Calculus/Improper Integrals

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Improper Integrals

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.)

Contents

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   .

DefinitionEdit

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

 
Example: Convergent Improper Integral

We claim that

 

To do this we calculate

   
 
 
 
Example: Divergent Improper Integral

We claim that the integral

  diverges.

This follows as

   
 
 
 

Therefore

  diverges.
Example: Improper Integral

Find   .

To calculate the integral use integration by parts twice to get

   
 
 
 

Now   and because exponentials overpower polynomials, we see that   and   as well. Hence,

 
Example: Powers

Show  

If   then

   
 
 
 

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

 = 
Example 1

Show  

If   then

   
 
 
 

Notice that we had to assume that   do avoid dividing by 0. So instead we do the   case separately,

 

which diverges.


Example 2

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.

Comparison TestEdit

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   .

  1. Suppose   for all   . Then if   converges so does   .
  2. Suppose   for all   . Then if   diverges so does   .

A similar theorem holds for improper integrals of the form   and for improper integrals with discontinuities.

Example: Use of comparsion test to show convergence

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.

Example: Use of Comparsion Test to show divergence

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.


Example

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.

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Improper Integrals