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let $a>0,ac-b^2>0,\alpha>\dfrac{1}{2}$

show that $$I=\int_{-\infty}^{\infty}\dfrac{\mathrm dx}{(ax^2+2bx+c)^{\alpha}}=\dfrac{(ac-b^2)^{\frac{1}{2}-\alpha}}{a^{1-\alpha}}\dfrac{\Gamma{(\alpha-\dfrac{1}{2})}}{\Gamma{(\alpha)}}\sqrt{\pi}$$

This problem is from the 2013 China university of science and technology of mathematical analysis examination questions,and this problem is last problem

My try:since $ac-b^2>0,a>0$ so $$ax^2+2bx+c>0$$ then I think $$(ax^2+2bx+c)^a=\left(a\left(x+\dfrac{b}{a}\right)^2+\dfrac{ac-b^2}{a}\right)^{\alpha}$$ so let $$A=\dfrac{ac-b^2}{a}>0,\sqrt{a}\left(x+\dfrac{b}{a}\right)=t$$

then $$I=\dfrac{1}{\sqrt{a}}\int_{-\infty}^{\infty}\dfrac{dt}{(t^2+A)^{\alpha}}=\dfrac{\sqrt{A}}{\sqrt{a}A^{\alpha}}\int_{-\infty}^{\infty}\dfrac{du}{(u^2+1)^{\alpha}}=\dfrac{2}{\sqrt{a}A^{\alpha-\frac{1}{2}}}\int_{0}^{\infty}\dfrac{1}{(u^2+1)^{\alpha}}du$$ where $u=\dfrac{t}{\sqrt{A}}$

Lemma: $$I_{1}=\int_{0}^{\infty}\dfrac{1}{(x^2+1)^{\alpha}}dx=\dfrac{\sqrt{\pi}\Gamma{(\alpha-\frac{1}{2})}}{2\Gamma{(\alpha)}}$$ proof:let $x^2=u$,then $$I_{1}=\dfrac{1}{2}\int_{0}^{\infty}\dfrac{t^{-\frac{1}{2}}}{(t+1)^a}dt=\dfrac{1}{2}B(\dfrac{1}{2},\alpha-\dfrac{1}{2})=\dfrac{1}{2}\dfrac{\Gamma{(\dfrac{1}{2})}\Gamma{(\alpha-\dfrac{1}{2})}}{\Gamma{(\alpha)}}=\dfrac{\sqrt{\pi}\Gamma{(\alpha-\frac{1}{2})}}{2\Gamma{(\alpha)}}$$ so

$$I=\int_{-\infty}^{\infty}\dfrac{\mathrm dx}{(ax^2+2bx+c)^{\alpha}}=\dfrac{(ac-b^2)^{\frac{1}{2}-\alpha}}{a^{1-\alpha}}\dfrac{\Gamma{(\alpha-\dfrac{1}{2})}}{\Gamma{(\alpha)}}\sqrt{\pi}$$

this problem have other methods? Thank you

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    $\begingroup$ It's one of the Beta function (en.wikipedia.org/wiki/Beta_function) representations. If you substitute $p = u^2$ it will become more apparent. $\endgroup$ – qoqosz Nov 6 '13 at 7:23
  • $\begingroup$ Yes,I have solution,But have other nice methods? $\endgroup$ – user94270 Nov 6 '13 at 7:30
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ \begin{align} &\int_{-\infty}^{\infty}{\dd x \over \pars{ax^{2} + 2bx + c}^{\alpha}} = {1 \over a^{\alpha}}\int_{-\infty}^{\infty} {\dd x \over \bracks{\pars{x + b/a}^{2} + \mu^{2}}^{\alpha}}\,, \qquad \mu \equiv {\root{ac - b^{2}} \over a} \end{align} Notice that $\mu \in {\mathbb R}.\quad$ $\mu > 0$.

\begin{align} &\int_{-\infty}^{\infty}{\dd x \over \pars{ax^{2} + 2bx + c}^{\alpha}} = {1 \over a^{\alpha}}\int_{-\infty}^{\infty} {\dd x \over \pars{x^{2} + \mu^{2}}^{\alpha}} = {2\mu^{1 - 2\alpha} \over a^{\alpha}}\int_{0}^{\infty} {\dd x \over \pars{x^{2} + 1}^{\alpha}} \\[3mm]&= {\mu^{1 - 2\alpha} \over a^{\alpha}}\int_{0}^{\infty} {x^{-1/2} \over \pars{x + 1}^{\alpha}}\,\dd x = {\mu^{1 - 2\alpha} \over a^{\alpha}}\, {\overbrace{\Gamma\pars{1/2}}^{=\ \root{\pi}}\,\,\,\Gamma\pars{\alpha - 1/2} \over \Gamma\pars{\alpha}} \end{align}

$$ \color{#0000ff}{\large% \int_{-\infty}^{\infty}{\dd x \over \pars{ax^{2} + 2bx + c}^{\alpha}} = {\pars{ac - b^{2}}^{1/2 - \alpha} \over a^{1 - \alpha}}\, {\Gamma\pars{\alpha - 1/2} \over \Gamma\pars{\alpha}}\,\root{\pi}} $$

See ${\tt http://dlmf.nist.gov/5.12.E3}\quad$ and $\quad{\tt http://dlmf.nist.gov/5.12.E1}$

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