Improper IntegralsThe integrals we have studied so far represent signed areas of bounded regions.

33/ 4 5/3

1

213 .200x x dx

There are two ways an integral can be improper: (1) The interval of integration may be infinite.(2) The integrand may tend to infinity.

We deal first with improper integrals over infinite intervals. One or both endpoints may be infinite:

Improper IntegralsHowever, areas of unbounded regions also arise in applications and are represented by improper integrals.

, , a

a

f x dx f x dx f x dx

Bell-shaped curve. The region extends infinitely far in both directions, but the total area is finite.

2

0

dx

x

How can an unbounded region have finite area? To answer this question, we must specify what we mean by the area of an unbounded region. Consider the area under the graph of f (x) = e−x over the finite interval [0,R]:

0

00

1R

Rx x R Re dx e e e e As ,R this area approaches a finite value :

0 0

lim lim 1 1R

x x R

R Re dx e dx e

It seems reasonable to take this limit as the definition of the area under the graph over the infinite interval [0, ). Thus, the unbounded region has area 1.

DEFINITION Improper Integral Fix a number a and assume that f (x) is integrable over [a, b] for all b > a. The improper integral of f (x) over [ , )a is defined as the following limit (if it exists):

We say that the improper integral converges if the limit exists (and is finite) and that it diverges if the limit does not exist.

0

limR

Ra

f x dx f x dx

Similarly, we define

lima a

RR

f x dx f x dx

A doubly infinite improper integral is defined as a sum (provided that both integrals on the right converge):

0

0

f x dx f x dx f x dx

32

Show that converges and compute its value.dx

x

Step 1. Integrate over a finite interval [2, R].

22

23 2

2

1 1 1 1

2 2 24 8

1RRdx

x Rx R

Step 2. Compute the limit as R → ∞.

3 22

1

8

1 1lim lim

8 2

R

R R

dx

x R

1

Determine whether converges.dx

x

11

n n0 l ll nR

R

dxx

xRR

1

lim lim lnR R

R

dxR

x

CONCEPTUAL INSIGHT If you compare the unbounded shaded regions in the last two examples, you may wonder why one has finite area and the other has infinite area. Convergence of an improper integral depends on how rapidly the function f (x) tends to zero as

or .x x Our calculations show that x−3 decreases rapidly enough for convergence, whereas x−1 does not.

1

limR

R

dx

x

32

1l

8im

R

R

dx

x

An improper integral of a power function f (x) = x−p is called a p-integral. Note that f (x) = x−p decreases more rapidly as p gets larger. Interestingly, our next theorem shows that the exponent p = −1 is the dividing line between convergence and divergence.

THEOREM 1 The p-Integral over [ , )a For a > 0,

1

, if 11

diverges, if 1{

pap

ppp

a

dx

x

32

1l

8im

R

R

dx

x

2

32

23 lim converges to

1

82

R

R

dxp

x

0

00

1 1 0 1 1 111R

Rx xR

R Rxe dx x e R e Re

e eR

0

?xxe dx

'

' 1

x

x

u x x

v x e

u x

v x e

* Choose so that ' is simpler than

* Choose ' so that ' can be evaluated.

u u u

v v v dx ' 'u x v x dx u x v x u x v x dx

Try writing the two factors of the integrand as '.uv

x x xxe dx xe e dx x x

u x du dx

e dx e C

1x xxxe e C x e C

0

1lim 1 lim 1 l 1i

1m

Rx

RR R RR

Rxe dx

ee

' ' L Hopital s Rule

' ' 'IbP s u x v x dx u x v x u x v x dx

Improper integrals arise in applications when it makes sense to treat certain large quantities as if they were infinite. For example, an object launched with escape velocity never falls back to earth but rather, travels “infinitely far” into space.

Escape Velocity The earth exerts a gravitational force of magnitude F (r) = GMem/r2 on an object of mass m at distance r from the center of the earth.

(a) Find the work required to move the object infinitely far from the earth.

REMINDER Me ≈ 5.98 · 1024 kg, re ≈ 6.37 · 106 m

The universal gravitational constant is… G ≈ 6.67 · 10−11 N-m2/kg2

A newton is… 1 kg-m/s2 A joule is… 1 N-m.

The work required to move an object from the earth’s surface (r = re) to a distance R from the center is

2

2

1

1 1 1 joules

lim

ee

e

e

R

e

e

R

err

r

e

e

eR e

R

eR

r

r dr GM mr

GM m GM mr r R

GM mr dr

GM m

GM m

r

is force in terms of distance from the center of the earth.F r r

Work is the antiderivative of force.

Escape Velocity The earth exerts a gravitational force of magnitude F (r) = GMem/r2 on an object of mass m at distance r from the center of the earth.(b) Calculate the escape velocity υesc on the earth’s surface.

The work required to move the object infinitely far from the earth.

By the principle of Conservation of Energy, an object launched with velocity υ0 will escape the earth’s gravitational field if its kinetic energy

2

0

1

2m v is at least as large as the work required to move the object to

infinity—that is, if

1/ 2

2

0 0

21

2e e

e e

GM m GMm v v

r r

2lime

eR

e

r eR

GM mr dGM m

rr

0 11,200 m/ 11,200 m/s sescvv

0.04 0.04

0 0

00

$

6000 6000

6000

150,00

150,000 lim0.04

150,000 00 1

t t

uu u

u

PV e dt e dt

e du e

Perpetual Annuity An investment pays a dividend continuously at a rate of $6000/year. Compute the present value of the income stream if the interest rate is 4% and the dividends continue forever.

