Laplace transform of $\frac{1-e^{-t}}{t}$ How to calculate the Laplace Transform of such a function like this:
$$\frac{1-e^{-t}}{t}$$
I try to separate, got the $\text{Ei}$ function, try to evaluate using Residue, got $0$. This function seems not to be on $L^1$ class, but the Wolfram Math calculates nevertheless, so how do I? Thanks!
 A: Hint.
By setting 
$$f(t):=\frac{1-e^{-t}}{t}
$$ one has 
$$
\mathcal{L}\{1-e^{-t} \}(s)=\mathcal{L} \{tf(t) \}(s)=-F'(s)
$$
where $F(s)$ is the Laplace transform of $f(t)$. Then use
$$
\mathcal{L}\{1-e^{-t} \}(s)=\frac{1}{s}-\frac{1}{s+1},\quad s>0.
$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\int_{0}^{\infty}{1 - \expo{-t} \over t}\,\expo{-st}\,\dd t & =
\int_{0}^{\infty}\bracks{\expo{-st} - \expo{-\pars{s + 1}t}}
\int_{0}^{\infty}\expo{-tx}\,\dd x\,\dd t
\\[5mm] & =
\int_{0}^{\infty}\int_{0}^{\infty}
\bracks{\expo{-\pars{x + s}t} - \expo{-\pars{x + s + 1}t}}\,\dd t\,\dd x
\\[5mm] & =
\int_{0}^{\infty}\pars{{1 \over x + s} - {1 \over x + s + 1}}\,\dd x =
\left.\ln\pars{x + s \over x + s + 1}\right\vert_{\ x\ =\ 0}^{\ x\ \to\ \infty} \\[5mm] &=
\bbx{\ds{\ln\pars{1 + {1 \over s}}}}
\end{align}
