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A few days ago, a poster named @Ryan posted the following integral:

$$\int_0^\infty dx \: \frac{x^3}{\left( x^4+7x^2+1 \right)^{\large\frac{5}{2}}} $$

I posted an answer to a different integral having an integral power of a polynomial in the denominator using the Residue Theorem, and glibly stated that the other integrals would follow suit similarly. On second thought, I believe I am wrong. I have been trying to evaluate this integral without success. Mathematica does provide an exact answer, which looks plausible. Can anyone provide an outline of how one evaluates such integrals? Is there a generalization of residue theory with which I am familiar (i.e., graduate-level complex analysis)?

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up vote 4 down vote accepted


  • Let $u = x^2$ $\quad du = 2 \,dx$

$$\int_0^\infty \frac{x^3\,dx}{\left( x^4+7x^2+1 \right)^{5/2}} ={1\over 2} \int_0^\infty {u\, du\over (u^2 + 7u + 1)^{5/2}} $$

  • Complete the square (manipulate denominator to do so).

$$ \int_{0}^{\infty} {u\,{\rm d}u \over \left[\left(u + 7/2\right)^{2} - 45/4\right]^{5/2}} $$

  • Use another substitution $\;t = u +\large\frac72$

Shortly, you can use a substitution involving the $\,\sec\,$ function.

There is a LOT of trigonometric manipulation/substitution required from this point.

Perhaps there is a more efficient route to integration, but this route/outline is what came to mind.

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You can cut down the ugliness here. Do an $x^2$ substitution to get $$\int_0^\infty \frac{x^3\,dx}{\left( x^4+7x^2+1 \right)^{\frac{5}{2}}} ={1\over 2} \int_0^\infty {x\, dx\over (x^2 + 7x + 1)^{5/2}} $$ This integral will yield (with some computational rigamaroll) to conventional complete the square elementary calculus methods.

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I think we were both composing at roughly the same time. – ncmathsadist Jan 6 '13 at 20:21
@amWhy: thanks, I do, and can work them out. (I am right now.) But I was wondering if the methods we employ using residues work here as well. – Ron Gordon Jan 6 '13 at 20:21

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