It's been a while since the last fractional-calculus question, so here's my question for all of you.
It can be found from the Riemann-Liouville definition of the fractional derivative that whenever $\Re(s)>1$,
$$\begin{align}I_x^\alpha\zeta(x)&={\frac {1}{\Gamma (\alpha )}}\int _{a}^{x}\zeta(t)(x-t)^{\alpha -1}\ dt\\&=\frac1{\Gamma(\alpha)}\int_a^x\sum_{n=1}^\infty\frac{(x-t)^{\alpha-1}}{n^t}\ dt\\&=\frac1{\Gamma(\alpha)}\sum_{n=1}^\infty\int_a^x\frac{(x-t)^{\alpha-1}}{n^t}\ dt\\&=\sum_{n=1}^\infty\frac{(-\ln n)^{-\alpha}}{n^x}\end{align}$$
Thanks to WolframAlpha for that last step. I imagine the interchange between integral and sum can be made with rigor, but I can't see how at the moment. Setting this into it's derivative form, I end up with
$$D_x^\alpha\zeta(x)=\sum_{n=1}^\infty\frac{(-\ln n)^\alpha}{n^x}$$
which shall be my fractional derivative of the Riemann zeta function.
I then wish to take the following: (is differentiation with respect to the fractional derivative nonsensical?)
$$D_\alpha^\beta D_x^\alpha\zeta(x)=D_\alpha^\beta\sum_{n=1}^\infty\frac{(-\ln n)^\alpha}{n^x}$$
Again, using the Riemann-Liouville definition, I end up with
$$D_\alpha^\beta\sum_{n=1}^\infty\frac{(-\ln n)^\alpha}{n^x}=\sum_{n=1}^\infty\frac{(\ln(-\ln n))^\beta(-\ln n)^\alpha}{n^x}$$
Which is a little weird since the summand is undefined at $n=1$. What should I do about this?
Also, my end goal is to get some crazy derivative of derivatives of the zeta function into
$$\sum_{n=1?}^\infty\frac{\prod_{k=0}^p(\ \overbrace{\ln\ln\dots\ln\ln}^k\ n\ )^{a_k}}{n^x}$$
Any ideas?
Simple idea:
Rewriting the zeta function as
$$\zeta(x)=1+\sum_{n=2}^\infty\frac1{n^x}$$
now removes the problem, and with linearity, should give us
$$D_x^\alpha\zeta(x)=\frac1{\Gamma(1-\alpha)x^\alpha}+\sum_{n=2}^\infty\frac{(-\ln n)^\alpha}{n^x}$$
$$D_\alpha^\beta D_x^\alpha\zeta(x)=\left(D_\alpha^\beta\frac1{\Gamma(1-\alpha)x^\alpha}\right)+\sum_{n=2}^\infty\frac{(\ln(-\ln n))^\beta(-\ln n)^\alpha}{n^x}$$
Though this feels a tad bit unsafe.