1

dx.

x(1 ’ x)1/2

0

SOLUTION

The key idea is that we can only handle one singularity at a time. This

integrand is singular at both endpoints 0 and 1. Therefore we divide the domain

of integration somewhere in the middle”at 1/2 say (it does not really matter

where we divide)”and treat the two singularities separately.

First we treat the integral

1/2 1

dx.

x(1 ’ x)1/2

0

Since the integrand has a singularity at 0, we consider

1/2

Y

1

lim dx.

x(1 ’ x)1/2

’0+

FL

This is a tricky integral to evaluate directly. But notice that

1 1

AM

≥

x(1 ’ x)1/2 x · (1)1/2

¤ x ¤ 1/2. Thus

when 0 <

TE

1/2 1/2 1/2

1 1 1

dx ≥ dx = dx.

x(1 ’ x)1/2 x · (1)1/2 x

We evaluate the integral: it equals ln(1/2) ’ ln . Finally,

lim ’ ln = +∞.

’0+

The ¬rst of our integrals therefore diverges.

But the full integral

1 1

dx

x(1 ’ x)1/2

0

converges if and only if each of the component integrals

1/2 1

dx

x(1 ’ x)1/2

0

and

1 1

dx

x(1 ’ x)1/2

1/2

139

CHAPTER 5 Indeterminate Forms

converges. Since the ¬rst integral diverges, we conclude that the original

integral diverges as well.

’1/3 dx

3

You Try It: Calculate as an improper integral.

’2 (2x)

5.3.3 AN APPLICATION TO AREA

Suppose that f is a non-negative, continuous function on the interval (a, b] which

is unbounded as x ’ a + . Look at Fig. 5.5. Let us consider the area under the graph

of f and above the x-axis over the interval (a, b]. The area of the part of the region

over the interval [a + , b], > 0, is

b

f (x) dx.

a+

Fig. 5.5

Therefore it is natural to consider the area of the entire region, over the interval

(a, b], to be

b

lim f (x) dx.

’0+ a+

This is just the improper integral

b

Area = f (x) dx.

a

140 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.20

Calculate the area above the x-axis and under the curve

1

y= 0 < x ¤ 1/2.

,

x · ln4/3 x

SOLUTION

According to the preceding discussion, this area is equal to the value of the

improper integral

1/2 1/2

1 1

dx = lim dx.

x · ln4/3 x x · ln4/3 x

’0+

0

For clarity we let •(x) = ln x, • (x) = 1/x. Then the (inde¬nite) integral

becomes

3

• (x)

dx = ’ 1/3 .

• 4/3 (x) • (x)

Thus

1/2

1/2 1 3

dx = lim ’

lim

x · ln4/3 x ln1/3 x

’0+ ’0+

’3 ’3

= lim ’ .

[’ ln 2]1/3 [ln ]1/3

’0+

Now as ’ 0 then ln ’ ’∞ hence 1/[ln ]1/3 ’ 0. We conclude that our

improper integral converges and the area under the curve and above the x-axis

equals 3/[ln 2]1/3 .

5.4 More on Improper Integrals

5.4.1 INTRODUCTION

Suppose that we want to calculate the integral of a continuous function f (x) over

an unbounded interval of the form [A, +∞) or (’∞, B]. The theory of the integral

that we learned earlier does not cover this situation, and some new concepts are

needed. We treat improper integrals on in¬nite intervals in this section, and give

some applications at the end.

141

CHAPTER 5 Indeterminate Forms

5.4.2 THE INTEGRAL ON AN INFINITE INTERVAL

Let f be a continuous function whose domain contains an interval of the form

[A, +∞). The value of the improper integral

+∞

f (x) dx

A

is de¬ned to be

N

lim f (x) dx.

N ’+∞ A

Similarly, we have: Let g be a continuous function whose domain contains an

interval of the form (’∞, B]. The value of the improper integral

B

g(x) dx

’∞

is de¬ned to be

B

lim f (x) dx.

M’’∞ M

EXAMPLE 5.21

Calculate the improper integral

+∞

x ’3 dx.

1

SOLUTION

We do this problem by evaluating the limit

N N

x ’3 dx = ’(1/2)x ’2

lim lim

N ’+∞ 1 1

N’+∞

lim ’(1/2) N ’2 ’ 1’2

=

N’+∞

1

=.

2

We conclude that the integral converges and has value 1/2.

EXAMPLE 5.22

Evaluate the improper integral

’32

x ’1/5 dx.

’∞

142 CHAPTER 5 Indeterminate Forms

SOLUTION

We do this problem by evaluating the limit

5 4/5 ’32

’32

’1/5

dx = lim

lim x x

M’’∞ 4

M’’∞ M M

5

= lim (’32)4/5 ’ M 4/5

M’’∞ 4

5

= lim 16 ’ M 4/5 .

M’’∞ 4

This limit equals ’∞. Therefore the integral diverges.

∞ ’3 dx.

1 (1 + x)

You Try It: Evaluate

Sometimes we have occasion to evaluate a doubly in¬nite integral. We do so by

breaking the integral up into two separate improper integrals, each of which can be

evaluated with just one limit.

