aarjaneiro/planetmath huggingface dataset:
integration of Laplace transform with respect to parameter

integration of Laplace transform with respect to parameter

We use the curved from the Laplace-transformed

functions

to the corresponding initial functions.

If

f

⁢

(

t

,

x

)

↶

F

⁢

(

s

,

x

)

,

then one can

integrate

both functions with respect to the

parametre

x

between the same which may be also infinite provided that the

integrals

converge:

∫

a

b

f

⁢

(

t

,

x

)

⁢

𝑑

x

↶

∫

a

b

F

⁢

(

s

,

x

)

⁢

𝑑

x

(1)

(1) may be written as

ℒ

⁢

{

∫

a

b

f

⁢

(

t

,

x

)

⁢

𝑑

x

}

=

∫

a

b

ℒ

⁢

{

f

⁢

(

t

,

x

)

}

⁢

𝑑

x

.

(2)

Proof.

Using the

definition

of the

Laplace transform

, we can write

∫

a

b

f

⁢

(

t

,

x

)

⁢

𝑑

x

↶

∫

0

∞

(

e

-

s

⁢

t

⁢

∫

a

b

f

⁢

(

s

,

x

)

⁢

𝑑

x

)

⁢

𝑑

t

=

∫

0

∞

(

∫

a

b

e

-

s

⁢

t

⁢

f

⁢

(

s

,

x

)

⁢

𝑑

x

)

⁢

𝑑

t

.

We change the of integration in the last double integral and use again the definition, obtaining

∫

a

b

f

⁢

(

t

,

x

)

⁢

𝑑

x

↶

∫

a

b

(

∫

0

∞

e

-

s

⁢

t

⁢

f

⁢

(

s

,

x

)

⁢

𝑑

t

)

⁢

𝑑

x

=

∫

a

b

F

⁢

(

s

,

t

)

⁢

𝑑

x

,

Q.E.D.

Title

integration of Laplace transform with respect to parameter

Canonical

name

IntegrationOfLaplaceTransformWithRespectToParameter

Date of creation

2013-03-22 18:44:47

Last modified on

2013-03-22 18:44:47

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Theorem

Classification

msc 44A10

Related topic

TableOfLaplaceTransforms

Related topic

TermwiseDifferentiation

Related topic

MethodsOfEvaluatingImproperIntegrals

Related topic

UsingConvolutionToFindLaplaceTransform

Related topic

RelativeOfCosineIntegral

Related topic

RelativeOfExponentialIntegral

Generated on Fri Feb 9 21:31:48 2018 by

LaTeXML
inverse Laplace transform of derivatives

inverse Laplace transform of derivatives

It may be shown that the

Laplace transform

F

⁢

(

s

)

=

∫

0

∞

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

is always

differentiable

and that its

derivative

can be formed by differentiating under the integral sign (

http://planetmath.org/DifferentiationUnderIntegralSign

), i.e. one has

F

′

⁢

(

s

)

=

∫

0

∞

∂

⁡

(

e

-

s

⁢

t

⁢

f

⁢

(

t

)

)

∂

⁡

s

⁢

𝑑

t

=

∫

0

∞

e

-

s

⁢

t

⁢

(

-

t

)

⁢

f

⁢

(

t

)

⁢

𝑑

t

.

This gives the rule

ℒ

-

1

⁢

{

F

′

⁢

(

s

)

}

=

-

t

⁢

f

⁢

(

t

)

.

(1)

Applying (1) to

F

′

⁢

(

s

)

instead of

F

⁢

(

s

)

gives

ℒ

-

1

⁢

{

F

′′

⁢

(

s

)

}

=

t

2

⁢

f

⁢

(

t

)

.

