## Convergence and Divergence of Improper Integrals Consider a function f(x) which exhibits a Type I or Type II behavior on the interval [a,b] (in other words, the integral is improper). We saw before that the this integral is defined as a limit. Therefore we have two cases:

1
the limit exists (and is a number), in this case we say that the improper integral is convergent;
2
the limit does not exist or it is infinite, then we say that the improper integral is divergent.

If the improper integral is split into a sum of improper integrals (because f(x) presents more than one improper behavior on [a,b]), then the integral converges if and only if any single improper integral is convergent.

Example. Consider the function on [0,1]. We have Therefore the improper integral converges if and only if the improper integrals are convergent. In other words, if one of these integrals is divergent, the integral will be divergent.

The p-integrals Consider the function (where p > 0) for . Looking at this function closely we see that f(x) presents an improper behavior at 0 and only. In order to discuss convergence or divergence of we need to study the two improper integrals We have and For both limits, we need to evaluate the indefinite integral We have two cases:

if p=1, then we have if , then we have In order to decide on convergence or divergence of the above two improper integrals, we need to consider the cases: p<1, p=1 and p >1. If p <1, then we have and  If p=1, then we have and  If p > 1, we have and The p-Test: Regardless of the value of the number p, the improper integral is always divergent. Moreover, we have  is convergent if and only if p <1  is convergent if and only if p >1

In the next pages, we will see how some easy tests will help in deciding whether an improper integral is convergent or divergent. [Geometry] [Algebra] [Differential Equations]
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Do you need more help? Please post your question on our S.O.S. Mathematics CyberBoard. Mohamed A. Khamsi
Tue Dec 3 17:39:00 MST 1996