aarjaneiro/planetmath huggingface dataset:
Laplace transform of t^n⁢f⁢(t)

Laplace transform of

t

n

⁢

f

⁢

(

t

)

Let

F

⁢

(

s

)

:=

ℒ

⁢

{

f

⁢

(

t

)

}

=

∫

0

∞

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

.

A

differentiation

under the

integral sign

with respect to

s

yields

F

′

⁢

(

s

)

=

-

∫

0

∞

e

-

s

⁢

t

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

-

ℒ

⁢

{

t

⁢

f

⁢

(

t

)

}

.

Differentiating again under the integral sign gives

F

′′

⁢

(

s

)

=

+

∫

0

∞

e

-

s

⁢

t

⁢

t

2

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

ℒ

⁢

{

t

2

⁢

f

⁢

(

t

)

}

.

One can continue similarly, and then we apparently have

F

(

n

)

⁢

(

s

)

=

(

-

1

)

n

⁢

∫

0

∞

e

-

s

⁢

t

⁢

t

n

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

(

-

1

)

n

⁢

ℒ

⁢

{

t

n

⁢

f

⁢

(

t

)

}

.

(1)

If this equation is multiplied by

(

-

1

)

n

, it gives the

ℒ

⁢

{

t

n

⁢

f

⁢

(

t

)

}

=

(

-

1

)

n

⁢

F

(

n

)

⁢

(

s

)

(2)

which is true for

n

=

1

,

2

,

3

,

…

Application.

Evaluate the improper integral

I

:=

∫

0

∞

t

3

⁢

e

-

t

⁢

sin

⁡

t

⁢

d

⁢

t

.

By the parent entry (

http://planetmath.org/LaplaceTransform

), we have

ℒ

⁢

{

sin

⁡

t

}

=

1

1

+

s

2

.  Using this and (2), we may write

∫

0

∞

t

3

⁢

e

-

s

⁢

t

⁢

sin

⁡

t

⁢

d

⁢

t

=

ℒ

⁢

{

t

3

⁢

sin

⁡

t

}

=

(

-

1

)

3

⁢

d

3

d

⁢

s

3

⁢

(

1

s

2

+

1

)

=

24

⁢

(

s

-

s

3

)

(

1

+

s

2

)

4

.

The value of

I

is obtained by substituting here

s

=

1

:

I

=

24

⁢

(

1

-

1

3

)

(

1

+

1

2

)

4

=

0

.

Title

Laplace transform

of

t

n

⁢

f

⁢

(

t

)

Canonical

name

LaplaceTransformOfTnft

Date of creation

2013-03-22 18:05:49

Last modified on

2013-03-22 18:05:49

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Related topic

TableOfLaplaceTransforms

Generated on Fri Feb 9 21:32:30 2018 by

LaTeXML
Laplace transform of {f⁢(t)}/t

Laplace transform of

f

⁢

(

t

)

t

Suppose that the quotient

f

⁢

(

t

)

t

:=

g

⁢

(

t

)

is Laplace-transformable (

http://planetmath.org/LaplaceTransform

).  It follows easily that also

f

⁢

(

t

)

is such.  According to the parent entry (

http://planetmath.org/LaplaceTransformOfTnft

), we may write

ℒ

-

1

⁢

{

G

′

⁢

(

s

)

}

=

-

t

⁢

g

⁢

(

t

)

=

-

f

⁢

(

t

)

=

ℒ

-

1

⁢

{

-

F

⁢

(

s

)

}

.

Therefore

G

′

⁢

(

s

)

=

-

F

⁢

(

s

)

,

whence

G

⁢

(

s

)

=

-

F

(

-

1

)

⁢

(

s

)

+

C

(1)

where

F

(

-

1

)

⁢

(

s

)

means any

antiderivative

of

F

⁢

(

s

)

.  Since each Laplace transformed

function

vanishes

in the infinity

s

=

∞

and thus

G

⁢

(

∞

)

=

0

,  the equation (1) implies

C

=

F

(

-

1

)

⁢

(

∞

)

and therefore

G

⁢

(

s

)

=

F

(

-

1

)

⁢

(

∞

)

-

F

(

-

1

)

⁢

(

s

)

=

∫

s

∞

F

⁢

(

u

)

⁢

𝑑

u

.

