Bear H.S. Understanding Calculus

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Chapter22 • ImproperIntegrals

The calculation for (b) is

similar,

but this time the integral

converges.

1 fb 1

= lim

2dx

1 x

b~oo

I X

= lim

_~]b

b~oo

x 1

= lim

(-~

= 1.

b~oo

The limit exists,so the integral

converges

and has the valueone (1):

2 dx = 1.

1 X

129

Example 22.1 shows that fl

-}p

dx converges for some P (e.g., P = 2) and diverges

forsome

P (e.g., P =

4).

Since

gets smaller on

[1,00)

as p gets bigger, there must be a

critical point

such that the integral fl

-}p

dx converges if p >

and diverges if p <

Po.

Here is the calculation:

- dx = lim

I x

b-e-co I

1 I

= lim

b~oo-p+l

= lim I

[b-

+! - 1].

b~oo-p+l

If p > 1, then the exponent in b

is negative,

limb~oo

= 0, and the integral converges.

p < 1, then 1 - p > 0 and

limb~oo

= 00. Therefore, fl

dx converges if p > 1

and diverges if

p < 1. If p = 1, the integration formula is different and we check that case

separately:

- dx = lim - dx

I X b-s-o: I X

= lim IOgx]b

b~oo

= lim (log e - 0) =

00.

b~oo

The integral diverges if p = 1. Hence

- dx converges ifand only if p > 1.

I x

The lower limit in the integrals above was taken to be 1 for simplicity.

course, the

lower limit must be positive since

-}p

is unbounded at 0. However, for any a > 0, fa

f (x) dx

converges if and only if fl

f(x)

dx converges, since the area between x = 1 and x = a is

certainly finite.

Integrals over unbounded intervals

(-00,

b] are treated in exactly the same way:

f(x)

aJ!f!!oo

f(x)

dx.

Now consider integrals

the type

-dx,

o x

(22.5)

130

Understanding

Calculus

where a > 0 and p >

The convergence or divergence of (22.5) does not depend on how

big a is, so we let a = 1 for convenience. Consider the following two special cases:

(a) 1

dx; (b) t'

dx.

"IX

The same simple integration as in Example 22.1 shows that

= lim

= lim (2 -

2JQ)

= 2,

"IX

....

"IX

x....

~dx

= lim r

-;'dx

= lim

(-1

+~]

10 x 0

....

The functions .;pfor p > 0 are all unbounded at 0, and some integrals converge (e.g., p =

while others diverge (e.g., p = 2). The bigger p is, the faster

-h

grows as x

0+,

and the

same kind of calculation as we made for integrals ft

shows that

converges ifand only if p < 1.

10 x»

There is an obvious similarity between integrals fo

.tIp

and integrals fl

dx.

For

positive p, p > 1 if and only if *< 1, so

~dx

and t

~dx

I x»

both converge or both diverge. The geometry makes the situation even clearer. Consider, for

example, the functions

and

These functions are inverses of each other, since

1 1

------x

fFr-

();f-

so their graphs are symmetric about the line y = x. Thus, the two shaded areas in Figure 22.1

are equal; that is,

(_1

= 1.

I x

Similarly, the curves y = .j. and y =

are symmetric about the line y =x, and both of the

following integrals converge:

t (I ) 1

I I

4'x

- 1

= I x3

= 2'

Figure 22.1

Chapter 22 • Improper Integrals

EXAMPLE

22.2

sinx

(a)

-3

dx.

For 0

1,0

sinx

x. Therefore,

sinx

x 1

-3-:::3=-'

Since

converges (p = ! < 1), the given integral with a smaller integrand also converges.

EXAMPLE

22.3

131

~dx.

l+x

First notice that the convergence or divergence of the integral

doesn't

depend on the lower limit. Since

(l:x

)

behaves roughly like

-;r

for large x, we make this comparison:

x x 1

--<-=-.

1+ x

Since

-!r

converges (p = 2 > 1), the given smaller integral converges.

EXAMPLE

22.4

dx,

This integral is improper for two

reasons-the

integrand is unbounded at 0, and the interval [0,

(0)

unbounded. All such integrals must be broken up into integrals with a single impropriety, and convergence

of the integral requires convergence of all the pieces. Thus, we write

dx = {

dx +

JOO

dx,

and check the two integrals separately for convergence. For 0

x ::: 1,

1, so

:::

)X,

and the

first integral converges. For

:::

and fl

converges:

oo e-x

= lim

b~oo

= lim [-e-xJr

b-s

-b

-1

= hm

+ e ) = -.

b~oo

Since fl

converges, fl

also converges. Therefore,

converges.

