Improper Integrals (BC Only)
As we finish up integrals, we have a few more topics to review that show up only on the BC Exam. These topics cover limits of integration that are not finite and the integration of discontinuous functions. We will also look at tests to determine if an improper integral converges or diverges.
Integrals with Infinite Limits of Integration
Consider the function on the intervalwhere a > 0. We know this function is continuous and non-negative on the interval. As the graph shows, the function is asymptotic along the positive x-axis.
Figure 1
We would like to know if the area under the curve for has a finite value or an infinite value. One way to check this is to find the area using a definite integral. It should be . However, we cannot use infinity as a limit because we cannot evaluate the definite integral as we have previously done.
As we have done before, we will try to adjust the problem to be more like something we are familiar with. In this case, we use a temporary variable for the upper limit. t is the most common one to use. Our integral changes to . That may not seem like much of an improvement, but it really is. Now we have a parameter that is considered a constant. We can take the limit of the problem as the parameter increases without bound. We are able to write the results in this manner:
We can take the antiderivative of the integral and evaluate it at the limits.
Applying the limit, we get
Since the result was a finite value, we say the improper interval converged .
We can use this same reasoning on any function that has an infinite interval.
Example 1.
Determine if the integral converges or diverges. If it converges, find the value.
We use the definition
Since the limit exists, the integral converges. The value of the integral is.
Example 2.
Determine if the integral converges or diverges. If it converges, find the value.
Using the definition, we can rewrite the problem as
Since the limit does not exist, the integral is divergent.
Example 3.
Determine the area (if it exists) of the region that is bounded by
This time, we use Case 2 of the definition since we are looking at a limit to negative infinity.
This value is finite, so the integral is said to converge.
Now we will look at an example that has infinite limits in both directions.
Example 4.
Find the value of .
Since this is an infinite limit on both sides, we use Case 3 and split the integral into two definite integrals at some convenient value for x. Based on the graph of the function, it seems reasonable to make the split at x = 0.
Example 4
So we will write the problem as
Clearly, this integral is convergent.
Integrals with Infinite Discontinuities
If a function is continuous on a closed interval [a, b], the Fundamental Theorem of Calculus tells us that the definite integral exists. If f(x) has an infinite discontinuity which is indicated by a vertical asymptote, we may still be able to determine some finite number that is associated with the integral. We do that in much the same way we worked problems in the preceding section: we assign a temporary variable to the limit where the infinite discontinuity occurs. Then we take the one-sided limit as x approaches the value of x that we replaced with the temporary variable.
The following rules explain how such integrals should be evaluated.
We will work a few examples to show how these rules apply.
Example 5.
Evaluate
The integrand has an infinite discontinuity at x = 4which is a right side limit. We will use Case 2.
Again, the integral converges..
Example 6.
Does converge or diverge? Show your reasoning.
goes infinite as x approaches zero from the right. We will use Case 1 for this example.
The integral diverges.
Example 7.
Determine whether diverges or converges.
The integrand is not defined at x = 4 where a vertical asymptote exists. This situation fits Case 3. We need to rewrite the problem as the sum of two separate integrals.
which does not exist. Therefore, the integral diverges.
Another point to remember is that the Fundamental Theorem does not apply in this example because the integrand is not continuous on the closed interval [1, 5]. If we had applied the Fundamental Theorem, we would have obtained
.
This result is clearly not possible because the answer is negative but the integrand was never negative.
If an integral is divided into two separate integrals as in Example 7, both parts must converge in order for the entire integral to be convergent. Otherwise, the integral will be divergent.
Example 8.
Evaluate
The integrand has an infinite discontinuity inside the interval at x = 2,which means that we follow Case 3 and write the integral as the sum of two separate definite integrals.
Remember, the integral is convergent because both parts were convergent.
The p-test for Convergence of Integrals
Through much experimentation and analysis, mathematicians have found that is the boundary that divides convergent and divergent integrals written in the form . This is important concept will be used extensively in the next chapter on sequences and series. We will examine this integral to develop the rules of evaluation. There are three cases to consider. When p <1, p = 1, and p > 1.
Case 1. p =1. The problem becomes ..
This case is divergent.
