Gamma Function as Integral of Natural Logarithm

Theorem
Let $x \in \R_{>0}$ be a strictly positive real number.

Then:
 * $\displaystyle \Gamma \left({x}\right) = \int_{\to 0}^1 \left({\ln \frac 1 t}\right)^{x - 1} \ \mathrm d t$

where $\Gamma$ denotes the Gamma function.

Proof
By definition of the Gamma function:
 * $\displaystyle \Gamma \left({x}\right) = \int_0^{\to \infty} t^{x-1} e^{-t} \ \mathrm d t$

In order to allow the limits to be evaluated, this is to be expressed as:
 * $\displaystyle \Gamma \left({x}\right) = \lim_{\delta \mathop \to 0^+, \ \Delta \mathrel \to +\infty} \int_\delta^\Delta t^{x-1} e^{-t} \ \mathrm d t$

where $0 < \delta < \Delta$.

Let $-t = \ln u$.

Then by Derivative of Logarithm Function and the Chain Rule:
 * $-1 = \dfrac 1 u \dfrac {\mathrm d u}{\mathrm d t}$

Also:

Hence:

Next:

Thus:

The result follows by renaming the dummy variable $u$ to $t$.