Continuing this way we can obtain the general rule

ℒ

-

1

⁢

{

F

(

n

)

⁢

(

s

)

}

=

(

-

1

)

n

⁢

t

n

⁢

f

⁢

(

t

)

,

(2)

or equivalently

ℒ

⁢

{

t

n

⁢

f

⁢

(

t

)

}

=

(

-

1

)

n

⋅

d

n

⁢

ℒ

⁢

{

f

⁢

(

t

)

}

d

⁢

s

n

,

(3)

for any

n

=

1

,

2

,

3

,

…

(and of course for

n

=

0

).

Example.

Let’s find the Laplace transform of the

first kind

and 0th

Bessel function

J

0

⁢

(

t

)

:=

∑

m

=

0

∞

(

-

1

)

m

(

m

!

)

2

⁢

(

t

2

)

2

⁢

m

,

which is the solution

y

⁢

(

t

)

of the Bessel’s equation

t

⁢

y

′′

⁢

(

t

)

+

y

′

⁢

(

t

)

+

t

⁢

y

⁢

(

t

)

=

0

(4)

satisfying the

initial condition

y

⁢

(

0

)

=

1

.  The equation implies that

y

′

⁢

(

0

)

=

0

.

By (3), the Laplace transform of the

differential equation

(4) is

-

d

⁢

ℒ

⁢

{

y

′′

⁢

(

t

)

}

d

⁢

s

+

ℒ

⁢

{

y

′

⁢

(

t

)

}

-

d

⁢

ℒ

⁢

{

y

⁢

(

t

)

}

d

⁢

s

=

0

.

Using here twice the rule 5 in the parent (

http://planetmath.org/LaplaceTransform

) entry gives us

-

d

⁢

(

s

2

⁢

Y

⁢

(

s

)

-

s

)

d

⁢

s

+

s

⁢

Y

⁢

(

s

)

-

1

-

d

⁢

Y

⁢

(

s

)

d

⁢

s

=

0

,

which is simplified to

(

s

2

+

1

)

⁢

d

⁢

Y

d

⁢

s

+

s

⁢

Y

=

0

,

i.e. to

d

⁢

Y

Y

=

-

s

⁢

d

⁢

s

s

2

+

1

.

Integrating this gives

ln

⁡

Y

=

-

1

2

⁢

ln

⁡

(

s

2

+

1

)

+

ln

⁡

C

=

ln

⁡

C

s

2

+

1

,

i.e.

Y

⁢

(

s

)

=

C

s

2

+

1

.

The initial condition enables to justify that the integration constant

C

must be 1.  Thus we have the result

ℒ

⁢

{

J

0

⁢

(

t

)

}

=

1

s

2

+

1

.

References

1

K. Väisälä:

Laplace-muunnos

.  Handout Nr. 163. Teknillisen korkeakoulun ylioppilaskunta, Otaniemi, Finland (1968).

Title

inverse Laplace transform of derivatives

Canonical name

InverseLaplaceTransformOfDerivatives

Date of creation

2013-03-22 16:46:27

Last modified on

2013-03-22 16:46:27

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

11

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Synonym

differentiation of Laplace transform

Related topic

MellinsInverseFormula

Related topic

SeparationOfVariables

Related topic

KalleVaisala

Related topic

TableOfLaplaceTransforms

Generated on Fri Feb 9 21:31:52 2018 by

LaTeXML
Kac’s theorem

Kac’s theorem

Let

f

:

M

→

M

be a

transformation

and

μ

a finite invariant measure for

f

. Let

E

be a subset of

M

with

positive measure

. We define the first return map for

E

:

ρ

E

⁢

(

x

)

=

min

⁡

{

n

≥

1

:

f

n

⁢

(

x

)

∈

E

}

If the set on the right is empty, then we define

ρ

E

⁢

(

x

)

=

∞

. The PoincarÃ© recurrence

theorem

asserts that

ρ

E

is finite for almost every

x

∈

R

.
We define the following sets:

E

0

=

{

x

∈

E

:

f

n

⁢

(

x

)

∉

E

,

n

≥

1

}

E

0

*

=

{

x

∈

M

:

f

n

⁢

(

x

)

∉

E

,

n

≥

0

}

By PoincarÃ© recurrence theorem,

μ

⁢

(

E

0

)

