These calculations are correct ? About  $\int\frac{e^{-x}}{x}dx$ Was trying to calculate $$\int_{0}^{\infty}e^{-x}\ln x dx=-\gamma$$ and I found this question:
I want to analyze $$\int\frac{e^{-x}}{x}dx$$
With $u=\displaystyle\frac{1}{x} \Rightarrow du = \displaystyle\frac{-1}{x^{2}} dx  $,  and $dv=e^{-x} \Rightarrow v=-e^{-x}$
Then 
$$\int\frac{e^{-x}}{x}dx = \displaystyle\frac{1}{x}\cdot-e^{-x}-\int-e^{-x}\cdot\displaystyle\frac{-1}{x^{2}} dx = -\displaystyle\frac{e^{-x}}{x}-\int \displaystyle\frac{e^{-x}}{x^{2}} dx$$   
Integrating from the same form gives:
$$\int\frac{e^{-x}}{x}dx = -\displaystyle\frac{e^{-x}}{x} + \displaystyle\frac{e^{-x}}{x^{2}} + 2\int\frac{e^{-x}}{x^{3}}dx$$
Are these calculations are correct?, and more is  valid say : 
$$\int\frac{e^{-x}}{x}dx = \displaystyle\sum\limits_{n=0}^\infty (-1)^{n+1}n!\frac{e^{-x}}{x^{n+1}}\ ?$$  

$\bf{EDIT}$: This series  helps me to
  calculate it ? :
  $$\int_{0}^{\infty}e^{-x}\ln xdx=-\gamma$$ I don't know  how to turn
  this series in something harmonic. If
  not, is this the way to calculate that this
  integral converges to $-\gamma$, which
  is the form ?

Thanks
 A: The series diverges, but converges to your integral asymptotically: If you add up the first $n$ terms the ratio of the error to the $n$th term goes to zero as $x$ goes to infinity
A: $\newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
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$$
\int_{0}^{\infty}x^{\mu}\,\expo{-x}\,\dd x = \Gamma\pars{\mu + 1}
$$
We take the derivative respect $\mu$:
$$
\int_{0}^{\infty}x^{\mu}\ln\pars{x}\,\expo{-x}\,\dd x = \Gamma\,'\pars{\mu + 1}
=
\Psi\pars{\mu + 1}\Gamma\pars{\mu + 1}
$$
We take the limit $\mu \to 0^{+}$:
$$
\color{#0000ff}{\Large\int_{0}^{\infty}\ln\pars{x}\expo{-x}\,\dd x}
=
\overbrace{\ \Psi\pars{1}\ }^{\ds{-\gamma}}
\quad
\overbrace{\ \Gamma\pars{1}\ }^{\ds{1}}
=
\color{#0000ff}{\Large -\,\gamma}
$$
