# Unilateral Laplace transform

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If $\mathbb{T}$ is a time scale, $s \in \mathbb{T}$, and $f$ is rd-continuous, then we define the unilateral Laplace transform of $f$ about $s$ by the formula $$\mathscr{L}_{\mathbb{T}}\{f\}(z;s) = \displaystyle\int_s^{\infty} f(t) e_{\ominus z}(\sigma(t),s) \Delta t,$$ for all $z$ for which the integral converges, where $\displaystyle\int \ldots \Delta t$ denotes the delta integral, $e_{\ominus z}$ denotes a delta exponential whose subscript is the forward circle minus of the constant $z$, and $\sigma$ is the forward jump.
 $\mathbb{T}=$ Unilateral Laplace transform $\mathbb{R}$ $\mathscr{L}_{\mathbb{R}}\{f\}(z;s)=\displaystyle\int_s^{\infty} f(\tau) e^{-z\tau} \mathrm{d}\tau$ $\mathbb{Z}$ $h\mathbb{Z}$ $\mathbb{Z}^2$ $\overline{q^{\mathbb{Z}}}, q > 1$ $\overline{q^{\mathbb{Z}}}, q < 1$ $\mathbb{H}$
 $f(t;s)$ $\mathscr{L}\{f(\cdot;s)\}(z)$ $e_{\alpha}(t;s)$ $\dfrac{1}{z-\alpha}$ $h_n(t;s)$ $\dfrac{1}{z^{n+1}}$ $\sinh_{\alpha}(t;s)$ $\dfrac{\alpha}{z^2-\alpha^2}$ $\cosh_{\alpha}(t;s)$ $\dfrac{z}{z^2-\alpha^2}$ $\sin_{\alpha}(t;s)$ $\dfrac{\alpha}{z^2+\alpha^2}$ $\cos_{\alpha}(t;s)$ $\dfrac{z}{z^2+\alpha^2}$