I need to find the Laplace transform of the product of a function $f(t)$ with the Unit Heaviside Step Function $H(t-c)$, i.e., $\text{L}[H(t-c)f(t)]$.
Given that
$$\text{L}[H(t)f(t-c)] = e^{-sc}F(s)\hspace{5cm} \text{EQ. 1}$$
is a well known result, I made the substitution $y=t-c$ to get
$$\text{L}[H(t-c)f(t)] = \text{L}[H(y)f(y+c)] =\text{L}[H(y)f(y-(-c))]$$
I then made what I'm afraid is the questionable claim that
$$\text{L}[H(y)f(y-(-c))]=e^{-s(-c)}F(s)=e^{sc}F(s)$$
The reason that I feel this is questionable is because I suspect that EQ. 1 above is only valid for $c\ge0$.
Could someone comment on whether my approach is valid, and if not, what the correct Laplace transform of $H(t-c)f(t)$ actually is?
Thanks
Your Eq. (1) is not true; it should be $\mathscr{L}[H(t-c)f(t-c)](s) = e^{-sc}F(s)$ instead, while $$ \begin{align} \mathscr{L}[H(t-c)f(t)](s) &= \int_0^\infty H(t-c)f(t)e^{-st} \,\mathrm{d}t \\ &= \int_c^\infty f(t)e^{-st} \,\mathrm{d}t \\ &= \int_0^\infty f(t+c)e^{-s(t+c)} \,\mathrm{d}t \\ &= e^{-sc}\mathscr{L}[f(t+c)](s) \end{align} $$ which cannot be expressed in function of $F(s)$ in general. However, alternatively, you may write from the second line : $$ \mathscr{L}[H(t-c)f(t)](s) = F(s) - \int_0^c f(t)e^{-st} \,\mathrm{d}t $$