6

So I was messing around with Bernoulli numbers and values of $\zeta'$ at integers $-$ and suddenly I came about a non trivial identity which can be written in terms of the logarithmic derivative of the zeta function, $z(x):=\frac{\zeta'(x)}{\zeta(x)}$.
This is the kind of identities I would have typically expected to find, maybe in some variation, on the Wolfram functions site, but I didn't. It states:

$$z(n)-z(n-2)+z(1-n)-z(3-n)+\frac1{n-1}+\frac1{n-2}=0.$$

The intriguing part are of course the two fractions.

This identity seems to hold not only for positive even $n$, but for random complex numbers as well, except obviously the points where one of the zeta values is $0$.
There are several ways to write it as a reflection formula. The most simple and beautiful one, as for me, is as follows (replacing $n$ by $x+1$): if we define a function $$f(x):=\frac{\zeta'(x+1)}{\zeta(x+1)}-\frac{\zeta'(x-1)}{\zeta(x-1)}+\frac1{x},$$ then the identity simply becomes $$f(x)=f(1-x),$$ as also illustrated by the graph. enter image description here I just cannot believe this result should be new...

Something about motivation: I noticed the occurring of expressions of the type $\color{red}{z(n)-z(n-2)}$ (always with even $n\in\mathbb N$) in closed forms of integrals like this one: $$\int_{0}^{\infty}\frac{1}{e^x-1}\left[\frac{12\,e^x}{(e^x-1)^2}-\frac{12}{x^2}+1\right]\,dx=\frac{5}{2}+\color{red}{\frac{\zeta'(2)}{\zeta(2)}-\frac{\zeta'(0)}{\zeta(0)}}$$ and this one: $$ \int\limits_0^\infty\dfrac{\tanh^3(x)}{x^2}dx=\frac{5}{6} - \frac{4}{15} \log 2 + \color{red}{\frac{\zeta'(4)}{\zeta(4)} -\frac{\zeta'(2)}{\zeta(2)}},$$ which further had lead me to $$\int\limits_0^\infty\dfrac{\tanh^5(x)}{x^2}dx= \frac{8}{15}-\frac{44}{189}\log2 + \color{red}{\frac{\zeta'(4)}{\zeta(4)} -\frac{\zeta'(2)}{\zeta(2)}} -\frac23\Bigl(\color{red}{\frac{\zeta'(6)}{\zeta(6)}-\frac{\zeta'(4)}{\zeta(4)}} \Bigr) $$ etc.

Wolfgang
  • 13,193
  • 3
    Your identity follows from the functional equation $\frac{\zeta'(s)}{\zeta (s)}+\frac{\zeta '(1-s)}{\zeta (1-s)}=\log (\pi)-\frac{1}{2}\psi^{(0)}\left(\frac{s}{2}\right)-\frac{1}{2} \psi^{(0)}\left(\frac{1-s}{2}\right)$. What is your question? – Steven Clark Nov 28 '21 at 20:20
  • @StevenClark Good, I expected something easy like that — and of course it is mainly about getting rid of the polygamma functions and other more "involved" constants. Between the lines (and part of my motivation) is in fact the question whether there are similar integrals with a closed form that "separates" $\frac{\zeta'}{\zeta}(s)$ from $\frac{\zeta'}{\zeta}(s-2)$ — a question BTW that was also asked on math.se in relation with the first integral, but never answered. Otherwise stated, for which $s$ can $\frac{\zeta'}{\zeta}(s)$ alone occur as an exponential period? – Wolfgang Nov 29 '21 at 09:14

0 Answers0