Improper Integrals - Integral Calculus, CSIR-NET Mathematical Sciences | Mathematics for IIT JAM, GATE, CSIR NET, UGC NET PDF Download

 Page 1


1 Improper Integrals
1.1 Introduction
InCalculusII,studentsde?nedtheintegral

b
a
f (x)dxovera?niteinterval[a,b].
The function f was assumed to be continuous, or at least bounded, otherwise
the integral was not guaranteed to exist. Assuming an antiderivative of f could
be found,

b
a
f (x)dx always existed, and was a number. In this section, we
investigate what happens when these conditions are not met.
De?nition 1 (Improper Integral) An integral is an improper integral if ei-
ther the interval of integration is not ?nite (improper integral of type 1) or
if the function to integrate is not continuous (not bounded) in the interval of
integration (improper integral of type 2).
Example 2

8
0
e
-x
dx is an improper integral of type 1 since the upper limit
of integration is in?nite.
Example 3

1
0
dx
x
is an improper integral of type 2 because
1
x
is not continu-
ous at 0.
Example 4

8
0
dx
x-1
is an improper integral of types 1 since the upper limit
of integration is in?nite. It is also an improper integral of type 2 because
1
x-1
is not continuous at 1 and 1 is in the interval of integration.
Example 5

2
-2
dx
x
2
-1
is an improper integral of type 2 because
1
x
2
-1
is not
continuous at-1 and 1.
Page 2


1 Improper Integrals
1.1 Introduction
InCalculusII,studentsde?nedtheintegral

b
a
f (x)dxovera?niteinterval[a,b].
The function f was assumed to be continuous, or at least bounded, otherwise
the integral was not guaranteed to exist. Assuming an antiderivative of f could
be found,

b
a
f (x)dx always existed, and was a number. In this section, we
investigate what happens when these conditions are not met.
De?nition 1 (Improper Integral) An integral is an improper integral if ei-
ther the interval of integration is not ?nite (improper integral of type 1) or
if the function to integrate is not continuous (not bounded) in the interval of
integration (improper integral of type 2).
Example 2

8
0
e
-x
dx is an improper integral of type 1 since the upper limit
of integration is in?nite.
Example 3

1
0
dx
x
is an improper integral of type 2 because
1
x
is not continu-
ous at 0.
Example 4

8
0
dx
x-1
is an improper integral of types 1 since the upper limit
of integration is in?nite. It is also an improper integral of type 2 because
1
x-1
is not continuous at 1 and 1 is in the interval of integration.
Example 5

2
-2
dx
x
2
-1
is an improper integral of type 2 because
1
x
2
-1
is not
continuous at-1 and 1.
Example 6

p
0
tanxdx is an improper integral of type 2 because tanx is not
continuous at
p
2
.
We now look how to handle each type of improper integral.
1.2 Improper Integrals of Type 1
These are easy to identify. Simply look at the interval of integration. If either
the lower limit of integration, the upper limit of integration or both are not
?nite, it will be an improper integral of type 1.
De?nition 7 (improper integral of type 1) Improperintegralsoftype 1are
evaluated as follows:
1. If

t
a
f (x)dx exists for all t= a,thenwede?ne
8

a
f (x)dx = lim
t?8
t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

8
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

8
a
f (x)dx is said
to be divergent (or to diverge).
2. If

b
t
f (x)dx exists for all t= b,thenwede?ne
b

-8
f (x)dx = lim
t?-8
b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
-8
f (x)dx
is said to be convergent (or to converge). Otherwise,

b
-8
f (x)dx is
said to be divergent (or to diverge).
3. If both

8
a
f (x)dx and

b
-8
f (x)dx converge, then we de?ne
8

-8
f (x)dx =
a

-8
f (x)dx+
8

a
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
Page 3


1 Improper Integrals
1.1 Introduction
InCalculusII,studentsde?nedtheintegral

b
a
f (x)dxovera?niteinterval[a,b].
The function f was assumed to be continuous, or at least bounded, otherwise
the integral was not guaranteed to exist. Assuming an antiderivative of f could
be found,

b
a
f (x)dx always existed, and was a number. In this section, we
investigate what happens when these conditions are not met.
De?nition 1 (Improper Integral) An integral is an improper integral if ei-
ther the interval of integration is not ?nite (improper integral of type 1) or
if the function to integrate is not continuous (not bounded) in the interval of
integration (improper integral of type 2).
Example 2

8
0
e
-x
dx is an improper integral of type 1 since the upper limit
of integration is in?nite.
Example 3

1
0
dx
x
is an improper integral of type 2 because
1
x
is not continu-
ous at 0.
Example 4

