How to transform $- \int_{0}^{\infty} \ln(1 - x e^{-x}) \ \mathrm{d}x$ into $\sum_{n=1}^{\infty} \frac{n!}{n^{n+2}}$?

Question

In a previous question, I was helped to determine the value of an improper integral. Now, I would like to understand how to get this result, and my calculus is a bit rusty.

I would like to understand these steps: $$ \begin{align} - \int_{0}^{\infty} \ln(1 - x e^{-x}) \ \mathrm{d}x &= \sum_{n=1}^{\infty} \frac{1}{n} \, \int_{0}^{\infty} x^{n} \, e^{-n x} \ \mathrm{d}x \\ &= \sum_{n=1}^{\infty} \frac{n!}{n^{n+2}}. \end{align} $$

I think I recognize Taylor series, but I'm not able to reproduce these steps myself.

Answer 1

You have the reduction formula $$ \int_0^{\infty} x^k e^{-nx} \, dx = \left[ -\frac{1}{n} x^k e^{-nx} \right]_0^{\infty} + \frac{k}{n} \int_0^{\infty} x^{k-1} e^{-nx} \, dx = \frac{k}{n} \int_0^{\infty} x^{k-1} e^{-nx}. $$ Together with the formula $\int_0^{\infty} e^{-nx} \, dx$, this gives $$ \int_0^{\infty} x^n e^{-nx} \, dx = \frac{n!}{n^{n+1}}. $$

Answer 2

Since $$\frac1{1-u} = \sum_{n=0}^{\infty} u^n$$

and the convergence is uniform on $[0,\delta]$ for any $0 < \delta < 1$, we can integrate both sides and the series term by term to get

$$\ln(1-u) = -\sum_{n=0}^{\infty} \frac{u^{n+1}}{n+1} = -\sum_{n=1}^{\infty} \frac{u^n}{n}$$

Then use the fact $\int_0^{\infty} x^m e^{-x}dx = m!$