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Why does the Laplace transform of a function multiplied by time produce the derivative in the Laplace domain of the Laplace transform of the function? Similarly, why does division by time produce an integral?

$$ \mathcal{L}\{{t f(t)}\} = -\frac{d}{ds}\mathcal{L}\{f(t)\} $$ $$ \mathcal{L}\{{\frac{1}{t} f(t)}\} = \int_s^\infty{F(u)du} $$

As an example for multiplication by time:

$$ \mathcal{L}\{{t y’}\} = -\frac{d}{ds}\mathcal{L}\{y'\} = -sY'(s)-Y(s) $$

From the definition of the Laplace transform: $$ \mathcal{L}\{{t y’}\} = \int_0^\infty e^{-st}ty’(t)dt $$

I can see how this could be started with integration by parts:

$$ =te^{-st}y(t)\Big|_0^\infty - (1-s)\int_0^\infty e^{-st}y(t)dt $$

I can see that the term on the right is: $$ -(1-s)\int_0^\infty e^{-st}y(t)dt = -(1-s)\mathcal{L}\{y(t)\} = -(1-s)Y(s) $$

but the integral evaluation eludes me.

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Your "Why?"-questions won't lead anywhere. You just have to accept the simple formulas you quote. Unlike the Fourier transform of functions or signals the Laplace transform of transient phenomena $t\mapsto f(t)$ $\>(t\geq0)$ has no intuitive physical interpretation. It is a purely formal operation applied to given or as yet unknown function terms ("expressions") resulting in other function terms hopefully listed in a catalogue. – While functions $f$ available only in the form of a discrete time series are Fourier transformed all the time, and interesting information about $f$ is revealed in this way, nobody would consider Laplace transforming such a data set, nor would anybody look at the graph of a Laplace transform.

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  • $\begingroup$ Hi Christian. This answer is a bit confusing. It seems as if you're stating that the Laplace Transform has no physical meaning. In fact, the bilateral LT is the FT with a rotation of coordinates. Perhaps I am missing something here. $\endgroup$
    – Mark Viola
    May 24, 2018 at 19:21
  • $\begingroup$ @MarkViola: Going through the motions to prove the quoted formulas is not the answer to the questions of the OP. But never mind. $\endgroup$ May 24, 2018 at 19:31
  • $\begingroup$ Christian, the OP asked, Why does the Laplace transform of a function multiplied by time produce the derivative in the Laplace domain of the Laplace transform of the function? Similarly, why does division by time produce an integral?" It seems that the OP wants an explanation that is afforded by proofs of the stated formulae. $\endgroup$
    – Mark Viola
    May 24, 2018 at 19:53
  • $\begingroup$ Thanks for the comment Christian. I was merely looking for the proof (and perhaps some guidance on solving the example problem). I wasn't asking a philosophical "why" :) $\endgroup$
    – drpm
    May 24, 2018 at 21:15
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Differentiation under the integral reveals

$$\begin{align} \frac{d}{ds}\mathscr{L}\{f\}(s)&=\frac{d}{ds}\int_0^\infty f(t)e^{-st}\,dt\\\\ &=\int_0^\infty f(t)\frac{de^{-st}}{ds}\,dt\\\\ &=\int_0^\infty f(t)\left(-te^{-st}\right)\,dt\\\\ &=-\int_0^\infty tf(t)e^{-st}\,dt \end{align}$$


Changing the order of integration reveals

$$\begin{align} \int_s^\infty\mathscr{L}\{f\}(s')\,ds'&=\int_s^\infty\int_0^\infty f(t)e^{-s't}\,dt\,ds'\\\\ &=\int_0^\infty f(t)\int_s^\infty e^{-s't}\,ds'\,dt\\\\ &=\int_0^\infty f(t)\left(\frac1t e^{-st}\right)\,dt\\\\ &=\int_0^\infty \frac{f(t)}{t}e^{-st}\,dt \end{align}$$

Of course, if $\int_0^L \frac{f(t)}{t}\,dt$, $L>0$, fails to exist, then this formal manipulation leads to a nonsensical result.

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  • $\begingroup$ Elegant solution Mark. Thank you! Can you comment on the evaluation of the integral in the question? My guess is that I took a wrong turn during integration by parts. $\endgroup$
    – drpm
    May 24, 2018 at 18:59
  • $\begingroup$ You're welcome. My pleasure. As for the example in the OP, it is correct. $\endgroup$
    – Mark Viola
    May 24, 2018 at 19:17
  • $\begingroup$ It might be correct but it's incomplete: what can be done with the evaluation of the integral? It seems like it should evaluate to -sY'(s) + sY(s) $\endgroup$
    – drpm
    May 24, 2018 at 21:14

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