# Evaluate integral - Wolfram Alpha works with a summation in terms of $\omega$

I found this "monster"

$$\int_0^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx$$

and I would like to know how to evaluate it. If not a complete answer, at least some hints.

Wolfram Alpha (here) works with a summation in terms of $\omega$. Can somebody explain me the meaning of this?

With @Grant B.’s guide and Wolfram Alpha , I obtained the roots.

I will try to make the partial fractions during the week-end.

• That notation indicates the sum over all roots $\omega$ of the denominator – Grant B. Mar 11 '17 at 0:17
• What Wolfram Alpha and Grant B. is saying is that you need to use partial fractions to solve it. – Sentinel135 Mar 11 '17 at 0:29

$\lim_{\omega \to \infty}$, use partial fraction and instead of inserting infinity,insert $\omega$ , you will get same result.

• This is too cryptic for me to make sense of, but then the difficulty may well originate in the Question itself. Can you fill in the details of how to "use partial fraction"? – hardmath Mar 11 '17 at 2:42

This is too long for a comment.

As you wrote, this is a real monster and I wish you very pleasant days with partial fraction decomposition. In any manner, since the roots of $x^7+6x^4+9=0$ cannot be expressed but just numerically evaluated, you could not get any "exact" result.

What I would suggest is just numerical integration $$\int_0^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx=\int_0^{1}\frac{x^5−6x^2}{x^7+6x^4+9}dx+\int_1^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx$$ For the second integral, use $x=\frac 1y$ to get $$\int_1^{\infty}\frac{x^5−6x^2}{x^7+6x^4+9}dx=-\int_0^{1}\frac{6 y^3-1}{9 y^7+6 y^3+1}dy$$

Edit

By the end, we need to compute $$I=\int_0^{1}\left(\frac{x^5−6x^2}{x^7+6x^4+9}- \frac{6 x^3-1}{9 x^7+6 x^3+1}\right)\,dx$$ Using the trapezoidal method with $101$ points, we then end with $$I=0.0837499$$ while the "exact" answer would be $0.0837447$.

Since all of that can easily be done using Excel, repeat the calculations using $1001$ points and get $0.0837447$ which is the exact solution for six significant figures.

• i've begun to do partial fractions and yes, it is very laborious. Why did you split the integral in two and why from 0 to "1" to infinity? Why "1"? – cgiovanardi Mar 11 '17 at 17:38
• Don't do that ! You waste your time and you will face nightmares and not get any exact result. I did split the integral in order to have these $(0,1)$ bounds for both integrals. I should use the trpzoidal method (say with 100 points) and the result will be sufficiently accurate. Do that with Excel. – Claude Leibovici Mar 12 '17 at 3:22