laplace transform of unit step function $u(t+1)$

Question

what is the laplace transform of unit step function $u(t+1)$?According to my understanding it should be $ e^s/s$ but I am not sure about it.

Answer 1

It is indeed. More generally,

$$\mathcal{L}(u(t-a)) = \frac{e^{-as}}{s}$$

Take $a=-1$ for your desired transform. A number of identities such as this can be found on this page I have bookmarked.

Answer 2

When you have doubt , you can always use the integral definition of the Laplace Transform: $$f(s)=\int_0^\infty u(t+1)e^{-st}dt$$ $$f(s)=\int_{-1}^\infty e^{-st}dt$$ $$f(s)=\dfrac {e^{-st}}{-s}\bigg|_{-1}^\infty$$ $$f(s)=\dfrac {e^{s}}{s}$$ So yes you're right.