How to differentiate the Riemann-Siegel $Z$-function?

Question

Is it well-known/or is there literature for differentiating the Riemann-Siegel $Z$-function?

$$Z(t)=e^{i \theta(t)}\zeta(1/2 +it)$$

I have tried differentiating this according to the chain rule but in the process, whenever I end up differentiating $[\zeta(1/2+it)]’$ using its formula

$$\zeta(1/2it)=e^{-i\theta(t)}Z(t)$$

the terms all end up cancelling with each other so I end up with the trivial result

$$Z’(t)=Z’(t)$$

Answer 1

Wolfram's answer linked by Tyma Gaidash is otained with the logarithmic derivative of $Z(t)$ :

\begin{align} \log Z(t)&=i \theta(t) +\log \zeta(1/2 +it)\\ \tag{1}\frac{Z'(t)}{Z(t)}&=i \theta'(t) +\frac{\zeta'(1/2 +it)}{\zeta(1/2 +it)}\\ \end{align}

where the derivative of the theta function is well known : \begin{align} \theta(t) &= \arg \left( \Gamma\left(\frac{1}{4}+i\frac t2\right) \right) - \frac{\log \pi}{2} t\\ \tag{2}\theta'(t) &=\frac 14\left(\psi\left(\frac 14+i\frac t2\right)+\psi\left(\frac 14-i\frac t2\right)-2\log \pi\right)\\ \end{align} (since the digamma function $\psi$ is the logarithmic derivative of $\Gamma$)

Combining $(1)$ and $(2)$ gives : $$Z'(t)=Z(t)\left(\frac i4\left(\psi\left(\frac 14+i\frac t2\right)+\psi\left(\frac 14-i\frac t2\right)-2\log \pi\right) +\frac{\zeta'(1/2 +it)}{\zeta(1/2 +it)}\right)$$

Answer 2

Let’s actually finish @Raymond Manzoni’s answer, Their result was: $$Z'(t)=Z(t)\left(\frac i4\left(\psi\left(\frac 14+i\frac t2\right)+\psi\left(\frac 14-i\frac t2\right)-2\log \pi\right) +\frac{\zeta'(1/2 +it)}{\zeta(1/2 +it)}\right)$$ Here is a solution using the series definition of the Zeta function for $\text {Re}\left(it+\frac12\right)>1\implies \text{Im}(t)<-\frac12$: $$Z'(t)=Z(t)\left(\frac i4\left(\psi\left(\frac 14+i\frac t2\right)+\psi\left(\frac 14-i\frac t2\right)-2\log \pi\right) +\frac{\frac d{dt}\sum\limits_{n=1}^\infty n^{-it-\frac12}}{\zeta(1/2 +it)}\right)= Z(t)\left(\frac i4\left(\psi\left(\frac 14+i\frac t2\right)+\psi\left(\frac \pi4-i\frac t2\right)-2\log \pi\right) -\frac{\sum\limits_{n=1}^\infty \frac{\ln(n)}{n^{it+\frac12}}}{\zeta\left(\frac12+it\right)}\right) $$ For $\text{Im}(t) <-\frac12 $, use the analytic continuation: $$ζ\left(it+\frac12\right)=2^{it+\frac12} \pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac\pi4\right) Γ\left(\frac12-it\right)ζ \left(\frac12-it\right)= 2^{it+\frac12} \pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac\pi4\right) Γ\left(\frac12-it\right) \sum\limits_{n=1}^\infty n^{it-\frac12} $$ Therefore: $$ζ’\left(it+\frac12\right)=2^{it+\frac12} \pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac14\right) Γ\left(\frac12-it\right)ζ \left(\frac12-it\right)= 2^{it+\frac12} \pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac\pi4\right) Γ\left(\frac12-it\right) \sum\limits_{n=1}^\infty n^{it-\frac12}=\frac d{dt}\left( 2^{it+\frac12} \right) \pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac14\right) Γ\left(\frac12-it\right) \sum\limits_{n=1}^\infty n^{it-\frac12} + 2^{it+\frac12} \frac d{dt}\left(\pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac14\right) Γ\left(\frac12-it\right) \sum\limits_{n=1}^\infty n^{it-\frac12}\right)= \frac d{dt}\left( 2^{it+\frac12} \right) \pi^{it-\frac12} \sin\left(\frac{i t \pi}2+\frac14\right) Γ\left(\frac12-it\right) \sum\limits_{n=1}^\infty n^{it-\frac12} + 2^{it+\frac12} \frac d{dt}\left(\pi^{it-\frac12}\right) \sin\left(\frac{i t \pi}2+\frac14\right) Γ\left(\frac12-it\right) \sum\limits_{n=1}^\infty n^{it-\frac12}+ \pi^{it-\frac12} \frac d{dt}\left(\sin\left(\frac{i t \pi}2+\frac14\right) Γ\left(\frac12-it\right)ζ \left(\frac12-it\right)\right)=…$$

Here is our final answer as the derivative is complicated using @Raymond Manzoni’s identities for $\text{Im}(t) <-\frac12 $ Note there may a typo. It used a bit of machine help:

$$Z'(t)=Z(t)\left(\frac i4\left(\psi\left(\frac 14+i\frac t2\right)+\psi\left(\frac 14-i\frac t2\right)-2\log \pi\right) -\frac{\left[-i2^{\frac12 + i t} π^{-\frac12 + i t} Γ\big(\frac12 - i t\big) \sin\big(\frac\pi 4 -\frac {\pi i t}2\big) \sum_{n=1}^\infty \ln(n) n^{it-\frac12}+ i2^{-\frac12 + i t} π^{\frac12+ i t} \cos(\frac\pi 4+ \frac {\pi i t}2) Γ\big(\frac12 - i t\big) ζ\big(\frac12- i t\big) + i2^{-\frac12 + i t} π^{-\frac12 + i t} Γ \big(\frac12 - i t\big) \ln(π) \sin (\frac\pi 4+ \frac {\pi i t}2) ζ \big(\frac12 - i t\big) + i2^{\frac12 + i t} π^{-\frac12 + i t} Γ \big(\frac12 - i t\big) \ln(2) \sin (\frac\pi 4+ \frac {\pi i t}2) ζ \big(\frac12 - i t\big) - i2^{\frac12 + i t} π^{-\frac12 + i t} Γ \big(\frac12- i t\big) ψ \big(\frac12- i t\big) \sin \big(\frac\pi 4+ \frac {\pi i t}2\big) ζ \big(\frac12 - i t\big)\right]}{\zeta\left(\frac12+it\right)}\right)$$

Here is the sum representation for the analytic continuation and main sum definition of the zeta function in terms of the derivative of the Riemann Siegel Z function. Please correct me and give me feedback!