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)

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