8.4

ImproperIntegrals

ln 2

0

tan−1 x + C

−x−1 + C

(-2,2)

(4,∞)

Because -1≤ cos x ≤1 for all x

limx→ ∞

4ex +55ex −3

= 45

Until now we have been finding integrals of continuous functions over closed intervals.

Sometimes we can find integrals for functions where the function or the limits are infinite. These are called improper integrals.

→

Example 1:

1

0

1

1

xdx

x

+−∫ The function is

undefined at x = 1 .

0

1

2

3

4

1

Since x = 1 is an asymptote, the function has no maximum.

Can we find the area under an infinitely high curve?

We could define this integral as:

01

1lim

1

b

b

xdx

x−→

+−∫

(left hand limit)

We must approach the limit from inside the interval.

→

01

1lim

1

b

b

xdx

x−→

+−∫

1

1

1

1

x

x

xdx

x

+++−∫ Rationalize the numerator.

2

1+x

1dx

x−∫

2 2

1 x

1 1dx dx

x x+

− −∫ ∫

21u x= −

2 du x dx=−1

2

du x dx− =1

1 21

sin2

x u du−− − ∫

→

2 2

1 x

1 1dx dx

x x+

− −∫ ∫

21u x= −

2 du x dx=−1

2

du x dx− =1

1 21

sin2

x u du−− − ∫

11 2sin x u− −

1 2

1 0lim sin 1

b

bx x

−

−

→− −

( ) ( )1 2 1

1lim sin 1 sin 0 1b

b b−

− −

→− − − − 1

2

π= +

2

π0 0

This integral converges because it approaches a solution.

→

Example 2:

-1

0

1

2

3

4

-1 1

1

0

dx

x∫

1

0lim ln

bbx

+→

0lim ln1 lnb

b+→

−

0

1lim lnb b+→

=∞ This integral diverges.

(right hand limit)

We approach the limit from inside the interval.

1

0

1lim

bbdx

x+→ ∫

→

Example 3:

( )

3

2031

dx

x −∫

The function approacheswhen .

∞1x →

0

1

2

3

4

1 2 3

( )233

01 x dx

−−∫

( ) ( )2 233 3

01 1lim 1 lim 1

b

cb cx dx x dx

− +

− −

→ →− + −∫ ∫

( ) ( )31 1

3 31 1

0

lim 3 1 lim 3 1b

b cc

x x− +→ →

− + −

→

( ) ( )2 233 3

01 1lim 1 lim 1

b

cb cx dx x dx

− +

− −

→ →− + −∫ ∫

( ) ( )31 1

3 31 1

0

lim 3 1 lim 3 1b

b cc

x x− +→ →

− + −

( ) ( ) ( )11 1 133 3 3

1 1lim 3 1 3 1 lim 3 2 3 1b c

b c− +→ →

⎡ ⎤⎡ ⎤− − − + ⋅ − −⎢ ⎥⎢ ⎥⎣ ⎦ ⎣ ⎦

0 0

33 3 2+

→

Example 4:

1 P

dx

x

∞

∫ 0P >

1 Px dx

∞ −∫

1lim

b P

bx dx−

→∞∫

1

1

1lim

1

bP

bx

P− +

→ ∞ − +

1 11lim

1 1

P P

b

b

P P

− + − +

→ ∞−

− + − +

What happens here?

If then gets bigger and bigger as , therefore the integral diverges.

1P ≤ 1Pb− +

b → ∞

If then b has a negative exponent and ,therefore the integral converges.

1P >1 0Pb− + →

(P is a constant.)

π

8.4 Tests for Convergence

1

xe dx∞ −∫

1lim

b x

be dx−

→∞∫

1lim

bx

be−

→∞−

( )1lim b

be e− −

→∞− − −

1 1lim

bb e e→∞− +

0 1

e=

Converges

→

Does converge?2

1

xe dx∞ −∫

Compare:

to for positive values of x.2

1xe

1xe

For2

2x

1 11,

ex x

xx e e

e> > ∴ <

→

0

1

1 2 3

2

1xe

1xe

For2

2x

1 11,

ex x

xx e e

e> > ∴ <

Since is always below , we say that it is

“bounded above” by .

2

1xe

1xe1

xe

Since converges to a finite number, must also converge!1

xe2

1xe →

Direct Comparison Test:

Let f and g be continuous on with

for all , then:

[ ),a ∞ ( ) ( )0 f x g x≤ ≤

x a≥

2 ( ) a

f x dx∞

∫diverges if diverges.( ) a

g x dx∞

∫

1 ( ) a

f x dx∞

∫ converges if converges.( ) a

g x dx∞

∫

→

Example 7:

2

21

sin

xdx

x

∞

∫The maximum value of so:sin 1x =

[ )2

2 2

sin 10 on 1,

x

x x≤ ≤ ∞on

Since converges, converges.2

1

x

2

2

sin x

x

→

Example 7:

21

1

0.1dx

x

∞

−∫

for positive values of x, so:2 0.1x x− <

Since diverges, diverges.1

x 2

1

0.1x −

2

1 1

0.1 xx≥

−on [ )1,∞

→

If functions grow at the same rate, then either they both converge or both diverge.

