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.

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