What Happens to the Laplace Transform of Asymptotic Functions?

[mhill]

let be f(x) \sim g(x) , in the sense that for big x f(x) is asymptotic to g(x) , my question if what happens to their Laplace transform ??

i believe that \int _{0}^{\infty}dt f(t)exp(-st) \approx \int _{0}^{\infty}dt g(t)exp(-st)

in first approximation the Laplace transform of f(x) and the Laplace transform of g(x) must be equal.

another question if we had a Linear operator L so we can define its inverse L^{-1} is it true that f(x) \sim L(g(x)) \rightarrow L^{-1} f(x)= g(x)

 

[HallsofIvy]

The answer to your first question is "what do you mean by \approx?". The difference between f and g may be extremely large for small x: and that difference will show up in the integral.

As for your second question- isn't that the definition of "inverse"?

 

[mhill]

Yes Hallsoftivy L^{-1} means the inverse operator, but is it valid even for asymptotic relations ?? , and as for the Laplace integral if f(x) is asymptotic to g(x) and the Laplace transform of f(x) and g(x) are denoted by F(s) G(s) then would be true that F(s) \sim G(s) ?, but i believe that having a kernel inside the integral transform exp(-st) for big values of 's' the remainder of the asymptotic relation could be make smoother