Integration Techniques, L’Hôpital’s Rule, and Improper Integrals8

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Improper Integrals8.8

Copyright © Cengage Learning. All rights reserved.

3

Evaluate an improper integral that has an infinite limit of integration.

Evaluate an improper integral that has an infinite discontinuity.

Objectives

4

Improper Integrals with Infinite Limits of Integration

5

Improper Integrals with Infinite Limits of Integration

The definition of a definite integral

requires that the interval [a, b] be finite.

And Fundamental Theorem of Calculus requires continuous f(x).

Here we develop a procedure for evaluating integrals that do not satisfy these requirements—usually because either one or both of the limits of integration are infinite, or f has a finite number of infinite discontinuities in the interval [a, b].

Integrals that possess either property are improper integrals.

A function f is said to have an infinite discontinuity at c if, from the right or left,

6

Improper Integrals with Infinite Limits of Integration

7

Geometrically, this says that although the curves 1/x and 1/x^2 and look very similar for x>0, the region under to the right of 1/x^2has finite area whereas the corresponding region under 1/x has infinite area. Note that both 1/x and 1/x^2 approach 0 as x-> infinity, but 1/x^2approaches 0 faster than 1/x. The values of 1/x don’t decrease fast enough for its integral to have a finite value.

8

Generalization:

1

1)1(

11

:

)1(

1

1

11)1

)ln(1

1)1

1

1

1

1

pdiverges

pconvergesapdx

x

Therefore

xpp

xdxxdx

xp

divergesxdxx

p

dxx

p

ap

ap

a

p

a

p

ap

aa

ap

9

My example

313

1

3

]313[

11

0

13

eedue

dxduxudxe

uu

x

10

22)arctan(

1

12

xdxx

11

eeedxeexdxex xxxx 1

)1()1( 1

11

11

12

My examples

xas 0 todecreasingeven NOT is f that Note

diverges --)2ln(ln)ln(1

]1[1

22

0

uduu

dxedueudxe

e xxx

x

422

1)arctan(

2

1

21

1

]2[

1

1

)2cosh(2

1

00

2

22

4

2

22

udu

u

dxedueu

dxe

edx

eedx

xxx

x

x

xx

Instead integrand approaches finite limit.

13

14

Improper Integrals with Infinite Discontinuities

15

Improper Integrals with Infinite Discontinuities

The 2nd type of improper integral is one that has an infinite discontinuity at or betweenthe limits of integration.

16

Example 6 – An Improper Integral with an Infinite Discontinuity

Evaluate

Solution:

The integrand has an infinite discontinuity

at x = 0, as shown in Figure 8.24.

You can evaluate this integral as shown

below.

Figure 8.24

2

3

3/2

11

0

3/21

0

3/11

03/1

xdxxdx

x

17

divergesx

dxx

0

1

2

1112

0

2

02

My Examples

dxx

dxx

dxx

2

02

0

32

2

32

111 Just showed that 2nd integral diverges => diverges

WRONGx

dxx 6

1

3

1

2

1112

3

2

32

NOTE that failing to recognize that there is interior singular point leads to wrong answer:

18

My generalization:

1

1)1(

11

:

)1(

1

1

11)1

)ln(1

1)1

1

1

0

01

0

1

00

00

0

pdiverges

pconvergesapdx

x

Therefore

xpp

xdxxdx

xp

divergesxdxx

p

dxx

pa

p

a

p

apap

a

p

aa

a

p

19

22)arctan(2

)1(

12

]2

1[

)1(

1

00

2

0

uduu

dxx

duxudxxx

20

24

2)arcsin(

1

1)'(1

1

1

1

1

11)'(1

112

2'1

1

0

1

02

1

0

2

22

22

2

22

22

2

sL

xdxx

dxys

xx

xx

x

xy

x

x

x

xyxy

21

3221

],2[2

1

3

0

3

0

5

2

uduu

dxduxudxx

3

1

1

0

3

1

1

0

3

0

|)1ln(||)1ln(|

111

xx

x

dx

x

dxdx

x

dx

22

11001

)ln()ln(1

0

1

0

1

0

dxxxxxdxx

23

24

25