AHHH!!!
So I tracked down the exact theorem.
Let's assume that \(A = \mathbb{C}^{\infty \times \infty}\) is an infinite square matrix. Which acts on an infinite vector \(\bf{x} \in \mathbb{C}^\infty\). If all the eigenvalues \(\lambda\) of \(A\) satisfy \(\Re(\lambda) > 0\)--then:
\[
\begin{align}
\left(\Gamma(1-z) A^{z-1}\right) {\bf x} &= \left(\int_0^\infty e^{-At}t^{-z}\,dt\right){\bf x}\\
\Gamma(1-z) A^{z-1}&=\int_0^\infty e^{-At}t^{-z}\,dt\\
\end{align}
\]
You can perform this on sectors as well, and arbitrary sectors in any complex direction. So you can make the exact same formula, upto a change of variables, so long as \(\theta < \arg(\lambda) < \theta'\).
This is equivalent to Von Neumann's iteration of operators on hilbert spaces, but is just written differently. Von Neumann focuses on the Fourier transform, and I spent most of my time using Mellin--but I did prove novel results about Von Neumann's iteration; but it was mostly just Ramanujan + Von Neumann added together to write straightforward obvious results, lol. I mostly just justified some results Ramanujan just fucking knew somehow, but I did it for linear operators acting on hilbert spaces.
Lmao, and if I'm bragging, I hope you know it's okay to brag a bit, Daniel
Also! You can use an analytic continuation formula by Euler to put it alllllllll together:
\[
\Gamma(1-z) A^{z-1} = \sum_{n=0}^\infty A^n \frac{(-1)^n}{(z-1-n)n!} + \int_1^\infty e^{-At}t^{-z}\,dt\\
\]
This constructs a holomorphic iteration \(A^z\) for \(\Re(z) > -1\). This is valid if we assume the spectrum \(\sigma(A) \subset \Re(z) > 0\).
Essentially if the spectrum \(\sigma(A)\) is in a sector of \(\mathbb{C}\), meaning \(\lambda \in \sigma(A)\) and \(\theta < arg(\lambda) < \theta'\) while additionally \(\theta' - \theta \le \pi\). Then we can fractionally iterate the linear operator \(A\) on the hilbert space it acts on. This is just an overly complicaed version of Ramanujan's Master Theorem, lol.
So I tracked down the exact theorem.
Let's assume that \(A = \mathbb{C}^{\infty \times \infty}\) is an infinite square matrix. Which acts on an infinite vector \(\bf{x} \in \mathbb{C}^\infty\). If all the eigenvalues \(\lambda\) of \(A\) satisfy \(\Re(\lambda) > 0\)--then:
\[
\begin{align}
\left(\Gamma(1-z) A^{z-1}\right) {\bf x} &= \left(\int_0^\infty e^{-At}t^{-z}\,dt\right){\bf x}\\
\Gamma(1-z) A^{z-1}&=\int_0^\infty e^{-At}t^{-z}\,dt\\
\end{align}
\]
You can perform this on sectors as well, and arbitrary sectors in any complex direction. So you can make the exact same formula, upto a change of variables, so long as \(\theta < \arg(\lambda) < \theta'\).
This is equivalent to Von Neumann's iteration of operators on hilbert spaces, but is just written differently. Von Neumann focuses on the Fourier transform, and I spent most of my time using Mellin--but I did prove novel results about Von Neumann's iteration; but it was mostly just Ramanujan + Von Neumann added together to write straightforward obvious results, lol. I mostly just justified some results Ramanujan just fucking knew somehow, but I did it for linear operators acting on hilbert spaces.
Lmao, and if I'm bragging, I hope you know it's okay to brag a bit, Daniel
Also! You can use an analytic continuation formula by Euler to put it alllllllll together:
\[
\Gamma(1-z) A^{z-1} = \sum_{n=0}^\infty A^n \frac{(-1)^n}{(z-1-n)n!} + \int_1^\infty e^{-At}t^{-z}\,dt\\
\]
This constructs a holomorphic iteration \(A^z\) for \(\Re(z) > -1\). This is valid if we assume the spectrum \(\sigma(A) \subset \Re(z) > 0\).
Essentially if the spectrum \(\sigma(A)\) is in a sector of \(\mathbb{C}\), meaning \(\lambda \in \sigma(A)\) and \(\theta < arg(\lambda) < \theta'\) while additionally \(\theta' - \theta \le \pi\). Then we can fractionally iterate the linear operator \(A\) on the hilbert space it acts on. This is just an overly complicaed version of Ramanujan's Master Theorem, lol.