PROBLEMS

Evaluate the integral, or show that it diverges.

22.1 1

dxs

22.2

22.31

o 1 + x

(Integrateby parts with u = logx.)

132

00 d

22.4

e X log x

22.5

1'''''

xe-

00 dx

22.6

2 x

22.7 2

22.81

x-I

1 dx

22.9

--2

o 1 - X

1 dx

22.10

r;--::

y1-x

22.11 1

logx dx

1 x

22.12 r dx

v'f=X2

22.

IOg2

e x

22.14

roo

9 +4x

22.15

-2-

x-x

1 1

22.16

~)dx.

yx+1

Hint:

vX+T

JX=d+JX.

Understanding Calculus

Use the comparisontest to tell whetherthe following convergeor diverge.

22.17

vT+?

1 1

22.18

roo

2 dx

l+x

22.19 1

3)-4

22.20 /00 1 +

~osx

1 X

Find the volumeobtainedby rotatingaboutthe x-axis the followingareas:

22.21 The area under

y =

for 1

x <

.22.22

The area under y = e-

for 1

x <

00.

22.23 The area under y = h for 0

x <

00.

yl+x

22.24 The area under y =

xe-

for 0

x <

00.

22.25 (i) Show that

xe:"

dx = 1.

(ii) Use integrationby parts with u =x

to showthat

1e-

dx =

(n + 1)

x't

e:"

dx.

(iii) Use (i) and (ii) to find

xne-

dx for n = 2, 3, 4,

....

Series

We extend the operation of addition from a finite number of terms to an infinite number of

terms by taking a limit; that is, we define the infinite sum

+a2 +a3 +...

+...

to be a limit of the finite sums

Sn,

where

(23.1)

(23.2)

Any indicated infinite sum like (1) is called a

series, and the sum of the series (23.1) is the

limit of the sequence

{sn}:

al + a2 +

...

= lim

Sn.

n-HX)

(23.3)

The numbers

{sn},

with each s; defined by (23.2), form the sequence of partial sums of the

series (1). The number

s; is the nth partial sum of the series.

The following notation for series is convenient:

Lan

+a2 + ...

+ ....

n=l

This E-notation can also be used for finite sums. For example,

Lan

+ a: +

+ a4 + a5,

n=l

2+2+2+2+2+2

= 12,

n=l

Lak

+...+

Sn.

k=l

We say the series E:1

converges if the sequence

{sn}

of its partial sums converges.

If the sequence

{sn}

does not converge, the series diverges. We will frequently omit the range

on the indices when dealing with infinite series, and write E

instead of

E:l

an.

133

134

Understanding Calculus

If L an is a convergent series, with

{sn}

its sequence of partial sums, then Sn ----+ s, and

Sn-l

----+

an =Sn -

Sn-l

----+ S - S =

A series cannot converge unless the terms tend to zero. The condition an ----+ 0 is NOT

sufficient for convergence, only necessary. We will see many divergent series whose terms

tend to zero.

EXAMPLE

23.1

Show that the following series diverge.

(a) L

2n:

1; (b)

L(-lt.

In series (a), an =

2n~1

!. The terms approach a limit, but the limit is not zero, so the series

diverges. In (b), the terms

(-I)n

oscillate between 1 and

-1,

so the sequence

{an}

does not approach a

limit, and the series diverges.

It is easy to see that if L an converges, then L

can

also converges and

Similarly, you can add the corresponding terms

two convergent series, so that if L an and

L b

both converge, then

L(a

)

converges and

L(a

)

Lan

The convergence or divergence

a series has nothing to do with the first 100 terms,

or the first 100 million terms. It is only the tail

the series that determines convergence.

Therefore, to determine convergence or divergence, we need only consider how the terms

behave for all sufficiently large n.

The following two series are instructive:

111

+ 2+ 3+ 4+

...

+ ;; +

...

, (23.4)

111

1 -

2+ 3- 4+... ± ;;

. . . . (23.5)

The first series, called the

harmonic series, diverges, and the second series converges. To see

that the harmonic series (23.4) diverges, consider the following partial sums:

= 1 2: 2'

1 1

s2=1+->2·-

2 -

S4 = 1 +

> 3 . !

2 3 4 -

Sg = 1 +

(~+~)

(~+

+~)

4·

2345678-

Continuing this way, we see that

S16

5.!' S32

6 . !' and so on. Clearly, Sn

00,

diverges.

