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starting from reflection formula and digamma function

I obtain \begin{align} \frac{\zeta'(s)}{\zeta(s)} + \frac{\zeta'(1-s)}{\zeta(1-s)} = \log(2\pi) + \frac{\pi}{2} \cot\left(\frac{\pi s}{2}\right) - \psi(1-s) \end{align} taking limit goes 1, reference says that \begin{align} \frac{\zeta'(0)}{\zeta(0)} = \log(2\pi) \end{align}

It seems that $\zeta'(1)=0$ and $\psi(0)=0$. I wonder how to prove this

Edit

due to @Mindlack I got $\frac{\zeta'(1)}{\zeta(1)}=0$, but still got problem with $\psi(0)=0$....

From Mathematica I see that $\psi(0) = -\infty$....

The reference I follow is Nicolas M Robles master thesis on zeta function regularization page 28

phy_math
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    You are aware that $\zeta(z)=\frac{1}{z-1}+O(1)$ as $z \rightarrow 1$? – Aphelli Jul 31 '19 at 14:31
  • @Mindlack, I see what you mean. zeta has a simple pole at $z=1$ so $\zeta'(1)=0$. I get it. – phy_math Jul 31 '19 at 14:32
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    No, since there is a pole $\zeta’(1)$ is undefined! However $\frac{\zeta’}{\zeta}$ has a limit in $1$ which is $0$. – Aphelli Jul 31 '19 at 14:35
  • @Mindlack, can you show me some details? I am confused that how you obtain $\frac{\zeta'}{\zeta} =0$ at $z=1$. – phy_math Jul 31 '19 at 14:38
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    $\zeta(z) = \frac{1}{z-1} f(z)$ where $f$ is analytic and $f(1)=1$ thus $f'/f$ is analytic and $\frac{\zeta'(z)}{\zeta(z)} = -\frac{1}{z-1}+\frac{f'(z)}{f(z)}=-\frac{1}{z-1}+O(1)$ – reuns Jul 31 '19 at 21:33

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A delicate calculation that requires justification \begin{align} \frac{\mathrm{d} \zeta}{\mathrm{d}z} &= \frac{\mathrm{d} }{\mathrm{d}z}\left( \sum_{n=1}^{\infty}n^{-z}\right) \\ &= \sum_{n=1}^{\infty}\left(- \ln(n) n^{-z} \right) \\ &= -\sum_{n=1}^{\infty}\left(\frac{\ln(n)}{n^{z}} \right) \\ &= -\sum_{n=2}^{\infty}\left(\frac{\ln(n)}{n^{z}} \right) \end{align} As $\ln 1 = 0$. Your result then follows. it is delicate but there is also a need to show that the sum converges uniformly and also that the sum of the derivatives converge uniformly.