Integral $\int_{-\infty}^{\infty}\frac{\mathrm dx}{(ax^2+2bx+c)^{\alpha}}$ 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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\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
$\quad{\tt http://dlmf.nist.gov/5.12.E1}$
