aarjaneiro/planetmath huggingface dataset:
integration of sqrt(x^2+1)

integration of

x

2

+

1

The

integral

I

:=

∫

x

2

+

1

⁢

𝑑

x

can be found by using the first Euler’s substitution (

http://planetmath.org/EulersSubstitutionsForIntegration

)

x

2

+

1

:=

-

x

+

t

,

but another possibility is to use partial integration (

http://planetmath.org/ASpecialCaseOfPartialIntegration

) if one knows the integral

∫

d

⁢

x

x

2

+

1

.  The corresponding may be said of the more general

∫

x

2

+

c

⁢

𝑑

x

.

We think that the integrand of

I

has the other factor 1 and integrate partially:

I

=

∫

1

⋅

x

2

+

1

⁢

𝑑

x

=

x

⁢

x

2

+

1

-

∫

x

⋅

1

2

⁢

x

2

+

1

⋅

2

⁢

x

⁢

𝑑

x

+

C

′

=

x

⁢

x

2

+

1

-

∫

x

2

x

2

+

1

⁢

𝑑

x

+

C

′

.

Writing the numerator as

(

x

2

+

1

)

-

1

and dividing its minuend and subtrahend separately, we can write

I

=

x

⁢

x

2

+

1

-

(

∫

x

2

+

1

⁢

𝑑

x

-

∫

1

x

2

+

1

⁢

𝑑

x

)

+

C

′

=

x

⁢

x

2

+

1

-

I

+

∫

d

⁢

x

x

2

+

1

+

C

′

.

Having

I

in two , we solve it from these equalities, obtaining

I

=

x

2

⁢

x

2

+

1

+

1

2

⁢

∫

d

⁢

x

x

2

+

1

+

C

,

i.e.,

∫

x

2

+

1

⁢

𝑑

x

=

x

2

⁢

x

2

+

1

+

1

2

⁢

arsinh

⁡

x

+

C

Title

integration of

x

2

+

1

Canonical name

IntegrationOfsqrtx21

Date of creation

2013-03-22 18:06:58

Last modified on

2013-03-22 18:06:58

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Derivation

Classification

msc 26A36

Synonym

antiderivative

of

x

2

+

1

Related topic

IntegrationByParts

Related topic

AreaFunctions

Related topic

DerivativeOfInverseFunction

Generated on Fri Feb 9 19:52:17 2018 by

LaTeXML
integration techniques

integration techniques

The following is an of

techniques of integration

:

•

general formulas for integration

•

integration by substitution

•

integration by parts

•

partial fraction decomposition

(

http://planetmath.org/ALectureOnThePartialFractionDecompositionMethod

)

•

integration of rational function of sine and cosine

•

integration of differential binomial

•

Euler’s substitutions for integration

•

integration of fraction power expressions

•

a

special case of partial integration

Title

integration techniques

Canonical name

IntegrationTechniques

Date of creation

2013-03-22 17:50:13

Last modified on

2013-03-22 17:50:13

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

7

Author

Wkbj79 (1863)

Entry type

Topic

Classification

msc 26A36

Synonym

techniques of integration

Related topic

AntiderivativeOfRationalFunction

Generated on Fri Feb 9 19:52:20 2018 by

LaTeXML
ostensibly discontinuous antiderivative

ostensibly discontinuous antiderivative

The

real function

x

↦

1

5

-

3

⁢

cos

⁡

x

(1)

is

continuous

for any

x

(the denominator is always positive) and therefore it has an

antiderivative

, defined for all

x

.  Using the

universal trigonometric substitution

cos

⁡

x

:=

1

-

t

2

1

+

t

2

,

d

⁢

x

=

2

⁢

d

⁢

t

1

+

t

2

,

t

=

tan

⁡

x

2

,

we obtain

5

-

3

⁢

cos

⁡

x

=

5

⁢

(

1

+

t

2

)

-

3

⁢

(

1

-

t

2

)

1

+

t

2

=

2

⁢

(

1

+

4

⁢

t

2

)

1

+

t

2

,

whence

∫

d

⁢

x

5

-

3

⁢

cos

⁡

x

=

∫

d

⁢

t

1

+

4

⁢

t

2

=

1

2

⁢

arctan

⁡

2

⁢

t

+

C

=

1

2

⁢

arctan

⁡

(

2

⁢

tan

⁡

x

2

)

+

C

.

