Name:_____________________ Improper Integrals Done Properly



Math 2414 Activity 8 (Due by August 1)

Determine if the following improper integrals converge or diverge by using the improper

integral toolbox. Give the reason for your conclusion!!!

1. [pic] {comparison} 2. [pic]{infinite limit comparison}

3. [pic] {comparison} 4. [pic] {comparison or limit comparison}

5. [pic] {comparison} 6. [pic] {finite limit comparison}

7. [pic] {comparison} 8. [pic] {infinite limit comparison}

9. [pic] {absolute test or Abel’s test} 10. [pic] {integrate by parts and absolute

test or Abel’s test}

11. [pic] {ratio or root test} 12. [pic] {ratio or root test}

13. [pic] {ratio or root test} 14. [pic] {root test}

15. [pic] {finite limit comparison test} 16. [pic] {finite limit comparison test}

17. For which values of [pic] does [pic] converge? {Hint: Let [pic].}

18. For which values of [pic] does [pic] converge? {Hint: Let [pic].}

19. Find [pic] so that [pic].

{Hint: Integrate by parts}

20. Find all the values of C so that the improper integral converges.

[pic] {Hint: Find a common denominator.}

21. Find all the values of C so that the improper integral converges.

[pic] {Hint: Find a common denominator.}

22. Find all the values of C so that the improper integral converges.

[pic] {Hint: Find a common denominator.}

23. [pic] may not equal [pic]. Show that [pic] diverges, and hence that [pic] diverges. Find the value of [pic].

24. Show that [pic] converges, and find its value.

{Hint: [pic] and [pic], and let [pic] in [pic].}

25. For the improper integral [pic], let’s look at the two cases: [pic] and [pic]. For [pic], [pic], so it’s convergent.

For [pic], [pic], but [pic] is convergent. So [pic] is convergent.

Use the substitution [pic] to find its value.

26. Use integration by parts and L’Hopital’s rule to find a simple formula for [pic] for n a nonnegative integer. [pic]

27. In the improper integral [pic] make the substitution [pic] , and use the result of problem #26 to find a simple formula for [pic] with n a nonnegative integer.

28. The improper integral [pic] converges. [pic], so on the interval [pic], [pic]. This gives us an upper bound on the integral as follows:

[pic].

Use the fact that [pic] to find a lower bound on the value of the integral.

29. Determine if the integral [pic] makes sense by making the substitution [pic] and examining the resulting improper integral.

30. Using [pic], find the values of

a) [pic] b)[pic].

{Hint: See problems 1 and 2 on Activity 6.}

c) Find the value of [pic]?

{Hint: [pic] is a rather odd function.}

31. Evaluate [pic] using the substitution [pic].

{Hint: [pic].}

Prove the following inequalities by comparing integrals with integrals: {Like in #28}

32. [pic] 33. [pic]

34. [pic] 35. [pic]

36. If [pic] is continuous on [pic], [pic], and [pic], where A is a number, then determine if [pic] is convergent or divergent. If it’s convergent, find its value.

{Hint: Use the definition of improper integral.}

37. a) If f is continuously differentiable with [pic] for all [pic] in [pic], then must [pic] exist as a number? How large could the limit be?

{Hint: [pic]}

b) If f is continuously differentiable with [pic] for all [pic] in [pic], then what must [pic] equal?

{Hint: [pic]}

38. Suppose that f is a continuous function with [pic] and [pic] for all [pic]. Determine if the improper integral [pic] is convergent or divergent.

{Hint: [pic] for [pic] implies that [pic].}

39. Suppose that f is continuously differentiable with [pic] for all x. Determine if the improper integral [pic] is convergent or divergent.

{Hint: [pic], so now apply the absolute convergence test.}

40. Consider the improper integral [pic].

a) Show that it is convergent.

{Hint: [pic].}

b) Find its value.

{Hint: [pic].}

41. In problem #10, you showed that the improper integral [pic] is convergent. Since [pic] is finite, [pic] is convergent. Let’s see if we can find its value. For [pic], let [pic], and let [pic].

a) Show that [pic] for [pic].

{Hint: [pic], using the identity [pic]with [pic]and [pic], we get

[pic]

using the identity [pic]with[pic]and[pic], we get

[pic].}

b) Show that [pic] for [pic].

{Hint: [pic]

[pic].}

c) Show that [pic].

{Hint: Let [pic] in [pic] to get [pic], and let [pic].}

d) Show that [pic].

{Hint: [pic] which appears to be an improper integral, but [pic], so it’s alright.

[pic][pic]

[pic]

But [pic], and [pic] is bounded on [pic]. So let [pic].}

e) Determine [pic].