Now consider (23.5), and more generally any series of the form

(23.6)

Chapter 23 • Series

135

where

...

and

---+

The partial sums

of (23.6) start at

= a1,

and then jump successively to the left and right as a2 is subtracted, a3 is added, a4 subtracted,

and so on. The jumps get smaller since the

an decrease, and all partial sums beyond

lie

in a fixed interval of length

a.: These intervals collapse to a single point since an

---+

0, so

{sn}

converges to that point. Any series converges

the signs alternate, the terms decrease

in magnitude, and the terms tend to

zero.

We will call such a series a proper alternating

series,

where the word "proper" indicates that not only do the signs alternate, but the other two

conditions are also satisfied, so

any proper alternating series converges. Moreover, because

of the way the s;

jump

back and forth in a proper alternating series, it is clear that s; is always

within a distance a

+1 of the limiting sum.

EXAMPLE 23.2

The following seriesis a properalternatingseries, and it is knownthat the sum is

1 1 I n

1 - - + - - - + ... =

357

Howmany terms must you add to get an approximation to

accurateto within .05?

Weknow the error between

5" in this alternatingseriesand the sum,

is less than the first term

omitted. If the firstterm omittedis

-ft'

the error will be less than .05. Yourcalculatorwill show

I I 1 I 1 1 1 I I

1 - - + - - - + - - - + - - - + - - -

.760.

3 5 7 9

13 15 17 19

Yourcalculator will also show that

.785, so .760 is indeed accurate within .05. Notice that since

the partial sumsjump back and forth overthe limit,the numberhalfwaybetweenthe sum to -

and the

sum to

+it is a much better approximation; that is,

(

1 1

1 -

- ... -

+ 2. 21

.760 + .024 = .784.

EXAMPLE 23.3

Tellwhetherthe seriesconverges or diverges, and why:

(a)

L(-l)"

(logn);

(b)

L(_I)"_n_

_ .

n 3n

+ 1

The first series, (a), converges because the signs alternate,and

(logn)

decreases, and decreasesto zero.

The series(a) is thereforea properalternatingseries. Theseries

(b)"

diverges because

(3"2~

-+>

Don't

be misledby alternatingsigns. The terms must decreaseto zero or it isn't a properalternatingseries.

A series L an such that L an converges but L

Ian

I diverges is called conditionally

convergent.

The series (23.5) is conditionally convergent. If L an and L

Ian

Iboth converge,

then

is absolutely convergent. It is a theorem that if the series L

Ian

I of absolute

values converges, then the series

necessarily also converges. Any cancellation because

of differing signs of the

only helps the convergence.

A very simple and very important series is the

geometric series

ax"

= a + ax + ax

...

+ ax" +

....

n=O

(23.7)

136 Understanding Calculus

If x

1, we find a formula for

as follows:

=a + ax + ax

+ .. ·+ ax",

=ax

+ax

+...

+ax

(1 - x)sn = a - ax

(23.8)

a - ax

Sn=----

I-x

The formula (23.8) works for any geometric series with x

1, but the series converges only

Ixl

< 1. If [x] < 1, then x

0, so

~:~:;

0, and

(1~x).

Hence, for

-1<x<l,

n 2 n a

ax =a + ax + ax + ... + ax +

...

--.

I-x

n=O

EXAMPLE

23.4

Find the sums of the geometric series.

00 1

(a) L 2

;

(b) 0.333

....

n=1

(a) The common ratio is ! < 1, so the series converges. The first term is a = !'so

00 1 !

L-=_2_

=1.

n=1 2

1 - 2

(b) Repeating decimals are simply a way of indicating a convergent geometric series. Here we

have

3 3 3

o333 . = - + - + - + ...

. .. 10 100 1000

3 1. 1

1 =

=-.

10 10 3

A conditionally convergent series has both positive and negative terms; the positive

terms add up to

+00,

and the negative terms to -00. Conditional convergence therefore

depends on a delicate cancellation between the positive and negative terms. It can be shown

that a conditionally convergent series can be rearranged to converge to anything you like, or

to diverge. Since the order in which the terms are added is all important in a conditionally

convergent series, this kind of summation is not an entirely satisfactory generalization of

a finite sum. On the other hand, if a series converges absolutely, then any rearrangement

will also converge, and converge to the same number. Moreover, if

L an and L b; both

converge absolutely, then you can add up all the products

anbk,

in any order, and the result

will be the product of

(L an) and (L bk) as it ought to be. No such statement can be made

about conditionally convergent series. Absolute convergence is the property that allows us

to treat infinite series pretty much like finite sums in terms of rearrangement, grouping, and

multiplying.

