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

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- Dec 19th 2005, 06:28 AMRed_FoxLaplace 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 - Dec 19th 2005, 08:23 AMCaptainBlackQuote:

Originally Posted by**Red_Fox**

RonL - Dec 19th 2005, 09:00 AMRed_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 (: - Dec 19th 2005, 09:20 AMCaptainBlackQuote:

Originally Posted by**Red_Fox**

let $\displaystyle t'=st$ then:

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

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

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

Now if I recall correctly the first integral on the RHS is $\displaystyle -\gamma$,

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 $\displaystyle -\frac{(\gamma+ln(s))}{s}$.

RonL - Dec 19th 2005, 11:14 AMRed_FoxThanks!
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!!! - Dec 22nd 2005, 06:52 AMmooshazzQuote:

Originally Posted by**CaptainBlack**

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? - Dec 22nd 2005, 07:51 AMCaptainBlackQuote:

Originally Posted by**mooshazz**

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

$\displaystyle \mathbb{C}$ when required.

See attachment for a definition restricted to s-real (I know its

intended for engineers but here it is anyway)

RonL - Dec 22nd 2005, 07:55 AMmooshazzQuote:

Originally Posted by**CaptainBlack**

but as a mathmatician i need to prove that for complex s

how can i do it? - Dec 22nd 2005, 10:30 AMCaptainBlackQuote:

Originally Posted by**mooshazz**

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 - Dec 22nd 2005, 02:50 PMmooshazzQuote:

Originally Posted by**CaptainBlack**

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 - Dec 22nd 2005, 11:10 PMCaptainBlackQuote:

Originally Posted by**mooshazz**

is uniquely determined by its values on a curve $\displaystyle \Gamma \subset A$. 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 $\displaystyle s.(0, \infty)$ and see if that works.

RonL - Dec 24th 2005, 08:15 AMmooshazz
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?