(not on the BC Course DescriIn 5.8 ption) it was shared that the present value (PV) after T years at interest rate r = 0.04 is

0.04

0

6000T

tPV e dt Over an infinite time interval,

0.04

0.04

u t

du dt

As , T u

Although an infinite number of dollars are paid out during the infinite time interval, their total present value is finite.

In practice, the word “forever” means “a long but unspecified length of time.” For example, if the investment pays out dividends for 100 years, then its present value is

The improper integral ($150,000) gives a useful and convenient approximation to this value.

1000.04

0

6000 $147,253te dt

An integral over a finite interval [a, b] is improper if the integrand becomes infinite at one or both of the endpoints of the interval. In this case, the region in question is unbounded in the vertical direction. For example,

Infinite Discontinuities at the Endpoints

9

0

dx

x is improper because the integrand f (x) = x−1/2 tends to

as x → 0+. Improper integrals of this type are defined as one-sided limits.

DEFINITION Integrands with Infinite Discontinuities If f (x) is continuous on [a, b) but discontinuous at x = b, we define

lim b R

R ba a

f x dx f x dx

Similarly, if f (x) is continuous on (a, b] but discontinuous at x = a,

lim b b

R aa R

f x dx f x dx

In both cases, we say that the improper integral converges if the limit exists and that it diverges otherwise.

9

0

?dx

x 1

f xx

The integral is improper because the integrand has an infinite discontinuity at x = 0.

lim b b

R aa R

f x dx f x dx

9

91/ 2 1/ 2

0 0 06

The integral converge

lim lim 2 lim 2

s

3

.

RR R RR

x dx x R

1/ 2

1/ 2

0 0 0lim lim ln lim l

ln 0.5 The integral diverg

n 0.5 ln

es.

RR R RR

dxx R

x

1/ 2

0

?dx

x

The integral is improper because the integrand has an infinite discontinuity at x = 0.

lim b b

R aa R

f x dx f x dx

1f x

x

THEOREM 2 The p-Integral over [0,a] For a > 0,

Theorem 2 is valid for all exponents p. However, the integral is not improper if p < 0.

1

, if 11diverges, if 1

0

{

paa pp

pp

dx

x

0 0

The -integrals and have

opposite behavior for 1. The first converges only for 1,

and the second converges only for 1 (both diverge for 1).

This

Graphical Ins

is reflect

ig t

ed

ha

p pp x dx x dx

p p

p p

1 0

in the graphs of and , which

switch places at 1. We see that a large value of helps

to converge but causes to diverge.

p q

ap p

y x y x

x p

x dx x dx

1

20

?1

dx

x

In Section 9.1, we will compute the length of a curve as an integral. It turns out that the improper integral in our next example represents the length of one-quarter of a unit circle. Thus, we can expect its value to be 1

24 2

1

2

1sin

1

dx

dx x

1

2 210 0

1 1

1

lim1 1

lim sin sin 02

R

R

R

dx dx

x x

R

Comparing IntegralsSometimes we are interested in determining whether an improper integral converges, even if we cannot find its exact value. For instance, the integral

1

xedx

x

cannot be evaluated explicitly. However, if x ≥ 1, then

10 1 0

xxe

ex x

In other words, the graph of y = e−x/x lies underneath the graph of y = e−x for x ≥ 1. Therefore

1

1 1

converges by direct computat n o0 ix

xedx e dx e

x

Multiply our compond

inequality .xe

1 1

1

1 1 1

lim

lim lim

u u u

RR

u R

RR R

e du e du e du

e e e e

1

1

xe dx e

u x du dx

Since the larger integral converges, we can expect that the smaller integral also converges (and that its value is some positive number less than e−1). This type of conclusion is stated in the next theorem.

THEOREM 3 Comparison Test for Improper Integrals Assume that f (x) ≥ g (x) ≥ 0 for x ≥ a.

a a

a a

* converges also converges.

* diverges also diverges.

f x dx g x dx

g x dx f x dx

The Comparison Test is also valid for improper integrals with infinite discontinuities at the endpoints.

1

1 1

converges by direct computat n o0 ix

xedx e dx e

x

What the Comparison Test says (for nonnegative functions):• If the integral of the bigger function converges, then the integral of the smaller function also converges.• If the integral of the smaller function diverges, then the integral of the larger function also diverges.

31

Show that converges.1

dx

x

We cannot evaluate this integral, but we can use the Comparison Test. To show convergence, we must compare the integrand (x3 + 1)−1/2 with a larger function whose integral we can compute.

3/ 2

3 3

1 1

1x

x x

3/ 21

31

converge converge s1

sdxx

x

d

x

31

Does converge?x

dx

x e

1/ 231

1 1We cannot use because diverges.

x

dx

xx e x

3 333

1 1x xxx

x e eex e

13 33

11

3 33

1 1lim lim

3 3

1 1lim

3 3

Rx x

x RR R

R

R

dxe e

e

e ee

converges

3 3u x du dx

31

convergesx

dx

x e

1

1

(for 1),

but (for 1)

p

p

x dx p

x dx p

0.5

8 20

Does converge?dx

jx x

0.5 0.5

8 20 0

2 diverg diveres ge

2s

ddx

x

x

x x

Notice that if 0 < x < 0.5, then x8 < x2, and therefore

8 2 28 2 2

1 12

2x x x

x x x