=

0

.
Kac’s theorem asserts that the

function

ρ

E

is integrable and

∫

E

ρ

E

⁢

𝑑

μ

=

μ

⁢

(

M

)

-

μ

⁢

(

E

0

*

)

When the system is

ergodic

, then

μ

⁢

(

E

0

*

)

=

0

, and Kac’s theorem implies:

1

μ

⁢

(

E

)

⁢

∫

E

ρ

E

⁢

𝑑

μ

=

μ

⁢

(

M

)

μ

⁢

(

E

)

This equality can be interpreted as: the mean return time to

E

s inversely proportional to the

measure

of

E

.

Title

Kac’s theorem

Canonical name

KacsTheorem

Date of creation

2014-03-19 22:18:04

Last modified on

2014-03-19 22:18:04

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

4

Author

Filipe (28191)

Entry type

Theorem

Related topic

PoincarÃ© Recurrence theorem

Generated on Fri Feb 9 21:32:05 2018 by

LaTeXML
Kingman’s subadditive ergodic theorem

Kingman’s subadditive ergodic theorem

Let

(

M

,

𝒜

,

μ

)

be a

probability space

, and

f

:

M

→

M

be a

measure preserving

dynamical system

. let

ϕ

n

:

M

→

𝐑

,

n

≥

1

be a

subadditive

sequence of

measurable functions

, such that

ϕ

1

+

is integrable, where

ϕ

1

+

=

max

⁡

{

ϕ

,

0

}

. Then, the sequence

(

ϕ

n

n

)

n

converges

μ

almost everywhere to a function

ϕ

:

M

→

[

-

∞

,

∞

)

such that:

ϕ

+

is integrable

ϕ

is

f

invariant, that is,

ϕ

⁢

(

f

⁢

(

x

)

)

=

ϕ

⁢

(

x

)

for

μ

almost all

x

, and

∫

ϕ

⁢

𝑑

μ

=

lim

n

⁡

1

n

⁢

∫

ϕ

n

⁢

𝑑

μ

=

inf

n

⁡

1

n

⁢

∫

ϕ

n

⁢

𝑑

μ

∈

[

-

∞

,

∞

)

The fact that the limit equals the infimum is a consequence of the fact that the sequence

∫

ϕ

n

⁢

𝑑

μ

is a subadditive sequence and Fekete’s subadditive lemma.

A superadditive version of the theorem also exists. Given a superadditive sequence

φ

n

, then the symmetric sequence is subadditive and we may apply the original version of the theorem.

Every additive sequence is subadditive. As a consequence, one can prove the

Birkhoff ergodic theorem

from Kingman’s subadditive ergodic theorem.

Title

Kingman’s subadditive ergodic theorem

Canonical name

KingmansSubadditiveErgodicTheorem

Date of creation

2014-03-18 14:34:03

Last modified on

2014-03-18 14:34:03

Owner

Filipe (28191)

Last modified by

Filipe (28191)

Numerical id

5

Author

Filipe (28191)

Entry type

Theorem

Related topic

birkhoff ergodic theorem

Generated on Fri Feb 9 21:32:07 2018 by

LaTeXML
Laplace transform

Laplace transform

Let

f

⁢

(

t

)

be a

function

defined on the interval

[

0

,

∞

)

. The

Laplace transform

of

f

⁢

(

t

)

is the function

F

⁢

(

s

)

defined by

F

⁢

(

s

)

=

∫

0

∞

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

,

provided that the

integral

converges.

1

1

Depending on the

definition

of
integral one is using, one may prefer to define the Laplace transform as

lim

x

→

0

+

⁡

∫

x

∞

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

It suffices that

f

be defined when

t

>

0

and

s

can be complex. We will
usually denote the Laplace transform of

f

by

ℒ

⁢

{

f

}

. Some
of the most common Laplace transforms are:

1.

ℒ

⁢

{

e

a

⁢

t

}

=

1

s

-

a

,

s

>

a

2.

