Answering this question, I've tried to alert the OP about the misleading definition of $\delta(t)$ - used in one of the answers - as:
$$ \delta(x) = \left\{\begin{array}{cc} \infty & x = 0 \\ 0 & x\neq 0 \end{array}\right. $$
And gave an example when this definition fails:
I'll show one example that illustrates why:
What is $\mathscr{L}\left[\,\delta(t)\,\right]$? And $\mathscr{L}\left[\,\delta(3t)\,\right]$?
Using the above interpretation, we have: $$ \begin{alignat}{0} \delta(t) = \left\{\begin{array}{cc} \infty & t = 0 \\ 0 & t\neq 0 \end{array}\right. &&\delta(3t) = \left\{\begin{array}{cc} \infty & 3t = 0 \\ 0 & 3t\neq 0 \end{array}\right. = \left\{\begin{array}{cc} \infty & t = 0 \\ 0 & t\neq 0 \end{array}\right. =\delta(t) \end{alignat} $$ So, since we have the same "function"... $\mathscr{L}\left[\,\delta(t)\,\right]$ must be equal to $\mathscr{L}\left[\,\delta(3t)\,\right]$, right? WRONG! $$ \begin{alignat}{1} \mathscr{L}\left[\,\delta(t)\,\right]&=1 \\ \mathscr{L}\left[\,\delta(3t)\,\right]&=\frac 1 3 \end{alignat} $$ But Laplace Transform Uniqueness Theorem states that:
If $f_1$ and $f_2$ are continuous on $[0,∞)$, then $\mathscr{L}[\,f_1\,]\neq\mathscr{L}[\,f_2\,] \iff f_1 \neq f_2$
Then: $$ \mathscr{L}\left[\,\delta(t)\,\right] \neq \mathscr{L}[\,\delta(3t)\,] \iff \delta(t) \neq \delta(3t) $$
I argued that $\delta(t)$ must be different of $\delta(3t)$ because of Laplace Uniqueness Theorem. But now I realized that my argument fails, because Dirac Delta "function" isn't continuous. So, my questions are:
Is this theorem valid for Delta "function", even if it isn't continuous in the traditional sense? If yes, why?
If no, is there a stronger theorem about uniqueness of Laplace Transform, which allows Dirac Delta "function" to be included?