Why does multiplication (division) by time of a function equal the derivative (integral) of the Laplace transform of the function? 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.
 A: 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. 
A: 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.
