# Laplace transform

• mhill
In summary, the conversation is discussing the use of big-O notation to prove that a function f(t) is equal to O(t). The speaker mentions using the Laplace transform of f(t) to perform an integral and shows that if f(t) is equal to O(t), then the transformed function F(s) is equal to O(s-2). However, it is noted that the converse has not been proven and the speaker asks if there is a method to prove that f(t)= O(t) if it is the solution of a certain integral equation.

#### mhill

let be a function f(t) , and i want to prove that $$f(t)=O(t)$$ in big-O notation.

i know that Laplace transform of f(t) is F(s) then i perform the integral

$$F(s)= \int_{0}^{\infty} dt f(t) e^{-st}$$ if we assume f(t)=O(t) then

$$F(s)= \int_{0}^{\infty} dt f(t) e^{-st} \le \int_{0}^{\infty} dt e^{-st}t$$

so it would be enough that $$F(s) \le Cs^{-2}$$ for a positive constant 'C'

is this enough ?

It is not clear what it is you are trying to prove. You have shown that if f(t)= O(t)[/itex] then F(t)= O(s-2). You have NOT shown the converse: that if F(t)= O(s-2) then f(t)= O(t).

sorry perhaps i did not explain myself well, the general idea y had is this given a function f(t) that is the solution to an integral equation

$$g(s)= \int_{0}^{\infty} K(s,t)f(t)$$

g(s) and K(s,t) are known , my question (more general than the previous one) is , would be a method to prove that $$f(t)= O(t)$$ if we know that f(t) is the solution of certain integral equation ? , thanks.