# Proving one of the laplace transform properties

• December 31st 2010, 07:28 AM
Liverpool
Proving one of the laplace transform properties
What is the proof of the following property? :

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

$\displaystyle L \left( \dfrac{F(t)}{t} \right) = \int_s^{\infty} f(u) \, du$
• December 31st 2010, 10:43 AM
chisigma
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

$\chi$ $\sigma$
• December 31st 2010, 01:10 PM
Krizalid
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 $s\to\infty.$
• December 31st 2010, 01:32 PM
Liverpool
I did not get it ;|
Can you give me a full proof?
Am lost :/
• December 31st 2010, 02:09 PM
Drexel28
Quote:

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

• December 31st 2010, 02:15 PM
Liverpool
Yes this one:

Quote:

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??
• January 1st 2011, 12:42 PM
chisigma
I apologize for the fact that a [stupid...] 'competition stress' often produces incomplete or 'impossible to undestand' answers(Thinking)...

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

$\chi$ $\sigma$
• January 3rd 2011, 03:09 PM
Liverpool
Thanks!
Now it is too clear :)