Improper Integrals - Comparison Test

[ISITIEIW]

Hey, not too sure about what function i would compare this integral from 1 to infinity of (3x^3 -2)/(x^6 +2) dx. I also have to show that it converges.

Thanks!

 

[polygamma]

For large $x$, the integrand behaves like $ \displaystyle \frac{3x^{3}}{x^{6}} = \frac{3}{x^{3}}$.

Thus the integral converges by the p-test.

 

[ISITIEIW]

thanks, i didn't know that you could have 3/x^p for the p test

 

[polygamma]

It's also important to note that $ \displaystyle \frac{3x^{3}-2}{x^{6}+2}$ is continuous on $[1,\infty)$.

 

[Krizalid1]

Hey, not too sure about what function i would compare this integral from 1 to infinity of (3x^3 -2)/(x^6 +2) dx. I also have to show that it converges.

You're taking the integral of such function for all $x\ge1$ and within this range, the function is continuous as said above which is important to bound the integrand the way we want. So for example $\dfrac{1}{{{x}^{6}}+2}<\dfrac{1}{{{x}^{6}}}$ holds always for $x\ge1$ and besides $3x^2-2<3x^2+2$ holds always, then actually for all $x\ge1$ you have $$\displaystyle\frac{3{{x}^{3}}-2}{{{x}^{6}}+2}<\frac{3{{x}^{3}}+2}{{{x}^{6}}}= \frac{3}{x^3}+\frac{2}{{{x}^{6}}},$$ implying $\displaystyle\int_{1}^{\infty }{\frac{3{{x}^{3}}-2}{{{x}^{6}}+2}\,dx}<\infty $

 

[Deveno]

Just a side-note on a "general tip" that may prove useful:

1. If you are integrating a function $\dfrac{f(x)}{g(x)}$ it is often helpful to find a bounding function:

$\dfrac{h(x)}{k(x)}$ where:

$f(x) \leq h(x)$ for all $x > N$

$g(x) \geq k(x)$ for all $x > M$ and

$h(x),k(x)$ share some common factor so we have some nice cancellation occurring.

Close examination shows this is exactly what is happening in Krizalid's post (caveat: finding the "right" bounding functions can often take some algebraic ingenuity).