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 $q$ be a constant complex number with $\operatorname{Re}(q) > -1$.

Let:


 * $t^q: \left({0\,.\,.\,\to}\right) \to \C$

be a branch of the complex power multifunction chosen such that $f$ is continuous on the half-plane $\operatorname{Re}(s) > 0$.

Then $f$ has a Laplace transform given by:


 * $\mathcal L\left\{{t^q}\right\}(s) = \dfrac{\Gamma\left({q+1}\right)}{s^{q+1}}$

where $\Gamma$ denotes the gamma function.

Proof
By the definition of Laplace transform for a function not continuous at zero,


 * $\displaystyle \mathcal L\left\{{t^q}\right\}(s) = \lim_{\varepsilon \to 0^+}\lim_{L \to +\infty}I\left({\varepsilon,L}\right)$

where:


 * $\displaystyle I\left({\varepsilon,L}\right) = \int_{\varepsilon}^{L} t^q e^{-s t} \rd t$.

The integrand of $I\left({\varepsilon,L}\right)$ is analytic.

Recall the case where $q$ is a positive integer:


 * $\displaystyle \mathcal L \left\{ {t^n} \right\} = \frac {n!} { s^{n+1} }$

Because the gamma function extends the factorial, a reasonable Ansatz is:


 * $\displaystyle \mathcal L \left\{ {t^q} \right\} = \frac {\Gamma\left({q+1}\right)} { s^{1+1} }$

To the end of expressing $\displaystyle I\left({\epsilon,L}\right)$ to a form similar to the integral defining $\Gamma$, we view $I$ as a contour integral.

Write $s = \sigma + i \omega$ for $\sigma, \omega \in \mathbb R$ and $\sigma > 0$.

The integrand for $I\left({\epsilon,L}\right)$ is analytic.

Thus the conditions for integration by substitution are satisfied for $\operatorname{Re}(s) > 0)$.

Substitute:

Definition:Ansatz

Denote:


 * $I_C = \displaystyle \oint_C \left({\frac u s}\right)^q e^{u} \rd \frac{u}{s}$

where:


 * $C = C_1 + C_2 - C_3 - C_4$

where:


 * $C_1$ is a line segment connecting $\sigma \varepsilon$ to $L\omega$


 * $C_2$ is a line segment connecting $L\omega$ to $L\omega + i L \omega$


 * $C_3$ is a line segment connecting $L\omega + i L \omega$ to $i\varepsilon$


 * $C_4$ is a circular arc connecting $i\varepsilon$ to $\varepsilon$

Note that $C_1$ and $C_2$ are positively oriented while $C_3$ and $C_4$ are negatively oriented.

The contours are illustrated in the following graph:

Picture here

To do

 * $\digamma$

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