TITLE: Laplace transform of $f(t)/t$
QUESTION [6 upvotes]: For a function $f(t)$ Laplace transform is defined as $F(s)=\int_0^{\infty} f(t)e^{-st}dt$.  
I have to show the property that the Laplace transform of $f(t)\over t$ is $\int _s^\infty F(s')ds'$.  
I tried the substitution $k={1\over t}$ but then I end up with $\int_0^\infty {e^{-sk} f(1/k)\over k} dk$.  
In  $\int _s^\infty F(s')ds'$ is  $s'$ a dummy variable or is it  the derivative  of $s$ in$ F(s)$.
Can someone please help me to show this

REPLY [8 votes]: $s'$ is a dummy variable, not the derivative of $s$ with respect to anything.  In any case, it can sometimes be helpful to start at the far side of the equals sign.  We have
\begin{align}
 \int_s^{\infty}F(s')ds' & = \int_s^{\infty}\int_0^{\infty}e^{-s' t}f(t)dtds'\\
 & = \int_0^{\infty}\int_s^{\infty}e^{-s't}f(t)ds'dt\\
 & = \int_0^{\infty}\left.-\frac{1}{t}e^{-s't}f(t)\right|_s^{\infty}dt\\
 & = \int_0^{\infty}e^{-st}\frac{f(t)}{t}dt\\
 & = \mathcal{L}\left[\frac{f(t)}{t}\right]
\end{align}