U(x- 4) is the "unit step function" that is 0 for x< 4, 1 for $\displaystyle x\ge 4$. The Laplace transform of that is $\displaystyle \int_0^\infty e^{-st}u(s- 4)ds= \int_4^\infty e^{-st}ds$. If you let x= s- 4, that becomes $\displaystyle \int_0^\infty e^{-(x+ 4)t}dx= e^{-4t}\int_0^\infty e^{-xt}dx= e^{-4t}L(1).$