LESSON X - Mathematics & Statistics
LESSON 12 INTEGRALS WITH INFINITE LIMITS OF INTEGRATION
Definition If the function f is continuous on the closed interval [pic], then
[pic].
NOTE: Recall that a definite integral exists if the integrand is continuous on the closed interval of integration. Since the integrand f is continuous on the closed interval [pic], then it is continuous on the closed interval [pic] for all [pic]. Thus, [pic] exists for all [pic]. Of course, if [pic], then since the function f is continuous at [pic], then the function f is defined at [pic] and [pic].
Definition If the function f is continuous on the closed interval [pic], then
[pic].
NOTE: Recall that a definite integral exists if the integrand is continuous on the closed interval of integration. Since the integrand f is continuous on the closed interval [pic], then it is continuous on the closed interval [pic] for all [pic]. Thus, [pic] exists for all [pic]. Of course, if [pic], then [pic].
The integrals [pic] and [pic] in the definitions above are called improper integrals. An improper integral is said to converge if the limit exists. Otherwise, it is said to diverge.
Definition If the function f is continuous on the real number line [pic] and a is any real number, then
[pic] = [pic] + [pic].
NOTE: The improper integral [pic] will converge if and only if the two improper integrals [pic] and [pic] converge. If one of these improper integrals diverges, then the improper integral [pic] diverges.
Examples Determine whether the following improper integrals converge or diverge. If the integral converges, then give its value.
1. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the interval [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] = [pic] =
[pic]
[pic] = [pic] = [pic] =
[pic] since [pic]
Answer: Diverges
2. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] =
[pic]
[pic] = [pic] =
[pic] = [pic] =
[pic] = [pic] = [pic]
NOTE: [pic]
Answer: Converges; [pic]
3. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] =
[pic]
[pic] = [pic] = [pic] =
[pic] since [pic]
Answer: Diverges
4. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
Using the integral formula [pic], where [pic], we have that [pic]. Thus,
[pic] = [pic] = [pic] =
[pic] = [pic]
since [pic] = [pic]
Recall that the horizontal line [pic] is a horizontal asymptote for the inverse tangent function [pic] as [pic]. Thus, [pic]. For future reference, the horizontal line [pic] is a horizontal asymptote for the inverse tangent function [pic] as [pic]. Thus, [pic].
Thus, [pic]. Thus, [pic] =
[pic]
Answer: Converges; [pic]
5. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic] = [pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] = [pic]
using the integral formula [pic], where [pic].
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] =
[pic] since [pic]
Answer: Converges; [pic]
6. [pic]
[pic] = [pic] + [pic] =
[pic] + [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed intervals [pic] and [pic]. The function is also continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic] = [pic] = [pic] = [pic]
using the integral formula [pic], where [pic].
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] = [pic]
since [pic]
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] since [pic]
Thus,
[pic] = [pic] + [pic] =
[pic] + [pic] = [pic] + [pic] = [pic]
Answer: Converges; [pic]
7. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers given by [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
We will find the partial fraction decomposition for [pic] = [pic]. Thus,
[pic] = [pic].
Multiplying both sides of this equation by [pic], we obtain the following equation.
[pic]
To solve for A, choose [pic]: [pic] [pic]
To solve for B, choose [pic]: [pic] [pic]
Thus, [pic] = [pic] = [pic] =
[pic] = [pic] = [pic]. Thus,
[pic] = [pic] =
[pic] =
[pic] = [pic]
[pic] = [pic] = [pic] =
[pic] = [pic]
[pic] = [pic] = [pic] = [pic] = [pic]
NOTE: Since [pic], then using L’Hopital’s Rule, we have that [pic] = [pic].
Thus, [pic] = [pic] = [pic]
Answer: Converges; [pic]
8. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers given by [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
Using the integral formula [pic], where [pic], we have that [pic]. Thus,
[pic] = [pic] = [pic] =
[pic]
Since the horizontal line [pic] is a horizontal asymptote for the inverse secant function [pic] as [pic], then [pic]. For future reference, the horizontal line [pic] is a horizontal asymptote for the inverse secant function [pic] as [pic]. Thus, [pic].