ℒ

⁢

{

cos

⁡

(

b

⁢

t

)

}

=

s

s

2

+

b

2

,

s

>

0

3.

ℒ

⁢

{

sin

⁡

(

b

⁢

t

)

}

=

b

s

2

+

b

2

,

s

>

0

4.

ℒ

⁢

{

t

n

}

=

Γ

⁢

(

n

+

1

)

s

n

+

1

,

s

>

0

,

n

>

-

1

.

5.

ℒ

⁢

{

f

′

}

=

s

⁢

ℒ

⁢

{

f

}

-

lim

x

→

0

+

⁡

f

⁢

(

x

)

For more particular Laplace transforms, see the table of Laplace
transforms.

Notice the Laplace transform is a linear transformation. It is worth noting that, if

∫

0

∞

e

-

s

⁢

t

⁢

|

f

⁢

(

t

)

|

⁢

𝑑

t

<

∞

for some

s

∈

ℝ

, then

ℒ

⁢

{

f

}

is an analytic function in the complex half-plane

{

z

⁢

∣

ℜ

⁡

z

>

⁢

s

}

.

Much like the

Fourier transform

, the Laplace transform has a

convolution

. However, the form of the convolution used is different.

ℒ

⁢

{

f

*

g

}

=

ℒ

⁢

{

f

}

⁢

ℒ

⁢

{

g

}

where

(

f

*

g

)

⁢

(

t

)

=

∫

0

t

f

⁢

(

t

-

s

)

⁢

g

⁢

(

s

)

⁢

𝑑

s

and

ℒ

⁢

{

f

⁢

g

}

⁢

(

s

)

=

∫

c

-

i

⁢

∞

c

+

i

⁢

∞

ℒ

⁢

{

f

}

⁢

(

z

)

⁢

ℒ

⁢

{

g

}

⁢

(

s

-

z

)

⁢

𝑑

z

The most popular usage of the Laplace transform is to solve

initial value problems

by taking the Laplace transform of both
sides of an

ordinary differential equation

; see the entry
“

image equation

(

http://planetmath.org/ImageEquation

)”.

Title

Laplace transform

Canonical

name

LaplaceTransform

Date of creation

2014-03-10 10:50:28

Last modified on

2014-03-10 10:50:28

Owner

rspuzio (6075)

Last modified by

pahio (2872)

Numerical id

26

Author

rspuzio (2872)

Entry type

Definition

Classification

msc 44A10

Related topic

DiscreteFourierTransform

Related topic

UsingLaplaceTransformToInitialValueProblems

Related topic

UsingLaplaceTransformToSolveHeatEquation

Generated on Fri Feb 9 21:32:09 2018 by

LaTeXML
Laplace transform of cosine and sine

Laplace transform of cosine and sine

We start from the easily formula

e

α

⁢

t

↶

1

s

-

α

(

s

>

α

)

,

(1)

where the curved from the Laplace-transformed

function

to the original function.  Replacing

α

by

-

α

we can write the second formula

e

-

α

⁢

t

↶

1

s

+

α

(

s

>

-

α

)

.

(2)

Adding (1) and (2) and dividing by 2 we obtain (remembering the linearity of the

Laplace transform

)

e

α

⁢

t

+

e

-

α

⁢

t

2

↶

1

2

⁢

(

1

s

-

α

+

1

s

+

α

)

,

i.e.

ℒ

⁢

{

cosh

⁡

α

⁢

t

}

=

s

s

2

-

α

2

.

(3)

Similarly, subtracting (1) and (2) and dividing by 2 give

ℒ

⁢

{

sinh

⁡

α

⁢

t

}

=

a

s

2

-

α

2

.

(4)

The formulae (3) and (4) are valid for

s

>

|

α

|

.

