Let $\varepsilon \in \mathbb{R}$, the following integral
$$ \int_{\varepsilon}^\infty \delta(t)e^{-st}dt. $$
converges to $1$ if $\varepsilon \to -0$ and $0$ if $\varepsilon \to +0$. This shows Laplace Trasform of delta function never converges, while Laplace Transform table says $\delta(t)$ is transformed into 1.
Why is Laplace Transform of $\delta(t)$ $F(s)=1$? Is there any proof or rationale?