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 bound variable $u$ to $t$.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.7 \ (1)$