There are the

hyperbolic identities

cosh

⁡

i

⁢

t

=

cos

⁡

t

,

1

i

⁢

sinh

⁡

i

⁢

t

=

sin

⁡

t

which enable the transition from hyperbolic to

trigonometric functions

.  If we choose

α

:=

i

⁢

a

in (3), we may calculate

cos

⁡

a

⁢

t

=

cosh

⁡

i

⁢

a

⁢

t

↶

s

s

2

-

(

i

⁢

a

)

2

=

s

s

2

+

a

2

,

the formula (4) analogously gives

sin

⁡

a

⁢

t

=

1

i

⁢

sinh

⁡

i

⁢

a

⁢

t

↶

1

i

⋅

i

⁢

a

s

2

-

(

i

⁢

a

)

2

=

a

s

2

+

a

2

.

Accordingly, we have derived the Laplace transforms

ℒ

⁢

{

cos

⁡

a

⁢

t

}

=

s

s

2

+

a

2

,

(5)

ℒ

⁢

{

sin

⁡

a

⁢

t

}

=

a

s

2

+

a

2

,

(6)

which are true for

s

>

0

.

Title

Laplace transform of cosine and sine

Canonical name

LaplaceTransformOfCosineAndSine

Date of creation

2013-03-22 18:18:27

Last modified on

2013-03-22 18:18:27

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Synonym

Laplace transform of sine and cosine

Generated on Fri Feb 9 21:32:12 2018 by

LaTeXML
Laplace transform of derivative

Laplace transform of derivative

Theorem

.

If the

real function

t

↦

f

⁢

(

t

)

and its

derivative

are Laplace-transformable and

f

is

continuous

for

t

>

0

,  then

ℒ

⁢

{

f

′

⁢

(

t

)

}

=

s

⁢

F

⁢

(

s

)

-

lim

t

→

0

+

⁡

f

⁢

(

t

)

.

(1)

Proof.

By the

definition

of

Laplace transform

and using

integration by parts

, the left hand side of (1) may be written

∫

0

∞

e

-

s

⁢

t

⁢

f

′

⁢

(

t

)

⁢

𝑑

t

=

/

t

=

0

∞

⁡

e

-

s

⁢

t

⁢

f

⁢

(

t

)

+

s

⁢

∫

0

∞

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

=

lim

t

→

∞

⁡

e

-

s

⁢

t

⁢

f

⁢

(

t

)

-

lim

t

→

0

⁡

e

-

s

⁢

t

⁢

f

⁢

(

t

)

+

s

⁢

F

⁢

(

s

)

.

The Laplace-transformability of

f

implies that

e

-

s

⁢

t

⁢

f

⁢

(

t

)

tends to zero as

t

increases boundlessly.  Thus the last

expression

leads to the right hand side of (1).

Title

Laplace transform of derivative

Canonical

name

LaplaceTransformOfDerivative

Date of creation

2013-03-22 18:24:54

Last modified on

2013-03-22 18:24:54

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Theorem

Classification

msc 44A10

Related topic

SubstitutionNotation

Generated on Fri Feb 9 21:32:15 2018 by

LaTeXML
Laplace transform of integral

Laplace transform of integral

On can show that if a

real function

t

↦

f

⁢

(

t

)

is
Laplace-transformable (

http://planetmath.org/LaplaceTransform

), as well
is

∫

0

t

f

⁢

(

τ

)

⁢

𝑑

τ

.  The latter is also

continuous

for

t

>

0

and by the
Newton–Leibniz formula (

http://planetmath.org/FundamentalTheoremOfCalculus

),
has the

derivative

equal

f

⁢

(

t

)

.  Hence we may apply the
formula for

Laplace transform of derivative

, obtaining

F

⁢

(

s

)

=

ℒ

⁢

{

f

⁢

(

t

)

}

=

s

⁢

ℒ

⁢

{

∫

0

t

f

⁢

(

τ

)

⁢

𝑑

τ

}

-

∫

0

0

f

⁢

(

t

)

⁢

𝑑

t

=

s

⁢

ℒ

⁢

{

∫

0

t

f

⁢

(

τ

)

⁢

𝑑

τ

}

,

i.e.

