Question

\[g(x)= \{t~~~~~t>1\]\[~~~~~~~~~~~~~\{1~~~~0 14 years ago

Answer 1

Laplace transform

Answer 2

\[\mathcal{L}\{g(t)\}=\int\limits_{0}^{1}1\cdot e^{-st}dt+\int\limits_{1}^{+\infty}t\cdot e^{-st}dt=\]\[=\frac{1-e^{-s}}{s}+\frac{(1+s)e^{-s}}{s^2}=\frac{s+e^{-s}}{s^2}=\frac{1}{s}+\frac{1}{s^2}e^{-s}\]

Answer 3

thanks nikvist