For what values of $ p \in (0, \infty] $ do we have $ f \in L^p (\mathbb{R}^3) $?

Question

I have presented a solution to the following problem. However by inspection of the result is clearly wrong. But I can't find where the error is in my proof.

Problem. Let $f(x)=\frac{1}{|x|^2}\frac{1}{(1+|x|)^2}$, for $x\in\mathbb{R}^3$, and $\mathbb{R}^3$ equipped with the usual Lesbesgue measure. What values of $p\in [0,\infty)$ do we have $f\in L^p(\mathbb{R}^3)$?

Solution. The function $x\mapsto \frac{1}{|x|^2}\frac{1}{(1+|x|)^2}$ belongs in $L^p(\mathbb{R}^3)$ if, only if, $$ \int_{\mathbb{R}^3}\left( \frac{1}{|x|^2}\frac{1}{(1+|x|)^2}\right)^p dx = \int_{\mathbb{R}^3} \frac{1}{|x|^{2p}}\frac{1}{(1+|x|)^{2p}}dx \\ =\int_{\mathbb{R}^3} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx < \infty $$ By definition of improper integral we have $$ \int_{\mathbb{R}^3} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx = \lim_{r\to \infty} \int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx $$ For all $\epsilon>0$ there is a $r>0$ such that $|x|>r$ implies $1-\epsilon< \frac{1}{(1+|x|)^2} <1$. Then $$ (1-\epsilon)\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx < \int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx < \int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx $$ Thus the integral $\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}\frac{1}{(1/|x|+1)^{2p}}dx$ converges if and only if the integral $\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx$ converges. Thus, it is sufficient to analyze the integral $\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx$. In fact,

\begin{align} \int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx =& \int_{1/r}^{r}\left(\int_{|x|=s} \frac{1}{|x|^{4p}} dS \right)ds \\ =& \int_{1/r}^{r}\left(\int_{|x|=s} \frac{1}{s^{4p}} dS \right)ds \\ =& \int_{1/r}^{r}\frac{1}{s^{4p}} \left(\int_{|x|=s} dS \right)ds \\ =& \int_{1/r}^{r}\frac{1}{s^{4p}} 4\pi s^2 ds \\ =& 4\pi \int_{1/r}^{r}s^{-4p+2} ds \\ =& \frac{4\pi}{-4p+3} \left[(r)^{-4p+3} -(1/r)^{-4p+3}\right] \\ \end{align} And $$ \lim_{r\to \infty} \frac{4\pi}{-4p+3} \left[(r)^{-4p+3} -(1/r)^{-4p+3}\right] = \left\{ \begin{array}{ccc} \mbox{does not exist} & \mbox{if} & p=3/4\\ \\ \infty & \mbox{if} & p< 3/4\\ \\ -\infty & \mbox{if} & p> 3/4 \end{array} \right. $$ But we know that something is wrong because if $p>3/4$ then $ \int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx>0 $ and $$ \lim_{r\to \infty}\int_{B[0,r]-B[0,1/r]} \frac{1}{|x|^{4p}}dx>0 $$

Answer 1

The issue is that the behavior near $x=0$ is not like $|x|^{-4p}$. You write that for $x\approx 0$, $$ 1-\epsilon< \frac{1}{(1+|x|)^2} <1$$ which is correct, but you use $$ 1-\epsilon< \frac{1}{(1/|x|+1)^2} <1$$ which is wrong. In fact, if $|x|\to 0$, then $\frac{1}{(1/|x|+1)^2} = \frac{|x|^2}{(1+|x|)^2} \le \frac{|x|^2}{(1+0)^2} \to 0$.

A sketch of the correct answer. The behavior near $0$ is like $|x|^{-2p}$. This is integrable if it doesn't explode too quickly; you should see $-2p>-d$, where $d=3$ is the dimension.

The behavior at infinity is like $|x|^{-4p}$. This is integrable if it decays fast enough: you should see $-4p < -d$. Together this gives the range $$ -d < -2p < - d/2 \iff d/4<p< d/2$$

Answer 2

HINT: using the change to spherical coordinates the integral reduces to $$ \int_{\Bbb R ^3}\frac1{|x|^{2p}(1+|x|^2)^p}\,\mathrm d x=4\pi\int_0^\infty \frac1{r^{2(p-1)}(1+r^2)^p}\,\mathrm d r $$ what makes the analysis simpler. In fact everything can be reduced to study the convergence of the improper integrals $$ \int_{0}^1\frac1{r^{2p-2}}\,\mathrm d r\qquad \text{ and }\qquad \int_1^{\infty }\frac1{r^{4p-2}}\,\mathrm d r $$ for $p>0$.