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
If anyone would like to answer my question about Laplace transforms, that would be nice:

http://math.stackexchange.com/questions/2021954/laplace-transform-of-power


 * Someone answered it! If I were to put it on PW, how much explicit justification is needed for the indentation of the contour around the branch point to not contribute to the integral? --GFauxPas (talk)


 * Enough to make it rigorous. :-) The boilerplate should of course be in place in Complex Analysis -- if not, then it is recommended that this is the case. Most of the reason for the fact that a lot of the Complex Analysis work has not been filled in is purely the technical detail has not been completed. (For some reason the Signature button is not working for me.) -- Prime.mover

Theorem
If $\phi = 0$, then $t^\phi$ is a constant function, and  is of exponential zero.

Let $\phi \ne 0$.

I lost my $t$! :( Silly mistake, I'll fix it later. --GFauxPas (talk) 11:59, 25 November 2016 (EST)

Theorem
Ansatz:


 * $\displaystyle \int_{L\sigma}^{L\sigma + iL\omega} z^a e^{-z} \, \mathrm dz \to 0 \text{ as } L \to +\infty$

Then we seek an integral of the form $I\left({L}\right)$ such that:


 * $0 \le \displaystyle \left \vert {\int_{L\sigma}^{L\sigma + iL\omega} z^a e^{-z} \, \mathrm dz  }\right \vert \le \left \vert {I\left({L}\right) }\right \vert$

Such that $\displaystyle \lim_{L \to +\infty} \left \vert { I\left({L}\right) }\right \vert = 0$

And use the squeeze theorem.

Parametrization

$\phi_2 \left({t}\right)= x\left({t}\right) + i y\left({t}\right)$

$ \phi_2 \left({t}\right) \, : \, \begin{cases} x\left({t}\right) = L\sigma \\ y\left({t}\right) = \omega t \\ t \in \left[{0 \,. \,. \, L}\right] \\ \end{cases}$

$\phi'\left({t}\right) \, : \, \begin{cases} x'\left({t}\right) = 0 \\ y'\left({t}\right) = \omega \end{cases}$

$\left \vert \phi'\left({t}\right) \right \vert = \sqrt{\omega^2} = \omega$

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