Since checking a series L an for absolute convergence involves checking the positive

series L

Ian

we now develop some tests for convergence of positive series; that is, series

L an with

2::

0 for all n. If

is the nth partial sum of a positive series, then

{sn}

is an

increasing sequence since each new term is obtained by adding a positive number; that is,

Sn+l

=s; +

an+l,

and

an+l

2::

It is a basic property of numbers that an increasing sequence

is either bounded, and converges, or is unbounded and diverges to

+00.

Therefore, a positive

Chapter 23 • Series

137

(23.9)

series converges if and only if its partial sums remain bounded. This observation immediately

gives us the following comparison test:

Comparison Test: /fO s all

and L b; converges, then L an converges; if

Lan

diverges, then L

bit

diverges.

EXAMPLE

23.5

(2)n

logn

(a) L n + I

;

()

-n-'

(a) This series converges by comparison with the geometric series L 3 .

(~)n,

since for all n,

(~)1l

< 3 .

(~)1l

n+l

3 - 3

(b)

Since"!

diverges and

> ! for n > 3, the series "

diverges.

L-

n-lI

L-

Our next test makes a comparison between the partial sums s; of a positive series, and

integrals

fIn

! (x)

a positive function. The improper integral ft

! (x)

converges if

and only if the integrals

ft ! (x )

remain bounded. We see from Figure 23.1 that if ! is

positive and decreasing,

(It

f(1)

f(2)

+ ... +

f(n

- 1)

f(x)dx,

and

f(2)

f(3)

+ ... +

f(n)

f(x)

dx.

(23.10)

The first inequality shows that if L ! (n) converges, then fl

! (x)

converges. The second

inequality shows that if

!(x)

converges, then the series L

!(n)

converges. Therefore,

f(l)

+ ... + f(n

-I)

ftnf(X)dx

~--

fen - 1)

f(2) +f(3) + ...+

f(n):S;

fti(x)dx

fen)

~./

U--r-r-1---

Figure 23.1

f(2)

n-l

138

Understanding Calculus

we havethe following convergence test:

Integral Test:

f (x) is a positive decreasing function, then L f (n)

converges

ifand

only if fl

f(x)

converges.

EXAMPLE

23.6

x 1

n log n

The function f

(x)

= x

I~g

x is positive and decreasing for x

::::

2. We can therefore compare the series

with

f,':>o

dx.

.. x ogx

The integrand has the form

du, with u =

logx,

du =

~,so

00 1

--dx

= lim

--dx

2 x log x

Il~oo

2 X

logx

= lim log IIOgxl]n

n~oo

= lim log

Ilognl-Iog

[log Z] =

00.

n~oc

Since the integral diverges, the series diverges.

Series of the form L

;!r

are called

p-series.

If p

;!r

-A

°and the series surely

diverges, so we consideronly

p-series

for p > 0. The divergent harmonic series is a p-series

with p = 1. Wesaw in Chapter22 that fl

-fp

dx converges if and only if p > 1, so a p-series

converges

ifand only if p > 1.

EXAMPLE

23.7

(a) L

n~;

(b) L n

: 1;

,,2+ sinn

(a) This is a p-series with p =

> 1, so the series converges.

(b) Clearly,

(11

: I) :::;

-!r,

and L

is a convergent p-series . The given series therefore converges.

+ sin 11 is always greater than or equal to 1, so

(2+jW

n) ::::

In.

Since L

is a divergent p-series (p =

< 1), the given series diverges.

An easy way to make the kind of comparison in (b) and (c) is given in the following test.

LimitComparisonTest: If L

and L b; arepositive

series,

andlimn~oo

i- 0,

then L

and L b; both

converge,

or both

diverge.

The limit comparisontest works because if

--+

£ i- 0, then

and b; are roughly

multiples of each other for large

n. Specifically, for all large n,

must be close enough to

£ so that

and

2£. Hence, b

and

ui,

for all large n. The first

inequalityshowsthat if

converges, so that L

~an

converges, then L b; also converges.

The secondinequalityshowsthat if

L b; converges, then L

converges.

--+

0, then

b; for all large n and consequently L

converges if L b

converges, but not conversely. If

--+

00,

then b

for all large

nand

L b; converges

converges, but not conversely.