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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.

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  • $\begingroup$ If $n$ is odd, we could let $u=1+r^2$ and apply binomial expansion. $\endgroup$ – Simply Beautiful Art Mar 3 '17 at 0:34
  • $\begingroup$ $\, n \gt 0 \,$. $\endgroup$ – Hazem Orabi Mar 3 '17 at 0:49
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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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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 $$ \frac{1}{2} \int_0^1 (1-y)^{n/2-1} y^{-1/2} \, dy = \frac{1}{2}B(1/2,n/2) = \frac{\sqrt{\pi}\Gamma(n/2)}{2\Gamma((n+1)/2)}. $$

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  • $\begingroup$ All too easy. (+1) $\endgroup$ – Mark Viola Mar 3 '17 at 0:51
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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.

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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

$$n!!= \begin{cases} n(n-2)\cdots 5\times 3 \times 1 & n \in \mathbb{2N}+1 \\ n(n-2)\cdots 6\times 4 \times 2 & n \in \mathbb{2N} \\ 1 & n \in \{-1,0\} \end{cases} $$

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