Talk:Initial value theorem

The printable version is no longer supported and may have rendering errors. Please update your browser bookmarks and please use the default browser print function instead.

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)[reply]

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

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

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)