{Hint: [pic].}

42. [pic] is convergent because [pic] is finite and [pic] on [pic]. Use integration by parts to find its exact value.

{Hint: [pic]. But [pic]. Now let [pic] and use the result from problem #41}

43. Consider the improper integral [pic]. Let [pic], to get [pic]. Now [pic]. Now let’s do integration by parts on the second

integral: [pic]. Since [pic] is absolutely convergent, we have that [pic] is convergent. This can also be shown using Abel’s Test.

a) Try something similar to show that [pic] is convergent.

b) Now let’s see if we can find the values of [pic] and [pic].

Let [pic] and differentiate with respect to [pic], to get [pic]. Now let [pic] to get

[pic]. Now using the identity [pic], we get

[pic]. Since [pic] and [pic], we can conclude that [pic]. Now let [pic] to get [pic]

[pic], where

[pic] and [pic].

For [pic], see if you can show that [pic].

c) Now we know that [pic], so let’s see what happens as [pic]. Let [pic] to get [pic], but for [pic], we also have

[pic]

[pic]

[pic]

[pic]

And [pic], [pic], [pic], and

[pic]. So

[pic]. Similarly, you can show that

[pic]. So see if you can solve for the values of [pic] and [pic] from the system of equations

[pic], and by the way it can be shown that [pic].

44. Convergent improper integrals of the form [pic] can be approximated using one of our approximation methods by converting it into an equivalent integral over a finite interval. To do it, let [pic], to get [pic]. The right endpoint and midpoint methods are no problem, but the left endpoint, trapezoid, and Simpson’s will require that [pic] exists so that we can define [pic].

Convert the following improper integrals into the form [pic] and determine which approximation methods can be used.

a) [pic] b) [pic] c) [pic]

45. Convergent improper integrals of the form [pic] can be approximated using one of our approximation methods by converting it into an equivalent integral. To do it, let [pic], to get [pic]. The right endpoint and midpoint methods are no problem, but the left endpoint, trapezoid, and Simpson’s will require that [pic] exists so that we can define [pic].

Convert the following improper integrals into the form [pic] and determine which approximation methods can be used.

a) [pic] b) [pic] c) [pic]

46. What would happen if you tried the original definition of the definite integral on the convergent improper integral [pic]?

a) If you partition the interval [pic] into n equal subintervals and construct a Riemann sum by choosing a number from each subinterval(excluding [pic]), you would get a sum of the form [pic]. If the number [pic], that you choose in the interval [pic] is [pic], then you can show that [pic], so what would be the limit of the Riemann sums in this case?

b) The result in part a) is a little strange since we know that [pic] is convergent. The problem is that in the interval [pic] where the function blows-up we can select a number so that the area of the approximating rectangle doesn’t approach zero as [pic], in fact, it might approach [pic]. What if we select [pic], then the area of the approximating

rectangle on [pic] would be less than [pic], and hence approaches zero. What is the smallest value of the approximating rectangle on [pic] for the improper integral [pic]?

47. Show that

a)[pic] b) [pic],

and hence

c) find the value of [pic].

48. a) Find [pic] {Hint: [pic], so let [pic].}

b) What happens if you try the same method to determine [pic]?

49. Let [pic].

a) Determine convergence or divergence of the improper integral [pic], and if convergent, determine its value.

b) Suppose that for [pic], [pic]. Determine convergence or divergence of the improper integral [pic], and if convergent, determine its value.

50. Consider the improper integral [pic] for [pic].

a) Show that [pic] converges since the Mean Value Theorem implies that [pic], for some [pic], so [pic].

b) Show that [pic] converges since the Mean Value Theorem implies that [pic], for some [pic], so [pic].

c) Since parts a) and b) imply that [pic] converges for [pic], we can consider the function [pic]. It can be shown that [pic]

[pic].

Find a simple formula for f by solving the differential equation [pic] with [pic].

d) Use the result of part c) to find the exact value of [pic].

51. Be careful about what you read, even if it’s in a popular high school textbook. The following statement is from the book Calculus: Concepts and Applications:

Note: An improper integral with an infinite limit of integration always diverges if the integrand has a limit other than zero as the variable of integration approaches infinity.

If the note is interpreted as [pic] means that [pic] diverges, then it’s incorrect!

a) Show that the improper integral [pic] is convergent.

{Hint: Let [pic], to get [pic], and look at problem 43.}

b) Show that the function [pic] is unbounded on every interval of the form [pic] for [pic], and certainly [pic].

