Integral Form of Gamma Function equivalent to Euler Form/Proof 2

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Theorem

The following definitions of the concept of Gamma Function are equivalent:

Integral Form

The gamma function $\Gamma: \C \setminus \Z_{\le 0} \to \C$ is defined, for the open right half-plane, as:

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

where $\MM$ is the Mellin transform.


For all other values of $z$ except the non-positive integers, $\map \Gamma z$ is defined as:

$\map \Gamma {z + 1} = z \map \Gamma z$

Euler Form

The Euler form of the gamma function is:

$\ds \map \Gamma z = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1} } = \lim_{m \mathop \to \infty} \frac {m^z m!} {z \paren {z + 1} \paren {z + 2} \cdots \paren {z + m} }$

which is valid except for $z \in \set {0, -1, -2, \ldots}$.


Proof

First we present a lemma:

Let $0 \le t \le m$.

Then:

$0 \le e^{-t} - \paren {1 - \dfrac t m}^m \le t^2 \dfrac {e^{-t} } m$

$\Box$


Recall the definition of the partial Gamma function:

$\ds \map {\Gamma_m} x := \frac {m^x m!} {x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + m} }$

We have that:

\(\ds \) \(\) \(\ds \int_0^\infty e^{-t} t^{x - 1} \rd t - \map {\Gamma_m} x\)
\(\ds \) \(=\) \(\ds \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \map {\Gamma_m} x\)
\(\ds \) \(=\) \(\ds \int_0^m e^{-t} t^{x - 1} \rd t + \int_m^\infty e^{-t} t^{x - 1} \rd t - \int_0^m \paren {1 - \frac t m}^m t^{x - 1} \rd t\) Partial Gamma Function expressed as Integral
\(\ds \) \(=\) \(\ds \int_m^\infty e^{-t} t^{x - 1} \rd t + \int_0^m \paren {e^t - \paren {1 - \frac t m}^m t^{x - 1} } \rd t\)


We have that for large $t$:

$t^{x - 1} < t^{t / 2}$

and so as $m \to \infty$:

$\ds \int_m^\infty e^{-t} t^{x - 1} \rd t \to 0$

Then:

\(\ds \) \(\) \(\ds \int_0^m \paren {e^t - \paren {1 - \frac t m}^m t^{x - 1} } \rd t\)
\(\ds \) \(\le\) \(\ds \int_0^m \frac {t^2 e^{-t} } m e^{-t} t^{x - 1} \rd t\) Lemma
\(\ds \) \(=\) \(\ds \frac 1 m \int_0^m t^{x + 1} e^{-t} \rd t\)
\(\ds \) \(<\) \(\ds \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t\)


Now we have that as $m \to \infty$:

$\ds \frac 1 m \int_0^\infty t^{x + 1} e^{-t} \rd t \to 0$

so:

$\ds \int_0^\infty e^{-t} t^{x - 1} \rd t - \map {\Gamma_m} x = 0$

leading to:

$\ds \int_0^\infty e^{-t} t^{x - 1} \rd t = \lim_{m \mathop \to \infty} \dfrac {m^x m!} {x \paren {x + 1} \paren {x + 2} \dotsm \paren {x + m} }$

as was to be demonstrated.

$\blacksquare$


Sources