Answer:
[tex]\mathcal{L}\{f(t)\} = \frac{f(0)}{2\cdot s}[/tex]
Step-by-step explanation:
The definition of the Laplace Transform lead to the following form:
[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = -\frac{1}{s}\cdot e^{-s\cdot t}\cdot f(t)|_{0}^{\infty}-\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dx[/tex]
[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = -\frac{1}{2\cdot s}\cdot e^{-s\cdot t}\cdot f(t)|_{0}^{\infty}[/tex]
[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = -\frac{1}{2\cdot s}\cdot e^{-s\cdot t}\cdot f(\infty)+\frac{1}{2\cdot s}\cdot f(0)[/tex]
[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = 0+\frac{1}{2\cdot s}\cdot f(0)[/tex]
[tex]\int\limits^{\infty}_{0} {e^{-s\cdot t}\cdot f(t)} \, dt = \frac{1}{2\cdot s}\cdot f(0)[/tex]
[tex]\mathcal{L}\{f(t)\} = \frac{f(0)}{2\cdot s}[/tex]
