Gamma function integral

Question

How does the derivation of this integral from the Gamma function is equivalent to (n-1)!?$$\Gamma(n)=\displaystyle\int_0^1\Bigg(\ln\frac1x\Bigg)^{n-1}dx$$

Answer 1

Using Gamma definition $$\Gamma(n)=\displaystyle\int_0^\infty e^{-u}u^{n-1}\ du$$ with substitution $e^{-u}=x$ $$\Gamma(n)=\displaystyle\int_0^1(\ln\frac1x)^{n-1}dx$$

Answer 2

I assume you want the result only for positive integer $n$.

Integrate by parts to get a reduction formula. $$ \int_0^1\left(\log\frac{1}{x}\right)^{n-1} = -\left.\frac{x}{n}\left(\log\frac{1}{x}\right)^n\;\right|_{x=0}^{x=1} +\int_0^1\frac{1}{n}\left(\log\frac{1}{x}\right)^{n} =\frac{1}{n}\int_0^1\left(\log\frac{1}{x}\right)^{n} $$ And do the easy initial case $n=1$ $$ \int_0^1\left(\log\frac{1}{x}\right)^{0}dx = 1. $$ So by induction, $$ \int_0^1\left(\log\frac{1}{x}\right)^{n-1} = (n-1)! $$

added
Computation for $$ \left.\frac{x}{n}\left(\log\frac{1}{x}\right)^n\;\right|_{x=0}^{x=1} . $$ For $x=1$, we have $\log(1) = 0$ so $$ \frac{1}{n}\left(\log\frac{1}{1}\right)^n = 0. $$ (Recall $n \ge 1$.)

For $x=0$, we have $$ \lim_{x \to 0^+}\frac{x}{n}\left(\log\frac{1}{x}\right)^n = 0 $$ since the $x$ goes to zero much, much faster than the $\log$ goes to $\infty$.