On a derivative involving the Riemann zeta function

Question

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.

Answer 1

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 $$c_n = \frac{1}{n!}\dfrac{d^{n}}{ds^{n}}\Big[\log\Big((s-1)\zeta(s)\Big)\Big]\Bigg|_{s=1}$$