Evaluate $\int_{0}^{\infty}\frac{1-e^{-t}}{t}e^{-st}\;dt$ 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!
 A: A last way similar to both Ron Gordon and Daniel Fischer is to use an usefull proposition related to the laplace transform. 

$$\mathcal{L}\left( \frac{f(t)}{t}\right) = \int_s^\infty F(\sigma) \,\mathrm{d}\sigma$$

Given that the integral exists, eg that there exists some $M$ such that $f(t)e^{-t} < M$ for all $t$. 
Here we have $$f(t) = 1-e^{-t} \quad \text{ and } \quad F(t) = \frac1\sigma - \frac1{1+\sigma}$$
Which you should know how to compute by now, or look it up in a standard laplace table. Hence
$$
\begin{align*}
\int_0^{\infty} \frac{1-e^{-t}}{t}e^{-st}\,\mathrm{d}t
& = \mathcal{L}\left(\frac{1-e^{-t}}{t}\right) \\
& = \int_s^\infty \mathcal{L}\left( 1-e^{-t}\right) \mathrm{d}\sigma \\
& = \int_s^\infty \frac1\sigma - \frac1{1+\sigma}\mathrm{d}\sigma \\
& = \log (1+s) - \log s
 = \log \left( 1 + \frac{1}{s} \right)
\end{align*}$$
And we are done $\square$
A: An easy way to evaluate the integral is using Frullani's theorem
$$\int_0^\infty \frac{f(at)-f(bt)}{t}\,dt=\bigg[f(0)-f(\infty)\bigg]\ln\left(\frac{b}{a}\right)$$
Taking $f(t)=e^{-t},\, a=s,$ and $b=1+s$ then the integral is simply evaluated to
$$\ln\left(\frac{1+s}{s}\right)$$
which matches up other's answers.
A: $\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}

A: Write
$$I(s) = \int_0^\infty \frac{1-e^{-t}}{t} e^{-st}\,dt$$
for $\operatorname{Re} s > 0$. Compute $I'(s)$ (by differentiating under the integral sign), and from that obtain $I(s)$ by noting that $\lim\limits_{\operatorname{Re} s \to +\infty} I(s) = 0$ by the dominated convergence theorem.
A: Use the fact that
$$\frac{1-e^{-t}}{t} = \int_0^1 du \, e^{-t u}$$
So the integral in question is
$$\int_0^{\infty} dt \, \int_0^1 du \, e^{-t (u+s)} $$
Because all integrals here converge absolutely, we may switch the order of integration and get
$$\int_0^1 du \, \int_0^{\infty} dt \, e^{-t (u+s)}  = \int_0^1 \frac{du}{u+s} = \log{\left ( 1+\frac1{s} \right )}$$
