Proof of a closed form of $\int_0^1(-\ln x)^ndx$

Question

$$\int_0^1(-\ln x)^ndx$$ Is there a step-by-step solution to a closed form of this expression? I've tried using different representations to re-write the expression but I couldn't find anything I knew how to simplify.

Answer 1

Take the integral $$I(a) = \int _{0}^{1} x^a \rm{d}x= \frac{1}{a+1} $$ Now take the derivative $n$ times and obtain $$I(a)^{(n)}= \frac{(-1)^nn!}{(a+1)^{n+1}},$$ which gives $$I(0)^{(n)} =(-1)^nn!.$$ On the other hand differentiating under the integral sign gives $$I(0)^{(n)}= \int _{0}^{1} \ln (x)^n \rm{d}x$$ And so conclude the result.

Answer 2

Substitute $$x = e^{-s}, \qquad dx = -e^s \,ds ,$$ so that the integral becomes $$\int_0^{\infty} s^n e^{-s}\,ds.$$ But, this is a well known formula for the factorial $n!$, or for $\Gamma(n + 1)$, where $\Gamma$ is the usual Gamma function.

If one doesn't know this formula, then, at least when $n$ is a nonnegative integer, one can evaluate the integral by

1. applying integration by parts to produce a reduction formula that expresses it in terms of $$\int_0^{\infty} s^{n - 1} e^{-s} \, ds,$$ and
2. applying an induction argument.