We have obtained the result

ℒ

⁢

{

f

⁢

(

t

)

t

}

=

∫

s

∞

F

⁢

(

u

)

⁢

𝑑

u

.

(2)

Application.

By the table of

Laplace transforms

,

ℒ

⁢

{

sin

⁡

t

}

=

1

s

2

+

1

.

Accordingly the formula (2) yields

ℒ

⁢

{

sin

⁡

t

t

}

=

∫

s

∞

1

u

2

+

1

⁢

𝑑

u

=

/

s

∞

⁡

arctan

⁡

u

=

π

2

-

arctan

⁡

s

=

arccot

⁡

s

.

Thus we have

ℒ

⁢

{

sin

⁡

t

t

}

=

arccot

⁡

s

=

arctan

⁡

1

s

.

(3)

This result is derived in the entry

Laplace transform of sine integral

in two other ways.

Title

Laplace transform of

f

⁢

(

t

)

t

Canonical name

LaplaceTransformOffracftt

Date of creation

2014-03-08 15:45:15

Last modified on

2014-03-08 15:45:15

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Related topic

FundamentalTheoremOfCalculusClassicalVersion

Related topic

SubstitutionNotation

Related topic

CyclometricFunctions

Generated on Fri Feb 9 21:32:33 2018 by

LaTeXML
Laplace transforms of derivatives

Laplace transforms of derivatives

where

As shown in the parent entry (

http://planetmath.org/LaplaceTransformOfDerivative

),
the

Laplace transform

of the

first derivative

of a Laplace-transformable

function

f

⁢

(

t

)

is got from the formula

ℒ

⁢

{

f

′

⁢

(

t

)

}

=

s

⁢

F

⁢

(

s

)

-

lim

t

→

0

+

⁡

f

⁢

(

t

)

.

(1)

The rule can be applied also to the function

f

′

⁢

(

t

)

:

ℒ

⁢

{

f

′′

⁢

(

t

)

}

=

s

⁢

[

s

⁢

F

⁢

(

s

)

-

lim

t

→

0

+

⁡

f

⁢

(

t

)

]

-

lim

t

→

0

+

⁡

f

′

⁢

(

t

)

=

s

2

⁢

F

⁢

(

s

)

-

s

⁢

f

⁢

(

0

+

)

-

f

′

⁢

(

0

+

)

Here the short notation

0

+

has been used for the right limits.

Further, one can use the rule to

f

′′

⁢

(

t

)

, getting

ℒ

⁢

{

f

′′′

⁢

(

t

)

}

=

s

⁢

[

s

2

⁢

F

⁢

(

s

)

-

s

⁢

f

⁢

(

0

+

)

-

f

′

⁢

(

0

+

)

]

-

f

′′

⁢

(

0

+

)

=

s

3

⁢

F

⁢

(

s

)

-

s

2

⁢

f

⁢

(

0

+

)

-

s

⁢

f

′

⁢

(

0

+

)

-

f

′′

⁢

(

0

+

)

.

Continuing similarly, one comes to the general formula

ℒ

⁢

{

f

(

n

)

⁢

(

t

)

}

=

s

n

⁢

F

⁢

(

s

)

-

s

n

-

1

⁢

f

⁢

(

0

+

)

-

s

n

-

2

⁢

f

′

⁢

(

0

+

)

-

…

-

f

(

n

-

1

)

⁢

(

0

+

)

.