ℒ

⁢

{

∫

0

t

f

⁢

(

τ

)

⁢

𝑑

τ

}

=

F

⁢

(

s

)

s

.

(1)

Application.

We start from the easily derivable rule

1

s

↷

1

,

where the curved from the Laplace-transformed

function

to the original function.  The formula (1) thus yields successively

1

s

2

↷

∫

0

t

1

⁢

𝑑

τ

=

t

,

1

s

3

↷

∫

0

t

τ

⁢

𝑑

τ

=

t

2

2

!

,

1

s

4

↷

∫

0

t

τ

2

2

!

⁢

𝑑

τ

=

t

3

3

!

,

etc.  Generally, one has

1

s

n

↷

t

n

-

1

(

n

-

1

)

!

∀

n

∈

ℤ

+

.

(2)

Title

Laplace transform of integral

Canonical name

LaplaceTransformOfIntegral

Date of creation

2014-03-17 10:43:31

Last modified on

2014-03-17 10:43:31

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

8

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Generated on Fri Feb 9 21:32:17 2018 by

LaTeXML
Laplace transform of logarithm

Laplace transform of logarithm

Theorem.

The

Laplace transform

of the

natural logarithm

function

is

ℒ

⁢

{

ln

⁡

t

}

=

Γ

′

⁢

(

1

)

-

ln

⁡

s

s

where

Γ

is Euler’s

gamma function

.

Proof.

We use the Laplace transform of the

power function

(

http://planetmath.org/LaplaceTransformOfPowerFunction

)

∫

0

∞

e

-

s

⁢

t

⁢

t

a

⁢

𝑑

t

=

Γ

⁢

(

a

+

1

)

s

a

+

1

by differentiating it with respect to the parametre

a

:

∫

0

∞

e

-

s

⁢

t

⁢

t

a

⁢

ln

⁡

t

⁢

d

⁢

t

=

Γ

′

⁢

(

a

+

1

)

⁢

s

a

+

1

-

Γ

⁢

(

a

+

1

)

⁢

s

a

+

1

⁢

ln

⁡

s

(

s

a

+

1

)

2

=

Γ

′

⁢

(

a

+

1

)

-

Γ

⁢

(

a

+

1

)

⁢

ln

⁡

s

s

a

+

1

Setting here

a

=

0

,  we obtain

ℒ

⁢

{

ln

⁡

t

}

=

∫

0

∞

e

-

s

⁢

t

⁢

ln

⁡

t

⁢

d

⁢

t

=

Γ

′

⁢

(

1

)

-

1

⋅

ln

⁡

s

s

,

Q.E.D.

Note.

The number

Γ

′

⁢

(

1

)

is equal the of the Euler–Mascheroni constant (

http://planetmath.org/EulersConstant

), as is seen in the entry

digamma and polygamma functions

.

Title

Laplace transform of logarithm

Canonical name

LaplaceTransformOfLogarithm

Date of creation

2013-03-22 18:26:01

Last modified on

2013-03-22 18:26:01

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Theorem

Classification

msc 44A10

Synonym

Laplace transform of logarithm function

Related topic

PowerFunction

Generated on Fri Feb 9 21:32:19 2018 by

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Laplace transform of periodic functions

Laplace transform of periodic functions

Let

f

⁢

(

t

)

be

periodic

with the

positive

period (

http://planetmath.org/PeriodicFunctions

)

p

.  Denote by

H

⁢

(

t

)

the

Heaviside step function

.  If now

g

⁢

(

t

)

:=

f

⁢

(

t

)

⁢

H

⁢

(

t

)

-

f

⁢

(

t

-

p

)

⁢

H

⁢

(

t

-

p

)

,

then it follows

g

⁢

(

t

)

=

{

f

⁢

(

t

)

for

⁢

0

<

t

<

p

,

0

otherwise.

