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