Laplace transform

  • Thread starter mhill
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  • #1
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Main Question or Discussion Point

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

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

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

[tex] F(s)= \int_{0}^{\infty} dt f(t) e^{-st} \le \int_{0}^{\infty} dt e^{-st}t [/tex]

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

is this enough ?
 

Answers and Replies

  • #2
HallsofIvy
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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).
 
  • #3
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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

[tex] g(s)= \int_{0}^{\infty} K(s,t)f(t) [/tex]

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

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