TITLE: Show $\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta \leq \frac{C}{(1+\vert v \vert)^2}$
QUESTION [5 upvotes]: $\textbf{Problem}$
\begin{equation*}
\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta 
\leq \frac{C}{(1+\vert v \vert)^2}
\end{equation*}

For obtaining the above upper bound, I tried to change the variable $\eta \rightarrow v+ \sigma \rho $ for $\sigma \in \mathbb{S}^2, \rho \in \mathbb{R}^{+}$. 
However, I stuck to handle the part $\displaystyle{\frac{1}{(1+\vert{\eta\vert})^4}}$.

$\textbf{Attempt}$
\begin{align*}
      \frac{1}{(1+\vert \eta \vert)^4} &= \frac{1}{(1+\vert v+\sigma \rho\vert)^4}\\
      \int_{\mathbb{R}^3} \frac{1}{\vert \eta - v \vert^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta &= \int_0^{\infty} \int_{\mathbb{S}^2} \frac{1}{\rho^2} \frac{1}{(1+\vert v + \sigma \rho\vert)^4} \rho^2 d\sigma d\rho  
\end{align*}

I don't know how to derive $(1+\vert v \vert)^2$ from the last integral term in my attempt. 
Any help is appreciated. 
Thank you!

REPLY [3 votes]: For all $v \in  \mathbb{R}^3$, we have
$$f(v):=\int_{\mathbb{R}^3}d\eta\,\frac{1}{|\eta-v|^2}\frac{1}{(1+|\eta|)^4}\leq \int_{B(v,1)}d\eta\,\frac{1}{|\eta-v|^2}+\int_{\mathbb{R}^3}d\eta\frac{1}{(1+|\eta|)^4}\\
=4\pi+4\pi\int_0^\infty dr\,\frac{r^2}{(1+r)^4}<+\infty$$
so that $v \mapsto f(v)$ is (uniformly) bounded. It suffices henceforth to show that $f(v)\leq \frac{C}{|v|^2}$. For that purpose, note that
$$f(v)\leq \int_{\mathbb{R}^3\setminus B(0,|v|/2)}d\eta\,\frac{1}{|\eta-v|^2}\frac{1}{|\eta|^3}+\int_{B(0,|v|/2)}d\eta\,\frac{1}{|\eta-v|^2}\frac{1}{(1+|\eta|)^4}$$
$$\int_{\mathbb{R}^3\setminus B(0,|v|/2)}d\eta\,\frac{1}{|\eta-v|^2}\frac{1}{|\eta|^3}=\frac{1}{|v|^2}\underbrace{\int_{\mathbb{R}^3\setminus B(0,1/2)}dx \frac{1}{|x-e_v|^2|x|^3}}_{=\text{const. indep. of $v$}<+\infty}$$
$$\int_{B(0,|v|/2)}d\eta\,\frac{1}{|\eta-v|^2}\frac{1}{(1+|\eta|)^4}\leq \int_{B(0,|v|/2)}d\eta\,\frac{1}{|v/2|^2}\frac{1}{(1+|\eta|)^4}\leq \frac{4}{|v|^2}\underbrace{\int_{\mathbb{R}^3}d\eta \frac{1}{(1+|\eta|)^4}}_{<+\infty}$$
Combining these results finishes the proof.