User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/existence

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/integrand

User:GFauxPas/Sandbox/Zeta2/lnxln1-x/evaluation

User:GFauxPas/Sandbox/Zeta2/FourierSeries/

User:GFauxPas/Sandbox/Zeta2/Informal Proof

$\mathcal L \left\{{}\right\}$

Theorem
Let $f\left({t}\right): \R \to \R$ or $\R \to \C$ be a function of exponential order $a$ for some constant $a \in \R$.

Let $\mu_c\left({t}\right)$ be the Heaviside Step Function.

Let $\mathcal L\left\{{f\left({t}\right)}\right\} = F\left({s}\right)$ be the Laplace Transform of $f$.

Then:


 * $\displaystyle \mathcal L \left\{{\mu_c\left({t}\right) f\left({t-c}\right)}\right\} = e^{-sc}F\left({s}\right)$

everywhere that $\mathcal Lf$ exists, for $\operatorname{Re}\left({s}\right) > a$.

Proof
Let $u = t - c$.

Then $\dfrac {\mathrm du}{\mathrm dt} = 1$.

Then $u \to 0^-$ as $t \to c^+$.

Also, $u \to +\infty$ as $t \to +\infty$.

So:

What's the best way to justify tweaking the lower limit to be $u = 0$ instead of $u \to 0^-$? --GFauxPas (talk) 02:38, 23 May 2014 (UTC)