Integral Form of Gamma Function equivalent to Euler Form

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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 1

It is taken for granted that the Gamma function increases monotonically on $\R_{\ge 1}$.


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We begin with an inequality that can easily be verified using cross multiplication.


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In particular: ... so verify it.
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Let $x$ be a real number between $0$ and $1$.

Let $n$ is a positive integer.

Then:

$\ds \frac {\log \map \Gamma {n - 1} - \log \map \Gamma n} {\paren {n - 1} - n} \le \frac {\log \map \Gamma {x + n} - \log \map \Gamma n} {\paren {x + n} - n} \le \frac {\log \map \Gamma {n + 1} - \log \map \Gamma n} {\paren {n + 1} - n}$

Since n is a positive integer, we can make use of the identity:

$\map \Gamma n = \paren {n - 1}!$

Simplifying, we get:

$\map \log {n - 1} \le \dfrac {\log \map \Gamma {x + n} - \map \log {\paren {n - 1}!} } x \le \map \log n$

We now make use of the identity:

$\ds \map \Gamma {x + n} = \prod_{k \mathop = 1}^n \paren {x + n - k} \map \Gamma x$

along with the fact that the Gamma Function is Log-Convex, to simplify the inequality:

$\ds \paren {n - 1}^x \paren {n - 1}! \prod_{k \mathop = 1}^n \paren {x + n - k}^{-1} \le \map \Gamma x \le n^x \paren {n - 1}!\prod_{k \mathop = 1}^n \paren {x + n - k}^{-1}$

Taking the limit as $n$ goes to infinity and using the Squeeze Theorem:

$\ds \map \Gamma x = \lim_{n \mathop \to \infty} n^x n! \prod_{k \mathop = 0}^n \paren {x + n - k}^{-1}$

which is another representation of Euler's form.

This proves equivalence for $x$ between $0$ and $1$.

The result follows from the Gamma Difference Equation.

$\blacksquare$


Proof 2

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.

$\blacksquare$

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