This result is not defined in the odd multiples of

π

, and it seems that the

function

(1) does not have a continuous antiderivative.

However, one can check that the function

x

↦

x

4

+

1

2

⁢

arctan

⁡

sin

⁡

x

3

-

cos

⁡

x

+

C

(2)

is everywhere continuous and has as its

derivative

the function (1); one has

|

sin

⁡

x

3

-

cos

⁡

x

|

≦

1

3

-

1

=

1

2

<

π

2

.

References

1

Ernst Lindelöf

:

Johdatus korkeampaan analyysiin

. Fourth edition. Werner Söderström Osakeyhtiö, Porvoo ja Helsinki (1956).

Title

ostensibly discontinuous antiderivative

Canonical name

OstensiblyDiscontinuousAntiderivative

Date of creation

2013-03-22 18:37:08

Last modified on

2013-03-22 18:37:08

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 26A36

Related topic

CyclometricFunctions

Generated on Fri Feb 9 19:52:22 2018 by

LaTeXML
reduction formulas

reduction formulas

To obtain a

reduction formula

for

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

:

- Split off

sin

⁡

x

to integrate by parts

=

∫

sin

n

-

1

⁡

x

⁢

cos

m

⁡

x

⁢

sin

⁡

x

⁢

d

⁢

x

Take

u

=

sin

n

-

1

⁡

x

⁢

cos

m

⁡

x

,

and

⁢

d

⁢

v

=

sin

⁡

x

⁢

d

⁢

x

So

d

⁢

u

=

[

(

n

-

1

)

⁢

sin

n

-

2

⁡

x

⁢

cos

m

+

1

⁡

x

-

m

⁢

sin

n

⁡

x

⁢

cos

m

-

1

⁡

x

]

⁢

d

⁢

x

,

and

⁢

v

=

-

cos

⁡

x

- Then simplify to get

=

-

sin

n

-

1

⁡

x

⁢

cos

m

+

1

⁡

x

+

(

n

-

1

)

⁢

∫

sin

n

-

2

⁡

x

⁢

cos

m

+

2

⁡

x

⁢

d

⁢

x

-

m

⁢

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

- Now use the identity

sin

2

⁡

x

+

cos

2

⁡

x

=

1

in the middle term and simplify to get

=

-

sin

n

-

1

⁡

x

⁢

cos

m

+

1

⁡

x

+

(

n

-

1

)

⁢

∫

sin

n

-

2

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

-

(

n

-

1

)

⁢

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

-

m

⁢

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

- Take the last two

integrals

to the left side:

[

1

+

(

n

-

1

)

+

m

]

⁢

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

=

-

sin

n

-

1

⁡

x

⁢

cos

m

+

1

⁡

x

+

(

n

-

1

)

⁢

∫

sin

n

-

2

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

- Since

[

1

+

(

n

-

1

)

+

m

]

=

m

+

n

divide both sides by

m

+

n

and hence

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

=

-

sin

n

-

1

⁡

x

⁢

cos

m

+

1

⁡

x

m

+

n

+

n

-

1

m

+

n

⁢

∫

sin

n

-

2

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

Using the exact same method but instead of splitting off

sin

⁡

x

, one can split off

cos

⁡

x

and follow similar procedure to obtain another reduction formula:

∫

sin

n

⁡

x

⁢

cos

m

⁡

x

⁢

d

⁢

x

=

sin

n

+

1

⁡

x

⁢

cos

m

-

1

⁡

x

m

+

n

+

m

-

1

m

+

n

⁢

∫

sin

n

⁡

x

⁢

cos

m

-

2

⁡

x

⁢

d

⁢

x

Title

reduction formulas

Canonical

name

ReductionFormulas

Date of creation

2013-03-22 17:37:06

Last modified on

2013-03-22 17:37:06

Owner

curious (18562)

Last modified by

curious (18562)

Numerical id

9

Author

curious (18562)

Entry type

Definition

Classification

msc 26A36

Synonym

powers of sines and cosines

Synonym

integration of trigonometric functions

Related topic

TrigonometricFormulasFromDeMoivreIdentity

Generated on Fri Feb 9 19:52:25 2018 by

LaTeXML
reduction formulas for integration of powers

reduction formulas for integration of powers

The following

reduction formulas

, with

integer

n

and via

integration by parts

, may be used for lowing (

n

>

0

) or raising (

n

<

0

) the the powers:

•

∫

sin

n

x

d

x

=

-

1

n

sin

n

-

1

x

cos

x

+

n

-

1

n

∫

sin

n

-

2

x

d

x

(

n

≷

0

)

•

∫

cos

n

x

d

x

=

1

n

cos

n

-

1

x

sin

x

+

n

-

1

n

∫

cos

n

-

2

x

d

x

(

n

≷

0

)

•

∫

(

ln

x

)

n

d

x

=

x

(

ln

x

)

n

-

n

∫

(

ln

x

)

n

-

1

d

x

(

n

≷

0

)

•

∫

1

(

1

+

x

2

)

n

d

x

=

1

2

⁢

n

-

2

⋅

x

(

1

+

x

2

)

n

-

1

+

2

⁢

n

-

3

2

⁢

n

-

2

∫

1

(

1

+

x

2

)

n

-

1

d

x

(

n

>

1

)

Example.

For finding

∫

d

⁢

x

sin

3

⁡

x

, we apply the first formula with

n

:=

-

1

,  getting first

∫

d

⁢

x

sin

⁡

x

=

-

1

-

1

⋅

cos

⁡

x

sin

2

⁡

x

+

-

2

-

1

⁢

∫

d

⁢

x

sin

3

⁡

x

.

From this we solve

∫

d

⁢

x

sin

3

⁡

x

=

-

1

2

⁢

cos

⁡

x

sin

2

⁡

x

+

∫

d

⁢

x

sin

⁡

x

=

-

1

2

⁢

cos

⁡

x

sin

2

⁡

x

+

ln

⁡

|

tan

⁡

x

2

|

+

C

(see

integration of rational function of sine and cosine

).

Note 1.

Instead of the two first formulae, it is simpler in the cases when

n

is a

positive

odd or a negative

even number

to use the following

∫

sin

2

⁢

m

+

1

⁡

x

⁢

d

⁢

x

=

∫

sin

2

⁢

m

⁡

x

⁢

sin

⁡

x

⁢

d

⁢

x

=

-

∫

(

1

-

cos

2

⁡

x

)

m

⁢

(

-

sin

⁡

x

)

⁢

𝑑

x

,

∫

cos

2

⁢

m

+

1

⁡

x

⁢

d

⁢

x

=

∫

cos

2

⁢

m

⁡

x

⁢

cos

⁡

x

⁢

d

⁢

x

=

∫

(

1

-

sin

2

⁡

x

)

m

⁢

cos

⁡

x

⁢

d

⁢

x

,

∫

1

sin

2

⁢

m

⁡

x

⁢

𝑑

x

=

∫

1

sin

2

⁢

m

-

2

⁡

x

⋅

1

sin

2

⁡

x

⁢

𝑑

x

=

-

∫

(

1

+

cot

2

⁡

x

)

m

-

1

⁢

d

⁢

cot

⁡

x

,

∫

1

cos

2

⁢

m

⁡

x

⁢

𝑑

x

=

∫

1

cos

2

⁢

m

-

2

⁡

x

⋅

1

cos

2

⁡

x

⁢

𝑑

x

=

∫

(

1

+

tan

2

⁡

x

)

m

-

1

⁢

d

⁢

tan

⁡

x

,

which may be found after making the powers on the

right hand sides

to

polynomials

.

