# Is $\int_1^{\infty}\frac{x \cos(x)^2}{1+x^3}$ convergent or divergent?

For the integral $$I= \int_1^{\infty}\frac{x \cos^2(x)}{1+x^3},$$ how do I test this for convergence or divergence?

I know that this an improper integral- however it cannot be solved so would need to use a comparison test for this.

Would the comparison test consist of: If $\cos(x)<1$ then we can use $1/(1+x^3)$ to show that it converges?

However... How can i compare the equation where there is an $x$ on the numerator of the original equation? Would I need to use something else to compare it with instead? Thanks.

• Might be of some use, Mathematica gives an answer, but it looks terrible... Commented May 6, 2015 at 3:17
• The first thing to test for, always, is absolute convergence.
– zhw.
Commented May 6, 2015 at 3:18
• @zhw. this is an improper integral, not a series Commented May 6, 2015 at 3:19
• Why does that matter?
– zhw.
Commented May 6, 2015 at 3:19
• @zhw Absolute convergence is a topic in series, not in improper integration. Commented May 6, 2015 at 3:24

Notice that $\cos^2(x) \leq 1$, therefore $$\frac{x \cos^2(x)}{1+x^3} \leq \frac{x}{1+x^3}.$$ You also know that $$\frac{1}{1+x^3} \leq \frac{1}{x^3}$$ for $x \in (1, \infty)$. Therefore $$\frac{x \cos^2(x)}{1+x^3} \leq \frac{x}{x^3} = \frac{1}{x^2}.$$
Consider comparing like this: $$\frac{x\cos^2 x}{1+x^3} \le \frac{x}{1+x^3} \le \frac{1}{x^2}$$
You can also do limit comparison to $\frac{x}{x^3}$.
• Limit comparison with $\frac{x}{x^3}$ is also possible, I just didn't show it because it's pretty elementary. Commented May 6, 2015 at 3:23
• It depends on the exact statement of limit comparison. If we compare with $\frac{1}{x^2}$ the limit does not exist. If we say compare with $\frac{1}{x^{3/2}}$ the limit is $0$, and we are finished. Commented May 6, 2015 at 4:45