Scaling property of Laplace Transform not working

Question

I know that $L(f(t)) = F(s) \implies L(f(at))= \frac{1}{a}F(s/a)$ and if $L(1) = 1/s$

• $L(2 \cdot 1) = 1/2 \cdot F(s/2) = 1/2 \cdot 1/(s/2) = 1/s$ but
• $L(2 \cdot 1) = 2 \cdot L(1) = 2 \cdot 1/s = 2/s$

Why is there a contradiction? What am I missing?

Answer 1

In both calculations, you appear to seek the Laplace transform of a function that always returns $2$. Your second calculation is correct. To get $1/2\cdot1/(s/2)$ you'd want a function that multiplies $t$ by $2$ before returning $1$, which is just the original always-$1$ function.

It helps to consider the more general $L(cf(at))=ca^{-1}F(s/a)$, which includes rescaling of not only the argument but also the function's returned value. We must then take care not to confuse $c=2,\,a=1$ with $c=1,\,a=2$.