I am trying to solve the following complex form. I will really appreciate the way to solve it. I have tried and failed myself. My results do not even qualify to be mention here.
\begin{equation} P = e^{-ne^{- \sigma \lambda t}} \end{equation}
\begin{equation} E(t) = \int_{0}^{+\infty} t \frac{\partial P}{\partial t}\ dt \end{equation}
This is part of the paper I am trying to solve. The solution that the author has given is
\begin{equation} E(t) = \frac{1}{\sigma \lambda} \sum^n_{k=1} (-1)^{k+1} {{n} \choose {k}} \frac{1}{k} \end{equation}
As I commented earlier, I think that there is a problem.
Since Olivier Oloa answered your next question $$\int^\infty_0 t(n\lambda e^{-ne^{-\lambda t} - \lambda t})dt=-\frac{n}{\lambda}\int_0^1\log( u) \:e^{-nu}du=\frac{\log (n)+\Gamma (0,n)+\gamma }{\lambda }$$ making $$E(t) = \int_{0}^{+\infty} t\ \frac{d P}{dt}\ dt=\frac{\log (n)+\Gamma (0,n)+\gamma }{ \sigma \lambda}$$ where appears the announced incomplete gamma function.
But $$ \frac{1}{\sigma \lambda} \sum^n_{k=1} (-1)^{k+1} {{n} \choose {k}} \frac{1}{k} =\frac{\psi ^{(0)}(n+1)+\gamma}{\sigma \lambda}$$ where appears the digamma function.
These two expressions are not the same. Just for illustration, consider $$\Delta_n=\log (n)+\Gamma (0,n)-\psi ^{(0)}(n+1)$$ and let us compute a few values $$\left( \begin{array}{cc} n &\log (n)+\Gamma (0,n) & \psi ^{(0)}(n+1)& \Delta_n \\ 1 & 0.21938 & 0.42278 & -0.20340 \\ 2 & 0.74205 & 0.92278 & -0.18074 \\ 3 & 1.11166 & 1.25612 & -0.14446 \\ 4 & 1.39007 & 1.50612 & -0.11604 \\ 5 & 1.61059 & 1.70612 & -0.09553 \\ 6 & 1.79212 & 1.87278 & -0.08066 \\ 7 & 1.94603 & 2.01564 & -0.06962 \\ 8 & 2.07948 & 2.14064 & -0.06116 \\ 9 & 2.19724 & 2.25175 & -0.05452 \\ 10 & 2.30259 & 2.35175 & -0.04916 \end{array} \right)$$