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,


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

where:


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

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:


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

From the poles of the gamma function, $\Gamma\left({-1}\right)$ is undefined.

This suggests that having $q > -1$ would be a good idea for $q$ wholly real.

Reasons for insisting $\operatorname{Re} q > -1$ for complex $q$ will become apparent during the course of the proof.

To the end of expressing $\displaystyle I\left({\varepsilon,L}\right)$ 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 $\sigma > 0$.

By the lemma, the integrand for $I\left({\varepsilon,L}\right)$ is analytic.

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

Substitute:

Denote $I_C = \displaystyle \int_C u^q e^{-u} \rd u$

Consider the contour:


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

where:


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


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


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


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

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:
 * LaplaceTransformGamma.png

By the Cauchy-Goursat theorem:


 * $I_C = I_{C_1} + I_{C_2} - I_{C_3} - I_{C_4} = 0$

Consider $I_{C_2}$ as $L \to +\infty$:

Now consider $I_{C_4}$ for $\varepsilon \to 0^+$:

Thus $I_{C_4} \to 0$ as $\epsilon \to 0^+$ and $I_{C_2} \to 0$ as $L \to +\infty$.

Thus, taking limits on $I_C = 0$,


 * $I_C = I_{C_1} - I_{C_3} = 0$

... that is, $I_{C_1} = I_{C_3}$ in the limit:

Thus:

... and the Ansatz is proved correct.

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