{Hint: Consider the values [pic] for [pic].}

52. Here is an alternative way of finding the exact value of [pic].

a) Use integration by parts to show that [pic].

b) Integrate both sides of the previous equation with respect to a from 0 to b to show that

[pic].

{Hint: [pic].}

c) Let [pic] in the previous equation to find the value of [pic].

{Hint: [pic].}

53. Evaluate the following limits:

a) [pic] {Hint: [pic] is a rather odd function.}

b) [pic] {Hint: [pic], so apply L’Hopital’s Rule.}

54. Consider the improper integral [pic].

a) Use integration by parts to show that the improper integral is convergent:

[pic], and [pic] is a proper integral, so check-out [pic].

b) Using the substitution [pic], you can show that [pic]. Use the substitution [pic]in [pic] to show that [pic].

c) Use the substitution [pic] in [pic] to show that [pic].

d) [pic], so [pic]. Use the

substitution [pic] in [pic] to conclude that [pic].

e) From the identity [pic], you can conclude that

[pic]. Now find the exact value of I.

55. Suppose that the improper integral [pic] converges. Use the substitution [pic] to find its value.

56. Find the exact value of the improper integral [pic].

{Hint: [pic], so investigate [pic] and [pic] using integration by parts.}

57. a) Find the value of the improper integral [pic].

{Hint: Use the substitution [pic] to get [pic], and then solve for the integral that you want.}

b) Find the exact value of [pic].

{Hint: You may use the result of part a).}

c) Find the exact value of [pic].

{Hint: You may use the result of part a).}

58. a) Suppose that [pic]. Evaluate the proper integral [pic] by converting it into an equivalent improper integral. First divide the numerator and denominator by [pic], and then use the substitution [pic].

b) Evaluate [pic] {Use the result of part a).}

c) Evaluate [pic] {Use the result of part a).}

d) Evaluate [pic] {Use the result of part a).}

59. Find the value of the convergent improper integral [pic].

{Hint: Let [pic].}

60. Show that [pic] for [pic].

{Hint: [pic], and let [pic].}

61. Consider the improper integral [pic]. It’s convergent if [pic] and [pic] are both convergent. [pic] will be convergent if [pic] is convergent. We already know that [pic] converges, and since [pic], we have convergence of [pic]. For [pic], let [pic] to get [pic], and since [pic], we also have convergence of [pic]. Make the substitution [pic] along with the identity [pic] to find the exact value of [pic].

62. If the improper integral [pic] converges, but the improper integral [pic] diverges, then the improper integral [pic] is said to be conditionally convergent. Let’s look at an example: We have already shown that [pic] is convergent, but what about [pic]? Well, let’s look at its graph:

From the graph, you can see that

[pic].

So [pic] is divergent. This means that [pic] is conditionally convergent. An alternative method of showing that [pic] is divergent is to consider the inequality [pic].

a) Determine if [pic] is absolutely or conditionally convergent.

b) Determine if [pic] is absolutely or conditionally convergent.

c) Determine if [pic] is absolutely or conditionally convergent.

63. Let’s sketch a proof of the Ratio Test:

Ratio Test: If [pic] is continuous on [pic] and [pic], then [pic] is absolutely convergent. If [pic] and [pic], then [pic] is divergent.

For convenience, let’s assume that [pic], and decompose the integral [pic] as [pic]. Since [pic], for [pic], we know that [pic], and hence that [pic]. From this we get that [pic] , [pic], and in general [pic]. So now we have that

[pic]

If we can show that [pic] has a finite value, then [pic] will be absolutely convergent. Let [pic], and notice that [pic]. This means that [pic].

a) Since [pic], find the value of [pic].

If [pic] and [pic], then for [pic], we know that [pic], and hence that [pic]. From this we get that [pic] , [pic], and in general [pic]. So now we have that

[pic]

b) Since [pic], what can you say about [pic]?

64. Let’s sketch a proof of the Root Test:

Root Test: If [pic] is continuous on [pic] and [pic], then [pic] is absolutely convergent. If [pic] and [pic], then [pic] is divergent.

Since [pic], we know that for [pic] that [pic], and hence that [pic]. So we have that

[pic].

If we can show that [pic] is bounded, then we can conclude that [pic] is absolutely convergent.

a) Show that [pic] is bounded for [pic].

If [pic] and [pic], then for [pic] that [pic], and hence that [pic]. So we have that

[pic]

If we can show that [pic] is divergent, then we can conclude that [pic] is divergent.

b) Show that [pic] is divergent.

65. Let’s sketch a proof of Abel’s Integral Test:

Abel’s Integral Test: If [pic], [pic], and [pic] are continuous on [pic] with [pic], [pic], [pic], and [pic] is bounded for all [pic], i.e. [pic]for all [pic], then [pic] is convergent.