Thus, [pic]. Thus, [pic] =
[pic] = [pic] = [pic]
Answer: Converges; [pic]
9. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers given by [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic] = [pic] = [pic]
Using the integral formula [pic], where [pic], we have that [pic]. Thus,
[pic] = [pic] =
[pic]. Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] =
[pic] = [pic] = [pic] = [pic]
NOTE: Since [pic], then [pic]. Also,
[pic].
Answer: Converges; [pic]
10. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the open interval [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] =
[pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = DNE since [pic] = DNE
Answer: Diverges
11. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers x such that [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic] = [pic] = [pic]
Using Integration by Parts on the integral [pic]:
Let [pic] and [pic]
Then [pic] [pic]
Thus, [pic] = [pic] = [pic] = [pic]
Thus, [pic] = [pic] =
[pic] [pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic]
We will use the Sandwich Theorem to evaluate [pic]:
Since [pic] for all t and [pic] (since t is approaching negative infinity), then [pic] [pic] for all [pic].
Since [pic] and [pic], then [pic] by the Sandwich Theorem. Thus,
[pic] = [pic] = [pic] = [pic] since
[pic] since the sine function is odd
Answer: Converges; [pic]
12. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] = [pic]
NOTE: [pic] = [pic]
Answer: Converges; [pic]
13. [pic]
[pic] = [pic] + [pic] = [pic] + [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed intervals [pic] and [pic]. The function is also continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
Thus,
[pic] = [pic] = [pic] = [pic] =
[pic] = [pic] NOTE: [pic]
Thus, the improper integral [pic] converges.
Now, consider the improper integral [pic].
[pic] = [pic] = [pic] = [pic]
since [pic].
Thus, the improper integral [pic] diverges. Thus, the improper integral [pic] diverges.
Answer: Diverges
14. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Using Integration by Parts:
Let [pic] and [pic]
Then [pic] [pic]
Thus, [pic] = [pic] = [pic] = [pic] =
[pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic] =
[pic]
Consider [pic], which has the indeterminate form [pic].
Rewrite [pic] = [pic], which has the indeterminate form [pic]. Applying L’Hopital’s Rule, we have that
[pic] = [pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic]
Answer: Converges; [pic]
15. [pic]
[pic] = [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Using Integration by Parts:
D I
[pic] [pic] To obtain the value of the indefinite integral,
[pic] [pic] + add the product of [pic] and [pic],
2 [pic] [pic] subtract the product of [pic] and [pic],
0 [pic] + add the product of 2 and [pic],
and add the constant of integration.
Thus, we have that [pic] = [pic] =
[pic].
Thus,
[pic] = [pic] = [pic] =
[pic] =
[pic]
Consider [pic], which has the indeterminate form [pic].
Rewrite [pic] = [pic], which has the indeterminate form [pic]. Applying L’Hopital’s Rule, we have that
[pic] = [pic] = [pic] = [pic]
Since [pic], then applying L’Hopital’s Rule again, we have that
[pic] = [pic] = [pic] = [pic]
Thus,
[pic] = [pic] = [pic]
Answer: Converges; [pic]
16. [pic]
[pic] = [pic] + [pic] =
[pic] + [pic]
The integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed intervals [pic] and [pic]. The function is also continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
Let [pic]
Then [pic]
[pic] = [pic] = [pic] = [pic] =
[pic]
Thus,
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] NOTE: [pic] =
[pic]
Thus, the improper integral [pic] converges.
Now, consider the improper integral [pic].
[pic] = [pic] = [pic] =
[pic] = [pic] = [pic] NOTE: [pic] =
[pic]
Thus, the improper integral [pic] converges.
Thus, the improper integral [pic] converges and
[pic] = [pic] + [pic] =
[pic] + [pic] = [pic]
Answer: Converges; 0
17. [pic]
[pic] = [pic] =
[pic]
Since [pic], then the integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers given by [pic]. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
We will find the partial fraction decomposition for [pic] = [pic]. Thus,
[pic] = [pic].