8
0
dx
x-1
is an improper integral of types 1 since the upper limit
of integration is in?nite. It is also an improper integral of type 2 because
1
x-1
is not continuous at 1 and 1 is in the interval of integration.
Example 5

2
-2
dx
x
2
-1
is an improper integral of type 2 because
1
x
2
-1
is not
continuous at-1 and 1.
Example 6

p
0
tanxdx is an improper integral of type 2 because tanx is not
continuous at
p
2
.
We now look how to handle each type of improper integral.
1.2 Improper Integrals of Type 1
These are easy to identify. Simply look at the interval of integration. If either
the lower limit of integration, the upper limit of integration or both are not
?nite, it will be an improper integral of type 1.
De?nition 7 (improper integral of type 1) Improperintegralsoftype 1are
evaluated as follows:
1. If

t
a
f (x)dx exists for all t= a,thenwede?ne
8

a
f (x)dx = lim
t?8
t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

8
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

8
a
f (x)dx is said
to be divergent (or to diverge).
2. If

b
t
f (x)dx exists for all t= b,thenwede?ne
b

-8
f (x)dx = lim
t?-8
b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
-8
f (x)dx
is said to be convergent (or to converge). Otherwise,

b
-8
f (x)dx is
said to be divergent (or to diverge).
3. If both

8
a
f (x)dx and

b
-8
f (x)dx converge, then we de?ne
8

-8
f (x)dx =
a

-8
f (x)dx+
8

a
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
1.3 Improper Integrals of Type 2
These are more di?cult to identify. One needs to look at the interval of in-
tegration, and determine if the integrand is continuous or not in that interval.
Things to look for are fractions for which the denominator becomes 0 in the in-
terval of integration. Keep in mind that some functions do not contain fractions
explicitly like tanx, secx.
De?nition 8 (improper integral of type 2) Improper integrals of type 2
are evaluated as follows:
1. if f is continuous on [a,b) and not continuous at b then we de?ne
b

a
f (x)dx = lim
t?b
- t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

b
a
f (x)dx is said
to be divergent (or to diverge).
2. if f is continuous on (a,b] and not continuous at a then we de?ne
b

a
f (x)dx = lim
t?a
+ b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

b
a
f (x)dx is said
to be divergent (or to diverge).
3. If f is not continuous at c wherea<c<b and both

c
a
f (x)dx and

b
c
f (x)dx converge then we de?ne
b

a
f (x)dx =
c

a
f (x)dx+
b

c
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
We now look at some examples.
Page 4


1 Improper Integrals
1.1 Introduction
InCalculusII,studentsde?nedtheintegral

b
a
f (x)dxovera?niteinterval[a,b].
The function f was assumed to be continuous, or at least bounded, otherwise
the integral was not guaranteed to exist. Assuming an antiderivative of f could
be found,

b
a
f (x)dx always existed, and was a number. In this section, we
investigate what happens when these conditions are not met.
De?nition 1 (Improper Integral) An integral is an improper integral if ei-
ther the interval of integration is not ?nite (improper integral of type 1) or
if the function to integrate is not continuous (not bounded) in the interval of
integration (improper integral of type 2).
Example 2

8
0
e
-x
dx is an improper integral of type 1 since the upper limit
of integration is in?nite.
Example 3

1
0
dx
x
is an improper integral of type 2 because
1
x
is not continu-
ous at 0.
Example 4

8
0
dx
x-1
is an improper integral of types 1 since the upper limit
of integration is in?nite. It is also an improper integral of type 2 because
1
x-1
is not continuous at 1 and 1 is in the interval of integration.
Example 5

2
-2
dx
x
2
-1
is an improper integral of type 2 because
1
x
2
-1
is not
continuous at-1 and 1.
Example 6

p
0
tanxdx is an improper integral of type 2 because tanx is not
continuous at
p
2
.
We now look how to handle each type of improper integral.
1.2 Improper Integrals of Type 1
These are easy to identify. Simply look at the interval of integration. If either
the lower limit of integration, the upper limit of integration or both are not
?nite, it will be an improper integral of type 1.
De?nition 7 (improper integral of type 1) Improperintegralsoftype 1are
evaluated as follows:
1. If

t
a
f (x)dx exists for all t= a,thenwede?ne
8

a
f (x)dx = lim
t?8
t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

8
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

8
a
f (x)dx is said
to be divergent (or to diverge).
2. If

b
t
f (x)dx exists for all t= b,thenwede?ne
b

-8
f (x)dx = lim
t?-8
b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
-8
f (x)dx
is said to be convergent (or to converge). Otherwise,