Let’s do integration by parts on [pic] with [pic] to arrive at [pic].

Now let’s focus on [pic].

[pic], but this implies that

[pic]. If we can show that [pic] is convergent, then we can conclude that [pic] is absolutely convergent, and hence that [pic] is convergent.

Show that [pic] is convergent.

66. Let’s find the value of [pic] by doing the following:

a) Integrate by parts with [pic], to get

[pic].

b) Use the substitution [pic] to show that [pic].

c) Use the substitution [pic], to find the value of [pic].

67. For any real number a, let’s find the value of [pic].

[pic]. Using the substitution

[pic] in the second integral, leads to

[pic]. Combine the two integrals to get the result.

68. For [pic], integrate by parts twice to get [pic]. Use the previous result to find the maximum value of [pic] for [pic].

{Hint: [pic].}

69. In Planck’s Radiation Law, the improper integral [pic] is used. Show that it converges using the Limit Comparison Test with [pic].

70. A careless computation indicates that the derivative of [pic] is [pic]. This would lead you to believe that the behavior of [pic] could be determined from the behavior of [pic]. But clearly [pic] goes to infinity as b goes to infinity, while [pic] never exceeds 2. What went wrong?

71. Consider the improper integral [pic]. For [pic], [pic], and since [pic] converges, we get that [pic] converges. For [pic],[pic], and since [pic] converges, we get that [pic] converges. So now we know that [pic] converges. What does it equal?

{Hint: [pic], so let [pic] in the first integral, and see what happens.}

72. In Calculus III, you show that [pic]. Use the substitution [pic] to find the exact value of [pic].

73. Don’t believe everything you read, even if it’s in an edited textbook. The following statement is from the book Random Integral Equations with Applications to Life Sciences and Engineering:

Lemma 1.1.7 If

i) [pic] is a continuous function on [pic], and its derivative [pic] is bounded for [pic];

ii) [pic] is a continuous function on [pic], [pic] for any [pic], [pic]; and

iii) [pic];

then

[pic].

For [pic] and [pic], show that all three of the conditions are satisfied, but the conclusion is false.

74. Pappus was a Greek mathematician who lived in Alexandria in the fourth century. One of his famous theorems is as follows:

The volume, V, of a solid of revolution is given by [pic], where A is the area of the region being revolved, [pic] is the distance from the axis of revolution to the centroid of the region, and the region doesn’t lie on both sides of the axis.

For example, if the circle [pic] is revolved about the y-axis, it would generate a bagel. The centroid of the circle is the point [pic], and hence [pic], [pic], and the volume of the bagel is [pic].

Pappus’s Theorem applies to bounded regions with centroids and finite areas. What happens if we try to extend it to unbounded regions with possibly infinite areas?

a) (Revolution of an infinite area)Consider the region under the graph of [pic] over the interval [pic]. Its area is given by the improper integral [pic]. If we revolve this region of infinite area about the x-axis and try to use Pappus’s Theorem, we would get a resulting volume of [pic]. Unless [pic], there’s not a chance of getting a finite volume, and it doesn’t appear that the region would balance along the x-axis. The volume generated by revolving the region about the x-axis is finite and is given by the improper integral [pic]. One major problem with the region is that it doesn’t have a centroid. The centroid of the region, [pic], is found from the formulas [pic] and [pic]. If it had a centroid, its y-coordinate would be 0, and [pic]. Pappus’s Theorem does work in a limiting sense. If we truncate the region to the interval [pic], then the region has a well-defined centroid of [pic], and we get [pic]. Find [pic].

{Note: [pic], which is the definition of [pic].}

b) (Revolution of an unbounded finite area)Consider the region under the graph of [pic] over the interval [pic]. Its area is given by the improper integral [pic]. If we revolve this region of finite area about the y-axis and try to use Pappus’s Theorem, we would get a resulting volume of [pic]. Unless [pic], there’s not a chance of getting an infinite volume, and it doesn’t appear that the region would balance at infinity. The volume generated by revolving the region about the y-axis is infinite and is given by the improper integral [pic]. One major problem with the region is that it doesn’t have a centroid. The centroid of the region, [pic], is found from the formulas [pic] and [pic]. If it had a centroid, its x-coordinate would be [pic], and [pic]. Pappus’s Theorem does work in a limiting sense. If we truncate the region to the interval [pic], then the region has a well-defined centroid of [pic], and we get [pic]. Find [pic].

{Note: [pic], which is the definition of [pic].}

-----------------------

b

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

[pic]

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