Gamma Function as Integral of Natural Logarithm

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Theorem

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

Then:

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

where $\Gamma$ denotes the Gamma function.


Proof

By definition of the Gamma function:

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

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

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

where $0 < \delta < \Delta$.


Let $-t = \ln u$.

Then by Derivative of Logarithm Function and the Chain Rule for Derivatives:

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

Also:

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


Hence:

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


Next:

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


Thus:

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

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

$\blacksquare$


Sources