30. Improper Integrals 2 (Rigorous)

7/28/2019 30. Improper Integrals 2 (Rigorous)

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Improper Integrals, Part 2:

The Concept Reloaded (Rigorous)Definition of Improper Integrals

Singularities in the Interval of Integration

Basic Improper IntegralsComparison Theorem

The Gamma Function


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Definition of Improper Integrals

00 0

0

e lim e lim e

lim e ( e ) 1

b bx x x

b b

b

b

dx dx

Definition

Example

Assume that the function f takes finite values in the interval [ , ).

If the limit lim f( ) exists and is finite, then we say that the the

improper integral of the function f over the inte

b

ab

a

x dx

rval [ , )

In this case we set

f( ) lim f( ) .

The improper integral f( ) if the limit lim

conver

f( ) does

not exist or is not finit

ges.

diverge

e.

s

b

a ab

b

a ab

a

x dx x dx

x dx x dx


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Generalizations

1

21

1dxx

1

2

1dx

x

The previous definitions generalize to cases where One of the end-points of the interval is negative infinity,or

A singular point is contained in the interval.

This integralconverges.

This integraldiverges.

Examples

1

2

11

2

1 1 1lim lim lim 1 1.dx

x x

1

2 210 0

1 1

lim limdx dxx x

10 0

1

1 1lim lim

x x


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Basic Improper Integrals

1converges for 1 and diverges otherwisepx dx p

1

1converges for 1 and diverges otherwise

pdx p

x

1

0converges for 1 and diverges otherwisepx dx p

Clearly (1) (2) and (3) (4). Proving these results is astraightforward computation.

3

4

0 e converges for 0 and diverges otherwiseax

dx a

1

0

1converges for 1 and diverges otherwise

pdx p

x

1

2

5


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Convergence of Improper Integrals

Sometimes, we cannot compute the limit of a givenimproper integral directly.

To find out whether such an integral converges or not,

we can compare the integral to a known integral (weknow that it either converges or diverges).

2 20

3

is an(1 2sin (4 ))(1 )

example of such an improper integral.

The graph of the function to beintegrated is shown in the picture.

dxx x


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Idea of the Comparison Theorem

2 2 2

Observe that the function to be integrated satisfies

3 30

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

for all 0. The following graph illustrates this observation.

x x x

x

2

3The blue curve is the graph of the function while the red curve is the1

graph of the function to be integrated.

x

The improper integral converges if

the area under theredred

curve is finite. Show that the area under the blueblue curve is finite.

Since the area under the red curve is smaller than the area

under the blue curve, it must then also be finite.

2 20

3

(1 2sin (4 ))(1 )dx

x x


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Examples (1)

2 20 0

3 3lim

1 1

b

bdx dx

x x

20

2 20

3This means that the improper integral converges.

1

3Hence also the improper integral converges.(1 2sin (4 ))(1 )

dxx

dxx x

To show that the area under the blue curve is finite, compute as

follows:

0

3lim3arctan( ) .

2

b

bx


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Comparison Theorem

a

Let , , , . Assume that the functions f and g satisfy

0 f( ) g( ) for all , . Assume also that the integral f( )

is improper.

1) If the improper integral g( ) converges,

b

a

b

a b a b

x x x a x b x dx

x dx

a

a

then also the improper

integral f( ) converges, and 0 f( ) g( ) .

2) If the improper integral f( ) diverges, then also the improper integral

g( ) diverges.

b b b

a a

b

b

a

x dx x dx x dx

x dx

x dx

Theorem

Remark

The integral f( ) is improper if either , or the

function f has singularities in the interval of integration.

b

ax dx a b


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Normal Distribution Function2

The improper integral e has important applications. It is related to

the normal distribution, an important concept in statistics. We need to understand

why the improper intergal converges.

x dx

2

e x dx

21

e x dx

This was shown earlierThis was shown earlier. Hence, we conclude that the integral

converges. The same argument also proves that

converges. Hence the integral converges.

2

1e x dx

2

2

1

2

1

To study the convergence of the integral e observe that for

1, . Hence also 0 e e for 1. The improper integral

e converges.

x

x x

x

dx

x x x x

dx

2 2 2 2

2

1 1

1 1

1

1

To show the convergence write e = e + e e .

The integral e is an ordinary definite integral which has a finite value.

x x x x

x

dx dx dx dx

dx

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30. Improper Integrals 2 (Rigorous)