What is the proof of the following property? :

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

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

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- Dec 31st 2010, 07:28 AMLiverpoolProving one of the laplace transform properties
What is the proof of the following property? :

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

$\displaystyle \displaystyle L \left( \dfrac{F(t)}{t} \right) = \int_s^{\infty} f(u) \, du$ - Dec 31st 2010, 10:43 AMchisigma
Setting $\displaystyle G(t)=\frac{F(t)}{t} \implies F(t)= t\ G(t)$ You have...

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

Integrating (1) You obtain...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Dec 31st 2010, 01:10 PMKrizalid
another statement is if $\displaystyle \dfrac{f(t)}t$ has limit as $\displaystyle t\to0^+$ and $\displaystyle f$ is exponentially bounded, then $\displaystyle \mathcal L\left(\dfrac{f(t)}{t}\right)=\displaystyle\int_s^ \infty \widehat f(u)\,du.$

the proof is basically to write $\displaystyle f(t)=t\cdot\dfrac{f(t)}t$ since $\displaystyle f$ is well defined for $\displaystyle t=0,$ then use the Laplace transform for $\displaystyle t\cdot f(t),$ integrate between $\displaystyle [s,\infty)$ and use that the Laplace transform of a function tends to 0 as $\displaystyle s\to\infty.$ - Dec 31st 2010, 01:32 PMLiverpool
I did not get it ;|

Can you give me a full proof?

Am lost :/ - Dec 31st 2010, 02:09 PMDrexel28
- Dec 31st 2010, 02:15 PMLiverpool
- Jan 1st 2011, 12:42 PMchisigma
I apologize for the fact that a [stupid...] 'competition stress' often produces incomplete or 'impossible to undestand' answers(Thinking)...

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Jan 3rd 2011, 03:09 PMLiverpool
Thanks!

Now it is too clear :)