laplace transform U(t-1)

L[e^t U(t-1)]
=∫ (e^t)(e^(-st))dt [from 1 to infinity]
=∫ e^[-(s-1)t] dt [from 1 to infinity]
=-e^[-(s-1)]/(s-1)

所以最後負負得正。就我所知就一個函數和一個階梯函數的乘積﹐是沒有公式套用的﹐每次都要親自計。

L{ f(t-p) u(t-p) } = e^(-ps) F(s)

The positive sign is obviously a mistake.