Write the integral $\int_0^\infty \left(\frac{1}{1 + x^2} \right)^\alpha x^\beta dx$ in terms of the Euler Beta Function?

Question

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?

Answer 1

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.

Answer 2

$$I=\int_0^\infty(x^2+1)^{-\alpha}x^{\beta}dx$$ 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

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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.