Multiplying both sides of this equation by [pic], we obtain the following equation.
[pic]
To solve for A, choose [pic]: [pic] [pic]
To solve for B, choose [pic]: [pic] [pic] [pic]
Thus, [pic] = [pic] = [pic] =
[pic] = [pic]. Thus,
[pic] = [pic] =
[pic] = [pic] =
[pic]
[pic] = [pic] =
[pic] = [pic] =
[pic] = [pic]
[pic] = [pic] = [pic] = [pic] = [pic]
NOTE: Since [pic], then using L’Hopital’s Rule, we have that [pic] = [pic].
Thus, [pic] = [pic] = [pic]
Answer: Converges; [pic]
18. [pic]
[pic] = [pic] + [pic] =
[pic] + [pic]
Note that [pic] is quadratic in [pic] and we have that [pic] = [pic].
Since [pic] = [pic], then the integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed intervals [pic] and [pic]. The function is also continuous on the closed intervals [pic] for all [pic] and on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
We will find the partial fraction decomposition for [pic] =
[pic]. Thus, [pic] = [pic].
Multiplying both sides of this equation by [pic], we obtain the following equation.
[pic]
If we let [pic], we obtain the equation [pic] [pic].
Note that the coefficient of the [pic] term in the product [pic] is [pic]. The coefficient of the [pic] term in the product [pic] is [pic]. Thus, the coefficient of the [pic] term on the right side of the equation is [pic]. The coefficient of the [pic] term on the left side of the equation is 0. Equating the coefficients of [pic] on both sides of the equation, we obtain that [pic].
[pic]
Note that the coefficient of the [pic] term in the product [pic] is [pic]. The coefficient of the [pic] term in the product [pic] is [pic]. Thus, the coefficient of the [pic] term on the right side of the equation is [pic]. The coefficient of the [pic] term on the left side of the equation is 1. Equating the coefficients of [pic] on both sides of the equation, we obtain that [pic].
We have that [pic] from above and [pic]. Using the equation [pic] to solve for D, we have that [pic]. Substituting into the first equation of [pic], we obtain that [pic] [pic] [pic] [pic]. Since [pic] and [pic], then [pic].
Note that the coefficient of the y term in the product [pic] is [pic]. The coefficient of the y term in the product [pic] is [pic]. Thus, the coefficient of the y term on the right side of the equation is [pic]. The coefficient of the y term on the left side of the equation is 0. Equating the coefficients of y on both sides of the equation, we obtain that [pic] [pic].
We have that [pic] from above and [pic]. Using the equation [pic] to solve for C, we have that [pic]. Substituting into the second equation of [pic], we obtain that [pic] [pic] [pic]. Since [pic] and [pic], then [pic].
Thus, [pic] = [pic] = [pic] =
[pic] = [pic]. Thus,
[pic] = [pic] =
[pic] =
[pic] =
[pic] =
[pic]
Thus,
[pic] = [pic] =
[pic] =
[pic] =
[pic] =
[pic] = [pic] =
[pic] = [pic]
NOTE: [pic] and [pic] since [pic]
Thus, the improper integral [pic] converges to [pic].
Now, consider the improper integral [pic].
[pic] = [pic] =
[pic] =
[pic] =
[pic] =
[pic] = [pic]
[pic] = [pic]
NOTE: [pic] and [pic] since [pic]
Thus, the improper integral [pic] converges to [pic].
Thus, the improper integral [pic] converges and
[pic] = [pic] + [pic] =
[pic] + [pic] = [pic]
Answer: Converges; [pic]
19. [pic]
[pic] = [pic]
Note that [pic] is quadratic in [pic] and we have that [pic] = [pic].
Since [pic] = [pic], then the integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic] for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
We will find the partial fraction decomposition for [pic] =
[pic]. Thus, [pic] = [pic].
Multiplying both sides of this equation by [pic], we obtain the following equation.
