<s>[INST] Show $\int_{\mathbb{R}^3} \frac{1}{\vert{\eta -v\vert}^2} \frac{1}{(1+\vert \eta \vert)^4} d\eta \leq \frac{C}{(1+\vert v \vert)^2}$ 
$\textbf{Problem}$
\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!
 [/INST] We can compute the integral by rotating the $\sigma$-coordinate system in such a way that $v$ points towards the north pole of the sphere (i.e. in $e_3$-direction) and introducing polar coordinates $(\theta,\phi)$. Then the norm becomes $\lvert v + \rho \sigma \rvert = \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rvert \rho \cos(\theta)}$ and the integral turns into
$$ \int \limits_0^\infty \int \limits_0^\pi \int \limits_0^{2\pi} \frac{\sin(\theta) \, \mathrm{d} \phi \, \mathrm{d} \theta \, \mathrm{d} \rho}{\left(1 + \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rvert \rho \cos(\theta)}\right)^4} = 2 \pi \int \limits_0^\infty \int \limits_{-1}^1 \frac{\mathrm{d} t \, \mathrm{d} \rho }{\left(1 + \sqrt{\lvert v \rvert^2 + \rho^2 + 2 \lvert v \rvert \rho t}\right)^4} \equiv f(\lvert v \rvert) \, .$$
Clearly, $f(0) = \frac{4 \pi}{3}$. For $r > 0$ we can let $s = \sqrt{r^2 + \rho^2 + 2 r \rho t}$ to obtain
$$ f(r) = 2 \pi \int \limits_0^\infty \int \limits_{\lvert r - \rho \rvert}^{r + \rho} \frac{s}{r \rho (1+s)^4} \, \mathrm{d} s \, \mathrm{d} \rho \, .$$
The remaining integrals are not particularly hard, albeit slightly tedious, and the final result looks rather nice:
$$ f(r) = \frac{4\pi}{3} \begin{cases} 1, &  r = 0 \\ \frac{1}{3}, &  r = 1 \\ \frac{1-r^4 + 4 r^2 \log(r)}{(1-r^2)^3} , &  r \in (0,\infty) \setminus \{1\}\end{cases}  \, .$$
We want to find some $C > 0$ such that $f(r) \leq \frac{C}{(1+r)^2}$ holds for every $r \geq 0$. 


*

*If any constant is fine, we can simply use the inequality $\frac{-\log(r)}{1-r} \geq \frac{2}{1+r}\, ,r > 0,$ which follows from $\frac{\tanh(t)}{t} \leq 1\, , t \in \mathbb{R}$, with $t = \frac{1}{2} \log(r)$.
It leads to 
$$ \frac{3}{4\pi} f(r) = \frac{1 + r + r^2 + r^3 - 4 r^2 \frac{-\log(r)}{1-r}}{(1+r)^3 (1-r)^2} \leq \frac{1 + r + r^2 + r^3 - 4 r^2 \frac{2}{1+r}}{(1+r)^3 (1-r)^2} = \frac{1 + \frac{2 r}{(1+r)^2}}{(1+r)^2} \leq \frac{3}{2 (1+r)^2} $$
and therefore $C = 2 \pi$.

*In order to find the optimal constant we need to use the more complicated inequality $$\frac{-\log(r)}{1-r} \geq \frac{-1 + 7r + 7r^2-r^3}{12 r^2} \, , \, r > 0,$$
which is a consequence of
$$ \frac{\tanh(s)}{s} \leq \frac{1}{4} \left[\frac{\sinh(s)}{s} + \frac{3}{\cosh(s)}\right]\, , \, s \in \mathbb{R},$$
with $s = \log(r)$ (the latter can be proven by rewriting it as $1 - \frac{\sinh(s) [4- \cosh(s)]}{3 s} \geq 0$ and noting that the Maclaurin series of the left-hand side contains only non-negative terms). It follows that
$$ \frac{3}{4\pi} f(r) \leq \frac{1 + r + r^2 + r^3 - \frac{-1+7r+7r^2-r^3}{3}}{(1+r)^3 (1-r)^2} = \frac{4}{3 (1+r)^2} $$
and the corresponding constant $C = \frac{16 \pi}{9}$ is the smallest possible, since $f(1) = \frac{4\pi}{9} = \frac{16 \pi}{9 (1+1)^2}$.


It would be nice to obtain the estimate without evaluating the integral explicitly, but I have not found a way to do that yet.
</s>