(2)

Use of (2) requires that

f

⁢

(

t

)

,

f

′

⁢

(

t

)

,

f

′′

⁢

(

t

)

, …,

f

(

n

)

⁢

(

t

)

are
Laplace-transformable and that

f

⁢

(

t

)

,

f

′

⁢

(

t

)

,

f

′′

⁢

(

t

)

, …,

f

(

n

-

1

)

⁢

(

t

)

are

continuous

when

t

>

0

(not only

piecewise continuous

).

Remark.

Suppose that

f

⁢

(

t

)

and

f

′

⁢

(

t

)

are
Laplace-transformable and that

f

⁢

(

t

)

is continuous for

t

>

0

except the point

t

=

a

where the function has a
finite

jump discontinuity

.  Then

ℒ

⁢

{

f

′

⁢

(

t

)

}

=

s

⁢

F

⁢

(

s

)

-

f

⁢

(

0

+

)

-

e

-

a

⁢

s

⁢

(

lim

s

→

a

+

⁡

f

⁢

(

s

)

-

lim

s

→

a

-

⁡

f

⁢

(

s

)

)

.

Application.

Derive the Laplace transform of

sin

⁡

a

⁢

t

using the

derivatives

of sine (cf.

Laplace transform of cosine and sine

).

We have

f

⁢

(

t

)

:=

sin

⁡

a

⁢

t

,

f

′

⁢

(

t

)

=

a

⁢

cos

⁡

a

⁢

t

,

f

′′

⁢

(

t

)

=

-

a

2

⁢

sin

⁡

a

⁢

t

.

Using (2) with

n

=

2

we obtain

ℒ

⁢

{

-

a

2

⁢

sin

⁡

a

⁢

t

}

=

s

2

⁢

ℒ

⁢

{

sin

⁡

a

⁢

t

}

-

s

⁢

sin

⁡

0

-

a

⁢

cos

⁡

0

,

i.e.

-

a

2

⁢

ℒ

⁢

{

sin

⁡

a

⁢

t

}

=

s

2

⁢

ℒ

⁢

{

sin

⁡

a

⁢

t

}

-

a

,

which implies

ℒ

⁢

{

sin

⁡

a

⁢

t

}

=

a

s

2

+

a

2

.

Title

Laplace transforms of derivatives

Canonical name

LaplaceTransformsOfDerivatives

Date of creation

2014-04-06 8:24:33

Last modified on

2014-04-06 8:24:33

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Topic

Classification

msc 44A10

Generated on Fri Feb 9 21:32:37 2018 by

LaTeXML
Maximal ergodic theorem

Maximal ergodic theorem

Let

(

X

,

ℬ

,

μ

)

be a

probability space

and

T

:

X

→

X

a

measure preserving

transformation. Let

f

be a

L

1

⁢

(

μ

)

function.
Define the averages

f

*

⁢

(

x

)

=

sup

N

≥

1

⁡

1

N

⁢

∑

i

=

0

N

-

1

f

⁢

(

T

i

⁢

(

x

)

)

Then, for any

λ

∈

𝐑

, we have:

∫

f

*

>

λ

f

d

μ

≥

λ

μ

(

{

f

*

>

λ

}

)

This theorem may be used in the proof of the

ergodic theorem

(also known as

Birkhoff ergodic theorem

, or pointwise or strong ergodic theorem)

Title

Maximal ergodic theorem

Canonical name

MaximalErgodicTheorem

Date of creation

2014-03-19 22:15:48

Last modified on

2014-03-19 22:15:48

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

3

Author

Filipe (28191)

Entry type

Theorem

Related topic

birkhoff ergodic theorem

Generated on Fri Feb 9 21:32:39 2018 by

LaTeXML
Mellin’s inverse formula

Mellin’s inverse formula

It may be proven, that if a

function

F

⁢

(

s

)

has the

inverse Laplace transform

f

⁢

(

t

)

, i.e. a

piecewise continuous

and exponentially real function

f

satisfying the condition

ℒ

⁢

{

f

⁢

(

t

)

}

=

F

⁢

(

s

)

,

then

f

⁢

(

t

)

is uniquely determined when not regarded as different such functions which differ from each other only in a point set having

Lebesgue measure

zero.

