Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $\infty$ ?

share|cite|improve this question
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

3 Answers 3

up vote 0 down vote accepted

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

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

share|cite|improve this answer

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

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.

share|cite|improve this answer
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


Then, using partial fractions you can split the denominator


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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.