From the previous post "Clock Face And Life, Two Pies And Habits" dated 30 Jul 2015,
\(\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta }) }=-\int^{\infty}_{-\infty}{it.g(\omega t)e^{-i\omega t}}d\,\omega=-it.G(t)\)
if instead,
\(\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta }) }=-\int^{\infty}_{-\infty}{i\omega.g(\omega t)e^{-i\omega t}}d\,t=-i\omega.G(\omega)\)
which is exactly Fourier. However,
since, \(\theta=\omega t\), \(\omega=2\pi.\cfrac{1}{T}\)
\(t=\cfrac{T}{2\pi}\theta\)
and
\(\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta }) }=-i\cfrac{T}{2\pi}\theta.G(\cfrac{T}{2\pi}\theta)\)
If I go fishing,
\(\int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(e^{ -i\theta }+z_o) }=-i\cfrac{T}{2\pi}\theta.G(\cfrac{T}{2\pi}\theta)\) --- (*)
where \(f(z)=\cfrac{h(z)}{(z-z_o)}\), \(z_o\) a complex constant.
from the post "What Holomorphic?" dated 30 Jul 2015. When we consider \(-\infty\) to \(\infty\) as the biggest close loop possible. From Cauchy's integral formula,
\(h(z_o)=\cfrac{1}{2\pi i}\oint_c{\cfrac{h(z)}{(z-z_o)}}dz\)
Unfortunately on the complex plane, \(z_o\) cannot just disappear; is (*) still valid? If so,
\(\oint_c{\cfrac{h(z)}{(z-z_o)}}dz=\int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(e^{ -i\theta }+z_o) }\)
\(2\pi ih(z_o)=i\cfrac{T}{2\pi}\theta_{z_o}. g\left(\cfrac{T }{2\pi}\theta_{z_o} \right)\)
\(h(z_o)=\cfrac{T}{(2\pi)^2}\theta_{z_o}. g\left(\cfrac{T }{2\pi}\theta_{z_o} \right)\)
where \(z_o=R_{\theta_{zo}}e^{i\theta_{zo}}\)
\(h(z_o)\) depends on \(\theta_{z_o}\) only. Which is very odd, I loop around a pole with \(e^{-i\theta}\) but it does not matter how long my reach is.
This must be wrong. Previously, the dependence on \(R_\theta\) disappeared because I am always on a unit circle. If we change \(e^{-i\theta}\to R_{zo}e^{-i\theta}\) such that we loop around \(z_o\),
\(\int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(R_{zo}e^{- i\theta }) }=R_{zo}.i\cfrac{T}{2\pi}\theta. g\left(\cfrac{T }{2\pi}\theta \right)\)
then,
\(h(z_o)=R_{zo}.\cfrac{T}{(2\pi)^2}\theta_{z_o}. g\left(\cfrac{T }{2\pi}\theta_{z_o} \right)\)
provided \(z_o\) is the only pole within the circle \(R_{zo}e^{-i\theta}\). Still,
\(\oint_c{\cfrac{h(z)}{(z-z_o)}}dz\ne \int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(R_{zo}e^{- i\theta }) }\)
And this is all wrong. Unless \(\oint_c{...}dz\) allows us to go around in circle specifically.
\(\oint_c{\cfrac{h(z)}{(z-z_o)}}dz=\oint_{cR}{\cfrac{h(z)}{(z-z_o)}}d(R_{zo}e^{- i\theta })=R_{zo} \int ^{\infty}_{-\infty}{\cfrac{h(z)}{(z-z_o)}d(e^{ -i\theta }) }\)
which is unfortunately valid for any \(R\gt R_{zo}\). Which leads us to strict \(R_{zo}\lt R\lt R_{zo}-\Delta R\).
...
All these suggest that with,
\(\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta }) }=F(\omega)\)
where \(\theta=\omega t\),
we have just extracted the argument part of the function \(f(z)\).
\(\cfrac{d F(w)}{d(e^{-i\theta})}=f(z)\)
put back the argument part of the function.
