Upper bounds

• Jul 30th 2009, 06:57 PM
acosta0809
Upper bounds
the question is does this converge or diverge and if it converges give its upper bound..

$\int\limits_1^\infty \frac{\sin(3x)+3}{x^3+x}dx$

what i have is by comparison
$x^3+x>x^3$

$\frac{1}{x^3+x}<\frac{1}{x^3}$

and
$-1\leq\sin(3x)\leq1$

$\frac{2}{x^3+x}\leq\frac{\sin(3x)}{x^3+x}\leq\frac {4}{x^3+x}$

so
$\frac{\sin(3x)}{x^3+x}<\frac{4}{x^3}$

and i know that $\frac{4}{x^3}$ converges

but not sure how to get the upper bound, can someone explain to me how to get the upper bound please! thank you
• Jul 30th 2009, 07:00 PM
Krizalid
What you actually found, it's the upper bound.

What's this? An improper integral? A series? Also, you didn't say for what $x$ you're working on.
• Jul 30th 2009, 07:07 PM
acosta0809
Sorry it is an improper integral, but were currently in the series section.
• Jul 31st 2009, 06:10 AM
Krizalid
Okay, then you used $\frac{\sin (3x)+3}{x^{3}+x}\le \frac{4}{x^{3}+x}<\frac{4}{x^{3}},$ and that was enough to show its convergence.