# Gamma Function as Integral of Natural Logarithm

## 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 $t$.

$\blacksquare$