Improper Integrals - Part II

Improper Integrals

In the definition of the definite integral

( )b

f x dxa∫

Two things are essential - without them the integral cannot exist - they are

1. The interval of integration is a finite interval [a, b] (the definition would also work for (a, b), (a, b], or [a, b) ).

2. The function f is bounded on the interval.

If either of these conditions is violated, then the integral is not defined, and we call it an improper integral. Some improper integrals can be given meaning and some cannot. We will investigate first the possible meaning of integrals that violatecondition 2, that is integrals of functions over intervals on which the functions become infinite.

We begin with the case of a finite interval. Then we will look at cases where both of the integrability conditions are violated.

Consider the following integral.1

0

dxx∫

The function becomes infinite in this interval, as shown in the following picture.

As before, we may consider this expression to represent the area under the curve from 0 to 1, as shown below..

Again, the key to giving meaning to this expression is the concept of limit. Consider the expression

1dxxl

∫

where l is a variable that is assumed to be strictly between 0 and 1

l

Definition. The improper integral of f over (a, b], where fbecomes unbounded as we approach a, is defined as

( ) lim ( )b b

f x dx f x dxl aa l

+=∫ ∫

→

This leads us to the following definition.

Example. Evaluate 2

0

dxx∫

Solution. First compute where l is a variable point

greater than 0 and less than 2.

2dxxl

∫

Thus2

lim ln(2) ln( )00

dx lx l +

= − =∞∫→

This integral diverges to infinity. Thus the area under this curve is not finite but infinite.

22ln| | ln(2) ln( )dx x l

x ll= = −∫

Example. Evaluate 3

0

dxx∫

The graph shows a singularity at 0, where the function becomes infinite as we approach from the right.

Solution. First compute where l is a variable point

greater than 0 and less than 3.

2 dxxl

∫

Thus3

lim 2 3 2 2 300

dx lx l +

= − =∫→

332 2 3 2dx x l

x ll= = −∫

This integral converges to . In this case we can also say that the area under the curve is or that the value of the integral is .

2 32 3

2 3

The place where the integrand becomes infinite may also be at the right endpoint of the interval of integration. We then modify the approach in the obvious way and get the following definition.

Definition. The improper integral of f over [a, b), where fbecomes unbounded as we approach b, is defined as

( ) lim ( )b l

f x dx f x dxl ba a

−=∫ ∫

→

Example. Evaluate 23

2

0(2 )

dx

x∫

−

The graph clearly shows a singularity at 2, where the function becomes infinite as we approach from the left.

Solution. First compute where l is a variable point

greater than 0 and less than 2.

230(2 )

l dx

x∫

−

2 1 13 3 3

23

( ) 3 3(2 )(2 )

dx u du u xx

−= − =− =− −∫ ∫

−

We make the substitution u = 2 − x, du = −dx.

Then

so1 13 3

23

33(2 ) 3 2 3(2 )0(2 ) 0

ll dx x lx

=− − = − −∫

−

13

23

2 3 3lim 3 2 3(2 ) 3 220(2 )

dx llx

−= − − =∫

→−

Example. Evaluate 22(4 )2

dxx

∫−−

Solution. The graph of this function over the range in questionis shown below.

It is clear that this function becomes infinite at both ends of the interval of integration, and so is improper for two different reasons.

Fundamental Principle. If an improper integral is improper at two or more places, always break it into parts, each of which contains only one of the places at which it is improper.

Thus, when we had an integral we broke it into two

parts, one from minus infinity to an intermediate point c, and one from c to infinity.

( ) ( ) ( )c

f x dx f x dx f x dxc

∞ ∞= +∫ ∫ ∫

−∞ −∞

( )f x dx∞∫

−∞

We must now do the same thing with the current integral. We write 2 0 2

2 2 2(4 ) (4 ) (4 )2 2 0

dx dx dxx x x

= +∫ ∫ ∫− − −− −

2 0 22 2 2(4 ) (4 ) (4 )2 2 0

dx dx dxx x x

= +∫ ∫ ∫− − −− −

Each of the integrals in the right hand side of

has only one place at which it is improper.