EXAMPLE 5.23

Evaluate the improper integral

∞ 1

dx.

’∞ 1 + x

2

SOLUTION

The interval of integration is (’∞, +∞). To evaluate this integral, we break

the interval up into two pieces:

(’∞, +∞) = (’∞, 0] ∪ [0, +∞).

(The choice of zero as a place to break the interval is not important; any

other point would do in this example.) Thus we will evaluate separately the

integrals

+∞ 0

1 1

dx and dx.

1 + x2 ’∞ 1 + x

2

0

For the ¬rst one we consider the limit

N 1 N

dx = lim Tan’1 x

lim

1+x 2

N ’+∞ 0 N ’+∞ 0

Tan’1 N ’ Tan’1 0

= lim

N ’+∞

π

= .

2

143

CHAPTER 5 Indeterminate Forms

The second integral is evaluated similarly:

0 1 0

dx = lim Tan’1 x

lim

1+x 2

M’’∞ M M’’∞ M

Tan’1 0 ’ Tan’1 M

= lim

M’’∞

π

= .

2

Since each of the integrals on the half line is convergent, we conclude that the

original improper integral over the entire real line is convergent and that its

value is

π π

+ = π.

2 2

∞ ’1 dx.

1 (1 + x)

You Try It: Discuss

5.4.3 SOME APPLICATIONS

Now we may use improper integrals over in¬nite intervals to calculate area.

EXAMPLE 5.24

Calculate the area under the curve y = 1/[x · (ln x)4 ] and above the x-axis,

2 ¤ x < ∞.

SOLUTION

The area is given by the improper integral

+∞ N

1 1

dx = lim dx.

x · (ln x)4 x · (ln x)4

N ’+∞

2 2

For clarity, we let •(x) = ln x, • (x) = 1/x. Thus the (inde¬nite) integral

becomes

1/3

• (x)

dx = ’ 3 .

• 4 (x) • (x)

Thus

1/3 N

N 1

dx = lim ’ 3

lim

x · (ln x)4

N ’+∞ 2 ln x 2

N’+∞

1/3 1/3

= lim ’ ’3

ln3 N ln 2

N’+∞

1/3

= 3.

ln 2

Thus the area under the curve and above the x-axis is 1/(3 ln3 2).

144 CHAPTER 5 Indeterminate Forms

EXAMPLE 5.25

Because of in¬‚ation, the value of a dollar decreases as time goes on. Indeed,

this decrease in the value of money is directly related to the continuous

compounding of interest. For if one dollar today is invested at 6% contin-

uously compounded interest for ten years then that dollar will have grown

to e0.06·10 = $1.82 (see Section 6.5 for more detail on this matter). This

means that a dollar in the currency of ten years from now corresponds to

only e’0.06·10 = $0.55 in today™s currency.

Now suppose that a trust is established in your name which pays 2t + 50

dollars per year for every year in perpetuity, where t is time measured in

years (here the present corresponds to time t = 0). Assume a constant

interest rate of 6%, and that all interest is re-invested. What is the total

value, in today™s dollars, of all the money that will ever be earned by your

trust account?

SOLUTION

Over a short time increment [tj ’1 , tj ], the value in today™s currency of the

money earned is about

(2tj + 50) · e’0.06·tj · tj .

The corresponding sum over time increments is

(2tj + 50) · e’0.06·tj tj .

j

This in turn is a Riemann sum for the integral

(2t + 50)e’0.06t dt.

If we want to calculate the value in today™s dollars of all the money earned from

now on, in perpetuity, this would be the value of the improper integral

+∞

(2t + 50)e’0.06t dt.

0

This value is easily calculated to be $1388.89, rounded to the nearest cent.

You Try It: A trust is established in your name which pays t + 10 dollars per year

for every year in perpetuity, where t is time measured in years (here the present

corresponds to time t = 0). Assume a constant interest rate of 4%. What is the total

value, in today™s dollars, of all the money that will ever be earned by your trust

account?

145

CHAPTER 5 Indeterminate Forms

Exercises

1. If possible, use l™Hôpital™s Rule to evaluate each of the following limits. In

each case, check carefully that the hypotheses of l™Hôpital™s Rule apply.

cos x ’ 1

(a) lim

x’0 x 2 ’ x 3

e2x ’ 1 ’ 2x

(b) lim

x2 + x4

x’0

cos x

(c) lim

x’0 x 2

[ln x]2

(d) lim

x’1 (x ’ 1)

(x ’ 2)3

(e) lim

x’2 sin(x ’ 2) ’ (x ’ 2)

ex ’ 1

(f ) lim

x’1 x ’ 1

2. If possible, use l™Hôpital™s Rule to evaluate each of the following limits. In

each case, check carefully that the hypotheses of l™Hôpital™s Rule apply.

x3

(a) lim

x’+∞ ex ’ x 2

ln x

(b) lim

x’+∞ x

e’x

(c) lim

x’+∞ ln[x/(x + 1)]

sin x

(d) lim

x’+∞ e’x

ex

(e) lim

x’’∞ 1/x