Equivalence of Definitions of Gamma Function

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Theorem

The following definitions of the concept of the 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:

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

Weierstrass Form

The Weierstrass form of the gamma function is:

$\ds \frac 1 {\map \Gamma z} = z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z / n} }$

where $\gamma$ is the Euler-Mascheroni constant.


The Weierstrass form is valid for all $\C$.

Hankel Form

The Hankel form of the gamma function is:

$\ds \frac 1 {\map \Gamma z} = \dfrac 1 {2 \pi i} \oint_\HH \frac {e^t \rd t} {t^z}$

where $\HH$ is the contour starting at $-\infty$, circling the origin in an anticlockwise direction, and returning to $-\infty$.


The Hankel form is valid for all $\C$.

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 $z \in \set {0, -1, -2, \ldots}$.



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Weierstrass Form equivalent to Euler Form

First it is shown that the Weierstrass form is equivalent to the Euler form.


\(\ds \frac 1 {\map \Gamma z}\) \(=\) \(\ds z e^{\gamma z} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z/n} }\) Weierstrass Form of $\Gamma$ Function
\(\ds \) \(=\) \(\ds z \paren {\lim_{m \mathop \to \infty} \exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} } \paren {\lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^\infty \paren {\paren {1 + \frac z n} e^{-z/n} } }\) Definition of Euler-Mascheroni Constant


Combining the limits:

\(\ds \frac 1 {\map \Gamma z}\) \(=\) \(\ds z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} \prod_{n \mathop = 1}^m \paren {\paren {1 + \frac z n} e^{-z/n} } }\)
\(\ds \) \(=\) \(\ds z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 + \frac 1 2 + \cdots + \frac 1 m - \ln \paren m} z} \exp \paren {\frac {-z} 1 + \frac {-z} 2 + \cdots + \frac {-z} m} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }\)
\(\ds \) \(=\) \(\ds z \lim_{m \mathop \to \infty} \paren {\exp \paren {\paren {1 - 1 + \frac 1 2 - \frac 1 2 + \cdots + \frac 1 m - \frac 1 m - \ln \paren m} z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }\) Exponential of Sum
\(\ds \) \(=\) \(\ds z \lim_{m \mathop \to \infty} \paren {m^{-z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }\)


But:

$(1): \quad m = \dfrac {m!} {\paren {m - 1}!} = \dfrac 2 1 \cdot \dfrac 3 2 \cdots \dfrac {x + 1} x \cdots \dfrac m {m - 1}$

Each term in $(1)$ is just $\dfrac {x + 1} x = 1 + \dfrac 1 x$, so:

$\ds m = \prod_{n \mathop = 1}^{m - 1} \paren {1 + \frac 1 n}$

Thus the expression for $\dfrac 1 {\map \Gamma z}$ becomes:

\(\ds \) \(\) \(\ds z \lim_{m \mathop \to \infty} \paren {\prod_{n \mathop = 1}^{m - 1} \paren {1 + \frac 1 n}^{-z} \prod_{n \mathop = 1}^m \paren {1 + \frac z n} }\)
\(\ds \) \(=\) \(\ds z \lim_{m \mathop \to \infty} \paren {\paren {1 + \frac 1 m}^z \prod_{n \mathop = 1}^m \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n} }\)
\(\ds \) \(=\) \(\ds z \lim_{m \mathop \to \infty} \paren {1 + \frac 1 m}^z \lim_{m \mathop \to \infty} \prod_{n \mathop = 1}^m \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n}\) Product Rule for Complex Sequences
\(\ds \) \(=\) \(\ds z \prod_{n \mathop = 1}^\infty \paren {1 + \frac 1 n}^{-z} \paren {1 + \frac z n}\)

Hence:

$\ds \map \Gamma z = \frac 1 z \prod_{n \mathop = 1}^\infty \paren {1 + \frac 1 n}^z \paren {1 + \frac z n}^{-1}$

which is the Euler form of the Gamma function.

$\Box$


Integral Form equivalent to Euler Form

This is proved in the page:

Integral Form of Gamma Function equivalent to Euler Form

$\blacksquare$


Sources

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