Exploring Laplace Transforms: Understanding the Division by t Theorem

[Revan3]

Okay so I am brushing up my Laplace transforms as an independent study and I come across this proof for the "division by t theorem". The idea proof it self I have no problems with except for the limits of the first integration, It feels like they just arbitrarily choose the limits to be from <s,inf> to so convienently get rid of the negative sign. Is there any explanation for the limits of integration?

here is the problem:

http://www.flickr.com/photos/64771553@N02/5897331839/in/photostream/


here are the related theorems:

http://www.flickr.com/photos/64771553@N02/5897942368/

 

[Revan3]

Really like no answers =[?

 

[vela]

It's just an application of the fundamental theorem of calculus, which tells you that
\int_a^x h(u)\,du = H(x)-H(a)
where H(x) is a function that satisfies H'(x)=h(x). They chose a=\infty to get rid of H(a).

 

[Revan3]

I actually want to know why you can pick integration limits it must have some meaning

 

[vela]

If you don't use limits, an integral gives you a set of functions, where two elements of the set differ by a constant. In this problem, you only want one function so you need to use limits.

You might find this derivation (from Arfken) more satisfying.

Let f(s) = \mathcal{L}[F(t)]. Then
\begin{eqnarray*}
\int_s^b f(u)\,du &= \int_s^b \int_{0}^\infty F(t)e^{-ut}\,dt \, du \\
& = \int_{0}^\infty \int_s^b F(t)e^{-ut}\,du \, dt \\
& = \int_{0}^\infty F(t) \frac{e^{-st}-e^{-bt}}{t} \, dt
\end{eqnarray*}
Taking the limit as b \to \infty gives
\int_s^\infty f(u)\,du = <br /> \lim_{b \to \infty} \int_{0}^\infty F(t) \frac{e^{-st}-e^{-bt}}{t} \, dt = <br /> \int_{0}^\infty \frac{F(t)}{t} e^{-st} \, dt = \mathcal{L}\left[\frac{F(t)}{t}\right]