(1)

By the parent entry (

http://planetmath.org/DelayTheorem

), the

Laplace transform

of

g

is

G

⁢

(

s

)

=

F

⁢

(

s

)

-

e

-

p

⁢

s

⁢

F

⁢

(

s

)

,

whence

F

⁢

(

s

)

=

G

⁢

(

s

)

1

-

e

-

p

⁢

s

=

1

1

-

e

-

p

⁢

s

⁢

∫

0

∞

e

-

s

⁢

t

⁢

g

⁢

(

t

)

⁢

𝑑

t

=

1

1

-

e

-

p

⁢

s

⁢

∫

0

p

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

.

Thus we have the rule

ℒ

⁢

{

f

⁢

(

t

)

}

=

1

1

-

e

-

p

⁢

s

⁢

∫

0

p

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

(

period

⁢

p

)

.

(2)

On the contrary, if

f

⁢

(

t

)

is

antiperiodic

with positive antiperiod

p

, then the

function

g

⁢

(

t

)

:=

f

⁢

(

t

)

⁢

H

⁢

(

t

)

+

f

⁢

(

t

-

p

)

⁢

H

⁢

(

t

-

p

)

also has the property (1).  Analogically with the preceding procedure, one may derive the rule

ℒ

⁢

{

f

⁢

(

t

)

}

=

1

1

+

e

-

p

⁢

s

⁢

∫

0

p

e

-

s

⁢

t

⁢

f

⁢

(

t

)

⁢

𝑑

t

(

antiperiod

⁢

p

)

.

(3)

Title

Laplace transform of periodic functions

Canonical name

LaplaceTransformOfPeriodicFunctions

Date of creation

2013-03-22 18:58:24

Last modified on

2013-03-22 18:58:24

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

5

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Related topic

RectificationOfAntiperiodicFunction

Related topic

TableOfLaplaceTransforms

Generated on Fri Feb 9 21:32:22 2018 by

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Laplace transform of power function

Laplace transform of power function

In the defining

integral

(

http://planetmath.org/ImproperIntegral

)

ℒ

⁢

{

t

r

}

=

∫

0

∞

e

-

s

⁢

t

⁢

t

r

⁢

𝑑

t

of the

Laplace transform

of the

power function

t

↦

t

r

,  we make the substitution (

http://planetmath.org/SubstitutionForIntegration

)

u

:=

s

⁢

t

:

ℒ

⁢

{

t

r

}

=

∫

0

∞

e

-

u

⁢

(

u

s

)

r

⁢

d

⁢

u

s

=

1

s

n

+

1

⁢

∫

0

∞

e

-

u

⁢

u

r

+

1

-

1

⁢

𝑑

u

Here we have assumed that

r

>

-

1

and

s

>

0

.  According to the definition of the

gamma function

, the last integral is equal to

Γ

⁢

(

r

+

1

)

.  Thus we obtain

ℒ

⁢

{

t

r

}

=

Γ

⁢

(

r

+

1

)

s

r

+

1

.

(1)

The special case

r

=

-

1

2

gives the result

ℒ

⁢

{

1

t

}

=

π

s

.

(2)

Title

Laplace transform of power function

Canonical name

LaplaceTransformOfPowerFunction

Date of creation

2013-03-22 18:17:42

Last modified on

2013-03-22 18:17:42

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

6

Author

pahio (2872)

Entry type

Derivation

Classification

msc 44A10

Related topic

EvaluatingTheGammaFunctionAt12

Generated on Fri Feb 9 21:32:25 2018 by

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Laplace transform of sine integral

Laplace transform of sine integral

0.1

Derivation of

ℒ

⁢

{

Si

⁢

t

}

If one performs the change of integration variable

u

=

t

⁢

x

,

d

⁢

u

=

t

⁢

d

⁢

x

in the defining

integral

(

http://planetmath.org/DefiniteIntegral

)

Si

⁢

t

=

∫

0

t

sin

⁡

u

u

⁢

𝑑

u

,

of the

sine integral

function

, one obtains

Si

⁢

t

=

∫

0

1

sin

⁡

t

⁢

x

t

⁢

x

⁢

t

⁢

𝑑

x

=

∫

0

1

sin

⁡

t

⁢

x

x

⁢

𝑑

x

,

getting limits (

http://planetmath.org/UpperLimit

).  We know (see the entry

Laplace transform of sine and cosine

) that

ℒ

⁢

{

sin

⁡

t

⁢

x

x

}

=

1

s

2

+

x

2

.