The inverse Laplace transform is directly given by

Mellin’s

inverse

formula

f

⁢

(

t

)

=

1

2

⁢

π

⁢

i

⁢

∫

γ

-

i

⁢

∞

γ

+

i

⁢

∞

e

s

⁢

t

⁢

F

⁢

(

s

)

⁢

𝑑

s

,

by the Finn R. H. Mellin (1854—1933).  Here it must be integrated along a

straight line

parallel

to the

imaginary axis

and intersecting the

real axis

in the point

γ

which must be chosen so that it is

greater

than the

real parts

of all singularities of

F

⁢

(

s

)

.

In practice, computing the

complex integral

can be done by using the

Cauchy residue theorem

.

Title

Mellin’s inverse formula

Canonical name

MellinsInverseFormula

Date of creation

2013-03-22 14:23:02

Last modified on

2013-03-22 14:23:02

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

13

Author

pahio (2872)

Entry type

Result

Classification

msc 44A10

Synonym

inverse Laplace transformation

Synonym

Bromwich integral

Synonym

Fourier-Mellin integral

Related topic

InverseLaplaceTransformOfDerivatives

Related topic

HjalmarMellin

Related topic

TelegraphEquation

Generated on Fri Feb 9 21:32:40 2018 by

LaTeXML
Proof of Fekete’s subadditive lemma

Proof of Fekete’s subadditive lemma

If there is a

m

such that

a

m

=

-

∞

, then, by subadditivity, we have

a

n

=

-

∞

for all

n

>

m

. Then, both sides of the equality are

-

∞

, and the

theorem

holds.
So, we suppose that

a

n

∈

𝐑

for all

n

. Let

L

=

inf

n

⁡

a

n

n

and let

B

be any number greater than

L

. Choose

k

≥

1

such that

a

k

k

<

B

For

n

>

k

, we have, by the

division algorithm

there are

integers

p

n

and

q

n

such that

n

=

p

n

⁢

k

+

q

n

, and

0

≤

q

n

≤

k

-

1

.
Applying the

definition

of subadditivity many times we obtain:

a

n

=

a

p

n

⁢

k

+

q

n

≤

a

p

n

⁢

k

+

a

q

n

≤

p

n

⁢

a

k

+

a

q

n

So, dividing by

n

we obtain:

a

n

n

≤

p

n

⁢

k

n

⁢

a

k

k

+

a

q

n

n

When

n

goes to

infinity

,

p

n

⁢

k

n

converges to

1

and

a

q

n

n

converges to zero, because the

numerator

is

bounded

by the maximum of

a

i

with

0

≤

i

≤

k

-

1

. So, we have, for all

B

>

L

:

L

≤

lim

n

⁡

a

n

n

≤

a

k

k

<

B

Finally, let

B

go to

L

and we obtain

L

=

inf

n

⁡

a

n

n

=

lim

n

⁡

a

n

n

Title

Proof of Fekete’s subadditive lemma

Canonical

name

ProofOfFeketesSubadditiveLemma

Date of creation

2014-03-18 14:41:50

Last modified on

2014-03-18 14:41:50

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

3

Author

Filipe (28191)

Entry type

Proof

Generated on Fri Feb 9 21:32:43 2018 by

LaTeXML
rectification of antiperiodic function

rectification of antiperiodic function

Let the positive number

p

be the antiperiod of the

real function

f

and

f

⁢

(

t

)

≧

0

for

⁢

0

<

t

<

p

.

Then the

function

f

1

defined by

f

1

⁢

(

t

)

:=

max

⁡

{

f

⁢

(

t

)

,

0

}

=

{

f

⁢

(

t

)

for

⁢

f

⁢

(

t

)

>

0

,

0

for

⁢

f

⁢

(

t

)

≦

0

is the

half-wave rectification

of

f

and the function

f

2

defined by

f

2

⁢

(

t

)

:=

|

f

⁢

(

t

)

|

is the

full-wave rectification

of

f

.  They are periodic (

http://planetmath.org/PeriodicFunctions

), the former with period (

http://planetmath.org/PeriodicFunctions

)

2

⁢

p

and the latter with

p

.

