Write the integral $\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$ in terms of the Euler Beta Function? How can I write the integral $$\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$$
with $\alpha, \beta >0$, in terms of the Euler Beta Function? For which $\alpha, \beta$ does the integral actually converge? Maybe some change of variable can do the trick?
 A: With $x=\tan t$ the integral becomes$$\int_0^{\pi/2}\sin^\beta t\cos^{-2\alpha-\beta-2}t\mathrm dt=\tfrac12\operatorname{B}\left(\tfrac{\beta+1}{2},\,\alpha-\tfrac{\beta+1}{2}\right)$$(if this evaluation isn't obvious, use $u=\sin^2t$). I'll leave you to deduce convergence conditions.
A: $$I=\int_0^\infty(x^2+1)^{-\alpha}x^{\beta}dx$$
now if we split up our regions, it is fair to say that:
$$\int_0^\infty f(x)dx=\int_0^1f(x)dx+\int_1^\infty f(x)dx$$
now let $u=\frac 1x\Rightarrow dx=-x^2du=-\frac{du}{u^2}$ so:
$$\int_0^\infty f(x)dx=\int_0^1f(x)dx-\int_1^0f(1/x)\frac{dx}{x^2}=\int_0^1\left[f(x)+\frac{f(1/x)}{x^2}\right]dx$$
in your case:
$$f(x)=(x^2+1)^{-\alpha}x^\beta$$
so:
$$f(x)+\frac1{x^2}f(1/x)=(x^2+1)^{-\alpha}x^\beta+\frac1{x^2}(1/x^2+1)^{-\alpha}(1/x)^{\beta}$$
and since:
$$\frac{1}{\left(\frac1{x^2}+1\right)^\alpha}x^{-\beta}=\frac{x^{2\alpha-\beta}}{(x^2+1)^{\alpha}}=(x^2+1)^{-\alpha}x^{2\alpha-\beta}$$
so you now have:
$$I=\int_0^1(x^2+1)^{-\alpha}\left[x^\beta+x^{2\alpha-\beta}\right]dx$$
now try using the substitution $v=x^2$ and splitting it into two integrals
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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With $\ds{-1 < \Re\pars{\beta} < 2\,\Re\pars{\alpha} - 1}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
\pars{1 \over 1 + x^{2}}^{\alpha}x^{\beta}\,\dd x}
\,\,\,\stackrel{x^{2}\ \mapsto\ x}{=}\,\,\,
{1 \over 2}\int_{0}^{\infty}
{x^{\beta/2 - 1/2} \over \pars{1 + x}^{\alpha}}\,\dd x
\end{align}
Lets $\ds{x \equiv {1 \over t} - 1 \implies
t = {1 \over 1 + x}}$.
Then,
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}
\pars{1 \over 1 + x^{2}}^{\alpha}x^{\beta}\,\dd x} =
{1 \over 2}\int_{1}^{0}
{\pars{1/t - 1}^{\beta/2 - 1/2} \over t^{-\alpha}}\,\pars{-\,{1 \over t^{2}}}\dd t
\\[5mm] = &\ 
{1 \over 2}\int_{0}^{1}
t^{-\beta/2 + \alpha - 3/2}\,\,\,\,\,\pars{1 - t}^{\beta/2 - 1/2}\,\,\,\dd t
\\[5mm] = &\
\bbx{{1 \over 2}\on{B}\pars{\alpha - {\beta + 1 \over 2},{\beta + 1 \over 2}}} \\ &
\end{align}
$\ds{\on{B}}$: Beta Function.