This transformation formula can be integrated with respect to the parametre

x

:

ℒ

⁢

{

∫

0

1

sin

⁡

t

⁢

x

x

⁢

𝑑

x

}

=

∫

0

1

1

s

2

+

x

2

⁢

𝑑

x

=

1

s

⁢

/

x

=

0

1

⁡

arctan

⁡

x

s

=

1

s

⁢

arctan

⁡

1

s

.

Thus we have the transformation formula of the sinus integralis:

ℒ

⁢

{

Si

⁢

t

}

=

1

s

⁢

arctan

⁡

1

s

.

(1)

0.2

Laplace transform of sinc function

By the formula

ℒ

⁢

{

f

′

}

=

s

⁢

ℒ

⁢

{

f

}

-

lim

x

→

0

+

⁡

f

⁢

(

x

)

of the parent (

http://planetmath.org/LaplaceTransform

) entry,
we obtain as consequence of (1), that

ℒ

⁢

{

d

d

⁢

t

⁢

Si

⁢

t

}

=

s

⋅

1

s

⁢

arctan

⁡

1

s

-

Si

⁢

0

,

i.e.

ℒ

⁢

{

sin

⁡

t

t

}

=

arctan

⁡

1

s

.

(2)

The formula (2) may be determined also directly using the definition of

Laplace transform

.  Take an additional parametre

a

to the defining integral

ℒ

⁢

{

sin

⁡

t

t

}

=

∫

0

∞

e

-

s

⁢

t

⁢

sin

⁡

t

t

⁢

𝑑

t

by setting

∫

0

∞

e

-

s

⁢

t

⁢

sin

⁡

a

⁢

t

t

⁢

𝑑

t

:=

φ

⁢

(

a

)

.

Now we have the derivative

φ

′

⁢

(

a

)

=

∫

0

∞

e

-

s

⁢

t

⁢

cos

⁡

a

⁢

t

⁢

d

⁢

t

,  where one can partially integrate twice, getting

φ

′

⁢

(

a

)

=

∫

0

∞

e

-

s

⁢

t

⁢

cos

⁡

a

⁢

t

⁢

d

⁢

t

=

1

s

-

a

2

s

2

⁢

∫

0

∞

e

-

s

⁢

t

⁢

cos

⁡

a

⁢

t

⁢

d

⁢

t

.

Thus we solve

∫

0

∞

e

-

s

⁢

t

⁢

cos

⁡

a

⁢

t

⁢

d

⁢

t

=

1

s

1

+

(

a

s

)

2

=

φ

′

⁢

(

a

)

,

and since

φ

⁢

(

0

)

=

0

, we obtain

φ

⁢

(

a

)

=

arctan

⁡

a

s

.  This yields

∫

0

∞

e

-

s

⁢

t

⁢

sin

⁡

t

t

⁢

𝑑

t

=

φ

⁢

(

1

)

=

arctan

⁡

1

s

,

i.e. the formula (2).

Formula (2) is derived here (

http://planetmath.org/LaplaceTransformOfFracftt

) in a third way.

Title

Laplace transform of sine integral

Canonical name

LaplaceTransformOfSineIntegral

Date of creation

2014-11-07 15:36:06

Last modified on

2014-11-07 15:36:06

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

15

Author

pahio (2872)

Entry type

Example

Classification

msc 44A10

Synonym

Laplace transform of sinc function

Related topic

SubstitutionNotation

Related topic

SincFunction

Related topic

TableOfLaplaceTransforms

Related topic

SineIntegral

Generated on Fri Feb 9 21:32:28 2018 by

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