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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


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    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.
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