Laplace transform of $L({1-e^{-t}\over t})$

Question

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?

Answer 1

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 $\displaystyle\lim_{\omega\to\infty}\ln\left(\frac{\omega}{\omega+1}\right)=0$.

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \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}