I am interested in the integral over the log-zeta function$$I=\int_0^1 \ln\left(\zeta(s)\right)ds$$
My attempts:
Sub $u=1-s$ to get $$I=\int_0^1 \ln\left(\zeta(1-u)\right)du=\int_0^1 \ln(\zeta(u))-u\ln2-(u-1)\ln\pi-\ln\left(\sin(\frac{\pi u}{2})\right)-\ln\left(\Gamma(1-u)\right) du$$ This doesn't seem to lead anywhere unfortunately. I also tried to set up a parameter $a$ as
$$I(a)=\int_0^1 \ln\left(\zeta(s+a)\right)ds$$ $$I'(a)=\int_0^1 \frac{\zeta'(s+a)}{\zeta(s+a)}ds=\ln\left(\zeta(s+a)\right)|_{s=0}^{s=1}=\ln\left(\zeta(a+1)\right)-\ln\left(\zeta(a)\right)$$ From this point on I am not sure of what to do as integrating with respect to $a$ to recover $I(a)$ does not simplify the integral. If anybody has any idea of how to approach finding a closed form of this integral then please let me know. Thank you in advance.