Indefinite integral and proving convergence

[SteliosVas]

Homework Statement

okay so the equation goes:

∫(x*sin²(x))/(x³-1) over the terminals:
b= ∞ and a = 2

Homework Equations

Various rules applying to the convergence or divergence of integrals such as the p-test, ratio test, squeeze test etc

The Attempt at a Solution

Okay so I have tried factorising the bottom, I have tried the squeeze test and the ratio test to no success..

would method should I be applying, because I am unsure how to split up this integral...

Thanks

 

[axmls]

Here's a hint: first notice that \sin^2 x \leq 1 for all x. Then you can compare your integral with another one whose integrand is always greater. You can then prove the convergence of that integral.

 

[SteliosVas]

Therefore using the comparison test should i compare this integral to 1/x³ over the same interval?

 

[STEMucator]

Lets apply a useful theorem. If $\int_a^b |f(x)| \space dx$ converges, then $\int_a^b f(x) \space dx$ converges. Note the converse of this theorem is not true. So you should always determine the convergence of $\int_a^b |f(x)| \space dx$ to determine if $\int_a^b f(x) \space dx$ converges.

Applying this theorem to your given problem:

$$\int_{2}^{\infty} \left| \frac{x \sin^2(x)}{x^3 - 1} \right| \space dx = \int_{2}^{\infty} \frac{|x| |\sin^2(x)|}{|x^3 - 1|} \space dx \leq \int_{2}^{\infty} \frac{|x|}{|x - 1| |x^2 + x + 1|} \space dx$$

For $x \in [2, \infty)$, we can remove the absolute values in the expression because all of the terms would be positive anyway. So the new question is, does this integral converge:

$$\int_{2}^{\infty} \frac{x}{(x - 1) (x^2 + x + 1)} \space dx$$

Hint: Start with a partial fraction expansion, and then complete the square.

 

[SteliosVas]

Right now I understand. So the absolute of f(x) is always said to converge but the the opposite isn't always true ?

Right do we are performing a comparison test to x/(x^3-1)