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

• 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. – qoqosz Nov 6 '13 at 7:23
• Yes,I have solution,But have other nice methods? – user94270 Nov 6 '13 at 7:30

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