Laplace transform of $L({1-e^{-t}\over t})$ I have to find the Laplace transform of $$\mathcal{L}\left[\dfrac{1-e^{-t}}t\right],$$ then this is equivalent to $$\mathcal{L}\left[\dfrac{1}t\right]-\mathcal{L}\left[\dfrac{e^{-t}}t\right]$$
But $\mathcal{L}\left[\dfrac{1}t\right]$ doesn't exist right?  Is there any way I can do this?
 A: Consider Laplace transform
$$
\mathcal{L}\left[f(t)\right]=F(s)=\int_0^\infty f(t)\ e^{-st}\ dt
$$
and property of the unilateral Laplace transform
$$
\mathcal{L}\left[\frac{f(t)}{t}\right]=\int_s^\infty F(\omega)\ d\omega,
$$
where $F(\omega)$ is Laplace transform of $f(t)$. We choose $f(t)=(1-e^{-t})$ and it is easy to show that
$$
F(s)=\mathcal{L}\left[1-e^{-t}\right]=\frac1s-\frac1{s+1}
$$
then
\begin{align}
\mathcal{L}\left[\frac{1-e^{-t}}{t}\right]&=\int_s^\infty F(\omega)\ d\omega\\
&=\int_s^\infty \left(\frac1\omega-\frac1{\omega+1}\right)\ d\omega\\
&=\left.\left[\ln \omega-\ln(\omega+1)\right]\right|_s^\infty\\
&=\left.\ln\left(\frac{\omega}{\omega+1}\right)\right|_s^\infty\\
&=-\ln\left(\frac{s}{s+1}\right)\\
&=\large\color{blue}{\ln\left(\frac{s+1}{s}\right)},
\end{align}
where $\displaystyle\lim_{\omega\to\infty}\ln\left(\frac{\omega}{\omega+1}\right)=0$.
A: $\newcommand{\+}{^{\dagger}}
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\begin{align}
&\color{#66f}{\large\int_{0}^{\infty}{1 - \expo{-t} \over t}\,\expo{-st}\,\dd t}
=-\int_{0}^{\infty}
\ln\pars{t}\bracks{-s\expo{-st} + \pars{s + 1}\expo{-\pars{s + 1}t}}\,\dd t
\\[3mm]&=\int_{0}^{\infty}\ln\pars{t \over s}\expo{-t}\,\dd t
-\int_{0}^{\infty}\ln\pars{t \over s + 1}\expo{-t}\,\dd t
=\int_{0}^{\infty}\ln\pars{s + 1 \over s}\expo{-t}\,\dd t
\\[3mm]&=\color{#66f}{\large\ln\pars{1 + {1 \over s}}}
\end{align}
