On the fractional derivatives of the Riemann zeta function and the derivatives of the derivatives 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.
 A: When the summand is $1$ the numerator in the sum is $(\text{log}(-\text{log}(1)))^b \cdot (-\text{log}(1))^a=(-\infty)^b \cdot 0^a$. When $b>0$, $a>0$ then the numerator equals $(-1)^b \cdot \infty \cdot 0$, and ignoring the $(-1)^b$ term gives $\infty \cdot 0=0/(1/\infty)=0/0$. Also, when $b<0$, $a<0$ then the numerator equals $(-1)^b \cdot 0 \cdot \infty$, and ignoring the $(-1)^b$ term gives $0 \cdot \infty=\infty/1/0=\infty/\infty$.
In both these cases l'hopsital's rule can be used, which gives $0$ when $b>0$, $a>0$ and diverges to plus or minus infinity when $b<0$, $a<0$, however, when b approaches minus infinity and $a<0$ the limit is $0$ also. When $b>0$, $a<0$ the answer is plus or minus infinity, and when $b<0$, $a>0$ the answer is $0$. Furthermore, when $b=0$, $a=0$ the answer is $1$.
I don't know what the answers are for complex values or whether there are more singularities when introducing complex numbers. I'll leave those two questions for you or someone else to answer.
Hopefully that resolved the paradox of the first term in your series.
