Gamma function

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Let $\mathbb{T}$ be a time scale and define $p_f(t,s)=e_{\frac{f}{\mathrm{id}}}(t,s)$, where $\mathrm{id}$ denotes the identity map $\mathrm{id} \colon \mathbb{T} \rightarrow \mathbb{T}$ and $e_{\frac{f}{\mathrm{id}}}$ denotes the delta exponential. With these definitions, we define the gamma operator $$\Gamma_{\mathbb{T}}(f;s)=\mathscr{L}_{\mathbb{T}}\{p_{f \boxminus_{\mu} 1}(\cdot,s)\}(1)=\displaystyle\int_0^{\infty} p_{f \boxminus_{\mu}1}(\eta,s) e_{\ominus_{\mu}1}^{\sigma}(\eta,0) \Delta \eta,$$ where $\mathscr{L}_{\mathbb{T}}$ denotes the Laplace transform, $\boxminus_{\mu}$ denotes forward box minus, $\ominus_{\mu}$ denotes forward circle minus, and $\sigma$ denotes the forward jump.

Properties of gamma functions

Convergence of gamma function at positive values
Gamma function diverges at zero
Gamma function diverges at infinity
Gamma function equals one at one
Gamma function of x boxplus one

Define the bracket number operators (they are actually functions) $$[n]_{\mathbb{T}} = \left\{ \begin{array}{ll} 0 &; n=0 \\ [n-1]_{\mathbb{T}} \boxplus_{\mu} 1 &; n=1,2,\ldots \end{array} \right.$$ and the bracket factorial $$[n]_{\mathbb{T}}! = \left\{ \begin{array}{ll} 1&; n=0 \\ \displaystyle\prod_{j=1}^n [j]_{\mathbb{T}} &; n=1,2,\ldots \end{array} \right.$$

Theorem: Let $n \in \mathbb{Z}^+$ and assume that $[k]_{\mathbb{T}}$ is a constant function on $\mathbb{T}^+$ for all $k\in[1,n]\bigcap \mathbb{Z}^+$. Then $$\Gamma_{\mathbb{T}}\left( [n]_{\mathbb{T}};s \right) = \dfrac{[n-1]_{\mathbb{T}}!}{s^{n-1}}.$$

Proof:

Examples of gamma functions

We write formulas for gamma functions defined for $x \in \mathbb{R}^+$ and $s \in \mathbb{T}^+$.

$\mathbb{T}=$ $\Gamma_{\mathbb{T}}(x;s)=$
$\mathbb{R}$ $\displaystyle\int_0^{\infty} \left( \dfrac{\tau}{s} \right)^{x-1}e^{-\tau} d\tau$
$h\mathbb{Z};h>0$ $h \displaystyle\sum_{k=0}^{\infty} \left( \displaystyle\prod_{j=s}^{k-1} \dfrac{j+x}{j+1} \right) \dfrac{1}{(1+h)^{k+1}}$
$\overline{q^{\mathbb{Z}}}; q>1$ $\dfrac{(q-1)s}{(1+(q-1)x)^{\log_q(s)}} \displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(1+(q-1)x)^k}{\prod_{j=-\infty}^{k} (1+(q-1)q^k)}$

References