The

Laplace transforms

are

ℒ

⁢

{

f

1

⁢

(

t

)

}

=

1

1

-

e

-

p

⁢

s

⁢

F

⁢

(

s

)

,

ℒ

⁢

{

f

2

⁢

(

t

)

}

=

1

+

e

-

p

⁢

s

1

-

e

-

p

⁢

s

⁢

F

⁢

(

s

)

.

Title

rectification of antiperiodic function

Canonical name

RectificationOfAntiperiodicFunction

Date of creation

2013-03-22 18:58:19

Last modified on

2013-03-22 18:58:19

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Definition

Classification

msc 44A10

Classification

msc 26A99

Synonym

rectification

Related topic

MaximalNumber

Related topic

LaplaceTransformOfPeriodicFunctions

Related topic

MinimalAndMaximalNumber

Defines

half-wave rectification

Defines

full-wave rectification

Generated on Fri Feb 9 21:32:46 2018 by

LaTeXML
relative of cosine integral

relative of cosine integral

For determining of the value of the improper integral

I

(

a

)

:=

∫

0

∞

cos

⁡

a

⁢

x

2

-

cos

⁡

a

⁢

x

x

d

x

(

a

>

0

)

,

related to the

cosine integral

, we think it as a

function

of the parametre

a

which we denote by

t

.  Then we can take the

Laplace transform

(see the integration with respect to a parametre in the table of Laplace transforms):

ℒ

⁢

{

I

⁢

(

t

)

}

=

ℒ

⁢

{

∫

0

∞

(

cos

⁡

t

⁢

x

2

-

cos

⁡

t

⁢

x

)

⁢

d

⁢

x

x

}

=

∫

0

∞

(

s

s

2

+

x

4

-

s

s

2

+

x

2

)

⁢

d

⁢

x

x

Splitting the

fractional expressions

to

http://planetmath.org/node/5812

partial fractions and integrating give

ℒ

⁢

{

I

⁢

(

t

)

}

=

1

s

⁢

∫

0

∞

(

1

x

-

x

3

s

2

+

x

4

-

1

x

+

x

s

2

+

x

2

)

⁢

𝑑

x

=

1

s

⁢

/

x

=

0

∞

⁡

[

1

2

⁢

ln

⁡

(

s

2

+

x

2

)

-

1

4

⁢

ln

⁡

(

s

2

+

x

4

)

]

=

1

4

⁢

/

x

=

0

∞

⁡

ln

⁡

(

s

2

+

x

2

)

2

s

2

+

x

4

=

-

ln

⁡

s

2

⁢

s

.

As seen in the

http://planetmath.org/node/10588

table of Laplace transforms, the gotten expression is the Laplace transform of

γ

+

ln

⁡

t

2

=

I

⁢

(

t

)

(N.B.

ℒ

⁢

{

1

}

=

1

s

), and thus we have the result

I

⁢

(

a

)

=

γ

+

ln

⁡

a

2

.

Title

relative of cosine integral

Canonical name

RelativeOfCosineIntegral

Date of creation

2013-03-22 18:44:35

Last modified on

2013-03-22 18:44:35

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Example

Classification

msc 44A10

Related topic

SubstitutionNotation

Related topic

EulersConstant

Related topic

RelativeOfExponentialIntegral

Related topic

IntegrationOfLaplaceTransformWithRespectToParameter

Generated on Fri Feb 9 21:32:50 2018 by

LaTeXML
relative of exponential integral

relative of exponential integral

Let

a

and

b

be positive numbers.  We want to calculate the value of the improper integral

∫

0

∞

e

-

a

⁢

x

-

e

-

b

⁢

x

x

⁢

𝑑

x

(1)

related to the

exponential integral

.