Note 2.

∫

tan

n

⁡

x

⁢

d

⁢

x

(

n

∈

ℤ

+

)  is obtained easily by the substitution (

http://planetmath.org/IntegrationBySubstitution

)

tan

⁡

x

:=

t

,

d

⁢

x

=

d

⁢

t

t

2

+

1

and a

division

; e.g.

∫

tan

5

⁡

x

⁢

d

⁢

x

=

∫

t

5

t

2

+

1

⁢

𝑑

t

=

∫

(

t

3

-

t

+

t

t

2

+

1

)

⁢

𝑑

t

=

t

4

4

-

t

2

2

+

1

2

⁢

ln

⁡

(

t

2

+

1

)

+

C

=

tan

4

⁡

x

4

-

tan

2

⁡

x

2

+

ln

⁡

tan

2

⁡

x

+

1

+

C

.

Title

reduction formulas for integration of powers

Canonical name

ReductionFormulasForIntegrationOfPowers

Date of creation

2013-03-22 18:36:53

Last modified on

2013-03-22 18:36:53

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

10

Author

pahio (2872)

Entry type

Topic

Classification

msc 26A36

Classification

msc 26A09

Synonym

integration of powers

Related topic

GeneralFormulasForIntegration

Related topic

IntegralTables

Related topic

WallisFormulae

Generated on Fri Feb 9 19:52:28 2018 by

LaTeXML
substitution for integration

substitution for integration

For determining the

antiderivative

F

⁢

(

x

)

of a given

real function

f

⁢

(

x

)

in a “

closed form

”, i.e. for integrating

f

⁢

(

x

)

, the result is often obtained by using the

Theorem

.

If

∫

f

⁢

(

x

)

⁢

𝑑

x

=

F

⁢

(

x

)

+

C

and

x

=

x

⁢

(

t

)

is a

differentiable function

,
then

F

⁢

(

x

⁢

(

t

)

)

=

∫

f

⁢

(

x

⁢

(

t

)

)

⁢

x

′

⁢

(

t

)

⁢

𝑑

t

+

c

.

(1)

Proof.

By virtue of the

chain rule

,

d

d

⁢

t

⁢

F

⁢

(

x

⁢

(

t

)

)

=

F

′

⁢

(

x

⁢

(

t

)

)

⋅

x

′

⁢

(

t

)

,

and according to the supposition,

F

′

⁢

(

x

)

=

f

⁢

(

x

)

.  Thus we get the claimed equation (1).

Remarks.

•

The expression

x

′

⁢

(

t

)

⁢

d

⁢

t

in (1) may be understood as the

differential

of

x

⁢

(

t

)

.

•

For returning to the original variable

x

, the

inverse function

t

=

t

⁢

(

x

)

of

x

⁢

(

t

)

must be substituted to

F

⁢

(

x

⁢

(

t

)

)

.

Example.

For integrating

∫

x

⁢

d

⁢

x

1

+

x

4

we take

x

2

=

t

as a new variable.  Then,

2

⁢

x

⁢

d

⁢

x

=

d

⁢

t

,

x

⁢

d

⁢

x

=

d

⁢

t

2

, and we get

∫

x

⁢

d

⁢

x

1

+

x

4

=

1

2

⁢

∫

d

⁢

t

1

+

t

2

=

1

2

⁢

arctan

⁡

t

+

C

=

1

2

⁢

arctan

⁡

x

2

+

C

.

