Evaluating $\int_{0}^{1} \left ( \ln \frac{1}{x} \right)^ndx$.

Question

I have found this result on the Internet

$$\int_{0}^{1} \left ( \ln \frac{1}{x} \right)^ndx = \Gamma(n+1).$$

I know that if $n \in \mathbb{N}$, the proof is not complicated. However, if $n \in \mathbb{R}$, Do you have any idea to proof it?

Answer 1

Replace $x$ with $e^{-t}$. You get:

$$\int_{0}^{1}(-\log x)^n\,dx = \int_{0}^{+\infty} t^{n}e^{-t}\,dt $$ that is precisely the definition of $\Gamma(n+1)$, given that $\Re(n)>-1$.

Answer 2

Let $x=e^{-u}$. Then, our integral is $$ \displaystyle\int_0^1 \left( - \ln x \right)^n \, \mathrm{d}x = \displaystyle\int_0^{+\infty} u^n e^{-u} \, \mathrm{d}u = \Gamma(n+1), $$ if and only if $ \text{Re} \left( n \right) > -1 $, because of definition. $\Box$

Answer 3

Let $x=e^{-t}$, so $dx=-e^{-t}dt$. When $x=0$, $t=\infty$ and when $x=1, t=0$. Then $$\int_{0}^{1}\left(\ln\frac{1}{x}\right)^{n}dx=-\int_{\infty}^{0}t^{n}e^{-t}dt=\int_{0}^{\infty}t^{n}e^{-t}dt$$ which is the definition of $\Gamma(n+1)$.