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

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?

• you don't need the functional equation, and I don't think you know how to prove it. HINT (assuming you know that $\eta(0) = 1/2$) : $\eta(s) = \sum_{n= 1}^\infty (-1)^{n+1} n^{-s} = s \int_0^\infty S(x) x^{-s-1} dx$ with $S(x) = \sum_{n \le x} (-1)^{n+1}$, hence $\eta'(s) = \int_0^\infty S(x) x^{-s-1} dx-s\int_0^\infty \ln(x) S(x) x^{-s-1} dx$ Commented Apr 21, 2016 at 1:47
• sorry not sure my hint was a so good hint. if you'd like to use the functional equation, you have to look at the pole at $s=0$ of $\frac{\pi}{2}\tan(\pi s/2)$ and of $\frac{\zeta'(1-s)}{\zeta(1-s)}$, and estimate the residual constant term of their respective Laurent series Commented Apr 21, 2016 at 2:36
• For some reason the r.h.s also equals $\int_0^1\ln{\left(\frac{1}{\Gamma(x)}\right)}\,\mathrm{d}x$, an integral which is perhaps easier to evaluate. Commented May 13, 2017 at 6:13

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.
• My definition of $\eta(s)$ is slightly different from the usual one due to the $(-1)^n$ factor. Commented Apr 26, 2016 at 12:27
• The usual Dirichlet $\eta$-function is defined through $\sum_{n\geq 1}\frac{(-1)^{\color{red}{n-1}}}{n^s}$, sorry for the confusion. Commented Apr 26, 2016 at 12:28
• I believe that $\zeta(s)=(2^{1-s} - 1) \eta(s)$ should instead be $\eta(s)=(2^{1-s} - 1) \zeta(s)$ Commented Jan 3, 2022 at 9:47