Title

substitution for integration

Canonical name

SubstitutionForIntegration

Date of creation

2013-03-22 14:33:38

Last modified on

2013-03-22 14:33:38

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

21

Author

pahio (2872)

Entry type

Theorem

Classification

msc 26A36

Synonym

variable changing for integration

Synonym

integration by substitution

Synonym

substitution rule

Related topic

IntegrationOfRationalFunctionOfSineAndCosine

Related topic

IntegrationOfFractionPowerExpressions

Related topic

ChangeOfVariableInDefiniteIntegral

Generated on Fri Feb 9 19:52:31 2018 by

LaTeXML
Taylor series of arcus sine

Taylor series of arcus sine

We give an example of obtaining the

Taylor series

of an

elementary function

by integrating the Taylor series of its

derivative

.

For

-

1

<

x

<

1

we have the derivative of the principal of the arcus sine (

http://planetmath.org/CyclometricFunctions

)

function

:

d

⁢

arcsin

⁡

x

d

⁢

x

=

1

1

-

x

2

=

(

1

-

x

2

)

-

1

2

.

Using the

generalized binomial coefficients

(

-

1

2

r

)

we thus can form the Taylor series for it as Newton’s

binomial series

(

http://planetmath.org/BinomialFormula

):

(

1

-

x

2

)

-

1

2

=

∑

r

=

0

∞

(

-

1

2

r

)

⁢

(

-

x

2

)

r

=

1

+

(

-

1

2

1

)

⁢

(

-

x

2

)

+

(

-

1

2

2

)

⁢

(

-

x

2

)

2

+

(

-

1

2

3

)

⁢

(

-

x

2

)

3

+

⋯

=

=

1

-

-

1

2

1

!

x

2

+

-

1

2

⁢

(

-

1

2

-

1

)

2

!

x

4

-

-

1

2

⁢

(

-

1

2

-

1

)

⁢

(

-

1

2

-

2

)

3

!

x

6

+

-

⋯

=

=

1

+

1

2

⁢

x

2

+

1

⋅

3

2

⋅

4

⁢

x

4

+

1

⋅

3

⋅

5

2

⋅

4

⋅

6

⁢

x

6

+

⋯

for

-

1

<

x

<

1

Because

arcsin

⁡

0

=

0

for the

principal branch

(

http://planetmath.org/GeneralPower

) of the function, we get, by integrating the series termwise (

http://planetmath.org/SumFunctionOfSeries

), the

arcsin

⁡

x

=

∫

0

x

d

⁢

x

1

-

x

2

=

x

+

1

2

⋅

x

3

3

+

1

⋅

3

2

⋅

4

⋅

x

5

5

+

1

⋅

3

⋅

5

2

⋅

4

⋅

6

⋅

x

7

7

+

⋯

,

the

validity

of which is true for

|

x

|

<

1

.  It can be proved, in

addition

, that it is true also when

x

=

±

1

.

Title

Taylor series of arcus sine

Canonical name

TaylorSeriesOfArcusSine

Date of creation

2013-03-22 14:51:18

Last modified on

2013-03-22 14:51:18

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Example

Classification

msc 26A36

Classification

msc 26A09

Classification

msc 11B65

Classification

msc 05A10

Related topic

ExamplesOnHowToFindTaylorSeriesFromOtherKnownSeries

Related topic

TaylorSeriesOfArcusTangent

Related topic

CyclometricFunctions

Related topic

LogarithmSeries

Generated on Fri Feb 9 19:52:33 2018 by

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vector-valued function

vector-valued function

Let

n

be a positive integer

greater

than 1.  A

function

F

from a subset

T

of

ℝ

to the Cartesian product

ℝ

n

is called a

vector-valued function

of one real variable.  Such a function to any

real number

t

of

T

a

coordinate vector

F

⁢

(

t

)

=

(

f

1

⁢

(

t

)

,

…

,

f

n

⁢

(

t

)

)

.

