Talk:Initial value theorem

Untitled

Is F supposed to be the Laplace transform of ƒ? If so, it should say so. This article is written in a generally rather verbally challenged style. Michael Hardy (talk) 04:49, 6 February 2009 (UTC)

... nobody answered, so I've done some further editing. Michael Hardy (talk) 22:13, 2 March 2009 (UTC)

Does anyone think a proof or examples should be added? Krazyman (talk) 17:53, 27 February 2012 (UTC)

About the proof

I really don't see the necessity to complicate the proof that much. The initial value theorem is only makes sense for the one-sided Laplace transform, which means that $f(0^{-})$ does not make much sense (the derivative may not even exist at $0$, e.g., when using the Heaviside function).

Hence, one can pose $\lim _{s\to +\infty }sF(s)=\lim _{s\to +\infty }\left[\lim _{\epsilon \to 0^{+}}\int _{\epsilon }^{+\infty }{\frac {\partial }{\partial x}}f(x)e^{-sx}dx+f(\epsilon )\right]$. By exchanging the limit and the summation (integral), which is allowed because of the uniform convergence for $x>\epsilon >0$, one obtains that the integral vanishes whenever $\Re (s)>0$ (implied by $s\to +\infty$), and thus the required answer.

Did I miss something essential here ?

ikingut 16:59, 26 November 2014 (UTC)