Proving one of the laplace transform properties

**Liverpool** · December 31st 2010, 08:28 AM

What is the proof of the following property? :

If F of class A and $L \left( F(t) \right)=f(s)$ then :

$\displaystyle L \left( \dfrac{F(t)}{t} \right) = \int_s^{\infty} f(u) \, du$

**chisigma** · December 31st 2010, 11:43 AM

Setting $G(t)=\frac{F(t)}{t} \implies F(t)= t\ G(t)$ You have...

$\displaystyle \mathcal{L} \{F(t)\}= -\frac{d}{ds} \mathcal {L} \{g(t)}\} \implies f(s)= -\frac{d g(s)}{ds}$ (1)

Integrating (1) You obtain...

$\displaystyle g(s) = \mathcal{L} \{\frac{F(t)}{t}\} = \int_{s}^{\infty} f(u)\ du$ (2)

Kind regards

$\chi$ $\sigma$

**Krizalid** · December 31st 2010, 02:10 PM

another statement is if $\dfrac{f(t)}t$ has limit as $t\to0^+$ and $f$ is exponentially bounded, then $\mathcal L\left(\dfrac{f(t)}{t}\right)=\displaystyle\int_s^ \infty \widehat f(u)\,du.$

the proof is basically to write $f(t)=t\cdot\dfrac{f(t)}t$ since $f$ is well defined for $t=0,$ then use the Laplace transform for $t\cdot f(t),$ integrate between $[s,\infty)$ and use that the Laplace transform of a function tends to 0 as $s\to\infty.$

**Liverpool** · December 31st 2010, 02:32 PM

I did not get it ;|
Can you give me a full proof?
Am lost :/

**Drexel28** · December 31st 2010, 03:09 PM

Originally Posted by Liverpool

I did not get it ;|
Can you give me a full proof?
Am lost :/

Is there a particular part about this that you don't get?

**Liverpool** · December 31st 2010, 03:15 PM

Yes this one:

Originally Posted by chisigma

$\displaystyle \mathcal{L} \{F(t)\}= -\frac{d}{ds} \mathcal {L} \{g(t)}\} \implies f(s)= -\frac{d g(s)}{ds}$ (1)

Integrating (1) You obtain...

$\displaystyle g(s) = \mathcal{L} \{\frac{F(t)}{t}\} = \int_{s}^{\infty} f(u)\ du$ (2)

How did we obtain the limits of the last integral?
from s to infinity??

**chisigma** · January 1st 2011, 01:42 PM

I apologize for the fact that a [stupid...] 'competition stress' often produces incomplete or 'impossible to undestand' answers

...

Setting $G(t)= \frac{F(t)}{t} \implies F(t)= t\ G(t)$ we have...

$\displaystyle g^{'}(s) = \frac{d}{ds} \int_{0}^{\infty} G(t)\ e^{-s t}\ dt= - \int_{0}^{\infty} t\ G(t)\ e^{-s t}\ dt = - \mathcal{L} \{F(t)\}$ (1)

... and the first step is terminated. The (1) can be written as an ordinary first order DE as follows...

$\displaystyle g^{'}(s) = - f(s)$ (2)

... with the 'condition' $\lim_{s \rightarrow \infty} g(s)=0$ and its solution is...

$\displaystyle g(s)= - \int_{\infty}^{s} f(u)\ du = \int_{s}^{\infty} f(u)\ du$ (3)

Kind regards

$\chi$ $\sigma$

**Liverpool** · January 3rd 2011, 04:09 PM

Thanks!
Now it is too clear

Thread: Proving one of the laplace transform properties

LinkBack

Thread Tools

Display

Proving one of the laplace transform properties

Similar Math Help Forum Discussions

Fourier transform properties

Fourier transform properties

Explanations Of These Laplace Transforms and Properties

Laplace transform and Fourier transform what is the different?

Discrete Fourier Transform properties

Search Tags