Ngô Quốc Anh

March 10, 2012

An integral of 1/(1+|x|^2) over the whole Euclidean space

Filed under: Giải Tích 2, Giải Tích 3, Linh Tinh — Ngô Quốc Anh @ 2:46

Sometimes, we need a precise value for following $\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}}, \quad \alpha>\frac{n}{2}.$

As such, I am going to calculate it and place the result here for future works.

In order to evaluate the above integral, we need to use the so-called co-area formula. We first write $\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \int_0^{ + \infty } {\left( {\int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} } \right)dr}.$

Note that $\displaystyle \begin{gathered} \int_{\partial {B_0}(r)} {\frac{{dS}}{{{{(1 + {r^2})}^\alpha }}}} = \frac{1}{{{{(1 + {r^2})}^\alpha }}}\int_{\partial {B_0}(r)} {dS} \hfill \\ \qquad\qquad\qquad= \frac{1}{{{{(1 + {r^2})}^\alpha }}}\text{Area}({B_0}(r)) = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}\end{gathered}.$

Therefore, $\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^\alpha }}}} = \frac{{2{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2}} \right)}}\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)}}{{\Gamma (\alpha )}}$

where we have used $\displaystyle\int_0^{ + \infty } {\frac{{{r^{n - 1}}}}{{{{(1 + {r^2})}^\alpha }}}} dr = \frac{1}{2}\frac{{\Gamma \left( {\alpha - \frac{n}{2}} \right)\Gamma \left( {\frac{n}{2}} \right)}}{{\Gamma (\alpha )}}.$

In particular, by the duplication formula, we get that $\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^n}}}} = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( {\frac{n}{2}} \right)}}{{\Gamma (n)}} = \frac{1}{{{2^n}}}\frac{{2{\pi ^{\frac{{n + 1}}{2}}}}}{{\Gamma \left( {\frac{n}{2} + \frac{1}{2}} \right)}} = \frac{{{\sigma _n}}}{{{2^n}}}$

where $\sigma_n$ is the volume of $n$-sphere of radius 1 in $\mathbb R^{n+1}$.

Another particular formula occurs when $\alpha=\frac{n+2}{2}$. We then have $\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^{\frac{n}{2} + 1}}}}} = {\pi ^{\frac{n}{2}}}\frac{{\Gamma \left( 1 \right)}}{{\Gamma \left( {\frac{n}{2} + 1} \right)}} = \frac{{{\pi ^{\frac{n}{2}}}}}{{\Gamma \left( {\frac{n}{2} + 1} \right)}} = {\omega _n}$

where $\omega_n$ is the volume of $n$-ball of radius 1 in $\mathbb R^{n+1}$.

Keep in mind that the $n$-sphere of radius 1 is nothing but the boundary the $n+1$-ball of radius 1. Therefore, $\sigma_n$ is nothing but the surface area of the $n+1$-ball of radius 1. Interestingly, we have the following well-known formula $\displaystyle {\omega _n} = \frac{{{\sigma _{n - 1}}}}{n}.$

As such, the previous formula can be expressed as $\displaystyle\int_{{\mathbb{R}^n}} {\frac{{dx}}{{{{(1 + |x{|^2})}^{\frac{n}{2} + 1}}}}} = \frac{{{\sigma _{n - 1}}}}{n}.$

1. Hi Ngo, how can you do with the same integral above times log(1+|x|^2)? Give me an advice, please!
Txs

Comment by Fab — April 23, 2012 @ 21:21

• Unless there is some motivation, otherwise, it is not worth considering that integral. As you may know, the one I considered above comes from blow-up analysis for PDEs.

Comment by Ngô Quốc Anh — April 23, 2012 @ 21:47

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