Fractional Integration

[JmsNxn]

I've recently posted a question on mathoverflow which some of you who know fractional calculus may be able to answer: https://mathoverflow.net/questions/44033...-sum-fn-nx
I would appreciate any thoughts on the matter

Thanks, 
Caleb

Hey, Caleb!

What you have essentially detailed is the Exponential Differintegral--or the Riemann Liouville Differintegral. The standard way to right this, is to use the Mellin transform, but there are many different possible expansions for it.

In essence, the way I like to write it is as:

\[
\frac{d^{-s}}{dx^{-s}} f(x) = \frac{1}{\Gamma(s)} \int_0^\infty f(x-y)y^{s-1}\,dy\\
\]

You will notice instantly that:

\[
\frac{d^{-s}}{dx^{-s}} e^x = e^x\\
\]

Now, this is not the entire definition, as the full definition would be written using arcs in \(\mathbb{C}\). In which we write--assuming \(f\) is integrable on \(\gamma\) where \(\gamma(0) = 0\) and \(\gamma(\infty) = \infty\) on the Riemann sphere:

\[
\frac{d^{-s}}{dx^{-s}} f(x) = \frac{1}{\Gamma(s)} \int_\gamma f(x-y)y^{s-1}\,dy\\
\]

So for example, if we were to take \(f(x) = e^{-x}\), we want to integrate across \([-\infty,0]\), upon which we are given the formula:

\[
\frac{d^{-s}}{dx^{-s}} e^{-x} = \frac{e^{-\pi i s}}{\Gamma(s)} \int_0^\infty f(x+y)y^{s-1}\,dy = e^{-\pi i s} e^{-x}\\
\]

Which, is perfectly generalizable too:

\[
\frac{d^{-s}}{dx^{-s}} e^{\lambda x} = \lambda^{-s} e^{\lambda x}\\
\]

And this is a rigorous operator on a specific space of entire functions \(f\), those which have some sort of decay at \(\infty\).

---

The relation between this operators has been around for centuries; where we can rewrite Riemann's famous expression for the zeta function as:

\[
\frac{d^{-s}}{dx^{-s}}\Big{|}_{x=0} \frac{e^{-x}}{1+e^{-x}} = \frac{d^{-s}}{dx^{-s}}\Big{|}_{x=0} \sum_{n=1}^\infty e^{-nx} = \zeta(s)\\
\]

As to your question, I am a tad confused, but I believe this does exist, though you would prove it differently. I believe you are absolutely correct--but if memory serves me right, this is a rewording of a known result.

But just for fun, let's prove your result

Okay, so take the function:

\[
\frac{d^{s}}{dx^{s}}\Big{|}_{x=0} f(x)= F(s)\\
\]

And let's take your function:

\[
G(s) = f^{(s)}(0)\\
\]

Which we state that \(H(s) = F(s) - G(s)\), where \(H(n) = 0\) for all \(n\ge 0\). The function \(F(s)\) is exponentially bounded, in such a manner that: \(|F(s)| = O(e^{\rho \Re(s) + \tau |\Im(s)|})\). The value \(\tau \in (0,\pi/2)\) and \(\rho \in \mathbb{R}^+\). Your function \(G(s)\) also exists in this space--I'm a little too lazy to prove it right now, but trust me it is. Therefore \(H(s)\) is also in this space.

The thing is.... If \(H(s) = O(e^{\rho \Re(s) + \tau |\Im(s)|})\) and is holomorphic for \(\Re(s) > 0\), then if \(H(n) = 0\) then \(H=0\). This is what I call the Ramanujan Identity Theorem, as it's a direct corollary of Ramanujan's master theorem.

I can add more details if you like  

In short, from my brief analysis you are absolutely correct, and this is an expression for the Exponential Differintegral!

Great job!

EDIT: Also, please note this is just a rough walk through. You do need more elbow grease to iron everything out. Why I didn't answer the MO question, MO asks for a higher standard, and I'm too lazy to work out all the details right now!