Integral of the function $ (1+|x|^2)^{-k}$ 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?   
 A: $$
\begin{align}
\int_{\mathbb{R}^n}{\frac{\mathrm{d}x}{\left(1+|x|^2\right)^k}}
&=\omega_{n-1}\int_0^\infty\frac{r^{n-1}\,\mathrm{d}r}{\left(1+r^2\right)^k}\tag1\\
&=\frac{\omega_{n-1}}2\int_0^\infty\frac{r^{\frac n2-1}\,\mathrm{d}r}{\left(1+r\right)^k}\tag2\\
&=\frac{\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\frac{\Gamma\!\left(\frac n2\right)\Gamma\!\left(k-\frac n2\right)}{\Gamma(k)}\tag3\\
&=\bbox[5px,border:2px solid #C0A000]{\frac{\pi^{\frac n2}\Gamma\!\left(k-\frac n2\right)}{\Gamma(k)}}\tag4
\end{align}
$$
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 $k\gt\frac n2$.

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
$$
\omega_{n-1}=\frac{2\pi^{\frac n2}}{\Gamma\!\left(\frac n2\right)}\tag9
$$
A: 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.
A: 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 
$$\int_1^\infty\frac{r^{n-1}}{(1+r^2)^k} dr \le\int_1^\infty \frac{1}{r^{2k-n+1}} dr $$
