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How can one prove that $$I\left( a,b \right)= \int_{0}^{\infty }\frac{x^{a-\frac{3}{2}}dx}{\left[ x^2+\left( b^2-2 \right)x+1 \right]^a}=b^{1-2a}\frac{\Gamma \left( \frac{1}{2} \right)\Gamma \left( a-\frac{1}{2} \right)}{\Gamma \left( a \right)},\ $$ where $a>\frac12,\ b\in \mathbb{R}^+$?

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Bounds on $a$ and $b$? I'm guessing $a > 1/2$, nothing on $b$. – Ron Gordon Feb 1 '13 at 2:21
@rlgordonma Yes, I dont see the bounds on the question – Ryan Feb 1 '13 at 2:43
i think you should use beta function by changing variable $I\left( a\ ,\ b \right)=\frac{1}{2}{{b}^{1-2a}}\beta (a,b)$ – Maisam Hedyelloo Feb 1 '13 at 3:02
@rlgordonma : I just edit some errors, can you hav a look again :) – Ryan Feb 1 '13 at 10:26

Observation 1:

The change of variables $x\leftrightarrow x^{-1}$ yields the identity $$\int_0^1\frac{x^{a-\frac32}dx}{\left[x^2+(b^2-2)x+1\right]^a} =\int_1^{\infty}\frac{x^{a-\frac12}dx}{\left[x^2+(b^2-2)x+1\right]^a},$$ which in turn implies that \begin{align} I(a,b)=\int_0^{\infty}\frac{x^{a-\frac32}dx}{\left[x^2+(b^2-2)x+1\right]^a} &=\int_1^{\infty}\frac{2x^{a}\cdot\frac{1}{2}\left(x^{-\frac12}+x^{-\frac32}\right)dx}{\left[x^2+(b^2-2)x+1\right]^a}\tag{1} \end{align}

Observation 2:

We have $$x^2+(b^2-2)x+1=(x-1)^2+b^2 x=x\left[\left(x^{\frac12}-x^{-\frac12}\right)^2+b^2\right].\tag{2}$$ Observation 3:

We also have $$d\left(x^{\frac12}-x^{-\frac12}\right)=\frac{1}{2}\left(x^{-\frac12}+x^{-\frac32}\right)dx.\tag{3}$$

Now, making in (1) the change of variables $s=x^{\frac12}-x^{-\frac12}$ and using (2) and (3), one finds that $$I(a,b)=2\int_0^{\infty}\frac{ds}{\left(s^2+b^2\right)^a}=2b^{1-2a}\int_0^{\infty}\frac{dt}{\left(t^2+1\right)^a}=b^{1-2a}\int_0^{\infty}\frac{u^{-\frac12}du}{\left(u+1\right)^a}.$$ The last integral transforms into the standard beta function integral by the change of variables $v=\frac{u}{u+1}$. It gives \begin{align} I(a,b)=b^{1-2a}\int_0^1v^{-\frac12}(1-v)^{a-\frac32}dv=b^{1-2a}B\left(\frac12,a-\frac12\right)=b^{1-2a}\sqrt{\pi}\frac{\Gamma\left(a-\frac12\right)}{\Gamma(a)}. \end{align}

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