I found a very strange pattern when I was playing around in Wolfram Alpha, and getting it to compute
$$\int_0^\infty \left(\frac{1-{e^{-x}}}{x}\right)^n \mathbb{d}x$$
for different values of $n$. I found that for integer $n$ ($n>1$) the answer works out to be $\frac{n}{(n-1)!}$ multiplied by an integer combination of $\ln(r)$ for every integer $r \le n$. Moreover, the coefficients grow very rapidly, and the coefficient of $\ln(r)$ seems to always have a large power of $r$ as a factor, yet they always manage to cancel out to produce a small answer. I don't have a particularly clear idea of the pattern, as Wolfram Alpha was unable to compute the answer for $n>8$, and Wolfram Alpha splits the composite terms, so I can't tell what the coefficients are except for primes $p$ such that $p \le n < 2p$.
I found this a very strange kind of answer, and I don't have any idea how to go about evaluating this integral. I would like to know what the pattern is, and understand why this is the answer. So I would appreciate it if someone who knows more calculus than me could give me an exact solution to the integral (and explain how to get the solution).
Best Answer
Hint: Integrate by parts $n-1$ times to express $$I_n=\int_0^\infty \left(\frac{1-{e^{-x}}}{x}\right)^n {d}x =\frac1{(n-1)!}\int_0^\infty \frac{d^{n-1} [(1-{e^{-x}})^n ]}{dx^{n-1}}\frac{dx}x $$ which can be evaluated by expanding $(1-{e^{-x}})^n $ and integrating piecewise to arrive at $$I_n= \frac1{(n-1)!}\sum_{k=0}^n\binom nk (-1)^{n+k}k^{n-1}\ln k $$