Show that $\zeta'(0)=-\frac{1}{2}\ln(2\pi)$

Question

I started with the functional equation which was derived in class, $$ \zeta(s)=2^s\pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s) $$ and took the logarithmic derivative of both sides to get $$ \frac{\zeta'(s)}{\zeta(s)}=\ln2+\ln\pi+\frac{\pi}{2}\frac{\cos(\frac{\pi s}{2})}{\sin(\frac{\pi s}{2})}+\psi(1-s)-\frac{\zeta'(1-s)}{\zeta(1-s)}. $$ Now I see that if I evaluate at $1$ or $0$ I will have an equation for $\zeta'(0)$ but I cannot solve either of the resulting equations. Can someone show me how to do this?

Answer 1

We may use the $\eta $ function as suggested by user1952009. Given $s\in(0,1)$, $$ \eta(s) = \sum_{n\geq 1}\frac{(-1)^n}{n^s} $$ fulfills $\lim_{s\to 0^+}\eta(s)=-\frac{1}{2}$ by Abel's lemma. On the other hand, $$ \eta'(s) = \sum_{n\geq 1}\frac{(-1)^n \log n}{n^s}=\sum_{n\geq 1}\frac{(-1)^n}{n^s}\int_{0}^{+\infty}\frac{e^{-u}-e^{-nu}}{u}\,du $$ by differentiation under the $\sum$ sign and Frullani's theorem. So we have: $$ \lim_{s\to 0^+}\eta'(s) = \int_{0}^{+\infty}\frac{-\frac{1}{2}e^{-u}+\frac{1}{1+e^u}}{u}\,du=\frac{1}{2}\log\frac{\pi}{2}$$ by Wallis' product and Frullani's theorem again. On the other hand, $$ \zeta(s) = (2^{1-s}-1)\,\eta(s) $$ so $\zeta(0)=-\frac{1}{2}$ and by switching to logarithmic derivatives we get: $$ \frac{\zeta'(s)}{\zeta(s)} = \frac{\eta'(s)}{\eta(s)}-\frac{2^{1-s}\log(2)}{2^{1-s}-1}$$ from which $\zeta'(0)=-\log\sqrt{2\pi}$ easily follows.