Hence one may say that the vector-valued function

F

is composed of

n

real functions

t

↦

f

i

⁢

(

t

)

,  the values of which at

t

are the

components

of

F

⁢

(

t

)

.  Therefore the function

F

itself may be written in the component form

F

=

(

f

1

,

…

,

f

n

)

.

(1)

Example.

The

ellipse

{

(

a

⁢

cos

⁡

t

,

b

⁢

sin

⁡

t

)

⁢

⋮

⁢

t

∈

ℝ

}

is the value set of a vector-valued function

ℝ

→

ℝ

2

(

t

is the eccentric anomaly).

Limit,

derivative

and

integral

of the function (1) are defined componentwise through the equations

•

lim

t

→

t

0

⁡

F

⁢

(

t

)

:=

(

lim

t

→

t

0

⁡

f

1

⁢

(

t

)

,

…

,

lim

t

→

t

0

⁡

f

n

⁢

(

t

)

)

•

F

′

⁢

(

t

)

:=

(

f

1

′

⁢

(

t

)

,

…

,

f

n

′

⁢

(

t

)

)

•

∫

a

b

F

⁢

(

t

)

⁢

𝑑

t

:=

(

∫

a

b

f

1

⁢

(

t

)

⁢

𝑑

t

,

…

,

∫

a

b

f

n

⁢

(

t

)

⁢

𝑑

t

)

The function

F

is said to be

continuous

,

differentiable

or integrable

on an

interval

[

a

,

b

]

if every component of

F

has such a property.

Example.

If

F

is continuous on

[

a

,

b

]

,  the set

γ

:=

{

F

(

t

)

⋮

t

∈

[

a

,

b

]

}

(2)

is a (continuous) curve in

ℝ

n

.  It follows from the above definition of the derivative

F

′

⁢

(

t

)

that

F

′

⁢

(

t

)

is the limit of the expression

1

h

⁢

[

F

⁢

(

t

+

h

)

-

F

⁢

(

t

)

]

(3)

as

h

→

0

.  Geometrically, the vector (3) is

parallel

to the

line segment

connecting (the

end points

of the

position vectors

of) the points

F

⁢

(

t

+

h

)

and

F

⁢

(

t

)

.  If

F

is differentiable in

t

, the

direction

of this line segment then tends infinitely the direction of the

tangent line

of

γ

in the point

F

⁢

(

t

)

.  Accordingly, the direction of the tangent line is determined by the derivative vector

F

′

⁢

(

t

)

.

Title

vector-valued function

Canonical name

VectorvaluedFunction

Date of creation

2013-03-22 19:02:19

Last modified on

2013-03-22 19:02:19

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Definition

Classification

msc 26A36

Classification

msc 26A42

Classification

msc 26A24

Related topic

Component

Related topic

DifferenceOfVectors

Defines

integrable

Generated on Fri Feb 9 19:52:36 2018 by

LaTeXML
Weierstrass substitution formulas

Weierstrass substitution formulas

The

Weierstrass substitution formulas

for

-

π

<

x

<

π

are:

sin

⁡

x

=

2

⁢

t

1

+

t

2

cos

⁡

x

=

1

-

t

2

1

+

t

2

d

⁢

x

=

2

1

+

t

2

⁢

d

⁢

t

They can be obtained in the following manner:

Make the

Weierstrass substitution

t

=

tan

⁡

(

x

2

)

. (This substitution is also known as the

universal trigonometric substitution

.) Then we have

cos

⁡

(

x

2

)

=

1

sec

⁡

(

x

2

)

=

1

1

+

tan

2

⁡

(

x

2

)

=

1

1

+

t

2

and

sin

⁡

(

x

2

)

=

cos

⁡

(

x

2

)

⋅

tan

⁡

(

x

2

)

=

t

1

+

t

2

.

Note that these are just the “formulas involving

radicals

(

http://planetmath.org/Radical6

)” as designated in the entry

goniometric formulas

; however, due to the

restriction

on

x

, the

±

’s are unnecessary.

