100%(1)1 out of 1 people found this document helpful
This preview shows page 29 - 31 out of 63 pages.
The test says that ifR∞ag(x)dxconverges, then so doesR∞af(x)dx, and ifR∞af(x)dxdiverges then sodoesR∞ag(x)dx. This should make sense: ifg(x) has finite area, then so mustf(x) sincef(x) is smaller,and iff(x) has infinite area, then so doesg(x) sinceg(x) is bigger.In truth, we can apply this test to all sorts of improper integrals provided that there is only one improperbound and it happens at an endpoint. However, the functionsmust be greater than or equal to zero.12.2.1What to compare to: the p-testNotice that, from the statement of the theorem, if we want to prove an integral converges, we need to finda bigger function which we know converges.Similarly, if we want to prove an integral diverges, then wemust find a smaller function which we know diverges. This means that when you first encounter an integral,you have to make an educated guess as to whether you think the integral converges or diverges, becausethis determines which way you need to bound the function. To make these guesses, you want to keep thefollowing in mind:(1) Integrals of the formZ∞a1xpdx,wherea >0 (must be strictly positive), converge whenp >1 and diverge whenp≤1.(2) Integrals of the formZa01xpdx,wherea >0 is not infinity, converge whenp <1 and diverge whenp≥1. Notice nothing good everhappens withp= 1, and that everything here is the(3) Now considerZ∞aekxdx,whereais not-∞. This converges ifk <0 and diverges ifk≥0. Exponential decay is as good as itgets.(4) Similarly, if we considerZa-∞ekxdx,now withanot∞, this converges ifk >0 and diverges ifk≤0. With these last two, notice that allyou need is for the graph of the exponential to be tending to 0 as you approach the infinite bound.That should make it easy to remember.12.2.2ExamplesExample 12.11.Determine whetherR∞11x4+1dxconverges or diverges.29
Math 31B NotesSudesh KalyanswamySolution.This is not an integral we know how to compute, so we use the comparison test. First, we needto make a guess, as this will determine which way we want inequalities to go. Notice that asx→ ∞, thebottom just behaves likex4, since the extra 1 is insignificant. So the fraction behaves like1x4, andR∞11x4dxconverges by case (1) of the integrals above. So we can make the guess that the integral converges. Onething you should try and do is “remove” the terms you don’t care about, assuming the inequality goes inthe proper direction when you do. In this case, we’d like to remove the 1 from the denominator, so isZ∞11x4+ 1dx≤Z∞11x4dx?Well yes, sincex4+ 1≥x4, meaning1x4+1≤1x4, which is what we wanted. The other way to think about itis that the denominator of1x4+1is bigger than that of1x4, so the fraction is smaller. The integralR∞11x4dxconverges by thep-test, so by comparison, the original integral converges.