# Value of a certain integral

How to find the value of the integral $$\int_o^{\infty} \! \frac{x^8}{1+x^4+x^6+x^{10}} \, \mathrm{d}x$$ given to be $\frac{1}{12}(3\sqrt{2}-1) \pi$ by WolframAlpha and, in general, is there a procedure to find the value of the definite integral of a rational function of the form $\dfrac{x^l}{p(x)}$ where $deg(p) > l > 1$ from $0$ to $\infty$ ?

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What do you mean by "such integrals" - what would the general form be? – Sanath K. Devalapurkar Mar 26 '14 at 0:16
I think he means rational functions – enthdegree Mar 26 '14 at 0:19
Every real polynomial can be factored into at most quadratic terms, every such fraction can be split into partial fractions (uniquely) and every fraction with at most quadratic denominator can be integrated. – user2345215 Mar 26 '14 at 0:21

You can apply the residue theorem; in this case, the integrand is even and therefore may be extended to the entire real line. You may show, then that the integral over the real line is equal to

$$\oint_C dz \frac{z^8}{z^{10}+z^6+z^4+1}$$,

where $C$ is a semicircle of radius $R \to \infty$ in the upper half plane, which, in turn, is equal to $i 2 \pi$ times the sum of the residues of the poles inside $C$. The poles inside $C$ turn out to be

$$z_1=e^{i \pi/6}$$ $$z_2=e^{i \pi/4}$$ $$z_3=e^{i \pi/2}$$ $$z_4=e^{i 3 \pi/4}$$ $$z_5=e^{i 5 \pi/6}$$

The original integral is therefore

$$i \pi \sum_{k=1}^5 \frac{z_k^5}{10 z_k^6+6 z_k^2+4 }$$

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You can type it in here to find the answer to the problem,

http://www.wolframalpha.com/input/?i=int_0%5Einf+%5Cfrac%7Bx%5E8%7D%7B1%2Bx%5E4%2Bx%5E6%2Bx%5E10%7Ddx

In general, these types of integral problems are generally tackled with "the calculus of residues". Marsden - Basic Complex Analysis - Chapter 4 - Page 296, gives a lovely table for these kinds of problems. For example, if $\deg(Q(x))\ge 2+\deg(P(x))$ then,

$$\int_{-\infty}^\infty \frac{P(x)}{Q(x)} dx = 2\pi i \cdot \sum\left(\text{residues in H^+}\right)+\pi i \cdot\sum\left(\text{residues on \mathbb{R}}\right).$$ I don't know of any general methods that don't use residue theory.

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Thanks for the reference. What is $H^+$ though ? – Guest Mar 26 '14 at 0:29
$H^+$ is the upper half plane. That is, $$H^+=\{x+yi:y>0\}$$ – Bobby Ocean Mar 26 '14 at 0:31

For the given integral, the denominator can be factored so you get

$1+x^4+x^6+x^{10}=1+x^4+x^6(1+x^4)=(1+x^4)(1+x^6)$.

Then, using partial fractions you can split the denominator

$\dfrac{x^8}{(1+x^4)(1+x^6)}=\dfrac{x^2+1}{2(1+x^4)}+\dfrac{x^4-x^2-1}{2(1+x^6)}$

Which gives you two solvable (although challenging) integrals. In general, this would be my approach to solving any integral of the given form, but if it failed I would probably try using a series to approximate the functions and integrate term by term.

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