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

 * $\displaystyle \mathcal L \left\{{\sin at}\right\} = \frac a {s^2 + a^2}$

Proof
Without using Euler's formula.


 * $\displaystyle \mathcal L \left\{{\sin at}\right\} = \int_0^{\to +\infty}e^{-st}\sin at \, \mathrm dt$

This proof holds for real sine, where $\operatorname{Re}\left({s}\right) > a$.

From Integration by Parts:


 * $\displaystyle \int fg' \, \mathrm dt = fg - \int f'g \, \mathrm dt$

Here:

So:

Call the above equation $(A)$.

Consider:


 * $\displaystyle \int e^{-st} \cos at \, \mathrm dt$

Again, using Integration by Parts:


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

Here:

So:

Substituting this into $(A)$:

Evaluating at $t = 0$ and $t \to +\infty$:

Theorem
Let $a$ be a constant real number.

Let $\mathcal L$ be the Laplace Transform.

Then:


 * $\displaystyle \mathcal L \left\{{a}\right\} = \frac a s$

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

Proof
Perhaps better to consider a constant mapping than a constant? --GFauxPas (talk) 15:19, 9 May 2014 (UTC)


 * Or even make the page about $\mathcal L \left\{ {1}\right\}$ and make $\mathcal L \left\{ {a}\right\}$ a corollary... --GFauxPas (talk) 15:26, 9 May 2014 (UTC)