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$$\int_0^1(-\ln x)^ndx$$ 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.

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    $\begingroup$ Use integration by parts:$$I_n = \int_0^1(-\ln x)^ndx = \left[x(-\ln x)^n\right]_0^1 + n\int_0^1 (-\ln x)^{n-1}\,dx = nI_{n-1}$$ $\endgroup$
    – sciona
    Dec 27, 2014 at 5:04
  • $\begingroup$ That doesn't eliminate the $\ln(0)$ in the answer. $\endgroup$
    – Waffle
    Dec 27, 2014 at 5:10
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    $\begingroup$ There is no $\ln 0$ in the answer. The lower limit $\lim\limits_{x \to 0} x(\ln x) = 0$ ! $\endgroup$
    – sciona
    Dec 27, 2014 at 5:19
  • $\begingroup$ In that case, a recursive definition simplified to a verbatim definition of a factorial is definitely a short and simple closed-form solution. $\endgroup$
    – Waffle
    Dec 27, 2014 at 5:30

2 Answers 2

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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.

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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.
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  • $\begingroup$ Can't you do integration by parts to the original integral, without first doing that substitution? $\endgroup$ Dec 27, 2014 at 5:02
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    $\begingroup$ @GerryMyerson Yes, of course, I chose the substitution because the resulting form is (to me, anyway) more familiar. $\endgroup$ Dec 27, 2014 at 5:10
  • $\begingroup$ Could you explain how you change the integral in the first step? $\endgroup$
    – Waffle
    Dec 27, 2014 at 5:11
  • $\begingroup$ @Waffle This is just reverse substitution. The trickiest part is the limits: Since $\lim_{s \to -\infty} x = \lim_{s \to \infty} e^{-s} = 0$, $x = 0$ becomes $s = \infty$, and since $e^{-(0)} = 1$, $x = 1$ becomes $s = 0$. Does that help? $\endgroup$ Dec 27, 2014 at 5:22

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