The value may be found e.g. by utilising the

derivative

of the

integral

I

⁢

(

y

)

:=

∫

0

∞

e

-

x

⁢

y

⋅

e

-

a

⁢

x

-

e

-

b

⁢

x

x

⁢

𝑑

x

which can be formed by differentiating under the

integral sign

(

http://planetmath.org/DifferentiationUnderIntegralSign

):

I

′

⁢

(

y

)

=

∫

0

∞

e

-

x

⁢

y

⁢

(

-

x

)

⁢

e

-

a

⁢

x

-

e

-

b

⁢

x

x

⁢

𝑑

x

=

∫

0

∞

(

e

-

(

y

+

b

)

⁢

x

-

e

-

(

y

+

a

)

⁢

x

)

⁢

𝑑

x

=

/

x

=

0

∞

⁡

(

e

-

(

y

+

b

)

⁢

x

-

(

y

+

b

)

-

e

-

(

y

+

a

)

⁢

x

-

(

y

+

a

)

)

=

1

y

+

b

-

1

y

+

a

Thus,

I

⁢

(

y

)

=

ln

⁡

(

y

+

b

)

-

ln

⁡

(

y

+

a

)

=

ln

⁡

y

+

b

y

+

a

,

and the integral (1) has the value

I

⁢

(

0

)

=

ln

⁡

b

a

.

There is another method via

Laplace transforms

.  By the table of Laplace transforms, we have

ℒ

⁢

{

e

-

a

⁢

t

-

e

-

b

⁢

t

}

=

1

s

+

a

-

1

s

+

b

and therefore

ℒ

⁢

{

e

-

a

⁢

t

-

e

-

b

⁢

t

t

}

=

∫

s

∞

(

1

u

+

a

-

1

u

+

b

)

⁢

𝑑

u

=

/

u

=

s

∞

⁡

ln

⁡

u

+

a

u

+

b

=

ln

⁡

s

+

b

s

+

a

,

i.e.

∫

0

∞

e

-

s

⁢

t

⋅

e

-

a

⁢

t

-

e

-

b

⁢

t

t

⁢

𝑑

t

=

ln

⁡

s

+

b

s

+

a

.

Letting

s

→

0

+

,  this yields the equation

∫

0

∞

e

-

a

⁢

x

-

e

-

b

⁢

x

x

⁢

𝑑

x

=

ln

⁡

b

a

.

Title

relative of exponential integral

Canonical

name

RelativeOfExponentialIntegral

Date of creation

2013-03-22 18:44:20

Last modified on

2013-03-22 18:44:20

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

12

Author

pahio (2872)

Entry type

Example

Classification

msc 44A10

Classification

msc 26A36

Related topic

SubstitutionNotation

Related topic

RelativeOfCosineIntegral

Related topic

IntegrationOfLaplaceTransformWithRespectToParameter

Related topic

IntegrationUnderIntegralSign

Generated on Fri Feb 9 21:32:54 2018 by

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rules for Laplace transform

rules for Laplace transform

If

ℒ

⁢

{

f

⁢

(

t

)

}

=

F

⁢

(

s

)

,  then

•

ℒ

⁢

{

e

a

⁢

t

⁢

f

⁢

(

t

)

}

=

F

⁢

(

s

-

a

)

for

s

>

a

,

•

ℒ

⁢

{

f

⁢

(

t

a

)

}

=

a

⁢

F

⁢

(

a

⁢

s

)

for

a

>

0

.

For deriving these rules, we start from the definition of

Laplace transform

.  In the first case, we shall use the notation

s

-

a

=

r

:

ℒ

⁢

{

e

a

⁢

t

⁢

f

⁢

(

t

)

}

=

∫

0

∞

e

-

s

⁢

t

⁢

e

a

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

∫

0

∞

e

-

(

s

-

a

)

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

∫

0

∞

e

-

r

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

F

⁢

(

r

)

=

F

⁢

(

s

-

a

)

.

