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 tex2html_wrap_inline128 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 tex2html_wrap_inline128 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 tex2html_wrap_inline136 on [0,1]. We have

displaymath140

Therefore the improper integral

displaymath142

converges if and only if the improper integrals

displaymath144

are convergent. In other words, if one of these integrals is divergent, the integral tex2html_wrap_inline146 will be divergent.

The p-integrals Consider the function tex2html_wrap_inline148 (where p > 0) for tex2html_wrap_inline152 . Looking at this function closely we see that f(x) presents an improper behavior at 0 and tex2html_wrap_inline156 only. In order to discuss convergence or divergence of

displaymath158

we need to study the two improper integrals

displaymath160

We have

displaymath162

and

displaymath164

For both limits, we need to evaluate the indefinite integral

displaymath166

We have two cases:

if p=1, then we have

displaymath170

if tex2html_wrap_inline172 , then we have

displaymath174

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.

tex2html_wrap_inline182
If p <1, then we have

displaymath186

and

displaymath188

tex2html_wrap_inline182
If p=1, then we have

displaymath194

and

displaymath196

tex2html_wrap_inline182
If p > 1, we have

displaymath202

and

displaymath204

The p-Test: Regardless of the value of the number p, the improper integral

displaymath158

is always divergent. Moreover, we have

tex2html_wrap_inline182
tex2html_wrap_inline212 is convergent if and only if p <1
tex2html_wrap_inline182
tex2html_wrap_inline218 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]
[Calculus] [Complex Variables] [Matrix Algebra] [Trigonometry]

S.O.S MATHematics home page

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

Copyright 1999-2014 MathMedics, LLC. All rights reserved.
Contact us
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA
users online during the last hour