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I know this is a difficult integral to compute, and I know that the answer is: $\int_{-\infty}^{\infty} \frac{dx}{(ax^2 + b)^k} = \frac{\sqrt\pi \Gamma(k-\frac{1}{2})}{\Gamma(k)}\frac{1}{\sqrt a b^{k-\frac{1}{2}}}$.

However, I don't know how to work it out.

I would really appreciate any hints.

If anyone is able to work the answer, would it also be possible to compute it for different bounds for the integral? (e.g. once a suitable change in variable x is performed so that one integrates between $\frac{-\pi}{2}$ and $\frac{\pi}{2}$, could I integrate just between, let's say $\frac{-\pi}{2}$ and $\frac{-\pi}{4}$?).

Thanks.

D B
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  • Maybe residue theorem with poles at $\sqrt{-b/a}$? Are $a$ and $b$ real numbers? – Henricus V. Mar 26 '16 at 18:59
  • Yes, b and a are real. a and k are strictly positive. I have never used the residue theorem before, could I get a hint on how to use it? – D B Mar 26 '16 at 19:21
  • @D B Do you know how to calculate residue? Laurent series? – Henricus V. Mar 26 '16 at 19:25
  • @Henry W. , regretfully, complex analysis is not my strong point. However, if you give me an idea of the steps needed to solve the integral I will surely try to work it out on my own. – D B Mar 26 '16 at 19:43
  • @D B Use a semicircular contour on the upper half plane. By residue's theorem, the integral along that contour is $2\pi i$ times the total residue of enclosed poles. Then show that the integral on the semicircle vanishes as radius $\to \infty$. – Henricus V. Mar 26 '16 at 19:52

2 Answers2

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Let $x=\left(\frac ba\right)^{1/2}u$. Then $$\int_{-\infty}^{\infty}\frac{dx}{(ax^2+b)^k}=\frac1{\sqrt ab^{k-\frac12}}\int_{-\infty}^{\infty}\frac{du}{(u^2+1)^k}=\frac2{\sqrt ab^{k-\frac12}}\int_0^{\infty}\frac{du}{(u^2+1)^k}$$ Now let $t=\frac1{u^2+1}$. Then $$\int_{-\infty}^{\infty}\frac{dx}{(ax^2+b)^k}=\frac1{\sqrt ab^{k-\frac12}}\int_0^1t^{k-\frac32}(1-t)^{-\frac12}\,dt=\frac1{\sqrt ab^{k-\frac12}}\text{B}\left(k-\frac12,\frac12\right)$$ The key is to recognize the beta function in the answer and work towards that.

user5713492
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  • All too easy. +1 – Mark Viola Mar 26 '16 at 22:50
  • You made it look easy :). Thank you! But what if I wanted to work out the value of the integral between c1 and c2 instead of between $-\infty$ and $\infty$? Any idea? – D B Mar 26 '16 at 23:29
  • Yeah, if you really needed the integral from $c_1$ to $c_2$, then that scale factor of $\left(\frac ba\right)^{\frac12}$ would be reflected in the limits in the first line above and if $c_1<0<c_2$, the folding operation would instead result in two integrals. Then we might end up with 2 incomplete Beta functions at the end. If $2k$ were an integer, the substitution $t=\cos^2\theta$ would result in a feasible integral. – user5713492 Mar 27 '16 at 02:05
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If you need to compute $$\int_{\alpha}^{\beta} \frac{dx}{(ax^2 + b)^k}=\int_{0}^{\beta} \frac{dx}{(ax^2 + b)^k}-\int_{0}^{\alpha} \frac{dx}{(ax^2 + b)^k}$$ $$\int_{0}^{\gamma} \frac{dx}{(ax^2 + b)^k}=\frac{\gamma}{b^k}\,\, _2F_1\left(\frac{1}{2},k;\frac{3}{2} ;-\frac{a }{b}\gamma ^2\right)$$ where appears the Gaussian hypergeometric function.

Expanded as a series $$\,_2F_1\left(\frac{1}{2},k;\frac{3}{2} ;t\right)=\frac{1}{\Gamma (k)}\sum_{n=0}^\infty \frac{\Gamma (k+n)}{(2 n+1) \Gamma(n+1)}\,t^n$$

Claude Leibovici
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