Gamma Function as Integral of Natural Logarithm

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Theorem

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

Then:

$\displaystyle \map \Gamma x = \int_{\to 0}^1 \paren {\ln \frac 1 t}^{x - 1} \rd t$

where $\Gamma$ denotes the Gamma function.


Proof

By definition of the Gamma function:

$\displaystyle \map \Gamma x = \int_0^{\to \infty} t^{x - 1} e^{-t} \rd t$

In order to allow the limits to be evaluated, this is to be expressed as:

$\displaystyle \map \Gamma x = \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_\delta^\Delta t^{x - 1} e^{-t} \rd t$

where $0 < \delta < \Delta$.


Let $-t = \ln u$.

Then by Derivative of Logarithm Function and the Chain Rule:

$-1 = \dfrac 1 u \dfrac {\rd u} {\rd t}$

Also:

\(\displaystyle -t\) \(=\) \(\displaystyle \ln u\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle e^{-t}\) \(=\) \(\displaystyle u\) Exponential of Natural Logarithm


Hence:

\(\displaystyle \int_\delta^\Delta t^{x - 1} e^{-t} \rd t\) \(=\) \(\displaystyle \int_{e^{-\delta} }^{e^{-\Delta} } \paren {-\ln u}^{x - 1} e^{\ln u} \paren {-\frac 1 u \frac {\rd u} {\rd t} } \rd t\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \int_{e^{-\Delta} }^{e^{-\delta} } \paren {-\ln u}^{x - 1} e^{\ln u} \paren {\frac 1 u \frac {\rd u} {\rd t} } \rd t\) Reversal of Limits of Definite Integral
\(\displaystyle \) \(=\) \(\displaystyle \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} u \frac 1 u \rd u\) Exponential of Natural Logarithm
\((1):\quad\) \(\displaystyle \) \(=\) \(\displaystyle \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} \rd u\) simplification


Next:

\(\displaystyle \lim_{\delta \mathop \to 0^+} e^{-\delta}\) \(=\) \(\displaystyle 1\) Exponential of Zero
\(\displaystyle \lim_{\Delta \mathop \to +\infty} e^{-\Delta}\) \(=\) \(\displaystyle 0\) Exponential Tends to Zero and Infinity


Thus:

\(\displaystyle \lim_{\delta \mathop \to 0^+, \ \Delta \mathop \to +\infty} \int_{e^{-\Delta} }^{e^{-\delta} } \paren {\ln \frac 1 u}^{x - 1} \rd u\) \(=\) \(\displaystyle \int_{e^{-\Delta} }^1 \paren {\ln \frac 1 u}^{x - 1} \rd u\) substituting for the top limit
\(\displaystyle \) \(=\) \(\displaystyle \int_{\to 0}^1 \paren {\ln \frac 1 u}^{x - 1} \rd u\) as $\ln 0$ is undefined

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

$\blacksquare$


Sources