From the previous post "Clock Face And Life, Two Pies And Habits" dated 30 Jul 2015,
\(\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta }) }=-\int^{\infty}_{-\infty}{it.g(\omega t)e^{-i\omega t}}d\,\omega=-it.G(t)\)
\(\theta=\omega t\) and \(z=R_\theta e^{i\theta}\)
\(R_\theta\) being a function of \(\theta\).
\(\int { f(z)d(e^{ -i\theta }) }\)
takes a free ride with Fourier. Must it still be holomorphic?
\(f(z)=g(\omega t)\)
as long as the Fourier tansform of \(g(\omega t)\) exist, \(\int ^{\infty}_{-\infty}{ f(z)d(e^{- i\theta }) }\) is valid. In the case of simple periodicity, \(T\) a constant,
\(\int ^{\infty}_{-\infty}{ f(z)d(e^{ -i\theta }) }=\int ^{T}_{0}{ f(z)d(e^{ -i\theta }) }=-it.G(t)|^{T}_0=-i.TG(T)\)
where \(G(t)\) is the Fourier transform of \(f(z)\) with \(z\) replaced by \(\omega\). \(-i\) is in the direction of \(\theta=0\).
Most liberating.
Note: \(-i.tG(t)\) is a complex function (\(\theta=\omega t\) being real), with a dual \(f(z)\), running around in circles in the complex plane.
\(e^{i\theta}\) being replaced with \(e^{-i\theta}\).
