Computing the second derivative of Riemann zeta function at $0$

Question

WolframAlpha gives $$\zeta ''(0)=\frac{\gamma ^2}{2} - \frac{π^2}{24} - \frac{\log^2(2 π)}{2} + \gamma_1$$

My attempt

Consider the functional equation $$\zeta(z)=2^z\pi^{z-1}\sin\left(\frac{\pi z}{2}\right)\Gamma(1-z)\zeta(1-z)$$ Taking log and differentiating both sides gives $$\zeta'(z)=\zeta(z)\left(\log(2\pi)+\frac{\pi}{2}\cot\left(\frac{\pi z}{2}\right)-\psi(1-z)-\frac{\zeta'(1-z)}{\zeta(1-z)}\right)$$ Again $$\frac{\zeta''(z)}{\zeta'(z)}=\frac{\zeta'(z)}{\zeta(z)}+\frac{-\frac{\pi^2}{4}\csc^2\frac{\pi z}{2}+\psi_1(1-z)+\frac{\zeta(1-z)\zeta''(1-z)-\zeta'(1-z)^2}{\zeta^2(1-z)}}{\log(2\pi)+\frac{\pi}{2}\cot\frac{\pi z}{2}-\psi(1-z)-\frac{\zeta'(1-z)}{\zeta(1-z)}}$$

Where $\psi_1$ is the trigamma function

Take $z\to 0$ $$\frac{\zeta''(0)}{\zeta'(0)}=\frac{\zeta'(0)}{\zeta(0)}+L$$ Where $$L=\lim_{z\to 0}\frac{-\frac{\pi^2}{4}\csc^2\frac{\pi z}{2}+\psi_1(1-z)+\frac{\zeta(1-z)\zeta''(1-z)-\zeta'(1-z)^2}{\zeta^2(1-z)}}{\log(2\pi)+\frac{\pi}{2}\cot\frac{\pi z}{2}-\psi(1-z)-\frac{\zeta'(1-z)}{\zeta(1-z)}}$$

I multiplied top and bottom by $z^3\zeta(1-z)^2$

$$\lim_{z\to 0}\frac{{\overbrace{z^3\zeta(1-z)^2\psi_1(1-z)}^{=\epsilon_1}}-z^3\zeta(1-z)^2\frac{\pi^2}{4}\csc^2\frac{\pi z}{2}+z^3\zeta(1-z)\zeta''(1-z)-z^3\zeta'(1-z)^2}{{\underbrace{z^3\zeta(1-z)^2\log(2\pi)+z^3\zeta(1-z)^2\frac{\pi}{2}\cot\frac{\pi z}{2}-z^3\zeta(1-z)^2\psi(1-z)}_{=\epsilon_2}}-z^3\zeta(1-z)\zeta'(1-z)}$$ We have $\epsilon_1$ and $\epsilon_2$ tend to $0$ I used the Laurent series of $\zeta(1-s)$ $$\zeta(1-s)=-\frac{1}{s}+\gamma+\gamma_1s+\frac{\gamma_2}{2}s^2+O(s^3)$$ $$-\zeta'(1-s)=\frac{1}{s^2}+\gamma_1+\gamma_2 s+O(s^2)$$ $$\zeta''(1-s)=-\frac{2}{s^3}+\gamma_2 +O(s)$$ $$\zeta(1-z)^2=\frac{1}{z^2}-\frac{2\gamma}{z}+\gamma^2-2\gamma_1+O(z)$$ $$z^3\zeta'(1-z)^2=\frac{1}{z}+2\gamma_1z+O(z^2)$$ $$z^3\zeta(1-z)\zeta''(1-z)=\frac{2}{z}+\gamma+O(z)$$ $$z^3\zeta(1-z)\zeta'(1-z)=1-\gamma z+O(z^3)$$ $$\csc^2\frac{\pi z}{2}=\frac{4}{\pi^2 z^2}+\frac{1}{3}+\frac{\pi^2 z^2}{60}+O(z^4)$$ $$z^3\zeta(1-z)^2\frac{\pi^2}{4}\csc^2\frac{\pi z}{2}=\frac{1}{z}-2\gamma+O(z)$$

putting these in $$\lim_{z\to 0}\frac{{\epsilon_1}-(\frac{1}{z}-2\gamma+O(z))+(\frac{2}{z}-2\gamma+O(z))-(\frac{1}{z}+2\gamma_1z+O(z^2))}{{\epsilon_2}-(1-\gamma z+O(z^3))}=0$$

But $\zeta''(0)$ is not equal to $\zeta'(0)^2/\zeta(0)$, whats wrong with my computations?

I suspect that I missed something in the expansion of the last step, but I can't seem to find it.

Any help is appreciated

Answer 1

$\newcommand{\d}{\mathrm{d}}$This much is correct:

$$\frac{\zeta''(z)}{\zeta'(z)}=\frac{\zeta'(z)}{\zeta(z)}+\frac{-\frac{\pi^2}{4}\csc^2\frac{\pi z}{2}+\psi_1(1-z)+\frac{\zeta(1-z)\zeta''(1-z)-\zeta'(1-z)^2}{\zeta^2(1-z)}}{\log(2\pi)+\frac{\pi}{2}\cot\frac{\pi z}{2}-\psi(1-z)-\frac{\zeta'(1-z)}{\zeta(1-z)}}$$

But it is not true that $\epsilon_2$ goes to zero:

$$\epsilon_2:=z^3\zeta(1-z)^2\log(2\pi)+z^3\zeta(1-z)^2\frac{\pi}{2}\cot\frac{\pi z}{2}-z^3\zeta(1-z)^2\psi(1-z)$$

All the other terms vanish but the middle term is $z^3$ multiplied by a function with an order $3$ pole at zero. $$z^3\zeta(1-z)^2\cdot\frac{\pi}{2}\cot\frac{\pi z}{2}=z^3(z^{-2}+O(z^{-1}))(z^{-1}+O(z))\to1$$

Assuming the rest of your work is correct we'd have: $$\lim_{z\to 0}\frac{{\epsilon_1}-(\frac{1}{z}-2\gamma+O(z))+(\frac{2}{z}-2\gamma+O(z))-(\frac{1}{z}+2\gamma_1z+O(z^2))}{{\epsilon_2}-(1-\gamma z+O(z^3))}\\=\lim_{z\to0}\frac{-2\gamma_1z+O(z)}{\gamma z+O(z)}$$Which is inconclusive. So you need to take further expansions. We have: $$\epsilon_2=1+(\log(2\pi)-3\gamma)z+O(z^2),\,\epsilon_1=\frac{\pi^2}{6}z+O(z^2)$$

And you can handle the rest.