Using the above formulas along with the

double angle formulas

, we obtain

sin

⁡

x

=

2

⁢

sin

⁡

(

x

2

)

⋅

cos

⁡

(

x

2

)

=

2

⋅

t

1

+

t

2

⋅

1

1

+

t

2

=

2

⁢

t

1

+

t

2

and

cos

⁡

x

=

cos

2

⁡

(

x

2

)

-

sin

2

⁡

(

x

2

)

=

(

1

1

+

t

2

)

2

-

(

t

1

+

t

2

)

2

=

1

1

+

t

2

-

t

2

1

+

t

2

=

1

-

t

2

1

+

t

2

.

Finally, since

t

=

tan

⁡

(

x

2

)

, solving for

x

yields that

x

=

2

⁢

arctan

⁡

t

. Thus,

d

⁢

x

=

2

1

+

t

2

⁢

d

⁢

t

.

The Weierstrass substitution formulas are most useful for integrating

rational functions

of sine and cosine (

http://planetmath.org/IntegrationOfRationalFunctionOfSineAndCosine

).

Title

Weierstrass substitution formulas

Canonical name

WeierstrassSubstitutionFormulas

Date of creation

2013-03-22 17:05:25

Last modified on

2013-03-22 17:05:25

Owner

Wkbj79 (1863)

Last modified by

Wkbj79 (1863)

Numerical id

12

Author

Wkbj79 (1863)

Entry type

Definition

Classification

msc 26A36

Classification

msc 33B10

Synonym

Weierstraß substitution formulas

Related topic

GoniometricFormulae

Related topic

IntegrationOfRationalFunctionOfSineAndCosine

Related topic

PolynomialAnalogonForFermatsLastTheorem

Defines

Weierstrass substitution

Defines

Weierstaß substitution

Defines

universal trigonometric substitution

Generated on Fri Feb 9 19:52:39 2018 by

LaTeXML
alternative definition of Lebesgue integral, an

alternative definition of Lebesgue integral, an

The standard way of defining

Lebesgue integral

is first to define it for

simple functions

, and then to take limits for arbitrary positive

measurable functions

.

There is also another way which uses the

Riemann integral

[

1

]

.

Let

(

X

,

ℳ

,

μ

)

be a

measure space

. Let

f

:

x

→

ℝ

+

∪

{

0

}

be a nonnegative measurable function. We will define

∫

f

⁢

𝑑

μ

in

[

0

,

∞

]

and will call it as the

Lebesgue integral

of

f

.

If there exists a

t

>

0

such that

μ

⁢

(

{

x

:

f

⁢

(

x

)

>

t

}

)

=

∞

, then we define

∫

f

⁢

𝑑

μ

=

∞

.

Otherwise, assume

μ

⁢

(

{

x

:

f

⁢

(

x

)

>

t

}

)

<

∞

for all

t

∈

(

0

,

∞

)

and let

F

f

⁢

(

t

)

=

μ

⁢

(

{

x

:

f

⁢

(

x

)

>

t

}

)

.

F

f

⁢

(

t

)

is a

monotonically

non-increasing

function

on

(

0

,

∞

)

, therefore its Riemann integral is well defined on any

interval

[

a

,

b

]

⊂

(

0

,

∞

)

, so it exists as an improper Riemann integral on

(

0

,

∞

)

. We define

∫

f

⁢

𝑑

μ

=

∫

0

∞

F

f

⁢

(

t

)

⁢

𝑑

t

.

The definition can be extended first to real-valued functions, then complex valued functions as usual.

References

1

Lieb, E. H., Loss, M.,

Analysis

, AMS, 2001.

