0
$\begingroup$

I tried to find a derivation of the derivative of $\zeta(s)$ but I couldn't so I went ahead myself and computed it. The answer I obtain seems to be correct but I want to make sure that the computation is flawless. Can one of you confirm it for me? (or find the flaw(s))

\begin{align} \zeta(s)&=\sum_{n=1}^{\infty} \frac{1}{n^s}\nonumber\\ &=\sum_{n=1}^{\infty} \frac{1}{n^{x+iy}}\nonumber\\ &=\sum_{n=1}^{\infty} \frac{1}{n^xn^{iy}}\nonumber\\ \end{align} We can rewrite $n^{iy}$ as $e^{iy\text{log}(n)}$ and apply Euler's formula to obtain: \begin{align}\label{z4} \zeta(s)&=\sum_{n=1}^{\infty} \frac{1}{n^x\text{cos}(y\text{log}(n))+in^x\text{sin}(y\text{log}(n))} \end{align} \textbf{Corollary} Using definition of division in complex numbers, $$\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$$ Therefore, equation \ref{z4} becomes, \begin{align} \zeta(s)=\sum_{n=1}^{\infty} \frac{\text{cos}(y\text{log}(n))}{n^x}-i\sum_{n=1}^{\infty} \frac{\text{sin}(y\text{log}(n))}{n^x}. \end{align} Now, since we have the zeta function in the form $f=u+iv$, we can check if $f$ is holomorphic on our domain: \begin{align}\label{a9} \frac{\partial u}{\partial x}&=-\frac{\text{log}(n)\text{cos}(y\text{log}(n))}{n^x} \end{align} \begin{align}\label{a10} \frac{\partial u}{\partial y}&=-\frac{\text{log}(n)\text{sin}(y\text{log}(n))}{n^x} \end{align} \begin{align}\label{a11} \frac{\partial v}{\partial x}&=\frac{\text{log}(n)\text{sin}(y\text{log}(n))}{n^x} \end{align} \begin{align}\label{a12} \frac{\partial v}{\partial y}&=-\frac{\text{log}(n)\text{cos}(y\text{log}(n))}{n^x}, \end{align}

Clearly Cauchy-Riemann equations are satisfied, hence the function is differentiable on the entire domain, $Re(s)>1$. The actual process of differentiation of the series is much easier. Since zeta is just $$\zeta(s)=1+\frac{1}{2^s}+\frac{1}{3^s}+...$$ and $$\frac{d}{dx}\left(\frac{1}{n^x}\right)=-\frac{\text{log}(n)}{k^x},$$ differentiating term by term we obtain \begin{equation} \frac{d\zeta(s)}{ds}=-\sum_{n=2}^{\infty} \frac{\text{log}(n)}{n^s}\text{, }Re(s)>1. \end{equation} Additionally, we can provide a result for a general nth derivative, \begin{equation} \frac{d^k\zeta(s)}{ds}=\left(-1\right)^k\sum_{n=2}^{\infty} \frac{\text{log}^k(n)}{n^s}\text{, }Re(s)>1. \end{equation}

$\endgroup$
  • 3
    $\begingroup$ Why bother with $x$ and $y$? For each positive integer $n$, $1/n^s = \exp(-\log(n)s)$ is an analytic function of $s$ with derivative $-\log(n)/n^s$. In a domain where the series converges (uniformly on compact subsets), the derivative of the series can be taken term-by-term. $\endgroup$ – Robert Israel Nov 10 '17 at 0:08
  • $\begingroup$ For $\Re(s) > 1$ : $\frac{d}{ds} \zeta(s) = \frac{d}{ds} \sum_{n=1}^\infty n^{-s} = \sum_{n=1}^\infty \frac{d}{ds} n^{-s}= \sum_{n=1}^\infty (-\log n) n^{-s}$, see any course on complex analysis, holomorphic and analytic functions. $\endgroup$ – reuns Nov 10 '17 at 11:43
0
$\begingroup$

Your work seems fine but $\zeta(s)$ has other representations as well that you could differentiate.

$$\zeta(s)=\prod_{p=primes}\frac{1}{1-p^{-s}}, Re(s)>1$$ $$log(\zeta(s))=-\sum_{p}log(1-p^{-s})$$ Differentiating: $$\frac{\zeta'(s)}{\zeta(s)}=-\sum_{p}\frac{log(p)p^{-s}}{1-p^{-s}}$$

Another representation of the Zeta Function that was derived by Riemann:$$\zeta(s)=2^s\pi^{s-1}sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s), Re(s)<1$$ $$log(\zeta(s))=slog(2)+(s-1)log(\pi)+log(sin(\frac{\pi s}{2}))+log(\Gamma(1-s))+log(\zeta(1-s))$$ Differentiating: $$\frac{\zeta'(s)}{\zeta(s)}=log(2)+log(\pi)+\frac{\pi}{2}cot(\frac{\pi s}{2})-\frac{\Gamma'(1-s)}{\Gamma(1-s)}-\frac{\zeta'(1-s)}{\zeta(1-s)}$$ Replacing $s$ with $1-s$: $$\frac{\zeta'(1-s)}{\zeta(1-s)}=log(2)+log(\pi)+\frac{\pi}{2}tan(\frac{\pi s}{2})-\frac{\Gamma'(s)}{\Gamma(s)}-\frac{\zeta'(s)}{\zeta(s)}$$

Replacing $\frac{\zeta'(s)}{\zeta(s)}$ with the Euler Product derived earlier:$$\frac{\zeta'(1-s)}{\zeta(1-s)}=log(2)+log(\pi)+\frac{\pi}{2}tan(\frac{\pi s}{2})-\frac{\Gamma'(s)}{\Gamma(s)}+\sum_{p}\frac{log(p)p^{-s}}{1-p^{-s}}$$ This is valid for $Re(s)>0$ and we finally have a derivative for zeta valid for negative values of zeta.

EDIT: Another representation of Zeta in terms of Stieltjes Constants that might be easier to differentiate and is valid for $z\neq1$: $$\zeta(s)=\frac{1}{s-1}+ \sum_{n=0}^\infty \frac{(-1)^n}{n!}\gamma_{n}(z-1)^n$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.