Improper integral I'd like some help to evaluate this integral : 
$$I=\int^\infty_0 \frac{x-1}{\ln(x)}\,e^{-x} \,dx$$
I tried to use parameter then I've got an integral of gamma function which I don't know how to integrate it . 
Any help will be greatly appreciate .
 A: First let $x = \mathrm{e}^t$, changing variables $\mathrm{d} x = \mathrm{e}^t \mathrm{d} t$
$$
   I = \int_{-\infty}^\infty \exp( t - \mathrm{e}^t ) \frac{\mathrm{e}^t - 1}{t}  \mathrm{d} t
$$
In order to evaluate the integral, first evaluate
$$
  \mathcal{I}_s = \int_{-\infty}^\infty \exp( t - \mathrm{e}^t ) \mathrm{e}^{s t} \mathrm{d} t
   = \int_{-\infty}^\infty \exp( (s+1)t - \mathrm{e}^t ) \mathrm{d} t
$$
Changing variables back to $x$:
$$
  \mathcal{I}_s = \int_0^\infty x^{s} \mathrm{e}^{-x} \mathrm{d} x = \Gamma(s+1)
$$
Now use $\int_0^1 \mathrm{e}^{s t} \mathrm{d} s = \frac{\mathrm{e}^t - 1}{t}$ to get
$$
   I = \int_0^1 \mathcal{I}_s \, \mathrm{d}s = \int_0^1 \Gamma(1+s) \, \mathrm{d} s
$$
The integral $I$, thus, hardly has a closed form. Its approximate numerical value:
$$
I = 0.9227459506806306051438805
$$

Added: Rereading the answer, we can forgo changes of variables, observing $\int_0^1 x^s \mathrm{d} s = \frac{x-1}{\log x}$. Then 
$$
   I = \int_0^\infty \frac{x-1}{\log x} \mathrm{e}^{-x} \mathrm{d} x = \int_0^\infty \int_0^1 x^s \mathrm{e}^{-x} \mathrm{d} s \mathrm{d} x  = \int_0^1 \int_0^\infty  x^s \mathrm{e}^{-x}  \mathrm{d} x \mathrm{d} s = \int_0^1 \Gamma(1+s) \mathrm{d} s
$$
