Alternative form of the 2nd translation theorem proof

[8bitgrafix]

hi guys, this is literally my first post here on physicsforums so i apologize in advance that my latex formatting sucks.

Homework Statement

prove the alternative form of the 2nd translation theorem of the laplace transform:
$L[f(t)u(t-a)]=e^{-sa}L[f(t+a)]$

where $u(t-a)$ is the unit step function and $$L[f(t)]=\int^\infty_0e^{-st}f(t)dt$$ is the definition of the laplace transform of $$f(x)$$

Homework Equations

The Attempt at a Solution

$L[f(t)u(t-a)]=\int^\infty_0 e^{-st}f(t)u(t-a)\,dt$
$$=\int^a_0 e^{-st}f(t)u(t-a)\,dt + \int^\infty_a e^{-st}f(t)u(t-a)\,dt$$
the first integral is 0 since the unit step function 0 for any t < a, and $$u(t-a)=1$$ for t >= a so the 2nd integral becomes, $$\int^\infty_a e^{-st}f(t)\,dt$$
then since the laplace transform is an integral from 0 to $$\infty$$ we need to make the substitutions $$u = t-a, \ t = u+a,\ du = dt$$ to change the lower limit of integration from a to 0. then the integral becomes, $$\int^\infty_0 e^{-s(u+a)}f(u+a)\,du$$
then you pull the $$e^{-sa}$$ outside the integral to give $$e^{-sa}\int^\infty_0 e^{-su}f(u+a)\,du$$ and this is where I'm stuck. I don't know what to do here since the definition of the laplace transform is $$L[f(u)]=\int^\infty_0e^{-su}f(u)du$$ and i can't make any more substitutions without changing the limits of integration.

 

[clamtrox]

What's $L[f(t+a)]$ then?