1 1 ( 2) ( 2)2 ( 2)( 2) ( 2) ( 2) ( 2)( 2)(4 )

A B A x B xx x x x x xx

− + + −= = + =− + − + − +−

Let us first find the indefinite integral of the above expression.

so ( 2) ( 2) 1A x B x+ + − =− Substituting x = 2 and x = −2 gives:

A = −1/4 and B = 1/4, therefore

1 1 1 1 12 4( 2) 4( 2)(4 ) x xx

=− +− +−

Thus 1 1ln 2 ln 22 4 4(4 )

dx x xx

=− − + +∫−

so

0 0 1 1lim lim ln 2 ln 22 2 4 4 2 2(4 ) (4 )2

dx dx l ll lx xl

+ += = − − + =∞∫ ∫

→ − → −− −−

00 1 1 1 1ln 2 ln 2 ln 2 ln 22 4 4 4 4(4 )

dx x x l llxl

= − − + + = − − +∫ −

then

similarly

2 1 1lim lim ln 2 ln 22 2 4 4 2 2(4 ) (4 )0 0

ldx dx l ll lx x− −

= = − − + + =∞∫ ∫→ →− −

22(4 )2

dxx

=∞∫−−

Thus the integral does not converge. We can write 2

2(4 )2

dxx

∫−−

Example. Evaluate 1

211

dx

x∫

−−

Solution. The graph of this function over the range in question is shown below.

Again it is clear that this function becomes infinite at both ends of the interval of integration, and so is improper for two different reasons.

1 0 1

2 2 21 1 11 1 0

dx dx dx

x x x= +∫ ∫ ∫

− − −− −Here again we write

( )cos 1sincos21

dx d d xx

θ θ θ θθ

−= = = =∫ ∫ ∫−

First we find the indefinite integral by making a trigonometric substitution. Let x = sin(θ), dx = cos(θ)dθ.

Then

and we treat the right hand integrals separately.

0 01 1lim lim sin ( ) sin (1)

2 2 2 1 11 11

dx dx ll lx xl

π+ +

− −= = − = =∫ ∫→ − → −− −−

similarly1

1 1lim lim sin ( ) sin (1)2 2 2 1 11 10 0

ldx dx ll lx x

π− −

− −= = = =∫ ∫→ →− −

Thus1

.211

dx

xπ=∫

−−

Example. Evaluate 23

4

1( 2)

dx

x∫

−

Solution. The graph of this function over the range in question is shown below.

It is clear that this function becomes infinite at the point 2 that lies between the two endpoints. Thus we must again break it into smaller parts, where each part has only one place where the integrand becomes infinite.

2 2 23 3 3

4 2 4.

1 1 2( 2) ( 2) ( 2)

dx dx dx

x x x= +∫ ∫ ∫

− − −

1/ 3 1/ 32 23 3

2lim lim 3( 2) 3(1 2) 3

2 21 1( 2) ( 2)

ldx dx ll lx x

− − = = − − − =∫ ∫ → →− −

and

1/ 3 1/ 32 23 3

4 4 3lim lim 3(4 2) 3( 2) 3 22 22( 2) ( 2)

dx dx ll llx x

+ + = = − − − =∫ ∫ → →− −

Thus23

4 33 3 21( 2)

dx

x= +∫

−

Example. Evaluate ( 1)0

dxx x

∞∫ +

Solution. The graph of this function over the range in question is shown below.

It is clear that this function becomes infinite for two completely different reasons. First it becomes infinite as we near 0, and secondly, the range of integration is infinite. Again, we break it into two parts.

1

( 1) ( 1) ( 1)0 0 1

dx dx dxx x x x x x

∞ ∞= +∫ ∫ ∫+ + +

Break integral here

Let then2 or , 2u x x u dx udu= = =

2 1 12 2tan 2tan2 2( 1) ( 1) 1

dx udu du u xx x u u u

− −= = = =∫∫ ∫+ + +

Therefore

1 1 1lim 2 tan 1 2 tan 2( 1) 4 200

dx lx x l

π π+

− − = − = =∫ + →

and

1 1lim 2 tan 2 tan 1 22 4( 1) 21

dx lx x l

π π π∞ − − = − = − =∫ + →∞

therefore

( 1)0

dxx x

π∞

=∫ +