Does a simpler form exist for $\frac{\Re \left(\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}{\Im \left(\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}$?

Question

I am aware that:

$$\frac{\Re \zeta(\frac12+it)}{\Im\zeta(\frac12+it)}=\cot \left( \frac12\,t\ln \left( \pi \right)+\frac{i}{2}\ln \left( {\frac {\Gamma \left( \frac14+\frac{it}{2}\right) }{\Gamma \left( \frac14-\frac{it}{2} \right)}}\right)\right)$$

Does a similar expression exist for:

$$\dfrac{\Re \left(\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}{\Im \left(\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}$$

and/or:

$$\dfrac{\Re \left((-\frac12 + it)\,\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}{\Im \left((-\frac12 + it)\,\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}$$

?

Answer 1

The functional equation $$\zeta(s)=\chi(s)\zeta(1-s)$$ gives $\log \zeta(s)- \log \zeta(1-s) = \log \chi(s)$ ie. $$arg\ \zeta(1/2+it)) = \frac12 \Im(\log \zeta(1/2+it)- \log \zeta(1/2-it)) = \frac12 arg\ \chi(1/2+it)$$

$$arg\ \zeta(1/2+it))\Gamma(1/2+it)= \frac12 arg\ \chi(1/2+it)+arg \ \Gamma(1/2+it)$$

Answer 2

With great help from the accepted answer, I decided to post the solutions I derived as a separate answer.

For $t \in \mathbb{R}$, define:

$$\hat{\chi}(t)= \left( 2\,\pi \right) ^{-\frac12+it}\sin\big(\pi \left( \frac14+\frac{it}{2} \right)\big)\, \Gamma\left( \frac12+it \right)$$

and then we get:

$$\dfrac{\Re \left(\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}{\Im \left(\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}={\frac {i}{ \hat{\chi}(t) -\frac12}}+i$$

and also:

$$\dfrac{\Re \left((-\frac12 + it)\,\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}{\Im \left((-\frac12 + it)\,\zeta(\frac12+it)\,\Gamma(\frac12+it)\right)}={\frac {i}{\frac{i+2t}{i-2t} \hat{\chi}(t) -\frac12}}+i$$

It appears that, contrary to the simplified expression for $\frac{\Re \zeta(\frac12+it)}{\Im\zeta(\frac12+it)}$, the $\cot$- and $\log$-functions can be fully removed for both these cases.

ADDED: Just to falsify my remark in the last sentence, I now found that:

$$\frac{\Re \zeta(\frac12+it)}{\Im\zeta(\frac12+it)}=i\cdot\frac{\left( 2\,\pi \right) ^{\frac12+it}+2\cos\big(\pi \left( \frac14+\frac{it}{2} \right)\big)\, \Gamma\left( \frac12+it \right)}{\left( 2\,\pi \right) ^{\frac12+it}-2\cos\big(\pi \left( \frac14+\frac{it}{2} \right)\big)\, \Gamma\left( \frac12+it \right)}$$