How do you use the direct comparison test for improper integrals?

Oct 10, 2014

If an improper integral diverges, then we probably do not want to wast time trying to find its value.

Let us assume that we already know:

${\int}_{1}^{\infty} \frac{1}{x} \mathrm{dx} = \infty$

Let us look examine this uglier improper integral.

${\int}_{1}^{\infty} \frac{4 {x}^{2} + 5 x + 8}{3 {x}^{3} - x - 1} \mathrm{dx}$

By making the numerator smaller and the denominator bigger,

$\frac{4 {x}^{2} + 5 x + 8}{3 {x}^{3} - x - 1} \ge \frac{3 {x}^{2}}{3 {x}^{3}} = \frac{1}{x}$

By Comparison Test, we may conclude that

${\int}_{1}^{\infty} \frac{4 {x}^{2} + 5 x + 8}{3 {x}^{3} - x - 1} \mathrm{dx}$ diverges.

(Intuitively, if the smaller integral diverges, then the larger one has no chance to converge.)

I hope that this was helpful.