b
-8
f (x)dx is
said to be divergent (or to diverge).
3. If both

8
a
f (x)dx and

b
-8
f (x)dx converge, then we de?ne
8

-8
f (x)dx =
a

-8
f (x)dx+
8

a
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
1.3 Improper Integrals of Type 2
These are more di?cult to identify. One needs to look at the interval of in-
tegration, and determine if the integrand is continuous or not in that interval.
Things to look for are fractions for which the denominator becomes 0 in the in-
terval of integration. Keep in mind that some functions do not contain fractions
explicitly like tanx, secx.
De?nition 8 (improper integral of type 2) Improper integrals of type 2
are evaluated as follows:
1. if f is continuous on [a,b) and not continuous at b then we de?ne
b

a
f (x)dx = lim
t?b
- t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

b
a
f (x)dx is said
to be divergent (or to diverge).
2. if f is continuous on (a,b] and not continuous at a then we de?ne
b

a
f (x)dx = lim
t?a
+ b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

b
a
f (x)dx is said
to be divergent (or to diverge).
3. If f is not continuous at c wherea<c<b and both

c
a
f (x)dx and

b
c
f (x)dx converge then we de?ne
b

a
f (x)dx =
c

a
f (x)dx+
b

c
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
We now look at some examples.
1.4 Examples
• Evaluating an improper integral is really two problems. It is an integral
problem and a limit problem. It is best to do them separately.
• When breaking down an improper integral to evaluate it, make sure that
each integral is improper at only one place, that place should be either
the lower limit of integration, or the upper limit of integration.
Example 9

8
1
dx
x
2
This is an improper integral of type 1. We evaluate it by ?nding lim
t?8

t
1
dx
x
2
.
First,

t
1
dx
x
2
=

1-
1
t

and lim
t?8

1-
1
t

=1 hence

8
1
dx
x
2
=1.
Example 10

8
1
dx
x
This is an improper integral of type 1. We evaluate it by ?nding lim
t?8

t
1
dx
x
First,

t
1
dx
x
=lnt and lim
t?8
lnt =8 hence

8
1
dx
x
diverges.
Example 11

8
-8
dx
1+x
2
This is an improper integral of type 1. Since both limits of integration are
in?nite, we break it into two integrals.

8
-8
dx
1+x
2
=

0
-8
dx
1+x
2
+

8
0
dx
1+x
2
Note that since the function
1
1+x
2
is even, we have

0
-8
dx
1+x
2
=

8
0
dx
1+x
2
;
we only need to do

8
0
dx
1+x
2
.

8
0
dx
1+x
2
= lim
t?8

t
0
dx
1+x
2
and

t
0
dx
1+x
2
=tan
-1
x


t
0
=tan
-1
t-tan
-1
0
=tan
-1
t
Page 5


1 Improper Integrals
1.1 Introduction
InCalculusII,studentsde?nedtheintegral

b
a
f (x)dxovera?niteinterval[a,b].
The function f was assumed to be continuous, or at least bounded, otherwise
the integral was not guaranteed to exist. Assuming an antiderivative of f could
be found,

b
a
f (x)dx always existed, and was a number. In this section, we
investigate what happens when these conditions are not met.
De?nition 1 (Improper Integral) An integral is an improper integral if ei-
ther the interval of integration is not ?nite (improper integral of type 1) or
if the function to integrate is not continuous (not bounded) in the interval of
integration (improper integral of type 2).
Example 2

8
0
e
-x
dx is an improper integral of type 1 since the upper limit
of integration is in?nite.
Example 3

1
0
dx
x
is an improper integral of type 2 because
1
x
is not continu-
ous at 0.
Example 4

8
0
dx
x-1
is an improper integral of types 1 since the upper limit
of integration is in?nite. It is also an improper integral of type 2 because
1
x-1
is not continuous at 1 and 1 is in the interval of integration.
Example 5

