Say that I have some time signal $f(t)$ which is a real function of the real time variable $t$. I am interested in the consequences of the Laplace transform of $f$ if I multiply it by an Heaviside step function to mimic the fact that there is some hard cut in time at some point.
I thus want to compute ${\rm LT}[f(t)\times (1-H(t-t_0))]$. From what I understand of the convolution theorem for Laplace transforms, then I think that it should look something like $${\rm LT}[f(t) \times H(t-t_0)] = {\rm LT}[f] \circledast {\rm LT}[1-H[(t-t_0)].$$
First of all, is this correct ?
Then, intuitively, I would guess that when $t_0 \rightarrow +\infty$, then I would recover the Laplace transform of $f(t)$ itself, which would thus mean, if the above is true, that ${\rm LT}[1-H[(t-t_0)]$ should tend to a dirac delta function (i.e. the neutral element of the convolution). However, I also understand that this might not be so trivial because it seems that for this to be true, the limit and the integral within the definition of the Laplace transform would have to commute.
Indeed, I computed that ${\rm LT}[1-H[(t-t_0)] = e^{\lambda t_0}/\lambda$, which does not converges to a dirac delta function when $t_0 \rightarrow +\infty$.
Where am I wrong ? Is there a way to express what happens at the level of the function $f$ at the level of its laplace transform ?
Thanks a lot for your help !