Case 2. p<1. The problem becomes .
This case is divergent also.
Case 3. p > 1. The problem becomes .
This case is convergent. Since p is greater than 1, . As , the value of the fraction will get smaller and smaller until it converges to zero.
The rule summarizes as converges as long as.
Tests for Convergence and Divergence
Whenever an improper integral cannot be evaluated directly, we first need to check for divergence or convergence. If the integral diverges, clearly the integral will have no finite value. In this case, we can stop. However, if the improper integral converges, then we know that the definite integral has some finite value. We may not be able to find the exact value using definite integrals and the Fundamental Theorem of Calculus. We can, however, approximate the answer using some numerical method such as a Riemann Sum or the Trapezoidal Rule. We have two main tests that we can use to establish convergence or divergence -- the Direct Comparison Test and the Limit Comparison Test.
We are covering these tests for definite integrals now because they serve as a model for similar tests that check for convergence of sequences and series -- topics that we will cover in the next chapter.
Direct Comparison Test
The Direct Comparison Test method depends on our ability to determine whether the function we are checking is somehow bounded by some other function that we know is convergent or divergent.
Case 1 fits the squeeze principle. If we know that is bounded between , then, as approaches zero as , must be squeezed to zero also. Therefore, is also convergent. Although we may not know what value converges to, we do know that it is convergent.
Case 2 is the opposite of Case 1. If is always smaller than , we know that will always act as a "floor" to . If diverges, then must be "pushed away" from a converging situation and is therefore divergent.
Example 9.
Determine if converges or diverges.
Once again, we rewrite the problem using a temporary variable in place of .
We are stuck at this point because there is no convenient antiderivative for the integrand. We need to check out convergence or divergence using some other method. We remember that is always positive but is always less than So, we check the integral for convergence or divergence.
Since converges and we know that is a "ceiling" for , then we can conclude that must also converge.
Once again, we stress that although we know does converge, we do not know what value it converges to.
Example 10.
Check for convergence.
We rewrite the integral with a temporary variable so that does not appear in the integral.
Again, we are in a situation where we have a correct form for a definite integral but we have no convenient antiderivative. This is the time to use the Direct Comparison Test. We knowmust always lie between zero and 1. Therefore,. We can compare . Since , .
By the p-value rule we know that converges. Therefore, must converge also.
You are probably thinking, "The concepts make sense. But how do I know which basic functions to use as boundary functions?" Good question. There are some strategies to try.
1.If you have in an expression, try to replace it with 1.
2.If you have square roots with expressions that contain x and constants, eliminate the constants. See what the expression simplifies to.
3.If there is a constant in the denominator, eliminate the constant to see what similar function applies.
4.Otherwise, check the function against a parent function that you know diverges or converges such as.
Example 11.
Does converge?
Eliminating the constant in the radical sign of the integrand gives us since . So we need to compare. We see that. Consequently, . And we know that diverges, so must also diverge.
Limit Comparison Test
We won't attempt to prove this test, but the logic is easy to understand. From our study on limits we know that
.
If one of the functions converges, the other, which differs only by a constant of multiplication, must also converge. Similarly, if one function diverges, the other must diverge.
Example 12.
Check the convergence of .
We use our strategy of eliminating the constant in the denominator. That means we should compare our problem with. Remember, the Limit Comparison Test does not compare the definite integrals, only the integrands.
Since we have a finite limit, we know that both functions behave the same. Because of the p-value Rule, we know that converges. So, also converges.
Example 13.
Does converge? Explain.
Our line of thought should check whether we should use the Direct Comparison Test or the Limit Comparison Test. It is difficult to think of a bounding function that is similar to the integrand, which moves us away from the Direct Comparison Test. If we use, the Limit Comparison Test, we need to find a similar function. Our previous strategy of eliminating constants in the denominator is a reasonable guess to try. The similar integral is . Once again, we remind you to check the limit of the ratio of the integrands, not the definite integrals.
We have shown that both definite integrals behave the same way. We also know that converges because all definite integrals in the form converge as long as a > 1. We correctly conclude that converges.
Now You Try It
Determine if the integral converges or diverges. If it converges, give the value.
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Determine if the integral converges or diverges. Do not find its value.
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