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 $f:\R \to \R$ or $\R \to \C$ be a Definition:Continuous function, differentiable on any interval of the form $0 \le t \le A$.

Let $f$ be of exponential order $a$.

Let $f'$ be piecewise continuous on said intervals.

Then $\mathcal L \left\{{f}\right\}$ exists for $\operatorname{Re}\left({s}\right) > a$, and:


 * $\mathcal L \left\{{f'\left({t}\right)}\right\} = s \mathcal L \left\{{f\left({t}\right)}\right\} - f\left({0}\right)$.

Proof
Consider:


 * $\int_0^A e^{-st}f'\left({t}\right) \, \mathrm dt$

By hypothesis, $f'$ is piecewise continuous.

So by Piecewise Continuous Function is Riemann Integrable, this integral exists.

This means that integration by parts can be invoked:


 * $\displaystyle \int hj\,' \, \mathrm dt = hj - \int h'j \, \mathrm dt$

Here:

So:

Now, take the limit as $A \to +\infty$:

Recall that $f$ is of exponential order $A$:

Recall $\operatorname{Re}\left({s}\right) > a$.

We invoke Modulus of Exponential is Modulus of Real Part and Complex Exponential Tends to Zero

How to explain this? hmmmmmm