improper integral

improper integral

The

Riemann integral

of a

function

f

:

I

→

𝐑

∫

I

f

⁢

(

t

)

⁢

𝑑

t

is defined when

f

is bounded and

I

=

[

a

,

b

]

⊂

𝐑

is a compact

interval

.
If

I

is any interval of

𝐑

and

f

is bounded on every compact subset of

I

(for example

f

is

continuous

), then we can define the

concept

of

improper integral

of

f

on

I

by approximation of

I

with compact sets.

Let

I

=

(

a

,

b

)

with

a

∈

[

-

∞

,

+

∞

)

and

b

∈

(

-

∞

,

+

∞

]

and let

f

be a

continuous function

on

I

. Given

x

,

y

∈

I

we know that the function

f

is Riemann integrable on

[

a

,

b

]

; hence we can define the

improper integral of

f

on

I

as

∫

I

f

⁢

(

t

)

⁢

𝑑

t

:=

lim

(

x

,

y

)

→

(

a

,

b

)

,

x

>

a

,

y

<

b

⁡

∫

x

y

f

⁢

(

t

)

⁢

𝑑

t

.

Notice that the limit is taken in two

variables

. In the case when

I

is compact
this is the usual Riemann integral on

I

=

[

a

,

b

]

(because the integral function is continuous). So there is no ambiguity in using the same simbol for improper integrals and usual Riemann integrals (but we will see that there is an ambiguity when dealing with

Lebesgue integrals

).
Similarly, in the case when the interval

I

is semi-open, i.e. when

I

=

[

a

,

b

)

with

a

>

-

∞

, the

definition

clearly reduces to

∫

I

f

⁢

(

t

)

⁢

𝑑

t

:=

lim

y

→

b

-

⁡

∫

a

y

f

⁢

(

t

)

⁢

𝑑

t

which is a limit in one variable.

Notice also that an improper integral can be

infinite

or can possibly not exist (even thought

f

is continuous).

The definition may be extended to the case when

I

is the union of a finite number of intervals

I

=

I

1

∪

…

⁢

I

N

by

summing

up the improper integrals on every interval:

∫

I

f

⁢

(

t

)

⁢

𝑑

t

:=

∑

k

=

1

N

∫

I

k

f

⁢

(

t

)

⁢

𝑑

t

.

In the case when

I

=

𝐑

(or more generally when

I

has some

symmetry

), one can define the

symmetric

improper integral

as

∫

𝐑

f

⁢

(

x

)

⁢

𝑑

x

=

lim

x

→

+

∞

⁡

∫

-

x

x

f

⁢

(

t

)

⁢

𝑑

t

.

This limit can exist in some cases when the

improper integral

(not symmetric) fails to exist.

1

Examples

Given

I

=

(

0

,

1

]

,

f

⁢

(

t

)

=

1

/

t

one has:

∫

(

0

,

1

]

1

t

⁢

𝑑

t

=

lim

x

→

0

+

⁡

∫

x

1

1

t

⁢

𝑑

t

=

lim

x

→

0

+

⁡

[

2

⁢

t

1

2

]

x

1

=

lim

x

→

0

+

⁡

(

2

-

2

⁢

x

)

=

2

.

Given

I

=

[

1

,

+

∞

)

,

f

⁢

(

t

)

=

(

sin

⁡

t

)

/

t

one can check that the improper integral

∫

I

f

⁢

(

t

)

⁢

𝑑

t

exists and is finite but the improper integral

∫

I

|

f

⁢

(

t

)

|

⁢

𝑑

t

is infinite. In particular this function is not summable (in the sense of Lebesgue integrals) on the interval

I

.

The function

f

⁢

(

t

)

=

t

/

(

1

+

t

2

)

has no improper integral in

I

=

𝐑

. But since

f

⁢

(

t

)

=

-

f

⁢

(

-

t

)

one can easily check that the symmetric integral is zero:

lim

x

→

+

∞

⁡

∫

-

x

x

t

1

+

t

2

⁢

𝑑

t

=

0

.

Title

improper integral

Canonical

name

ImproperIntegral

Date of creation

2013-03-22 12:30:06

Last modified on

2013-03-22 12:30:06

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

8

Author

paolini (1187)

Entry type

Definition

Classification

msc 40A10

Related topic

CauchyPrinciplePartIntegral

Related topic

ExampleOfUsingResidueTheorem

Related topic

FunctionXx

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