-(1/2) i Exp^(-(s^2/4)) Sqrt[\[Pi]] + DawsonF[s/2]

There is not a certain function as a solution like F(s) for your question, since

By definition of Laplace transform we eventually obtain that

exp(-(s^2/4)) \int_{-s/2}^{\infinity} exp(u^2) du

but the term including of integral is divergent for all s. this implies that Laplace transform of exp(t^2) is divergent for any s in real line.

On the other hand, we recall that the necessary condition for existence of Laplace transform of a function (means F(s)) is that the limit F(s) when s goes into infinity is 0.

But about this example this limit is equal to

\int_{-\infinity ^{\infinity} exp(u^2) du

which is obviously equal to infinity.

I think you mean Laplace transform of exp(-t^2)??? that is probably a better question to consider.

Can you clarify your question?

Your function is not of exponential order, so the integral defining Laplace transform of this function does not exist.

See

michael a. B. Deaktn

laplace transforms for superexponential

functions

j. Austral. Math. Soc. Ser. B 36(1994), 17-25

michael a. B. Deaktn

Dear Hassan, what is the answer, that u want to add here?