f \in L^{p}(X)$ when $f(x)=|x|^{-1}$ and

$X=B(0,1)$
$X=\mathbb{R}^{n} \backslash B(0,1)$
$X=\mathbb{R}^{n}$.

My attempt:

Suppose $0\leq a<b\leq \infty$ and we consider the annulus $E_{a,b}:=\{x\in\Bbb{R}^n\,:\, a<|x|<b\}$. Then, for any $p\in\Bbb{R}, $ we have $\int_{E_{a,b}}\frac{1}{|x|^{p}}\,dx=\int_a^b\frac{1}{r^{p}}A_{n-1}r^{n-1}\,dr=A_{n-1}\int_a^b\frac{1}{r^{p+1-n}}\,dr$, where $A_{n-1}$ is the surface area of the unit sphere $S^{n-1}\subset\Bbb{R}^n$

$A_{n-1} \frac{-(p+1-n)}{r^{p+2-n}}\bigr\vert_{a}^{b}$

In the first case, $a = 0$ and $b = 1$, in the second case $a = 1$ and $b = \infty$, and in the third case $a = 0$ and $b = \infty$. In the first case the integral is finite when $p +2 ≤ n$, in the second case the integral is finite when $p + 2 ≥ n$, and in the third case $p + 2 = n$.

Is my attempt correct?

Solution 1:

For any $\lambda\in \Bbb{R}$,

$\int_0^1\frac{1}{x^{\lambda}}\,dx$ is finite if and only if $\lambda<1$.
$\int_1^{\infty}\frac{1}{x^{\lambda}}\,dx$ is finite if and only if $\lambda>1$.
clearly, $\int_0^{\infty}\frac{1}{x^{\lambda}}\,dx$ is finite if and only if $\int_0^1\frac{1}{x^{\lambda}}\,dx$ and $\int_1^{\infty}\frac{1}{x^{\lambda}}\,dx$ are both finite. This happens if and only if $\lambda<1$ and $\lambda>1$.... i.e this integral is never finite.