Exact Values of the integal $\int_0^\infty \frac{r^{n-1}}{(1+r^2)^{\frac{s}{2}}}\,dr$

Question

Does any one know the exact expression of the integral,

$$E_n(s)=\int_0^\infty \frac{r^{n-1}}{(1+r^2)^{\frac{s}{2}}}\,dr~~~~s>n, n\in \mathbb{N}$$ or more generally, $$E_a(s)=\int_0^\infty \frac{r^{a-1}}{(1+r^2)^{\frac{s}{2}}}\,dr~~~~s>a, a\in \mathbb{R}$$ For the special case $s=n, n+2$ I find out by induction that $$ E_{n-1}(n)=\frac{\omega_{n-1}}{2\omega_{n-2}}~~\text{and}~~E_{n-1}(n+2)=\frac{\omega_{n-1}}{2n\omega_{n-2}}. $$ where $\omega_{n-1} = \frac{\pi^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}$ is the surface measure of the n-dimensional sphere of $\mathbb{R}^n$. Further result is welcome

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{0}^{\infty}{r^{n - 1} \over \pars{1 + r^{2}}^{s/2}}\,\dd r & \stackrel{r^{2}\ \mapsto\ r}{=}\,\,\, {1 \over 2}\int_{0}^{\infty}{r^{n/2 - 1} \over \pars{1 + r}^{s/2}}\,\dd r \,\,\,\stackrel{\pars{r + 1}\ \mapsto\ r}{=}\,\,\, {1 \over 2}\int_{1}^{\infty}{\pars{r - 1}^{n/2 - 1} \over r^{s/2}}\,\dd r \\[5mm] & \stackrel{r\ \mapsto\ 1/r}{=}\,\,\, {1 \over 2}\int_{1}^{0}{\pars{1/r - 1}^{n/2 - 1} \over \pars{1/r}^{s/2}}\, {\dd r \over -r^{2}} = {1 \over 2}\int_{0}^{1}r^{s/2 - n/2 - 1}\,\pars{1 - r}^{n/2 - 1}\,\dd r \end{align} The integral converges whenever $$ \left.\begin{array}{lcl} \ds{\Re\pars{{s \over 2} - {n \over 2} - 1}} & \ds{>} & \ds{-1} \\[2mm] \ds{\Re\pars{{n \over 2} - 1}} & \ds{>} & \ds{-1} \end{array}\right\} \qquad\implies\qquad \bbx{\ds{0 < \Re\pars{n} < \Re\pars{s}}} $$ In such a case \begin{align} \int_{0}^{\infty}{r^{n - 1} \over \pars{1 + r^{2}}^{s/2}}\,\dd r & = {1 \over 2}\,\mrm{B}\pars{{s \over 2} - {n \over 2},{n \over 2}} = \bbx{\ds{{\Gamma\pars{s/2 - n/2}\Gamma\pars{n/2} \over 2\Gamma\pars{s/2}}}} \end{align}

$\ds{\mrm{B}}$: Beta Function. $\ds{\quad\Gamma}$: Gamma Function.

Answer 2

By setting $\frac{1}{1+r^2}=u$ we get that $E_n(s)$ depends on a value of the Beta function:

$$ E_n(s) = \frac{\Gamma\left(\frac{n}{2}\right)\,\Gamma\left(\frac{s-n}{2}\right)}{2\,\Gamma\left(\frac{s}{2}\right)}.\tag{1} $$