2
-2
dx
x
2
-1
is an improper integral of type 2 because
1
x
2
-1
is not
continuous at-1 and 1.
Example 6

p
0
tanxdx is an improper integral of type 2 because tanx is not
continuous at
p
2
.
We now look how to handle each type of improper integral.
1.2 Improper Integrals of Type 1
These are easy to identify. Simply look at the interval of integration. If either
the lower limit of integration, the upper limit of integration or both are not
?nite, it will be an improper integral of type 1.
De?nition 7 (improper integral of type 1) Improperintegralsoftype 1are
evaluated as follows:
1. If

t
a
f (x)dx exists for all t= a,thenwede?ne
8

a
f (x)dx = lim
t?8
t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

8
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

8
a
f (x)dx is said
to be divergent (or to diverge).
2. If

b
t
f (x)dx exists for all t= b,thenwede?ne
b

-8
f (x)dx = lim
t?-8
b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
-8
f (x)dx
is said to be convergent (or to converge). Otherwise,

b
-8
f (x)dx is
said to be divergent (or to diverge).
3. If both

8
a
f (x)dx and

b
-8
f (x)dx converge, then we de?ne
8

-8
f (x)dx =
a

-8
f (x)dx+
8

a
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
1.3 Improper Integrals of Type 2
These are more di?cult to identify. One needs to look at the interval of in-
tegration, and determine if the integrand is continuous or not in that interval.
Things to look for are fractions for which the denominator becomes 0 in the in-
terval of integration. Keep in mind that some functions do not contain fractions
explicitly like tanx, secx.
De?nition 8 (improper integral of type 2) Improper integrals of type 2
are evaluated as follows:
1. if f is continuous on [a,b) and not continuous at b then we de?ne
b

a
f (x)dx = lim
t?b
- t

a
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

b
a
f (x)dx is said
to be divergent (or to diverge).
2. if f is continuous on (a,b] and not continuous at a then we de?ne
b

a
f (x)dx = lim
t?a
+ b

t
f (x)dx
provided the limit exists as a ?nite number. In this case,

b
a
f (x)dx is
said to be convergent (or to converge). Otherwise,

b
a
f (x)dx is said
to be divergent (or to diverge).
3. If f is not continuous at c wherea<c<b and both

c
a
f (x)dx and

b
c
f (x)dx converge then we de?ne
b

a
f (x)dx =
c

a
f (x)dx+
b

c
f (x)dx
The integrals on the right are evaluated as shown in 1. and 2..
We now look at some examples.
1.4 Examples
• Evaluating an improper integral is really two problems. It is an integral
problem and a limit problem. It is best to do them separately.
• When breaking down an improper integral to evaluate it, make sure that
each integral is improper at only one place, that place should be either
the lower limit of integration, or the upper limit of integration.
Example 9

8
1
dx
x
2
This is an improper integral of type 1. We evaluate it by ?nding lim
t?8

t
1
dx
x
2
.
First,

t
1
dx
x
2
=

1-
1
t

and lim
t?8

1-
1
t

=1 hence

8
1
dx
x
2
=1.
Example 10

8
1
dx
x
This is an improper integral of type 1. We evaluate it by ?nding lim
t?8

t
1
dx
x
First,

t
1
dx
x
=lnt and lim
t?8
lnt =8 hence

8
1
dx
x
diverges.
Example 11

8
-8
dx
1+x
2
This is an improper integral of type 1. Since both limits of integration are
in?nite, we break it into two integrals.

8
-8
dx
1+x
2
=

0
-8
dx
1+x
2
+

8
0
dx
1+x
2
Note that since the function
1
1+x
2
is even, we have

0
-8
dx
1+x
2
=

8
0
dx
1+x
2
;
we only need to do

8
0
dx
1+x
2
.

8
0
dx
1+x
2
= lim
t?8

t
0
dx
1+x
2
and

t
0
dx
1+x
2
=tan
-1
x


t
0
=tan
-1
t-tan
-1
0
=tan
-1
t
Thus

8
0
dx
1+x
2
= lim
t?8

tan
-1
t

=
p
2
It follows that

8
-8
dx
1+x
2
=
p
2
+
p
2
= p
Example 12
 p 2 0
secxdx
This is an improper integral of type 2 because secx is not continuous at
p
2
.We
evaluate it by ?nding lim
t?
p 2 - 
t
0
secxdx.
First,
t

0
secxdx=ln|secx+tanx||
t
0
=ln|sect+tant|
and lim
t?
p 2 - (ln|sect+tant|)=8 hence
 p 2 0
secxdx diverges.
Example 13

p
0
sec
2
xdx
This is an improper integral of type 2, sec
2
x is not continuous at
p
2
.Thus,

p
0
sec
2
xdx =
 p 2 0
sec
2
xdx+

p
p 2 sec
2
xdx
First, we evaluate
 p 2 0
sec
2
xdx.
 p 2 0
sec
2
xdx=lim
t?
p 2 - 
t
0
sec
2
xdx

t
0
sec
2
xdx=tant-tan0
=tant
Therefore,
 p 2 0
sec
2
xdx= lim
t?
p 2 - (tant)
=
It follows that
 p 2 0
sec
2
xdx diverges, therefore,

p
0
sec
2
xdx also diverges.