Prove that $1/(1+|x|^2)^n$ is integrable and find its value I was studying the dual space $\mathscr{S}'(\mathbb{R}^n)$ of Schwarz space (tempered distributions). I was given a proof of the claim that $L^p(\mathbb{R^n})\subset \mathscr{S}'(\mathbb{R}^n)$, which goes like this: Suppose $f\in L^p$. We identify $f$ with $f(\phi)=\int f\phi d\mu$ where $\phi\in \mathscr{S}(\mathbb{R^n})$ is a Schwarz function of rapid decrease and $\mu$ is the Lebesgue measure on $\mathbb{R}^n$. Then we have the bound
$$
\left|\int_{\mathbb{R}^n} f(x)\phi(x) d\mu\right|\leq \|f\|_{L^p}\cdot\|(1+|x|^2)^{-n}\|_{L^{p'}}\cdot\|(1+|x|^2)^n \phi\|_{\infty}.
$$
This shows that $f\in \mathscr{S}'(\mathbb{R}^n)$. So it would seem that
$$
\int_{\mathbb{R}^n}\frac{1}{(1+|x|^2)^n}d\mu<\infty \qquad |x|^2=x_1^2+\cdots+x_n^2.
$$
How can one show that this is the case? i.e. $1/(1+|x|^2)^n$ is integrable on $\mathbb{R}^n$? If so can we evaluate this integral and give an explicit expression? (I tried for $n\leq 7$ on Mathematica and get no obvious pattern, but it seems that this integral is proportional to $\pi^{\lceil n/2\rceil}$). Finally, with this how can we show that
$$\|(1+|x|^2)^{-n}\|_{L^{p'}}<\infty$$
for any real $p'\geq 1$? Thanks!
 A: It is true that if $\psi(x) = (1+\vert x \vert^2)^{-n}$ and $p \geqslant 1$ then $\psi \in L^p(\mathbb R^n)$. It actually is very easy to show and follows from two simple observations:

*

*Since $\psi$ is smooth, in any bounded subset $\Omega$ of $\mathbb R^n$ it is in $L^p(\Omega)$.

*As $\vert x \vert \to \infty $, $$ \frac 1 {(1+\vert x \vert^2)^{np}} \sim \frac 1 {\vert x \vert^{2np}} $$ and $\vert x \vert^{-2np}$ is integrable away from zero since $2np >n$. If you are not familiar with this notation $\sim$ means asymptotic.


If you want to actually calculate this integral, you must use polar coordinates: \begin{align*}
\int_{\mathbb R^n} \frac 1 {(1+\vert x \vert^2)^{np}} \, dx &= \int_0^\infty \int_{\partial B_\rho} \frac 1 {(1+\rho^2)^{np}} \, d\mathcal H^{n-1}_x \, d\rho \\
&= n \omega_n \int_0^\infty  \frac {\rho^{n-1}} {(1+\rho^2)^{np}} \,  d\rho
\end{align*} where $d\mathcal H^{n-1}_x$ is the $(n-1)$-Hausdorff measure (think surface element), $\partial B_\rho$ is the $(n-1)$ sphere centred at zero with radius $\rho$, and $\omega_n$ is the volume of the unit ball. Via Mathematica, (or otherwise), we actually have $$\int_0^\infty  \frac {\rho^{n-1}} {(1+\rho^2)^{np}} \,  d\rho = \frac 1 2 \mathrm B (n/2, n(p -1)/2)  $$ where $\mathrm B$ is the Beta function provided $n>0$ and $p>1/2$. Hence, $$ \int_{\mathbb R^n} \frac 1 {(1+\vert x \vert^2)^{np}} \, dx = \frac {n \omega_n} 2 \mathrm B (n/2, n(p -1)/2).$$
