Computing the value of $\zeta'(0)$ Using the relation $$\zeta(s)=2^s \pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s),$$ in addition to the following identities $$\lim_{s\to1}\left[\frac{d}{ds}\Gamma(s)\right]=-\gamma
            \>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\>\lim_{s\to1}\left[\frac{d}{ds}(s-1)\zeta(s)\right]=\gamma,$$ where $\gamma$ is the Euler constant, show that $$\zeta'(0)=-\frac{1}{2}\log(2\pi).$$
I have tried differentiating both sides, differentiating the log of both sides, but to no avail. I suspect that there may be some nice way to incorporate laurent series, but I am just not seeing it.
Cheers
 A: $$
\zeta(s)=2^s \pi^{s-1}\sin(\frac{\pi s}{2})\Gamma(1-s)\zeta(1-s)
$$
$$
\zeta'(s)=\left(2^s \pi^{s-1}\Gamma(1-s)\right)'\sin(\frac{\pi s}{2})\zeta(1-s)+2^s \pi^{s-1}\Gamma(1-s)\left(\sin(\frac{\pi s}{2})\zeta(1-s)\right)'
$$For ease of presentation, let $G(s)= 2^s \pi^{s-1}\Gamma(1-s)$. We have $G(0)=\pi^{-1}$ and
$$
G'(s) = 2^s \pi^{s-1}\Gamma(1-s) \ln(2)+2^s \pi^{s-1}\Gamma(1-s)\ln(\pi)-2^s \pi^{s-1}\Gamma'(1-s);
$$
$$
G'(0)=\frac{1}{\pi}\left(\ln(2)+\ln(\pi)+\gamma\right)
$$Since $\lim_{s\to 0}s\cdot \zeta(1-s)=-1,$ using $\lim_{\theta\to 0}\sin(k\theta)/\theta=k$ and a change of variable we have:
$$
\zeta'(0)=G'(0)\cdot \frac{-\pi}{2}+G(0)\lim_{s\to 0}\left(\sin(\frac{\pi s}{2})\zeta(1-s)\right)'
$$
$$
=G'(0)\cdot \frac{-\pi}{2}+\pi^{-1}\lim_{s\to 1}\left(\sin\left(\frac{\pi (1-s)}{2}\right)\zeta(s)\right)'
$$
$$
=\color{red}{-\frac{1}{2}\ln(2\pi)}-\frac{\gamma}{2}+\frac{1}{2}\lim_{s\to 1}\left((1-s)\zeta(s)\right)'
$$
$$
={-\frac{1}{2}\ln(2\pi)}-\frac{\gamma}{2}+\frac{\gamma}{2}={-\frac{1}{2}\ln(2\pi)}
$$
