Laplace transform of $\frac{1-e^{-t}}{t}$

Question

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!

Answer 1

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 $$ \mathcal{L}\{1-e^{-t} \}(s)=\frac{1}{s}-\frac{1}{s+1},\quad s>0. $$

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}