Integers

The set $\mathbb{Z}=\{\ldots,-2,-1,0,1,2,\ldots\}$ of integers is a time scale.

$\mathbb{T}=\mathbb{Z}$

Forward jump:	$\sigma(t)=t+1$
Forward graininess:	$\mu(t)=1$
Backward jump:	$\rho(t)=t-1$
Backward graininess:	$\nu(t)=1$
$\Delta$-derivative	$f^{\Delta}(t)=f(t+1)-f(t)$
$\nabla$-derivative	$f^{\nabla}(t)=f(t)-f(t-1)$
$\Delta$-integral	$\displaystyle\int_s^t f(\tau) \Delta \tau = \left\{ \begin{array}{ll} \displaystyle\sum_{k=s}^{t-1} f(k) &; t \gt s \\ 0 &; t=s \\ -\displaystyle\sum_{k=t}^{s-1} f(k) &; t \lt s \end{array} \right.$
$\nabla$-integral	$\displaystyle\int_s^t f(\tau) \nabla \tau = \left\{ \begin{array}{ll} \displaystyle\sum_{k=s+1}^t f(k) &; t>s \\ 0 &; t=s \\ -\displaystyle\sum_{k=t+1}^s f(k) &; t\lt s \end{array} \right.$
$\Delta$-exponential	$e_p(t,s) = \left\{ \begin{array}{ll} \displaystyle\prod_{k=s}^{t-1} 1+p(k) &= t \gt s \\ 1 &= t=s \\ \displaystyle\prod_{k=t}^{s-1} \dfrac{1}{1+p(k)} &= t \gt s \\ \end{array} \right.$ (derivation)
$\nabla$-exponential
$\mathrm{sin}_p(t,s)$	$\sin_p(t,s) = \left\{ \begin{array}{ll} \dfrac{\displaystyle\prod_{k=s}^{t-1}1+ip(k) - \displaystyle\prod_{k=s}^{t-1}1-ip(k)}{2i} &; t>s \\ 0 &; t=s \\ \dfrac{\displaystyle\prod_{k=t}^{s-1} \frac{1}{1+ip(k)} - \displaystyle\prod_{k=t}^{s-1} \frac{1}{1-ip(k)}}{2i} &; t<s \end{array} \right.$ (derivation)
$\mathrm{sin}_1(t,0)$	$\begin{array}{ll} \sin_1(t,0) &= \dfrac{(1+i)^{t}-(1-i)^{t}}{2i} \\ &= \dfrac{\displaystyle\sum_{k=0}^{t} {t \choose k} i^k - \displaystyle\sum_{k=0}^{t} (-1)^k {t \choose k} i^k}{2i} \end{array}$
$\mathrm{cos}_p(t,t_0)$	$\begin{array}{ll} \cos_p(t,t_0) &= \dfrac{e_{ip}(t,t_0)+e_{-ip}(t,t_0)}{2} \\ &= \dfrac{\displaystyle\prod_{k=t_0}^{t-1}1+ip(k) + \displaystyle\prod_{k=t_0}^{t-1}1-ip(k)}{2} \end{array}$
$\mathrm{cos}_1(t,0)$	\begin{array}{ll} \cos_1(t,0) &= \dfrac{(1+i)^{t}+(1-i)^{t}}{2} \\ &= \dfrac{\displaystyle\sum_{k=0}^{t} {t \choose k} i^k + \displaystyle\sum_{k=0}^{t} (-1)^k {t \choose k} i^k}{2} \end{array}
Hilger circle
Gamma function:	$\Gamma_{\mathbb{Z}}(t;s)=\displaystyle\sum_{k=0}^{\infty} \left( \displaystyle\prod_{j=s}^{k-1} \dfrac{j+x}{j+1} \right) \dfrac{1}{2^{k+1}}$