$I = \int\limits_0^\infty {\frac{{{e^{ - nx}} - {e^{ - x}}}}{x}dx} = ?$$n \in \mathbb{N}$

Let, ${e^{ - x}} = t$

$ \Rightarrow  - {e^{ - x}}dx = dt$

$ \Rightarrow  - tdx = dt$

$I = \int\limits_1^0 {\frac{{{t^n} - t}}{{ - \ln t}}\frac{{dt}}{{ - t}}}  =  - \int\limits_0^1 {\frac{{{t^{n - 1}} - 1}}{{\ln t}}dt} $

But, $I(\lambda ) = \int\limits_0^1 {\frac{{{x^\lambda } - 1}}{{\ln x}}dx = \ln (\lambda  + 1)} $ (Proof can be seen in this problem)

So, $I =  - \ln (n - 1 + 1) =  - \ln n$