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First, according to

\begin{align*} \int_{0}^{\infty} x^{-m}(1-e^{-x})^{n} \, dx =\frac{n}{1-m}\int_{0}^{\infty} x^{1-m}(1-e^{-x})^{n} \, dx -\frac{n}{1-m}\int_{0}^{\infty} x^{1-m}(1-e^{-x})^{n-1} \, dx, \end{align*}

which can be denoted by

\begin{align*} I_{m,n} = \frac{n}{1-m}I_{m-1,n}-\frac{n}{1-m}I_{m-1,n-1}. \tag{1} \end{align*}

I also find

\begin{align*} \int_{0}^{\infty} e^{-nx}\ln x \, dx = -\frac{\gamma}{n}-\frac{\ln n}{n}. \end{align*}

So

\begin{align*} \int_{0}^{\infty} x^{-2} \left(1 - e^{-x}\right)^{n} \, dx &= -\int_{0}^{\infty} \left(1 - e^{-x} \right)^{n} \, d\left(\frac{1}{x}\right) = n \int_{0}^{\infty} \frac{\left( 1 - e^{-x} \right)^{n - 1} e^{-x}}{x} \, dx \\ &= n \int_{0}^{\infty} \frac{\sum_{k = 0}^{n - 1} \binom{n-1}{k} ( -1 )^{k} e^{-(k + 1)x} }{x} \, dx \\ &= n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} (k+1) \int_{0}^{\infty} e^{-(k+1)x} \ln x \, dx \\ &= n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} (k+1) \left( -\frac{\gamma}{k+1} - \frac{\ln (k + 1)}{k + 1} \right) \\ &= -n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} \left( \gamma + \ln (k + 1) \right) \\ &= -n\sum_{k = 0}^{n - 1} \binom{n-1}{k} (-1)^{k} \ln (k + 1). \end{align*}

Hence, it seems that we can solve this integral by recursive relation $\text{(1)}$, but how to get a accurate result about the integral.

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  • $\begingroup$ You should use double dollar signs instead of the [ ... ] commands to render equations. $\endgroup$ – Danu Jul 11 '14 at 3:09
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A basic idea is to use integration by parts, as you did for $I_{2,n}$.

Let $2 \leq m \leq n$ be integers and denote $f(x) = (1 - e^{-x})^{n}$. We have two observations:

  • $f^{(k)}(x) = O(x^{n-k})$ near $x = 0$ for $0 \leq k \leq n$.
  • $f^{(k)}(x) = O(e^{-x})$ near $x = \infty$ for $1 \leq k \leq n$.

Using this, applying integration by parts $m$ times, we obtain

\begin{align*} I_{m,n} = \int_{0}^{\infty} \frac{(1 - e^{-x})^{n}}{x^{m}} \, dx &= -\frac{1}{(m-1)!} \int_{0}^{\infty} f^{(m)}(x) \log x \, dx \\ &= \frac{(-1)^{m-1}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m} \int_{0}^{\infty} e^{-kx} \log x \, dx \\ &= \frac{(-1)^{m}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m-1} (\gamma + \log k) \\ &= \frac{(-1)^{m}}{(m-1)!} \sum_{k=1}^{n} \binom{n}{k} (-1)^{k} k^{m-1} \log k. \end{align*}

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