Calculation of derivative values of Riemann zeta function

Question

How the values of $\zeta'(0)=-1/2 \log⁡(2\pi)$ is calculated?

Answer 1

One elegant derivation of $\zeta'(0)=-\ln(2\pi)/2$ is as follows:

Consider the following representation for the Dirichlet eta function: $$\eta(s)=\frac{1}{2}+\frac{1}{2}\left[\left(\frac{1}{1^s}-\frac{1}{2^s}\right)-\left(\frac{1}{2^s}-\frac{1}{3^s}\right)+\left(\frac{1}{3^s}-\frac{1}{4^s}\right)-...\right],$$ which holds for $\text{Re}(s)>-1$.

Differentiating termwise (which, if we were more rigorous, would have to be motivated by showing uniform convergence) gives: \begin{align*} \eta'(s)&=\frac{1}{2}\left[\left(-\frac{\ln(1)}{1^s}+\frac{\ln(2)}{2^s}\right)-\left(-\frac{\ln(2)}{2^s}+\frac{\ln(3)}{3^s}\right)+\left(-\frac{\ln(3)}{3^s}+\frac{\ln(4)}{4^s}\right)-...\right]\\ &=\frac{1}{2}\left[\left(-\frac{\ln(1)}{1^s}+\frac{\ln(2)}{2^s}\right)+\left(\frac{\ln(2)}{2^s}-\frac{\ln(3)}{3^s}\right)+\left(-\frac{\ln(3)}{3^s}+\frac{\ln(4)}{4^s}\right)+...\right]\\ &=\frac{1}{2}\left[-\ln(1^{1^{-s}})+\ln(2^{2^{-s}})+\ln(2^{2^{-s}})-\ln(3^{3^{-s}})-\ln(3^{3^{-s}})+\ln(4^{4^{-s}})+...\right]\\ &=\frac{1}{2}\ln\left(\frac{2^{2^{-s}}}{1^{1^{-s}}}\cdot\frac{2^{2^{-s}}}{3^{3^{-s}}}\cdot\frac{4^{4^{-s}}}{5^{5^{-s}}}\cdot\frac{4^{4^{-s}}}{7^{7^{-s}}}\cdot...\right) \end{align*} Now, letting $s=0$, we can make use of the Wallis product to obtain $$\eta'(0)=\frac{1}{2}\ln\left(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{4}{7}\cdot...\right)=\frac{1}{2}\ln\left(\frac{\pi}{2}\right).$$ Now we differentiate the relation $$\eta(s)=(1-2^{1-s})\zeta(s),$$ which gives us $$\eta'(s)=2^{1-s}\ln(2)\zeta(s)+(1-2^{1-s})\zeta'(s).$$ Solving for $\zeta'(s)$ leads to $$\zeta'(s)=\frac{\eta'(s)-2^{1-s}\ln(2)\zeta(s)}{1-2^{1-s}}.$$ Setting $s=0$ and recalling that $\zeta(0)=-\frac{1}{2}$ finally leaves us with \begin{align*} \zeta'(0)&=\frac{\frac{1}{2}\ln(\frac{\pi}{2})-2\ln(2)(-\frac{1}{2})}{1-2}\\ &=-\left(\frac{\ln(\frac{\pi}{2})}{2}+\ln(2)\right)\\ &=-\left(\frac{\ln(\frac{\pi}{2})+\ln(4)}{2}\right)\\ &=-\frac{1}{2}\ln(2\pi). \end{align*}

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Here is a derivation of the above representation for $\eta(s)$ for those interested.

Recall the regular series for $\eta(s)$: $$\eta(s)=\frac{1}{1^s}-\frac{1}{2^s}+\frac{1}{3^s}-\frac{1}{4^s}+...$$ convergent for $\text{Re}(s)>0$. We can split each term as follows: \begin{align*} \eta(s)&=\left(\frac{1}{2\cdot 1^s}+\frac{1}{2\cdot1^s}\right)-\left(\frac{1}{2\cdot2^s}+\frac{1}{2\cdot2^s}\right)+\left(\frac{1}{2\cdot3^s}+\frac{1}{2\cdot3^s}\right)-...\\ &=\frac{1}{2}\left[\left(\frac{1}{1^s}+\frac{1}{1^s}\right)-\left(\frac{1}{2^s}+\frac{1}{2^s}\right)+\left(\frac{1}{3^s}+\frac{1}{3^s}\right)-...\right]. \end{align*} Regrouping consecutive terms (which does not require absolute convergence, only convergence) then gives: \begin{align*} \eta(s)&=\frac{1}{2}\left[\frac{1}{1^s}+\left(\frac{1}{1^s}-\frac{1}{2^s}\right)+\left(-\frac{1}{2^s}+\frac{1}{3^s}\right)+\left(\frac{1}{3^s}-\frac{1}{4^s}\right)+...\right]\\ &=\frac{1}{2}+\frac{1}{2}\left[\left(\frac{1}{1^s}-\frac{1}{2^s}\right)-\left(\frac{1}{2^s}-\frac{1}{3^s}\right)+\left(\frac{1}{3^s}-\frac{1}{4^s}\right)+...\right]. \end{align*} This shows that the representation at least holds for $\text{Re}(s)>0$ (since that is the convergence of the original series), but it in fact holds for all $\text{Re}(s)>-1$ aswell. One might wonder why it is the case that the values of the infinite series in the interval $-1<\text{Re}(s)\leq0$ (where the derived series converges but not the original) indeed equal $\eta(s)$. However, if we recall that there is a unique analytic continuation to the Dirichlet eta function (as with the Riemann zeta function), and observe that our series is indeed analytic, we can conclude that the series must equal $\eta(s)$ for all $\text{Re}(s)>-1$.