On a derivative involving the Riemann zeta function

Let $n$ be a positive real number. Can the equality

$$\dfrac{d^{n}}{ds^{n}}\Big[s^{n-1}\ln\Big(\pi^{-s/2}\Gamma\Big(1+\frac{s}{2}\Big)\Big)\Big]\Bigg|_{s=1} = - \dfrac{d^{n}}{ds^{n}}\Big[s^{n-1}\ln\Big(\ln(s-1)\zeta(s)\Big)\Big]\Bigg|_{s=1}$$

be possible for any positive real $n$, where $\zeta(s)$ is the Riemann zeta function and $\Gamma(s)$ is the usual gamma function in number theory ?

My approach was by fractional calculus (to accommodate all reals), but I did not complete it since it appeared terribly malicious too me. I'm wondering if there can be some shorter and more intuitive way? Even a long complete proof by fractional calculus will still be very much appreciated.

• Did you use the functional equation of $\zeta$? Sep 29 '16 at 14:43
• Maybe an expansion of all the functions involved in Taylor series could help. Sep 29 '16 at 14:50
• @SylvainJULIEN, wouldn't that make things extra messy ?
– SPD
Sep 29 '16 at 14:51
• The "usual gamma function in number theory" is usually denoted by $\Gamma(s)$, no? Sep 29 '16 at 23:08
• There are different approaches to fractional differentiation; could you indicate in your post which one you have in mind? Sep 30 '16 at 4:42

Many basic complex-analyis misunderstandings here.

• First of all, $\frac{d^n}{ds^n}$ means $n \in \mathbb{N}$ (otherwise you have to define it). Then you probably wanted to write the Laurent series at $s=1$ of $\log \zeta(s)$. Unfortunately, $s=1$ is a branch point of $\log \zeta(s)$ not an isolated singularity, so it doesn't have a Laurent series there, and the coefficients you wrote on the RHS don't exist.

• Then I can guess you wanted instead to consider $\dfrac{d^{n}}{ds^{n}}\Big[s^{n-1}\log\Big((s-1)\zeta(s)\Big)\Big]\Bigg|_{s=1}$ since $\log\Big((s-1)\zeta(s)\Big)$ is analytic at $s=1$, but it also means $\log\Big((s-1)\zeta(s)\Big)= \sum_{k=0}^\infty c_k (s-1)^k$ (for $|s-1| < 3$) and hence $\dfrac{d^{n}}{ds^{n}}\Big[s^{n-1}\log\Big((s-1)\zeta(s)\Big)\Big]\Bigg|_{s=1} = \dfrac{d^{n}}{ds^{n}}\Big[\sum_{k=0}^\infty c_k(s-1)^{n+k-1}\Big]\Bigg|_{s=1}$ $=\sum_{k=1}^\infty c_k\frac{(n+k-1)!}{n!}(s-1)^{k-1}\Bigg|_{s=1} = c_1$

where $c_1 = \frac{d}{ds}\Big[\log\Big((s-1)\zeta(s)\Big)\Big]\Bigg|_{s=1} = \lim_{s \to 1} \frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1} = \gamma$ (the Euler-Mascheroni constant)

• In the same way $\log\Big(\pi^{-s/2}\Gamma\Big(1+\frac{s}{2}\Big)\Big) = \sum_{n=0}^\infty a_n (s-1)^n$ and $\dfrac{d^{n}}{ds^{n}}\Big[(s-1)^{n-1}\log\Big(\pi^{-s/2}\Gamma\Big(1+\frac{s}{2}\Big)\Big)\Big]\Bigg|_{s=1} = a_{1} = -\frac{\log(\pi)}{2}+ \frac{\Gamma'(3/2)}{2\Gamma(3/2)}$ $-\frac{\log(\pi)}{2}+1+\frac{\Gamma'(1/2)}{2\Gamma(1/2)}=-\frac{\log(\pi)}{2}+1- \frac{2\log(2)+\gamma}{2} \ne -\gamma$

• Finally, note that the coefficients $c_n$ of the Taylor series of $\log\Big((s-1)\zeta(s)\Big)$ around $s=1$ are given by $$c_n = \frac{1}{n!}\dfrac{d^{n}}{ds^{n}}\Big[\log\Big((s-1)\zeta(s)\Big)\Big]\Bigg|_{s=1}$$

• actually, the coefficients on the right hand side are known to exist, hence i'm not sure what you mean. Anyway, does your answer suggest that the equality is impossible ?
– SPD
Sep 30 '16 at 7:29
• And please check that i clearly stated that $n$ is a positive real number, hence not necessarily any integer.
– SPD
Sep 30 '16 at 7:35
• @SPD you need to define $d^n/dn$ for $n \not \in \mathbb{N}$, and there is almost no interesting reason for looking at the fractional derivative of functions related to $\zeta(s)$, secondly I told you your formula is unclear, and one as to guess what you meant. Now I see that a possiblity is that you meant $\log(\log((s-1)\zeta(s))) = F(s)$ that is holomorphic at $s=1$. Hence (for $n \in \mathbb{N}$) your RHS is $F'(1)$. can you compute it ? Sep 30 '16 at 7:45
• it also seems that you're considering only the case when $n$ is a positive integer, which doesn't completely address the question.
– SPD
Sep 30 '16 at 7:45
• ''There is almost no interesting for looking at the fractional derivative involving $\zeta(s)$''. To me this sounds like your own perception which you shouldn't necessarily assume to hold for everyone.
– SPD
Sep 30 '16 at 7:49