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

1 Answer
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^infty1/x dx=infty#

Let us look examine this uglier improper integral.

#int_1^infty{4x^2+5x+8}/{3x^3-x-1} dx#

By making the numerator smaller and the denominator bigger,

#{4x^2+5x+8}/{3x^3-x-1} ge {3x^2}/{3x^3}=1/x#

By Comparison Test, we may conclude that

#int_1^infty{4x^2+5x+8}/{3x^3-x-1} dx# diverges.

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

I hope that this was helpful.