# A few fractional integral

\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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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

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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