About the integrabity of $f(x) = |x|^{-(n+1)}$ and $f(x) = |x|^{-n}$ on $\mathbb{R}^n$

Question

The function $f(x) = |x|^{-(n+1)}$ on $\mathbb{R}^n$ is obviously unbounded as we approach $0$. Thus, in order to prove that the function is not integrable on $\mathbb{R}^n$, I use the following evaluation

$$\int_{\mathbb{R}^n} f(x) d \mu (x) \ge \int_{|x| < \epsilon} f(x) d \mu (x) \ge \int_{|x| < \epsilon} \epsilon ^{-(n+1)} d \mu (x) = c\epsilon ^n \times\epsilon ^{-(n+1)} = \dfrac{c}{\epsilon} $$ and let $\epsilon \to 0$

Please help me to verify my argument.

Furthermore, can the power $-(n+1)$ be replaced by $-n$? I think the answer is negative, but I can't prove it.

Answer 1

Your argument is correct. To prove that the result is true with $n+1$ replaced by $n$ note that $\int \frac 1 {|x|^{n}} \geq \sum_k \int_{R_k \leq |x| \leq R_{k+1}} \geq \sum_k \frac {cR_{k+1}^{n}-cR_k^{n}} {R_{k+1}^{n}}$ for any increasing sequence $\{R_k\}$. Choose these numbers so that $\frac {R_{k+1}} {R_k} \to \infty$ to see that the general term of the series here does not tend to $0$ and hence the series is divergent.

Answer 2

I suggest to use polar coordinate and changes everything into 1-dim. Note that $$ \int_0^{\infty}\dfrac{1}{r^k}\mathtt{d} r=\infty,\quad\forall k\in\mathbb{Z} $$ When $k=1$, it becomes log-type and the integral also blows up.

So the result follows immediately. ;)