Laplace transform of $f(t)/t$

Question

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

Answer 1

Hint: compute both $$ \frac d{ds} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt \\ \lim _{s\to\infty} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt $$

Answer 2

$s'$ 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}