# 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.

• In one place you have an exponent of $-\alpha$ while another has a positive exponent...which is it? If the former, there are troubles at $n=1$ even before you bring up the additional trouble. Dec 10, 2016 at 2:01
• @Clayton Hm, good point. And I changed the $I$ to a $D$, integral to derivative, hence change in sign for $\alpha$. Dec 10, 2016 at 2:02
• @Clayton Perhaps I should shift the sum and factor out the first term. Then treat it like a polynomial? But it still doesn't let me escape most of my problems. Dec 10, 2016 at 2:03
• Yes, what are you doing? Dec 11, 2016 at 19:13
• @SkeletonBow Trying to get a different form for$$\sum_{n=1?}^\infty\frac{\prod_{k=0}^p(\ \overbrace{\ln\ln\dots\ln\ln}^k\ n\ )^{a_k}}{n^x}$$ Dec 11, 2016 at 22:19

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 $$1$$.