In the second case, we make the change of variable

t

a

=

u

and later use the notation

s

⁢

a

=

r

:

ℒ

⁢

{

f

⁢

(

t

a

)

}

=

∫

0

∞

e

-

s

⁢

t

⁢

f

⁢

(

t

a

)

⁢

𝑑

t

=

a

⁢

∫

0

∞

e

-

s

⁢

a

⁢

u

⁢

f

⁢

(

u

)

⁢

𝑑

u

=

a

⁢

∫

0

∞

e

-

r

⁢

u

⁢

f

⁢

(

u

)

⁢

𝑑

u

=

a

⁢

F

⁢

(

r

)

=

a

⁢

F

⁢

(

a

⁢

s

)

.

Title

rules for Laplace transform

Canonical name

RulesForLaplaceTransform

Date of creation

2013-03-22 18:31:08

Last modified on

2013-03-22 18:31:08

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Related topic

TableOfLaplaceTransforms

Generated on Fri Feb 9 21:32:56 2018 by

LaTeXML
sine integral at infinity

sine integral at infinity

The value of the improper integral (one of the

Dirichlet integrals

)

∫

0

∞

sin

⁡

x

x

⁢

𝑑

x

=

lim

x

→

∞

⁡

Si

⁡

x

,

where Si means the

sine integral

(

http://planetmath.org/SineIntegral

)

function

, is most simply determined by using

Laplace transform

which may be aimed to the integrand (see

integration of Laplace transform with respect to parameter

).  Therefore the integrand must be equipped with an additional parametre

t

:

ℒ

⁢

{

∫

0

∞

1

x

⁢

sin

⁡

t

⁢

x

⁢

d

⁢

x

}

=

∫

0

∞

1

x

⋅

x

s

2

+

x

2

⁢

𝑑

x

=

∫

0

∞

d

⁢

x

s

2

+

x

2

=

1

s

⁢

/

x

=

0

∞

⁡

arctan

⁡

x

s

=

π

2

⋅

1

s

The obtained

transform

π

2

⋅

1

s

corresponds (see the

inverse Laplace transformation

) to the function

t

↦

π

2

because

ℒ

⁢

{

1

}

=

1

s

.  Thus we have the result

∫

0

∞

sin

⁡

x

x

⁢

𝑑

x

=

π

2

.

(1)

Note 1.

Since

x

↦

sin

⁡

x

x

or

x

↦

sinc

⁡

x

is an

even function

, the result (1) may be written also

∫

-

∞

∞

sinc

⁡

x

⁢

d

⁢

x

=

π

;

see the

sinc

-function (

http://planetmath.org/SincFunction

).

Note 2.

The result (1) may be easily generalised to

∫

0

∞

sin

⁡

a

⁢

x

x

d

x

=

π

2

(

a

>

0

)

(2)

and to

∫

0

∞

sin

⁡

a

⁢

x

x

d

x

=

(

sgn

a

)

π

2

(

a

∈

ℝ

)

.

(3)

Title

sine integral at infinity

Canonical name

SineIntegralAtInfinity

Date of creation

2013-03-22 15:17:22

Last modified on

2013-03-22 15:17:22

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

18

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Classification

msc 30A99

Synonym

limit of sine integral

Related topic

SineIntegral

Related topic

SincFunction

Related topic

SubstitutionNotation

Related topic

IncompleteGammaFunction

Related topic

ExampleOfSummationByParts

Related topic

SignumFunction

Generated on Fri Feb 9 21:32:59 2018 by

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Structural stability theorem

Structural stability theorem

Let

M

be a compact

differentiable manifold

and let

f

:

M

→

M

be a

C

r

diffeomorphism

.

Then,

f

is

structurally stable

in the

C

1

topology

if and only if it is Axiom A and satisfies the strong transversality condition.
The question of knowing if this is valid for the

C

r

topology,

r

≥

2

is still open.

