Finding Laplace Transform of $(t-4)h(t-4)$

Question

Find the Laplace Transform of $(t-4)h(t-4)$, where

$$h(t) = \begin{cases} 1, & t > 0 \\ 0, & t < 0\text{.} \end{cases}$$

I am thinking about using the shifting theorem here. But coud not get to the core of it. Are there multiple ways to do this ?

Answer 1

The Laplace transform is $$ \int_0^\infty f(t)e^{-st}dt $$ Since you have a shifted unit step, the integral becomes $$ \int_0^\infty (t-4)h(t-4)e^{-st}dt=\int_0^4 0\cdot(t-4)e^{-st}dt+\int_4^\infty 1\cdot(t-4)e^{-st}dt=\int_4^\infty (t-4)e^{-st}dt $$ Can you integrate this?