Calculate $\int\limits_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^p}$

Question

I need to calculate the integral: $$\int\limits_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^p}$$ (That's all the question asks, so I believe that I also should determine the values of $p$ for which this integral converges).
My attempt is using the co-area formula with $\Phi(x)=|x|$ (which has $|\nabla\Phi|=1$):

$$\int\limits_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^p}= \int\limits_{0}^{\infty} \left ( \int\limits_{S_r} \frac{1}{(1+r^2)^p}dS(x) \right ) dr=\int\limits_{0}^{\infty} \frac{1}{(1+r^2)^p}vol_{n-1}(S_r) dr= \\ \omega \int\limits_{0}^{\infty} \frac{r^{n-1}}{(1+r^2)^p} dr$$ where $S_r$ is the $0$-centered $(n-1)$-dimentional sphere of radius $r$, and $\omega$ is the $(n-1)$-volume of $S_1$.

Now I'm stuck with that last improper integral. Any ideas/alternative methods will be welcomed. Thank you!

Answer 1

The integral of interest, $\int_0^\infty \frac{r^{n-1}}{(1+r^2)^p}\,dr$, converges for $n<2p$ and diverges otherwise.

Enforcing the substitution $r\to r^{1/2}$ reveals

$$\begin{align} \int_0^\infty \frac{r^{n-1}}{(1+r^2)^p}\,dr&=\frac12\int_0^\infty \frac{r^{n/2-1}}{(1+r)^p}\,dr\\\\ &=\frac12B\left(n/2,p-n/2\right)\\\\ &=\frac12\frac{\Gamma(n/2)\Gamma(p-n/2)}{\Gamma(p)} \end{align}$$

Answer 2

The surface area of the $n$-dimensional unit sphere is given by $A_n=\frac{\pi^{n/2}}{\Gamma\left(n/2\right)}$, hence by Cavalieri's principle the given integral equals:

$$ \frac{\pi^{n/2}}{\Gamma\left(n/2\right)}\int_{0}^{+\infty}\frac{r^{n-1}}{(1+r^2)^p}\,dr = \frac{\pi^{n/2}\,\Gamma\left(p-\frac{n}{2}\right)}{2\,\Gamma(p)}$$ under the assumption $p>\frac{n}{2}$ (otherwise, the integral is not convergent).

Answer 3

Not a complete answer, but too long to be posted as a comment.

The exact value of the integral was already given in the other answers. But it is worth to note that sometimes the value is not important; sometimes we only need to know that the integral converges. In this case, the following estimate (which does not invoke the gamma function) can help. $$\int\limits_{0}^{\infty} \frac{r^{n-1}}{(1+r^2)^p} dr=\int\limits_{0}^{1} \frac{r^{n-1}}{(1+r^2)^p} dr+\int\limits_{1}^{\infty} \frac{r^{n-1}}{(1+r^2)^p} dr:=I_1+I_2$$ where $I_1<\infty$ because $r\mapsto \frac{r^{n-1}}{(1+r^2)^p}$ is continuous on $[0,1]$ and, for $p>n/2$, $$I_2\leq \int\limits_{1}^{\infty} \frac{r^{n-1}}{r^{2p}} dr=\int\limits_{1}^{\infty} r^{n-1-2p}dr=\frac{r^{n-2p}}{n-2p}<\infty.$$ So, for $p>n/2$, $$\int\limits_{\mathbb{R}^n} \frac{dx}{(1+|x|^2)^p}=\omega(I_1+I_2)<\infty.$$

Remark. A similar calculation can be done with the integral $$\int\limits_{\mathbb{R}^n} \frac{dx}{(1+|x|^p)^2}$$ which appears in the proof of the embedding $H^p\hookrightarrow L^\infty$ for $n/2<p<\infty$. In this context, the estimate is useful because we only need to know that the integral converges.