[pic]
If we let [pic], we obtain the equation [pic].
Note that the coefficient of the [pic] term in the product [pic] is [pic]. The coefficient of the [pic] term in the product [pic] is [pic]. Thus, the coefficient of the [pic] term on the right side of the equation is [pic]. The coefficient of the [pic] term on the left side of the equation is 0. Equating the coefficients of [pic] on both sides of the equation, we obtain that [pic].
[pic]
Note that the coefficient of the [pic] term in the product [pic] is [pic]. The coefficient of the [pic] term in the product [pic] is [pic]. Thus, the coefficient of the [pic] term on the right side of the equation is [pic]. The coefficient of the [pic] term on the left side of the equation is 48. Equating the coefficients of [pic] on both sides of the equation, we obtain that [pic].
We have that [pic] from above and [pic]. Using the equation [pic] to solve for B, we have that [pic]. Substituting into the second equation of [pic], we obtain that [pic] [pic] [pic] [pic]. Since [pic] and [pic], then [pic].
Note that the coefficient of the x term in the product [pic] is [pic]. The coefficient of the x term in the product [pic] is [pic]. Thus, the coefficient of the x term on the right side of the equation is [pic]. The coefficient of the x term on the left side of the equation is 57. Equating the coefficients of x on both sides of the equation, we obtain that [pic].
We have that [pic] from above and [pic]. Using the equation [pic] to solve for C, we have that [pic]. Substituting into the second equation of [pic], we obtain that [pic] [pic] [pic]. Since [pic] and [pic], then [pic].
Thus, [pic] = [pic] = [pic] =
[pic]. Thus,
[pic] = [pic] =
[pic] =
[pic] =
[pic]
[pic]
[pic] =
[pic] =
[pic] =
[pic] =
[pic]
Thus,
[pic] = [pic] =
[pic] =
[pic] [pic]
[pic]
[pic] = [pic] = [pic] =
[pic] = [pic]
[pic] and [pic] since [pic]
[pic] [pic]
[pic] =
[pic] =
[pic] =
[pic] =
[pic]
Answer: Converges; [pic]
20. [pic]
[pic] = [pic]
Note that [pic] is quadratic in [pic] and we have that [pic] = [pic].
Since [pic] = [pic], then the integrand of [pic] is continuous on its domain of definition, which is the set of all real numbers. Thus, the function f is continuous on the closed interval [pic]. The function is also continuous on the closed intervals [pic]for all [pic]. Thus, the Fundamental Theorem of Calculus can be applied.
[pic]
We will find the partial fraction decomposition for [pic] =
[pic]. Thus, [pic] = [pic].
Multiplying both sides of this equation by [pic], we obtain the following equation.
[pic]
If we let [pic], we obtain the equation [pic].
Note that the coefficient of the [pic] term in the product [pic] is [pic]. Thus, the coefficient of the [pic] term on the right side of the equation is [pic]. The coefficient of the [pic] term on the left side of the equation is 0. Equating the coefficients of [pic] on both sides of the equation, we obtain that [pic].
[pic]
Note that the coefficient of the [pic] term in the product [pic] is [pic]. Thus, the coefficient of the [pic] term on the right side of the equation is [pic]. The coefficient of the [pic] term on the left side of the equation is 28. Equating the coefficients of [pic] on both sides of the equation, we obtain that [pic] [pic].
We have that [pic] from above. Since [pic] and [pic], then [pic] [pic].
[pic]
Note that the coefficient of the t term in the product [pic] is [pic]. Thus, the coefficient of the t term on the right side of the equation is [pic]. The coefficient of the t term on the left side of the equation is [pic]. Equating the coefficients of t on both sides of the equation, we obtain that [pic].
Since [pic] from above and [pic], then [pic].
[pic] = [pic] = [pic] =
[pic] = [pic] Thus,
[pic] = [pic] =
[pic] = [pic] =
[pic] = [pic]
Thus,
[pic] = [pic] =
[pic] =
[pic] =
[pic] =
[pic] = [pic] =
[pic]
Answer: Converges; [pic]
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