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