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Improper Integrals
Chapter 7.8
April 10, 2007
Improper Integrals
(We defined them in Day 6 materials)


An integral having at least one nonfinite limit or an integrand
that becomes infinite between the limits of integration.

Interval is infinite (easiest to identify)
 f (x)dx
1

Function “Blows” up! (down)
1
1
1 x 2 dx
Which of the following integrals are improper?
A.
B.
C.

0

9

1
1
3
4
r 2  3r  2
ln  3s  8  ds
2
tan t  1 dt
dr
Look where the
denominator is zero
Both -1 and -2 are in
the interval [-3,0] so
the integral is
IMPROPER!
Look where 3s - 8 is
zero
8/3 = 2 2/3 is NOT in
the interval [4,9] so the
integral is proper.
Look where the angle
“t+1” = π/2 +multiples of π
π/2 - 1 is about .57 and is
in the interval [-2,1] so
the integral is
IMPROPER!

Let’s look at the Improper Integral:


dx
1 x 2
The “problem” is that we can’t apply the Fundamental Theorem
of Calculus because we don’t have a finite interval.
So let’s get an idea of what is happening by using a finite
100
interval: 100 dx
1
1
99



1

1 x 2
x 1 100
100
1000

1
dx
x2
100,000

1
1
1
999


1 
x 1
1000
1000
1000
dx 1

2
x
x 1
100,000

1
99,999
1 
100,000
100,000

Let’s look at the Improper Integral:

Instead of continuing with different numbers,
let’s use a variable “n” and let “n” take on the different values.
n
dx
1 x 2

1
1


1
x 1
n
n
Look at the limit as n goes to infinity:
 1 
lim   1  0  1  1
n   n


dx
1 x 2
We say the integral converges to 1

Compare that with Improper Integral:
2
x
 dx
1

Using the same technique as in the previous example,
let’s use the variable “n” so that [1,n] is finite.
n
3 n
3
x
2
n
1
x
dx



1
3 1 3 3
Looking at the limit as n goes to infinity:

 n3 1 
lim     
n 
 3 3
We say the integral diverges.

Think about both integrals in terms of their area:

dx
1 x 2

2
x
 dx
1

Compare the Improper Integrals:

1
1 x 2 dx

and
n

1
1
2
1
x
dx 
1
x
2
 2x
1 n
2
1
1
Looking at the limit as n goes to infinity:
lim  2 n  2   
n

1
We already know the first converges to 1, what about the
second? Again, let the variable “n” replace the infinity so that the
interval [1,n] is finite.
n


1
dx
x
We say the integral diverges.
2 n 2
Think about both integrals in terms of their area:

dx
1 x 2


1
1
dx
x

What about

1
1 x dx
Let the variable “n” be so that the interval [1,n] is finite.
n
1
1 x dx  ln x

n
1
 ln n  ln 1
Looking at the limit as n goes to infinity:
lim ln n  
n

We say the integral diverges.
Compare the three graphs:

What about


1
1
dx
3
x
Let the variable “n” be so that the interval [1,n] is finite.
n

1


1
dx   1
3
x
2x 2
n
1

1
1

2n 2 2
Looking at the limit as n goes to infinity:
1
1
 1
lim   2   0 
n   2n
2
2
1
We say the integral converges to .
2
Other Examples:


1
dx
1  x2
1

0

cos(x) dx
ex
 ex  1 dx
Converges to π/4
diverges



1
ln x
dx
x2
diverges
Need better way to take limits….
Next time L’Hopital’s (4.4)
Method for Improper Integrals:

Define:



b
a



t
f (x) dx  lim  f (x) dx
t 
a
b
f (x) dx  lim  f (x) dx
t  t
Evaluate the Integral and then evaluate the limit
If there is an infinite integral in both directions, then we define
the integral to be:



a


a
f (x) dx   f (x) dx   f (x) dx
In groups,determine if the integral is convergent
or divergent. If convergent, evaluate.

1
0 z 2  3z  2 dz
n
1
0 z 2  3z  2 dz
Replace the infinity with n so that the
interval [0,n] is finite.
1 
 1
 

 dz

z  2 z 1
0
n
To integrate, we will use partial fractions:
1
A
B


z 2  3z  2 z  2 z  1
1  A z  1  B z  2 
when z  1 we B  1
when z  2 we A  1
   ln z  2  ln z  1
z 1
 ln
z2
n
 ln
0
n
0
n 1
1
 ln
n2
2
Look at the limit as n goes to infinity
1
1
 n 1
lim  ln
 ln   ln1  ln  0  ln1  ln 2   ln 2
n 
n2
2
2
We say the integral converges to ln2