1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why does Laplace(t) not diverge ?

  1. Nov 14, 2006 #1
    hello , anybody can explain me why Laplace(t) does not diverge ? when we perform integration by parts, the "uv" part contains the "t" variable and when we place the infinity bound here, t should cause the "uv" part to diverge as the exponential goes to zero.. so why doesnt it diverge? thanks in advance
     
  2. jcsd
  3. Nov 14, 2006 #2
    the question is that if [tex] f(t)=O(e^{at}) [/tex] or what is the same..

    [tex] f(t)e^{-at}) [/tex] for and a>0 then the F(s) Laplace transform exist for every s>a if you try to apply the Laplace transform of [tex] exp(ax^{2} [/tex] for a>0 you will say that this does not exist since for big x the quadratic term is dominant..another example of a function not having a Laplace trasnform is f(t)=1/(t-1) due to the singular point at t=1.
     
  4. Nov 15, 2006 #3

    thank you for your interest Karlisbad , i need a better explanation, can you or someone else explain it a bit better so that i can understand well.thank you.
     
  5. Nov 15, 2006 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    The point is that it still has e-st in it. That exponential goes to 0 as s goes to infinity and "dominates" any polynomial. That is, for any polynomial P(s), [itex]lim_{s\rightarrow \infty}P(s)e^{-st}= 0[/itex].

    In fact the Laplace transform is only defined for functions for which that is true.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Why does Laplace(t) not diverge ?
Loading...