I started with the basic question:
why does:
\( C e^s = \int_{-\infty}^{\infty} \frac{s^t}{\Gamma(t+1)}\partial t \)
Differentiate and we can see it. Disregarding convergence.
The fundamental theorem I proved is the following:
\( \mathcal{E} f = \frac{1}{\Gamma(-s)}\int_0^{\infty} f(-t)t^{-s-1} \partial t \)
\( \mathcal{L}f = \int_{-\infty}^{\infty} f(t) \frac{s^t}{\Gamma(t+1)}\, \partial t \)
\( \mathcal{E} \mathcal{L} = \mathcal{L}\mathcal{E} = 1 \)
We also note that:
\( \mathcal{E}f = \frac{\partial^s f(t)}{\partial t^s} |_{t=0} \)
if we think of complex iterations of the derivative are in agreement with the Riemann-Liouville differintegral where \( e^s \) is the fixpoint.
This is the thoerem that I spent about two weeks trying to prove. It's not actually that hard but I'd prefer keeping it to myself until I write a full paper with all the things I've found using this linear operator \( \mathcal{E} \)
By example I was able to deduce:
\( e^s = \int_{-\infty}^{\infty} \frac{s^t}{\Gamma(t+1)}\,\partial t \)
\( \sin(s) = \int_{-\infty}^{\infty}\sin(\frac{\pi}{2}t) \frac{s^t}{\Gamma(t+1)}\,\partial t \)
\( \cos(s) = \int_{-\infty}^{\infty}\cos(\frac{\pi}{2}t) \frac{s^t}{\Gamma(t+1)}\,\partial t \)
There are many more results that I found. I'm trying to compile them and make logical sense of all the connections. Lots of very interesting things happen. I'm calling these Taylor integral representations. Hopefully I'm not the only one who sees the parallel to Taylor series.
I analyzed \( \mathcal{E} f(g(s)) \) and doing some basic integral rearrangements and substituting \( g \) for \( e^s \) and \( \ln(s) \) I got the result at the top.
Also \( \bf{E} \neq \mathcal{E} \) in case you thought that...
why does:
\( C e^s = \int_{-\infty}^{\infty} \frac{s^t}{\Gamma(t+1)}\partial t \)
Differentiate and we can see it. Disregarding convergence.
The fundamental theorem I proved is the following:
\( \mathcal{E} f = \frac{1}{\Gamma(-s)}\int_0^{\infty} f(-t)t^{-s-1} \partial t \)
\( \mathcal{L}f = \int_{-\infty}^{\infty} f(t) \frac{s^t}{\Gamma(t+1)}\, \partial t \)
\( \mathcal{E} \mathcal{L} = \mathcal{L}\mathcal{E} = 1 \)
We also note that:
\( \mathcal{E}f = \frac{\partial^s f(t)}{\partial t^s} |_{t=0} \)
if we think of complex iterations of the derivative are in agreement with the Riemann-Liouville differintegral where \( e^s \) is the fixpoint.
This is the thoerem that I spent about two weeks trying to prove. It's not actually that hard but I'd prefer keeping it to myself until I write a full paper with all the things I've found using this linear operator \( \mathcal{E} \)
By example I was able to deduce:
\( e^s = \int_{-\infty}^{\infty} \frac{s^t}{\Gamma(t+1)}\,\partial t \)
\( \sin(s) = \int_{-\infty}^{\infty}\sin(\frac{\pi}{2}t) \frac{s^t}{\Gamma(t+1)}\,\partial t \)
\( \cos(s) = \int_{-\infty}^{\infty}\cos(\frac{\pi}{2}t) \frac{s^t}{\Gamma(t+1)}\,\partial t \)
There are many more results that I found. I'm trying to compile them and make logical sense of all the connections. Lots of very interesting things happen. I'm calling these Taylor integral representations. Hopefully I'm not the only one who sees the parallel to Taylor series.
I analyzed \( \mathcal{E} f(g(s)) \) and doing some basic integral rearrangements and substituting \( g \) for \( e^s \) and \( \ln(s) \) I got the result at the top.
Also \( \bf{E} \neq \mathcal{E} \) in case you thought that...