Laplace Transform of log(t)

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December 19th 2005, 06:28 AM
Red_Fox

Laplace Transform of log(t)

Hi everybody!
I'm trying to find the Laplace transform of log(t) (Our lecturer promised to give a lot of respect to the one who solve that :) ...)

Can you guys please help me and give me some direction?
Thanks!

Red_Fox
December 19th 2005, 08:23 AM
CaptainBlack

Quote:

Originally Posted by Red_Fox

Hi everybody!
I'm trying to find the Laplace transform of log(t) (Our lecturer promised to give a lot of respect to the one who solve that :) ...)

Can you guys please help me and give me some direction?
Thanks!

Red_Fox

Have you consided looking it up in a table of Laplace transforms? :D

RonL
December 19th 2005, 09:00 AM
Red_Fox

well... where can I find one? I mean, I didn't find it in the elementry Laplace Transform table!
And by the way, I would like to know the way to find it.

P.S I've found something: (-y+lns)/s , when y is Euler constant (limiting difference between the harmonic series and the natural logarithm).
Maybe you know how they reached this??

Thanks (:
December 19th 2005, 09:20 AM
CaptainBlack

Quote:

Originally Posted by Red_Fox

well... where can I find one? I mean, I didn't find it in the elementry Laplace Transform table!
And by the way, I would like to know the way to find it.

P.S I've found something: (-y+lns)/s , when y is Euler constant (limiting difference between the harmonic series and the natural logarithm).
Maybe you know how they reached this??

Thanks (:

${\cal{L}}\ ln(t)=\int_0^{\infty}e^{-st}ln(t)dt$

let $t'=st$ then:

${\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}ln(\frac{t'}{s}) \frac{1}{s} dt'$

${\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}(ln(t')-ln(s)) \frac{1}{s} dt'$

${\cal{L}}\ ln(t)= \frac{1}{s} \int_0^{\infty}e^{-t'}ln(t') dt' -\frac{ln(s)}{s} \int_0^{\infty} e^{-t'}dt'$

Now if I recall correctly the first integral on the RHS is $-\gamma$ ,
and you should be able to finish the derivation from here.

So not so difficult after all?

By the way my handbook says it is $-\frac{(\gamma+ln(s))}{s}$ .

RonL
December 19th 2005, 11:14 AM
Red_Fox

Thanks!

Wow (: it's so simple after t'=st!
By the way , it is very interesting how to reach from this integral (with the e and the ln) to the limiting difference between the harmonic series and the natural logarithm. see also [HTML]http://en.wikipedia.org/wiki/Euler-Mascheroni_constant[/HTML]

Thank You Very Much!!!
December 22nd 2005, 06:52 AM
mooshazz

Quote:

Originally Posted by CaptainBlack

${\cal{L}}\ ln(t)=\int_0^{\infty}e^{-st}ln(t)dt$

let $t'=st$ then:

${\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}ln(\frac{t'}{s}) \frac{1}{s} dt'$

${\cal{L}}\ ln(t)=\int_0^{\infty}e^{-t'}(ln(t')-ln(s)) \frac{1}{s} dt'$

${\cal{L}}\ ln(t)= \frac{1}{s} \int_0^{\infty}e^{-t'}ln(t') dt' -\frac{ln(s)}{s} \int_0^{\infty} e^{-t'}dt'$

Now if I recall correctly the first integral on the RHS is $-\gamma$ ,
and you should be able to finish the derivation from here.

So not so difficult after all?

By the way my handbook says it is $-\frac{(\gamma+ln(s))}{s}$ .

RonL

you have one little mistake when you put t'
you forgot s is a complex variable
so you changed the intire intigration
you need to prove it still works under complex variabels

can you help me with that?
December 22nd 2005, 07:51 AM
CaptainBlack

1 Attachment(s)

Quote:

Originally Posted by mooshazz

you have one little mistake when you put t'
you forgot s is a complex variable
so you changed the intire intigration
you need to prove it still works under complex variabels

can you help me with that?

Its not a mistake, I am using a definition of the Laplace transform
for s real.

At least in some places the Laplace transform is defined for s real,
with appropriate continuation being used to extend it to most of
$\mathbb{C}$ when required.

See attachment for a definition restricted to s-real (I know its
intended for engineers but here it is anyway)

RonL
December 22nd 2005, 07:55 AM
mooshazz

Quote:

Originally Posted by CaptainBlack

At least in some places the Laplace transform is defined for s real,
with appropriate continuation being used to extend it to most of
$\mathbb{C}$ when required.

See attachment for a definition restricted to s-real (I know its
intended for engineers but here it is anyway)

RonL

but as a mathmatician i need to prove that for complex s
how can i do it?
December 22nd 2005, 10:30 AM
CaptainBlack

Quote:

Originally Posted by mooshazz

but as a mathmatician i need to prove that for complex s
how can i do it?

This is a bit sketchy as I don't have the relevant reference books to
hand, but as you are a mathematician you should be able to fill in the
detail.

1. Show that in this case the Laplace transform is analytic in the right half
plane. You should have a set of conditions for a functions to guarantee
that its Laplace transform has this property.

2. Then the value of the Laplace transform at all points in the right half
plane is uniquely defined by its values on an arc contained in the right half
plane.

3. Any segment of the positive real axis supplies such an arc. The given
Laplace transform with s complex is a function analytic on the right
half plane, which agrees with the real Laplace transform on the real axis,
and so is the complex Laplace transform.

Does that make any sort of sense?

RonL
December 22nd 2005, 02:50 PM
mooshazz

Quote:

Originally Posted by CaptainBlack

This is a bit sketchy as I don't have the relevant reference books to
hand, but as you are a mathematician you should be able to fill in the
detail.

1. Show that in this case the Laplace transform is analytic in the right half
plane. You should have a set of conditions for a functions to guarantee
that its Laplace transform has this property.

2. Then the value of the Laplace transform at all points in the right half
plane is uniquely defined by its values on an arc contained in the right half
plane.

3. Any segment of the positive real axis supplies such an arc. The given
Laplace transform with s complex is a function analytic on the right
half plane, which agrees with the real Laplace transform on the real axis,
and so is the complex Laplace transform.

Does that make any sort of sense?

RonL

not even a bit
i didn't understand any of the stages

at st. 1 did you ment the result of the transform or the interior of the integral

at st. 2 i didn't get anything
and so in stage 3
December 22nd 2005, 11:10 PM
CaptainBlack

Quote:

Originally Posted by mooshazz

not even a bit
i didn't understand any of the stages

at st. 1 did you ment the result of the transform or the interior of the integral

at st. 2 i didn't get anything
and so in stage 3

Stage 2 is key. Roughtly it is the result that a function analytic on a region $A \subseteq \mathbb C$
is uniquely determined by its values on a curve $\Gamma \subset A$ . If you haven't covered something
like that in Complex Analysis this method is not available to you and you will
have to try evaluating the integral after the change of variable as a line
integral over the line $s.(0, \infty)$ and see if that works.

RonL
December 24th 2005, 08:15 AM
mooshazz

as a hebrew speaker it's a bit difficult for me with mathmatical english
can you explain me those stages in the language of signs and numbers?