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)

As a Limit
Consider the real function:


 * $y = x^x$.

This function is [[Definition:Power (Algebra)/Real Number/Definition 1|well defined for $x > 0$.

It is not obvious whether or not the right hand limit:


 * $\displaystyle \lim_{x \mathop \to 0^+} y$

exists.

If it does, it would be nice if:


 * $\displaystyle \lim_{x \mathop \to 0^+} x^x = 0^0$.

To that end,


 * Note $x^x = \exp(x \log x) \to e^0 = 1$. &mdash; Lord_Farin (talk) 15:24, 9 December 2016 (EST)

Theorem
Let $f: \R \to \R$ be a real function that has a limit at infinity.

Then $f$ is of exponential order $0$.

Proof
Define the constant mapping:


 * $\displaystyle C \left({t}\right) = - \lim_{t \mathop \to +\infty} f \left({t}\right)$

Further define:


 * $g \left({t}\right) = f \left({t}\right) + C \left({t}\right)$

From:


 * Constant Function is of Exponential Order Zero,


 * Sum of Functions of Exponential Order,

it is sufficient to prove that $g$ is of exponential order $0$.

To that end,

Theorem

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

Proof
Recall that for $n \in \N_{\ge 0}$:


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

for $\operatorname{Re}\left({s}\right) > 0$.

With Gamma Function Extends Factorial in mind, let our Ansatz be:


 * $\displaystyle \mathcal L \left\{ {t^a} \right\} \stackrel{?}{=} \dfrac{\Gamma\left({a+1}\right)}{s^{a+1}}$

for $\operatorname{Re}\left({s}\right) > 0$.

The conditions on $a$ for which this guess might be correct will be determined later.

Denote $s = \operatorname{Re}\left({s}\right) + i \operatorname{Im}\left({s}\right) = \sigma + i \omega$.

By the definition of Laplace Transform:

Denote:
 * $\displaystyle I_1 = \int_\epsilon^R t^a e^{-st} \, \mathrm d t$.

By the integral form of the gamma function:

Denote:
 * $\displaystyle I_3 = \int_{\epsilon}^R t^z e^{-t} \, \mathrm d t$.

By Complex Riemann Integral is Contour Integral, $I_1$ is a contour integral.

We perform the following substitution:

I'm not quite happy with how I'm using the template here, and am open to suggestions. I will actually need to do this sort of thing for a bunch of proofs with Laplace transforms. --GFauxPas (talk) 20:52, 29 November 2016 (EST)


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

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}$

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

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

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