Direct Comparison Test for Improper Integrals

Here is the Direct Comparison Test (in restricted form).

Example Exercise 2

Direct Comparison Test for Improper Integrals

We know e x

x



and e ^{ x} are positive, continuous functions over the interval  ^{1, }  . Moreover, 1

x   e ^{x x} x e

   . Thus, if

1

e dx x

 

 converges, then

1

e x

x dx

 

 also converges by the Direct Comparison Test.

  ^{1   1}

1 1

1

lim lim lim

1 1 1 1 1

lim 0



    

  

 

 

        

 

          



 

t

x x x t t

t t t

t t

e

e dx e dx e e e

e e e e e

1

e dx x



  converges 

1

e x

x dx

 

 converges.

Direct Comparison Test for Improper Integral:

Let  and T be continuous on  ^{a , }  such that ^{0  }   ^{x  T x}   ^{  x a .}

Then,  

a

T x dx

  converges  

a

 x dx



  converges

and

 

a

 x dx

  ^{diverges  }

a

T x dx



  ^diverges

Does

1

e x

x dx

 

 converge or diverge?

Example Exercise 3

Direct Comparison Test for Improper Integrals

Recall that e ^{ x}

²

is famously difficult to integrate and does not have an antiderivative in terms of any functions we know. Instead, compare e ^{ x}

²

to e ^{ x} . Note that both functions are always positive. Find where the two functions intersect.

 

2

2

2 2

ln ln

0

1 0

0, 1

x x

x x

e e

e e

x x

x x x x

x x

 

 



 

   

   

   

  

The two functions are equal only when x = 0 and 1, which implies one function is greater than the other over the interval   ^{1,  . Indeed, e  x  e  x}

²

^over   ^{1,  , so e  x  e  x}

²

 ^{1, }  . Thus, by the Direct Comparison Test, if e ^{ x} converges, e ^{ x}

²

converges.

  ^{ }

 

   

1 1

1 1

1

1

1 1

1

lim

lim lim

lim

lim lim

0 1

t

x x

t

x t t

t t

t t

t

t t

e

e dx e dx

e e e

e e

e e

e

e e





 



  

 

 



 

 







 

       

  

  

  

  

 

1

e dx x



  converges; therefore,

²

1

e x dx



  converges by the Direct Comparison Test.

Does

²

1

e x dx



  converge or diverge?

Think how all the output values of the sine-wave range from  1 to 1.

2

1 sin 1

0 sin 1

x x

  

  

Thus, the numerator of the integrand is always less than or equal to 1. Dividing all three sides of the inequality by x ² gives us the inequality below.

2

2 2

sin 1

0 x

x x

 

We know by the p-Test for Improper Integrals that ₂

1

1 dx x

  converges. We know, too, that the functions

2

2 2

sin 1

and x

x x are continuous. Thus, by Direct Comparison Test,

2 2 1

sin x x dx

 

converges.

Does ₂

1

x dx

 converge or diverge?

Example Exercise 5

Direct Comparison Test for Improper Integrals

Think of ₁₀₀ ¹ as insignificant for a moment. Then, the integrand would be ¹

₂

¹ _x

x  . By the p-Test for Improper Integrals, we know

1

1 dx x

  diverges. Since

₂

1 100

1

x  and ¹ _x are positive and continuous on  ^1,  , we know

2 1

1 100

1 dx

x



  will also diverge if

2 1

100

1 1

x x

  for all x  1 (by the Direct Comparison Test).

2

2 1

100

2 1

100

2 1

1 100

1,

1 1

1 diverges

x x x

x x

x x

dx x



  

  

 



  

Example Exercise 6

Direct Comparison Test for Improper Integrals

The functions ¹ ₂

1  x and 1 ₂

x are positive and continuous on  ^{1, }  . By the p-Test for Improper Integrals, we know ₂

1

1 dx x

  converges.

2 2

2 2

1, 1

1 1

1

x x x

x x

   

 



Thus, by the Direct Comparison Test, ₂

1

1

1 dx

x



  converges.

Does

2 1

1 100

1 dx

x



  converge or diverge?

Does ₂

1

1

1 dx

x



  converge or diverge?

The Limit Comparison Test, stated below, does not include inequalities in its hypotheses.

Example Exercise 8

Limit Comparison Test for Improper Integrals

On the interval  ^{1, }  ^, ₂ ¹

2

x  x and 1 ₂

x are both positive, continuous functions. By the p-Test for Improper Integrals, we know ₂

1

1 dx x

  converges.

2 H H

2

2 2

lim lim lim 1

2 2 2 2

x x x

x x

x x x

  

        

           

 

Thus, by the Limit Comparison Test, ₂

1

1 2 dx

x x



  must converge.

Limit Comparison Test for Improper Integrals: Let f and g be positive,

continuous functions on  ^{a , }  . Let L be a positive number less than infinity. Then,

       

lim and

x

a a

f x L f x dx g x dx

g x

 



 

   

    both converge or both diverge.

Does ₂

1

1 2 dx

x x



  converge or diverge?

#1 converges in direct comparison to

₂

1

1 1 dx

x



  #3 converges in direct comparison to

3 1

1





x ^dx

#5 diverges in direct comparison to

1

1 dx x





#7 converges in limit comparison with

₃

2 1

1 dx x





Practice Problems in Calculus: Concepts and Contexts by James Stewart 1st ed. problem set: Section 5.9 #1, #41-49 odd

2nd ed. problem set: Section 5.10 #1, #415-49 odd 3rd ed. problem set: Section 5.10 #1, #41-49 odd

Practice Problems in Calculus: Early Transcendentals by Briggs and Cochran

2nd ed. problem set: Not available.

Practice Problems Determine if the following intervals converge or diverge.

#1

2 2 1

cos 1

x dx x



  ^{#2 2}

1

1

x dx x e



 

#3 3

1

1 1

dx x



  ^#4

1

e x

x dx

 



#5 ⁴  

1

1 3sin x x dx

 

 ^{#6 6}

1

1 1

dx x



 

HINT: Use limit comparison with ¹

₃

x .

#7 5 3

1

x dx x x



  ^{#8 3 5 2 4}

1

2 5 4 3

3 2 1

x x x

x x x dx

   

  



HINT: Use limit comparison with

3 2

1 x

. HINT: Use limit comparison with ¹

₂

x .