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


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.

  • $\begingroup$ 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. $\endgroup$
    – Clayton
    Dec 10, 2016 at 2:01
  • $\begingroup$ @Clayton Hm, good point. And I changed the $I$ to a $D$, integral to derivative, hence change in sign for $\alpha$. $\endgroup$ Dec 10, 2016 at 2:02
  • $\begingroup$ @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. $\endgroup$ Dec 10, 2016 at 2:03
  • $\begingroup$ Yes, what are you doing? $\endgroup$ Dec 11, 2016 at 19:13
  • $\begingroup$ @SkeletonBow Trying to get a different form for$$\sum_{n=1?}^\infty\frac{\prod_{k=0}^p(\ \overbrace{\ln\ln\dots\ln\ln}^k\ n\ )^{a_k}}{n^x}$$ $\endgroup$ Dec 11, 2016 at 22:19

1 Answer 1


Sorry I don't know how to use MathJax.

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.


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