Does converge?21 1

dx

x

∞

+∫

As the “1” in the denominator becomes

insignificant, so we compare to .

x → ∞

2

1

x

2

2

1

lim1

1

x

x

x

→∞

+

2

2

1limx

x

x→∞

+=

2lim

2x

x

x→∞= 1= Since converges,

converges.

2

1

x2

1

1 x+ →

21 1

dx

x

∞

+∫Of course

21lim

1

b

b

dx

x→∞=

+∫1

1lim tan

b

bx−

→∞=

1 1lim tan tan 1b

b− −

→∞= −

2 4

π π= −

4

π=

tany x=

As , 2

y xπ

→ ∞ →

2

π

2

π4

π4

π

→

21 1

dx

x

∞

+∫

21lim

1

b

b

dx

x→∞=

+∫1

1lim tan

b

bx−

→∞=

1 1lim tan tan 1b

b− −

→∞= −

2 4

π π= −

4

π=

Of course 21

1 dx

x

∞

∫2

1lim

b

bx dx−

→∞= ∫

1

1lim

b

bx−

→∞= −

1 1lim

1b b→∞

⎛ ⎞= − − −⎜ ⎟⎝ ⎠

1=

π

0

QuickTime™ and a decompressor

are needed to see this picture.

Gabriel’s Horn:

Created by revolving y = 1

x, 1≤ x ≤∞, around x-axis.

Find the volume and the surface area.

QuickTime™ and a decompressor

are needed to see this picture.

Gabriel’s Horn: y = 1

x, 1≤ x ≤∞, around x-axis.

Find the volume and the surface area.

Volume = limb→ ∞

πx2

1

b

∫ dx = limb→ ∞

-πx 1

b

= 0 + π = π

Surface Area = limb→ ∞

2π 1 + -1x2

⎛⎝⎜

⎞⎠⎟

2

1

b

∫ dx = ∞

It’s a solid with infinite surface area wrapped around a finite volume!!

Evaluate the improper integral or state that it diverges.

41

dx

x

∞

∫

Evaluate the improper integral or state that it diverges.

b

4 4b1 1

b

3b1

3b

dx dx = lim

x x

1 = lim -

3x

-1 1 1 = lim + =

3b 3 3

∞

→ ∞

→ ∞

→∞

∫ ∫

Evaluate the improper integral:

-2x

1

x e dx∞

∫

Evaluate the improper integral:

b-2x -2x -2x

b1 1

-2x

b-2x -2x

b1

x e dx = lim x e dx u = x dv = e dx

-1 du = dx v = e

2

-x 1 = lim e + e dx

2 2

∞

→ ∞

→ ∞

⎛ ⎞⎜ ⎟⎝ ⎠

∫ ∫

∫b

b-2x -2x -2x

1b b1

-2b 2

b

-2

-x 1 -x 1 = lim e - e = lim - e

2 4 2 4

-b 1 -1 1 = lim - e - - e

2 4 2 43

= e4

→∞ →∞

−

→∞

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Evaluating an integral on ( , )−∞ ∞Evaluate the improper integral.

xe dx∞

−

−∞∫

Evaluate the improper integral.

( ) ( )

0x x -x

- 0

0 ax -x

a - aa 0

a -a

a - a

e dx = e dx + e dx

= lim e dx + lim e dx

= lim 1 - e + lim -e + 1

= 1 + 1 = 2

∞ ∞−

−∞ ∞

→ ∞ →∞

→ ∞ →∞

∫ ∫ ∫

∫ ∫

Evaluating an integral on ( , )−∞ ∞

Interior Infinite Discontinuity

State why the integral is improper. Then evaluate the integral or state that it diverges.

2

20

dx

1 x−∫

Interior Infinite Discontinuity2

20

1 2

2 20 1

c 2

2 2c 1- c 10 c

c 2

c 1- c 1+0 c

dx has an infinite discontinuity at x = 1

1 x

dx dx= +

1 x 1 x

dx dx lim + lim

1 x 1 x

1 x - 1 1 x - 1 lim ln + lim ln

2 x + 1 2 x + 1

, t

→ → +

→ →

−

− −

=− −

⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞= − −⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠

= ∞

∫

∫ ∫

∫ ∫

he integral diverges