Integral of the function $ (1+|x|^2)^{-k}$

Question

I want to prove that the $$\int _{ { R }^{ n } }{ \frac { 1 }{ { (1+{ |x| }^{ 2 }) }^{ k } } } <\infty $$ if $k>\frac n 2$; where |x| is the usual norm in ${R}^{n}$. I tried this: $$\int _{ { R }^{ n } }{ \frac { 1 }{ { (1+{ |x| }^{ 2 }) }^{ k } } }=\int _{ { R }^{ n }-{ B }_{ 1 }(0) }{ \frac { 1 }{ { (1+{ |x| }^{ 2 }) }^{ k } } } + \int _{ { B }_{ 1 }(0) }{ \frac { 1 }{ { (1+{ |x| }^{ 2 }) }^{ k } } } $$ where ${ B }_{ 1 }(0)$ is the unit ball, the second term of the sum is finite since $$\int _{ { B }_{ 1 }(0) }{ \frac { 1 }{ { (1+{ |x| }^{ 2 }) }^{ k } } } \le \int _{ { B }_{ 1 }(0) }{ \frac { 1 }{ { ({ |x| }^{ 2 }) }^{ k } } }<\infty $$ if $k>\frac n 2$ but i do not know how to estimate the first term of the sum, any idea?

Answer 1

By Polar coordinates we get

$$\int_{\Bbb R^n}\frac{dx}{(1+|x|^2)^k}\mathrm dr= c_n\int_0^\infty\frac{r^{n-1}}{(1+r^2)^k} dr = c_n\int_0^1\frac{r^{n-1}}{(1+r^2)^k} dr + c_n\int_1^\infty\frac{r^{n-1}}{(1+r^2)^k} dr $$

where $c_n $ is the area of the unit sphere. By Riemann integral the above integral converges if and only if $2k-n>0$ since

$$\int_1^\infty\frac{r^{n-1}}{(1+r^2)^k} dr \le\int_1^\infty \frac{1}{r^{2k-n+1}} dr $$

Answer 2

Note that converting to hyperspherical coordinates your integral is $$ \lim_{R\to \infty}\int_{\partial B(0,1)}\int_0^R\frac{r^{n-1}}{(1+r^2)^k}\mathrm dr\mathrm dS $$ where $\mathrm dS$ is the surface element depending only upon the angular variables. I.e the usual integrating over shells method. Now, if $n\alpha(n)^{n-1}$ denotes the surface area of the unit sphere, your integral is in fact $$ n\alpha(n)^{n-1}\lim_{R\to \infty}\int_0^R\frac{r^{n-1}}{(1+r^2)^k}\mathrm dr $$ a single variable integral, with the integrand asymptotic to $$ \frac{r^{n-1}}{r^{2k}} $$ at infinity. Now if $2k>n+\epsilon$ for an $\epsilon>0$, we have for $r>1$, $$ \frac{r^{n-1}}{r^{2k}}<\frac{r^{n-1}}{r^{n+\epsilon}}=\frac{1}{r^{1+\epsilon}} $$ which is integrable away from $0$. Since this is the only singularity you have to worry about, you are done.

Answer 3

$$ \begin{align} \int_{\mathbb{R}^n}{\frac{\mathrm{d}x}{\left(1+|x|^2\right)^k}} &=\omega_{n-1}\int_0^\infty\frac{r^{n-1}\,\mathrm{d}r}{\left(1+r^2\right)^k}\tag1\\ &=\frac{\omega_{n-1}}2\int_0^\infty\frac{r^{\frac n2-1}\,\mathrm{d}r}{\left(1+r\right)^k}\tag2\\ &=\frac{\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\frac{\Gamma\!\left(\frac n2\right)\Gamma\!\left(k-\frac n2\right)}{\Gamma(k)}\tag3\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^{\frac n2}\Gamma\!\left(k-\frac n2\right)}{\Gamma(k)}}\tag4 \end{align} $$ Explanation:
$(1)$: convert to polar coordinates
$(2)$: substitute $r\mapsto r^{\frac12}$
$(3)$: apply $(9)$ and the Beta Function
$(4)$: cancel common factors

Formula $(4)$ is finite for $k\gt\frac n2$.

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Computation of $\boldsymbol{\omega_{n-1}}$ $$ \begin{align} 1 &=\int_{\mathbb{R}^n} e^{-\pi|x|^2}\mathrm{d}x\tag5\\ &=\omega_{n-1}\int_0^\infty e^{-\pi r^2}r^{n-1}\,\mathrm{d}r\tag6\\ &=\frac{\omega_{n-1}}2\int_0^\infty e^{-\pi r}r^{\frac n2-1}\,\mathrm{d}r\tag7\\ &=\frac{\omega_{n-1}}2\pi^{-\frac n2}\Gamma\!\left(\frac n2\right)\tag8 \end{align} $$ Explanation:
$(5)$: integral is the product of $n$ copies of $\int_{-\infty}^\infty e^{-\pi x^2}\mathrm{d}x=1$
$(6)$: convert to polar coordinates
$(7)$: substitute $r\mapsto r^{\frac12}$
$(8)$: apply the Gamma Function

Equation $(8)$ implies $$ \omega_{n-1}=\frac{2\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\tag9 $$