I have to prove the next problem
$$\mathcal{L} \left(\int_{0}^{t}\frac{1-e^{-u}}{u}du,s\right) = \frac{1}{s}\log\left(1+\frac{1}{s}\right)$$
I'm quite new in the subject and I have troubles with this one. There's no need to put the step by step, just the initial ones to know that I'm going in the right way (but if you want to,that would be great)
Thanks!
Assuming $\text{Re}(s)>0$ we have: $$\begin{eqnarray*}\mathcal{L}\left(\int_{0}^{t}\frac{1-e^{-u}}{u}\,du\right)&=&\int_{0}^{+\infty}\int_{0}^{t}\frac{e^{-st}-e^{-st-u}}{u}\,du\,dt\\(\text{Sub.}\;u=tv)\quad&=&\int_{0}^{+\infty}\int_{0}^{1}\frac{e^{-st}-e^{-(s+v)t}}{v}\,du\,dt\\(\text{Fubini})\quad&=&\int_{0}^{1}\frac{1}{v}\int_{0}^{+\infty}\left(e^{-st}-e^{-(s+v)t}\right)\,dt\,dv\\&=&\int_{0}^{1}\frac{dv}{v}\left(\frac{1}{s}-\frac{1}{s+v}\right)\,dv\\&=&\frac{1}{s}\int_{0}^{1}\frac{dv}{s+v}=\frac{1}{s}\left[\log(s+1)-\log(s)\right]\\&=&\color{red}{\frac{1}{s}\,\log\left(1+\frac{1}{s}\right)}.\end{eqnarray*}$$