If $\mathcal{L}\{f(t)\} = F(s)$ , then

• $\mathcal{L}\{e^{at}f(t)\} \,=\, F(s\!-\!a)$     for $s > a$ ,
• $\mathcal{L}\{f(\frac{t}{a})\} \;=\; a\,F(as)$         for $a > 0$ .
  

For deriving these rules, we start from the definition of Laplace transform. In the first case, we shall use the notation $s\!-\!a = r$ : $$\mathcal{L}\{e^{at}f(t)\} = \int_0^\infty\!e^{-st}e^{at}f(t)\,dt = \int_0^\infty\!e^{-(s-a)t}f(t)\,dt = \int_0^\infty\!e^{-rt}f(t)\,dt = F(r) = F(s\!-\!a).$$ In the second case, we make the change of variable $\frac{t}{a} = u$ and later use the notation $sa = r$ : $$\mathcal{L}\{f(\frac{t}{a})\} = \int_0^\infty\!e^{-st}f(\frac{t}{a})\,dt = a\!\int_0^\infty\!e^{-sau}f(u)\,du = a\!\int_0^\infty\!e^{-ru}f(u)\,du = aF(r) = a\,F(as).$$