Why does Laplace(t) not diverge ?

[kkirtac]

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 doesn't it diverge? thanks in advance

 

[Karlisbad]

the question is that if $$f(t)=O(e^{at})$$ or what is the same..

$$f(t)e^{-at})$$ for and a>0 then the F(s) Laplace transform exist for every s>a if you try to apply the Laplace transform of $$exp(ax^{2}$$ 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.

 

[kkirtac]

the question is that if $$f(t)=O(e^{at})$$ or what is the same..

$$f(t)e^{-at})$$ for and a>0 then the F(s) Laplace transform exist for every s>a if you try to apply the Laplace transform of $$exp(ax^{2}$$ 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.

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.

 

[HallsofIvy]

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), $lim_{s\rightarrow \infty}P(s)e^{-st}= 0$.

In fact the Laplace transform is only defined for functions for which that is true.