Title

alternative definition of Lebesgue integral, an

Canonical name

AlternativeDefinitionOfLebesgueIntegralAn

Date of creation

2013-03-22 17:32:46

Last modified on

2013-03-22 17:32:46

Owner

Gorkem (3644)

Last modified by

Gorkem (3644)

Numerical id

6

Author

Gorkem (3644)

Entry type

Definition

Classification

msc 26A42

Classification

msc 28A25

Generated on Fri Feb 9 19:52:46 2018 by

LaTeXML
an integrable function which does not tend to zero

an integrable function which does not tend to zero

In this entry, we give an example of a

function

f

such that

f

is

Lebesgue
integrable

on

ℝ

but

f

⁢

(

x

)

does not tend to zero as

x

→

∞

.

Set

f

⁢

(

x

)

=

∑

k

=

1

∞

k

k

6

⁢

(

x

-

k

)

2

+

1

.

Note that every term in this series is positive, hence we may integrate
term-by-term, then make a change of variable

y

=

k

3

⁢

x

-

k

4

and
compute the answer:

∫

-

∞

+

∞

f

⁢

(

x

)

⁢

𝑑

x

=

∑

k

=

1

∞

∫

-

∞

+

∞

k

⁢

d

⁢

x

k

6

⁢

(

x

-

k

)

2

+

1

=

∑

k

=

1

∞

1

k

2

⁢

∫

-

∞

+

∞

d

⁢

y

y

2

+

1

=

π

2

6

⋅

π

=

π

3

6

However, when

k

is an integer,

f

⁢

(

k

)

>

k

, so not only does

f

⁢

(

x

)

not
tend to zero as

x

→

∞

, it gets arbitrarily large.
As we can see from the plot, we have a

sequence

of peaks which, as they get
taller, also get narrower in such a way that the total area under
the curve stays finite:

\includegraphics

integrable-no-tend-zero.gif

By a variation of our procedure, we can produce a function which is
defined almost everywhere on the

interval

(

0

,

1

)

, is Lebesgue integrable,
but is

unbounded

on any

subinterval

, no matter how small. For

instance

,
define

f

⁢

(

x

)

=

∑

m

=

1

∞

∑

n

=

1

∞

1

(

m

+

n

)

6

⁢

(

x

-

m

/

(

m

+

n

)

)

2

+

1

.

Making a computation

similar

to the one above, we find that

∫

-

∞

+

∞

f

⁢

(

x

)

⁢

𝑑

x

=

∑

m

=

1

∞

∑

n

=

1

∞

∫

-

∞

+

∞

d

⁢

x

(

m

+

n

)

6

⁢

(

x

-

m

/

(

m

+

n

)

)

2

+

1

=

∑

m

=

1

∞

∑

n

=

1

∞

1

(

m

+

n

)

3

⁢

∫

-

∞

+

∞

d

⁢

y

y

2

+

1

=

π

⁢

∑

k

=

1

∞

k

-

1

k

3

Hence the

integral

is finite.

Now, however, we find that

f

cannot be bounded in any interval,
however small. For, in any interval, we can find

rational
numbers

. Given a rational number

r

, there are an

infinite

number of ways to express it as a fraction

m

/

(

m

+

n

)

. For
each of these ways, we have a term in the series which equals 1
when

x

=

r

, hence

f

⁢

(

r

)

diverges

to

infinity

.

To help in understanding this function, we have made a slide show
which shows partial sums of the series. As before, the successive
peaks become narrower in such a way that the arae under the curve
stays finite but, this time, instead of marching off to infinity,
they become dense in the interval.

\includegraphics

integrable-everywhere-unbounded_3.gif

Title

an

integrable function which does not tend to zero

Canonical name

AnIntegrableFunctionWhichDoesNotTendToZero

Date of creation

2014-02-01 3:01:47

Last modified on

2014-02-01 3:01:47

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

15

Author

rspuzio (6075)

Entry type

Example

Classification

msc 26A42

Classification

msc 28A25

Generated on Fri Feb 9 19:52:49 2018 by

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