Title

Structural stability theorem

Canonical

name

StructuralStabilityTheorem

Date of creation

2014-03-19 22:14:32

Last modified on

2014-03-19 22:14:32

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

5

Author

Filipe (28191)

Entry type

Theorem

Synonym

stability conjecture

Related topic

Ω

-stability theorem

Generated on Fri Feb 9 21:33:01 2018 by

LaTeXML
Von Neumann’s ergodic theorem

Von Neumann’s ergodic theorem

Let

U

:

H

→

H

be an

isometry

in a Hilbert space

H

. Consider the

subspace

I

⁢

(

U

)

=

{

v

∈

H

:

U

⁢

v

=

v

}

, called the space of invariant vectors. Denote by

P

the

orthogonal projection

over the subspace

I

⁢

(

U

)

. Then,

lim

n

→

∞

⁡

1

n

⁢

∑

j

=

0

n

-

1

U

j

⁢

(

v

)

=

P

⁢

(

v

)

,

∀

v

∈

H

This general theorem for Hilbert spaces can be used to obtain an

ergodic theorem

for the

L

2

⁢

(

μ

)

space by taking

H

to be the

L

2

⁢

(

μ

)

space, and

U

to be the composition operator (also called Koopman operator) associated to a

transformation

f

:

M

→

M

that preserves a measure

μ

, i.e.,

U

f

⁢

(

ψ

)

=

ψ

∘

f

, where

ψ

:

M

→

𝐑

. The space of invariant functions is the set of functions

ψ

such that

ψ

∘

f

=

ψ

almost everywhere. For any

ψ

∈

L

2

⁢

(

μ

)

, the

sequence

:

lim

n

→

∞

⁡

1

n

⁢

∑

j

=

0

n

-

1

ψ

∘

f

j

converges in

L

2

⁢

(

μ

)

to the orthogonal projection

ψ

~

of the function

ψ

over the space of invariant functions.

The

L

2

⁢

(

μ

)

version of the ergodic theorem for Hilbert spaces can be derived directly from the more general

Birkhoff ergodic theorem

, which asserts pointwise convergence instead of convergence in

L

2

⁢

(

μ

)

. Actually, from Birkhoff ergodic theorem one can derive a version of the ergodic theorem where convergence in

L

p

⁢

(

μ

)

holds, for any

p

>

1

.

Title

Von Neumann’s ergodic theorem

Canonical name

VonNeumannsErgodicTheorem

Date of creation

2014-03-18 14:02:09

Last modified on

2014-03-18 14:02:09

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

6

Author

Filipe (28191)

Entry type

Theorem

Related topic

Birkhoff ergodic theorem

Generated on Fri Feb 9 21:33:13 2018 by

LaTeXML
Mellin transform

Mellin transform

The

Mellin transform

is an

integral transform

defined as follows:

F

⁢

(

s

)

=

∫

0

∞

f

⁢

(

t

)

⁢

t

s

-

1

⁢

𝑑

t

Intuitively, it may be viewed as a

continuous

analogue of a power series
— instead of synthetizing a

function

by summing multiples of integer
powers, we integrate over all real powers. This

transform

is closely
related to the

Laplace transform

— if we make a change of variables

t

=

e

-

r

and define

g

by

f

⁢

(

e

-

r

)

=

g

⁢

(

r

)

, then the above

integral

becomes

F

⁢

(

s

)

=

-

∫

-

∞

+

∞

g

⁢

(

r

)

⁢

e

-

r

⁢

s

⁢

𝑑

r

,

which is a

bilateral Laplace transform

.

(more to come)

Title

Mellin transform

Canonical name

MellinTransform

Date of creation

2015-02-17 15:10:57

Last modified on

2015-02-17 15:10:57

Owner

rspuzio (6075)

Last modified by

pahio (2872)

Numerical id

7

Author

rspuzio (2872)

Entry type

Definition

Classification

msc 44A15

Generated on Fri Feb 9 21:33:15 2018 by

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