The Laplace transform of the delta function

Question

$f(t) = 1$ must equal to delta function in the Laplace domain since "constant in one domain is delta in the other domain". On the other hand, table says that it must be $1/s$ in the Laplace domain. This also seems true indeed since differentiating the constant we'll delta-pulse in the time domain and constant 1 in the Laplace domain (s is differentiation operator in the Laplace domain). I can make two (likely fallacies) conclusions here: Laplace operator is the same as s (diff) operator since it translates between const-1 and delta pulse and, secondly, delta-pulse = hyperbola $1/s$. Where is the flaw?

Answer 1

The standard argument is: the Laplace transform of $f$ is $$F(s)= \int_0^\infty e^{-st} f(t)\,dt$$ For the delta-function at $0$ this gives $F(s)\equiv 1$.

However,

"constant in one domain is delta in the other domain"

does not work in the other direction. In contrast to the Fourier transform (which is very similar to its inverse), the Laplace transform is quite different from its inverse. The Laplace transform of the function that is identically equal to $1$ on $[0,\infty)$ is $1/s$, not a delta function.

$$F(s)= \int_0^\infty e^{-st} 1\,dt = \frac1s$$

differentiation = s in Laplace domain

True: if $f$ transforms to $F(s)$, then $f'$ transforms to $sF(s)$.

getting 1 means that we have got delta-pulse in the time domain

Yes, that's what I began with. The delta-function at $0$ transforms into $F(s)\equiv 1$.

We have concluded that delta-pulse function is equal to hyperbola 1/s.

I haven't, and I don't see how you did. My guess is that you confused two kinds of pulses: jump by $1$, like in the Heaviside function, and unit point mass. The Heaviside function $H$ does transform into $1/s$. The delta-function is the (distributional) derivative of $H$. By the derivative rule for Laplace transform, the delta-function transforms to $s(1/s) = 1$.