How can I calculate $\int_0^\infty\frac{r^{n-1}}{(1+r^2)^{(n+1)/2}}\,dr$? How can I calculate: 
$$ \displaystyle\int_0^\infty \frac{r^{n-1}}{(1+r^2)^\frac{n+1}{2}}\,dr $$
I encountered this integral in proving that a function is an approximation to the identity. But I don't know how to solve this integral. I would greatly appreciate any help.
 A: Let $u=r^2$, so $du/u = 2dr/r $, and the integral becomes
$$ I = \frac{1}{2}\int_0^{\infty} \frac{u^{n/2-1}}{(1+u)^{(n+1)/2}} \, du. $$
Now, this is an integral that can be written in terms of the Beta-function: in particular the transformation $u=1/y-1$ turns it into
$$ \frac{1}{2} \int_0^1 (1-y)^{n/2-1} y^{-1/2} \, dy = \frac{1}{2}B(1/2,n/2) = \frac{\sqrt{\pi}\Gamma(n/2)}{2\Gamma((n+1)/2)}. $$
A: Let $x=r^2>0$ then $x^{\frac{n-2}{2}}=r^{n-2}$ and we have $dx=2rdr$ so that we have,
$$\frac{1}{2} \int_{0}^{\infty} \frac{x^{\frac{n-2}{2}}}{(1+x)^{\frac{n+1}{2}}} dx $$
Now use the Beta Function properties. See this.
A: As shown we have 
$$I_n =\displaystyle\int_0^\infty \frac{r^{n-1}}{(1+r^2)^\frac{n+1}{2}}\,dr=  \frac{\sqrt{\pi}\Gamma(n/2)}{2\Gamma((n+1)/2)}$$
Now use the formula 
$$\Gamma \left(n+\frac{1}{2}\right) = \frac{(2n-1)!! \sqrt{\pi}}{2^n}$$
For even integers 
$$I_{2k} = \frac{ (k-1)!}{(2k-1)!!}2^{k-1}$$
and for odd integers 
$$I_{2k+1} =\frac{(2k-1)!!}{k!}\frac{\pi}{2^{k+1}}$$
where the double factorial 
$$n!!= \begin{cases} 
      n(n-2)\cdots 5\times 3 \times 1 & n \in \mathbb{2N}+1 \\
      n(n-2)\cdots 6\times 4 \times 2 & n \in \mathbb{2N} \\
      1 & n \in \{-1,0\}
   \end{cases}
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\int_{0}^{\infty}{r^{n - 1} \over \pars{1 + r^{2}}^{\pars{n + 1}/2}}\,\dd r & =
{1 \over 2}\int_{0}^{\infty}
{r^{n/2 - 1} \over \pars{1 + r}^{\pars{n + 1}/2}}\,\dd r
\\[5mm] & =
{1 \over 2}\int_{0}^{\infty}r^{n/2 - 1}\bracks{%
{1 \over \Gamma\pars{\bracks{n + 1}/2}}\int_{0}^{\infty}t^{\pars{n - 1}/2}
\expo{-\pars{1 + r}t}\dd t}\dd r
\\[5mm] & =
{1 \over 2\Gamma\pars{\bracks{n + 1}/2}}\int_{0}^{\infty}t^{\pars{n - 1}/2}\expo{-t}
\int_{0}^{\infty}r^{n/2 - 1}\expo{-t\,r}\dd r\,\dd t
\\[5mm] & =
{1 \over 2\Gamma\pars{\bracks{n + 1}/2}}\int_{0}^{\infty}t^{\pars{n - 1}/2}\expo{-t}
{1 \over t^{n/2}}\int_{0}^{\infty}r^{n/2 - 1}\expo{-r}\dd r\,\dd t
\\[5mm] & =
{1 \over 2\Gamma\pars{\bracks{n + 1}/2}}\pars{\int_{0}^{\infty}t^{-1/2}\expo{-t}\,\dd t}
\pars{\int_{0}^{\infty}r^{n/2 - 1}\expo{-r}\dd r}
\\[5mm] & =
{\Gamma\pars{1/2}\Gamma\pars{n/2} \over 2\,\Gamma\pars{\bracks{n + 1}/2}} =
\bbx{\ds{{1 \over 2}\,\root{\pi}\,
{\Gamma\pars{n/2} \over \Gamma\pars{\bracks{n + 1}/2}}}}
\end{align}
