Talk:Initial value theorem

Untitled
Is F supposed to be the Laplace transform of &fnof;? 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 $$ \lim_{s\to+\infty} sF(s) = \lim_{s\to+\infty}\left[ \lim_{\epsilon\to0^+}\int_{\epsilon}^{+\infty}\frac{\partial}{\partial x}f(x)e^{-sx}dx + f(\epsilon) \right] $$. 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)