What is the significance of 'sufficiently large' values in Laplace Transform?

[Miike012]

In the attachment that I added I highlighted the portion I am questioning.

I will define L[f(t)](s) to be the laplace transform of the function f(t).

f(t) = e^t

L[f(t)](s) = 1/(s-1). The laplace transform is defined for all values s≠1.

L[f(t)](2) = 1.

Question: "What do they mean by sufficiently large vales of s" because I don't consider s = 2 a large value.

 

[Ray Vickson]

In the attachment that I added I highlighted the portion I am questioning.

I will define L[f(t)](s) to be the laplace transform of the function f(t).

f(t) = e^t

L[f(t)](s) = 1/(s-1). The laplace transform is defined for all values s≠1.

L[f(t)](0) = -1.

Question: "What do they mean by sufficiently large vales of s" because I don't consider s = 0 a large value.

Sufficiently large means s ≥ a (or maybe s > a) for some finite number 'a'. Although 1/(s-1) exists for all s ≠ 1, it really only represents the Laplace integral for s > 1. Here, '1' is a sufficiently large number. You are right: '0' is not sufficiently large.

 

[Miike012]

I think it would be better if they said s can be any value so long the integral (in the attachment) converges.

Example:

L[e^(at)](s) = ∫e^-ste^atdt [from 0 to ∞] = e^[t(a-s)]/(a-s) [from 0 to ∞] = 0 + 1/(s-a)

where the term e^[R(a-s)]/(a-s) tends to 0 as R→∞ IF s>a

 

[Miike012]

Sufficiently large means s ≥ a (or maybe s > a) for some finite number 'a'. Although 1/(s-1) exists for all s ≠ 1, it really only represents the Laplace integral for s > 1. Here, '1' is a sufficiently large number. You are right: '0' is not sufficiently large.

Sorry I noticed that the laplace didn't converge for s = 0 so I changed it to s = 2