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)

Sandbox
Needed: Thm, Absolutely convergent complex integral converges.

Vretblad's Chapter 3:


 * Definition of Laplace Transform
 * to do: define convention that heaviside without subscript is $\mu(t) = \mu_0(t)$. Also denoted as $H(t)$.


 * Definition of Exponential order


 * $e^{iat} \in \mathcal{E}, a \in \R$
 * $e^{iat} \in \mathcal{E}, a \in \R$
 * $e^{zt} \in \mathcal{E}, z \in \C$


 * Theorem $3.1$: if $f \in \mathcal{E}_a$, then $\mathcal{L}\left\{{f}\right\}$ exists for $\operatorname{Re}\left({s}\right) > a$, and the integral converges absolutely.


 * Definition: The rate of growth of a function of exponential order is $\sigma_0 = \inf \left\{{a: f \in \mathcal{E}_a}\right\}$


 * Theorem: $f \in \mathcal{E}_{a} \implies f \in \mathcal{E}_{b}$ for $b > a$.


 * Theorems: The following functions are all of exponential order (and thus have Laplace transforms):
 * polynomials, sine, cosine, exponentials.


 * Theorem: sums and products of functions of exponential order are themselves of exponential order.

Theorem
Let $f \left({t}\right): \R \to \R$ or $\R \to \C$ be a function.

Let $f$ be continuous on the real interval $\left [{0 \,.\,.\, \to} \right)$, except possibly for some finite number of jump discontinuities in every finite subinterval of $\left [{0 \,.\,.\, \to} \right)$.

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

Then the Laplace transform $\mathcal L\left \{{f}\right\}(s)$ exists for $\operatorname{Re}\left({s}\right) > a$.

Proof
By the definition of exponential order, $f$ is piecewise continuous on any interval of the form $\left[{0 \,.\,.\, T}\right], T > 0$

By Piecewise Continuous Function is Riemann Integrable, the following integral exists:

$$\int_0^T f\left({t}\right)e^{-st} \, \mathrm dt$$

Recall $f$ is of exponential order $a$.

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

By the definition of exponential order, there exists some constant $M > 0$ such that:


 * $\forall t \ge M: \left\vert {f \left({t}\right)} \right \vert < K e^{a t}$

Thus:

Eventually
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

User:GFauxPas/Sandbox/NumberTheory