Write

\(I(s)=\int_0^{\infty} \frac{1-e^{-t}}{t}e^{-st}dt\)

for Re s>0. Compute I′(s) (by differentiating under the integral sign), and from that obtain I(s) by noting that \(\displaystyle\lim_{{{R}{e}{s}\Rightarrow+\infty}}{I}{\left({s}\right)}={0}\) by the dominated convergence theorem.

\(\displaystyle{I}{\left({s}\right)}={\log{{\left({1}+{\frac{{{1}}}{{{s}}}}\right)}}}\)

I am getting

\(\displaystyle{I}'{\left({s}\right)}=-{\frac{{{1}}}{{{s}+{1}}}}\)

integrating both sides gives

\(\displaystyle{I}{\left({s}\right)}={\ln{{\left({\frac{{{1}}}{{{s}+{1}}}}\right)}}}\)

You don't get a factor of s down by differentiating.

\(\displaystyle{\frac{{\partial}}{{\partial{s}}}}{e}^{{-{s}{t}}}=-{t}\cdot{e}^{{-{s}{t}}}\)

so it`s

\(I'(s)=\int_0^{\infty} (e^{-t}-1) \cdot e^{-st} dt\)