# Gamma function

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.
We write formulas for gamma functions defined for $x \in \mathbb{R}^+$ and $s \in \mathbb{T}^+$.
 $\mathbb{T}=$ $\Gamma_{\mathbb{T}}(x;s)=$ $\mathbb{R}$ $\displaystyle\int_0^{\infty} \left( \dfrac{\tau}{s} \right)^{x-1}e^{-\tau} \mathrm{d}\tau$ $h\mathbb{Z};h>0$ $h \displaystyle\sum_{k=0}^{\infty} \left( \displaystyle\prod_{j=s}^{k-1} \dfrac{j+x}{j+1} \right) \dfrac{1}{(1+h)^{k+1}}$ $\overline{q^{\mathbb{Z}}}; q>1$ $\dfrac{(q-1)s}{(1+(q-1)x)^{\log_q(s)}} \displaystyle\sum_{k=-\infty}^{\infty} \dfrac{(1+(q-1)x)^k}{\prod_{j=-\infty}^{k} (1+(q-1)q^k)}$