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

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\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.

