Laplace Transform of Shifted Function

Question

Why do we need to multiply the shifted function $f(t - a)$ by the shifted step function $u(t - a)$ to obtain the Laplace transform?

$$ \mathcal{L\{f(t - a)\}} = \int_0^\infty u(t - a)f(t - a)\mathrm{e}^{-st}\,\mathrm{d}t $$ instead of $$ \int_0^\infty f(t - a)\mathrm{e}^{-st}\,\mathrm{d}t $$

Thanks

Answer 1

That doesn't look right. For $a \geq 0$, the basic identity is $L(u(t-a) f(t-a))=e^{-as} L(f(t))$. From this identity it follows that $L(u(t-a) f(t)) = e^{-as} L(f(t+a))$ (the more commonly used version of the formula when transforming forward).

Just $L(f(t-a))$ has no simple formula in terms of $L(f(t))$, because it involves the values of $f$ on $[-a,0)$, which have no contribution to the Laplace transform of $f$ itself.