TITLE: Evaluate $\int_{0}^{\infty}\frac{1-e^{-t}}{t}e^{-st}\;dt$
QUESTION [9 upvotes]: This is laplace transform of $\dfrac{1-e^{-t}}{t}$ and the integral exists according to wolfram
Do i get any help/hints about how to work this ? I have been trying integration by parts with different combinations for u and dv but none of them are working. Any help is appreciated thanks!

REPLY [4 votes]: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#66f}{\large\int_{0}^{\infty}{1 - \expo{-t} \over t}\,\expo{-st}\,\dd t}
=\int_{0}^{\infty}\bracks{\expo{-st} - \expo{-\pars{s + 1}t}}\
\overbrace{\int_{0}^{\infty}\expo{-tx}\,\dd x\,\dd t}^{\ds{=\ \color{#c00000}{1 \over t}}}
\\[5mm]&=\int_{0}^{\infty}\int_{0}^{\infty}
\bracks{\expo{-\pars{s + x}t} - \expo{-\pars{s + 1 + x}t}}
\,\dd t\,\dd x
=\int_{0}^{\infty}\pars{{1 \over x + s} - {1 \over x + s + 1}}\,\dd x
\\[5mm]&=\left.\ln\pars{x + s \over x + s + 1}
\right\vert_{\, x\ =\ 0}^{\, x\ \to\ \infty}
=-\ln\pars{s \over s + 1}=\color{#66f}{\large\ln\pars{1 + {1 \over s}}}
\end{align}

Another way to evaluate the integral is:
  \begin{align}
&\color{#66f}{\large\int_{0}^{\infty}{1 - \expo{-t} \over t}\,\expo{-st}\,\dd t}
=s\int_{0}^{\infty}\ln\pars{t}\expo{-st}\,\dd t
-\pars{s + 1}\int_{0}^{\infty}\ln\pars{t}\expo{-\pars{s + 1}t}\,\dd t
\\[5mm]&=\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
=\ln\pars{s + 1 \over s}\
\underbrace{\int_{0}^{\infty}\expo{-t}\,\dd t}_{\ds{=\ \color{#c00000}{1}}}
\\[5mm]&=\color{#66f}{\large\ln\pars{1 + {1 \over s}}}
\end{align}