# A few fractional integral [closed]

\begin{align} & \int_0^\infty \frac{x^4}{\left( x^4+x^2+1 \right)^3}\text{d}x \\ & \int_0^\infty \frac{x^3}{\left( x^4+7x^2+1 \right)^{\frac{5}{2}}} \text{d}x \\ & \int_0^\infty \frac{\sqrt{x}}{\left( x^4+14x^2+1 \right)^{\frac{5}{4}}} \text{d}x \end{align}

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## closed as off-topic by Jonas Meyer, Newb, Michael Albanese, David K, Zev ChonolesApr 4 at 4:40

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Jonas Meyer, Newb, Michael Albanese, David K, Zev Chonoles
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Are these homework? What insights have you obtained in working on these? –  Ron Gordon Jan 5 '13 at 10:42
Yiiikes! Too many questions, too little (wero, in fact) self work, ideas, effort shown. This is important, in particular with homerwork-related questions, as yours seem to be. –  DonAntonio Jan 5 '13 at 10:57
I guess the title is misleading: you are talking about integrals of fractions and not about fractional integrals. –  Fabian Jan 5 '13 at 11:16
@Ryan : Your way of coding TeX is bizarre and offends common sense. See my edits. There's no need to write {{{x}^{2}}} where x^2 will do, etc. –  Michael Hardy Jan 5 '13 at 16:49

## 1 Answer

I will provide a rough outline of the first integral. It will be convenient to rewrite it as

$$\frac{1}{2} \int_{-\infty}^{\infty} dx \: \frac{x^4}{(x^4 + x^2 + 1)^3}$$

The simplest way to evaluate such an integral is through a common technique from complex analysis that is a result of something called the Residue Theorem, described here: http://en.wikipedia.org/wiki/Residue_theorem for example. To proceed, we consider an integral in the complex plane:

$$\int_C dz \: \frac{z^4}{(z^4 + z^2 + 1)^3}$$

where $C$ is a contour consisting of a path $C_1$ along the real $z$ axis (i.e., the $x$ axis) over the interval $[-R,R]$ and a path $C_2$ along a semicircle in the upper half plane from the point $[R,0]$ to the point $[-R,0]$ as illustrated below.

The Residue Theorem states that the latter integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrand contained within $C$. It should be noted that each pole has a multiplicity of 3, due to the polynomial in the denominator being cubed. The residue of such a pole $p$ in this case is

$$\mathrm{Res}_{z=p} \frac{z^4}{(z^4 + z^2 + 1)^3} = \frac{1}{2} \lim_{z \rightarrow p} \frac{d^2}{dz^2} \left [ (z-p)^3 \frac{z^4}{(z^4 + z^2 + 1)^3} \right ]$$

We then find the roots of the expression in the denominator, which are at $z = \pm \exp{(i \pi/3)}$ and $z = \pm \exp{(i 2 \pi/3)}$. We need only consider those roots inside the contour $C$, i.e., $z = \exp{(i \pi/3)}$ and $z = \exp{(i 2 \pi/3)}$. I will leave the work of computing the derivatives for the residue calculation to you (best done with the help of computer algebra such as Mathematica). The result is that

$$\mathrm{Res}_{z=\exp{(i \pi/3)}} \frac{z^4}{(z^4 + z^2 + 1)^3} = -\frac{27+i \sqrt{3}}{288}$$

$$\mathrm{Res}_{z=\exp{(i 2 \pi/3)}} \frac{z^4}{(z^4 + z^2 + 1)^3} = \frac{27-i \sqrt{3}}{288}$$

$$i 2 \pi \left [ \mathrm{Res}_{z=\exp{(i \pi/3)}} \frac{z^4}{(z^4 + z^2 + 1)^3} + \mathrm{Res}_{z=\exp{(i 2 \pi/3)}} \frac{z^4}{(z^4 + z^2 + 1)^3} \right ] = \frac{\pi}{24 \sqrt{3}}$$

The complex integral over the contour $C$ may be split into the components over $C_1$ and $C_2$. The integral over $C_1$ is equal to the integral we seek in the limit of the semicircle getting infinitely big. The integral over $C_2$ can be shown to vanish in this limit (as $1/R^7$ as $R \rightarrow \infty$).

Thus, we may write

$$\int_{-\infty}^{\infty} dx \: \frac{x^4}{(x^4 + x^2 + 1)^3} = \frac{\pi}{24 \sqrt{3}}$$

or

$$\int_{0}^{\infty} dx \: \frac{x^4}{(x^4 + x^2 + 1)^3} = \frac{\pi}{48\sqrt{3}}$$

The other integrals may be attacked similarly. Please note that I gave the full treatment of the integral over $C_2$ short shrift and there is much more detail involved.

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