The definition of a definite integral:
requires the interval be finite. The Fundamental Theorem of Calculus requires that be continuous on . In this section, you will be studying a method of evaluating integrals that fail these requirements—either because their limits of integration are infinite, or because a finite number of discontinuities exist on the interval . Integrals that fail either of these requirements are improper integrals. (If you are not familiar with L'Hôpital's rule, it is a good idea to review it before reading this section.)
- 1 Improper Integrals with Infinite Limits of Integration
- 2 Improper Integrals with a Finite Number Discontinuities
- 3 Comparison Test
Improper Integrals with Infinite Limits of IntegrationEdit
Consider the integral
Assigning a finite upper bound in place of infinity gives
This improper integral can be interpreted as the area of the unbounded region between , (the -axis), and .
1. Suppose exists for all . Then we define
- , as long as this limit exists and is finite.
If it does exist we say the integral is convergent and otherwise we say it is divergent.
2. Similarly if exists for all we define
3. Finally suppose is a fixed real number and that and are both convergent. Then we define
We claim that
To do this we calculate
We claim that the integral
This follows as
To calculate the integral use integration by parts twice to get
Now and because exponentials overpower polynomials, we see that and as well. Hence,
Notice that we had to assume that to avoid dividing by 0. However the case was done in a previous example.
Improper Integrals with a Finite Number DiscontinuitiesEdit
First we give a definition for the integral of functions which have a discontinuity at one point.
Definition of improper integrals with a single discontinuityEdit
If is continuous on the interval and is discontinuous at , we define
If the limit in question exists we say the integral converges and otherwise we say it diverges.
Similarly if is continuous on the interval and is discontinuous at , we define
Finally suppose has an discontinuity at a point and is continuous at all other points in . If and converge we define
Notice that we had to assume that do avoid dividing by 0. So instead we do the case separately,
The integral is improper because the integrand is not continuous at . However had we not noticed that we might have been tempted to apply the fundamental theorem of calculus and conclude that it equals
which is not correct. In fact the integral diverges since
and and both diverge.
We can also give a definition of the integral of a function with a finite number of discontinuities.
Definition: Improper integrals with finite number of discontinuitiesEdit
Suppose is continuous on except at points in . We define as long as each integral on the right converges.
Notice that by combining this definition with the definition for improper integrals with infinite endpoints, we can define the integral of a function with a finite number of discontinuities with one or more infinite endpoints.
There are integrals which cannot easily be evaluated. However it may still be possible to show they are convergent by comparing them to an integral we already know converges.
Theorem (Comparison Test) Let be continuous functions defined for all .
- Suppose for all . Then if converges so does .
- Suppose for all . Then if diverges so does .
A similar theorem holds for improper integrals of the form and for improper integrals with discontinuities.
Show that converges.
For all we know that so . This implies that
We have seen that converges. So putting and into the comparison test we get that the integral converges as well.
Show that diverges.
Just as in the previous example we know that for all . Thus
We have seen that diverges. So putting and into the comparison test we get that diverges as well.
An extension of the comparison theoremEdit
To apply the comparison theorem you do not really need for all . What we actually need is this inequality holds for sufficiently large (i.e. there is a number such that for all ). For then
so the first integral converges if and only if third does, and we can apply the comparison theorem to the piece.
Show that converges.
The reason that this integral converges is because for large the factor in the integrand is dominant. We could try comparing with , but as , the inequality
is the wrong way around to show convergence.
Instead we rewrite the integrand as .
Since the limit we know that for sufficiently large we have . So for large ,
Since the integral converges the comparison test tells us that converges as well.