On questions about the integral of the logarithm of the Riemann zeta function II With the help of Wolfram Alpha online calculator I know good approximations of the real part of the integral $$\int_0^1 \log \left(\zeta(x)\right)dx,$$
where $\zeta(x)$ denotes the Riemann zeta function (evaluated over real numbers).
See it as int log(zeta(x))dx, from x=0 to 1 or 15 digits of int log(zeta(x))dx, from x=0 to 1

Question. I'm curious to know if this constant $$\Re\int_0^1 \log \left(\zeta(x)\right)dx$$ is in the literature. Can you provide me the reference? Then answer this question as a reference request adding what is the article showing/defining the value of this real constant, and I am going to search and read such statement. Many thanks.

My problem here is that I can not find the numeric value of this constant in 
The On-Line Encyclopedia of Integer Sequences or in Internet. 
 A: It is possible to evaluate the integral in term of series over non-trivial zeros of $\zeta\left(s\right)$. From the product $$\zeta(s)=\pi^{\frac{s}{2}}\frac{\prod_{\rho}\left(1-\frac{s}{\rho}\right)}{2(s-1)\Gamma\left(1+\frac{s}{2}\right)}$$ where $\rho$ runs over the non-trivial zeros of $\zeta\left(s\right),$ we get $$\int_{0}^{1}\log\left(-\zeta\left(s\right)\right)ds=\log\left(\pi\right)\int_{0}^{1}\frac{s}{2}ds-\log\left(2\right)-\int_{0}^{1}\log\left(1-s\right)ds$$ $$-\int_{0}^{1}\log\left(\Gamma\left(1+\frac{s}{2}\right)\right)ds+\int_{0}^{1}\sum_{\rho}\log\left(1-\frac{s}{\rho}\right)ds.$$Obviously $$\log\left(\pi\right)\int_{0}^{1}\frac{s}{2}ds-\log\left(2\right)-\int_{0}^{1}\log\left(1-s\right)ds=\frac{\log\left(\pi\right)}{4}-\log\left(2\right)-1$$ and $$\int_{0}^{1}\log\left(\Gamma\left(1+\frac{s}{2}\right)\right)ds=2\int_{0}^{1/2}\log\left(\Gamma\left(1+x\right)\right)dx$$ $$=3\log\left(A\right)-1-\frac{7}{12}\log\left(2\right)+\frac{\log\left(\pi\right)}{2}$$ where $A$ is the Glaisher-Kinkelin Constant. So, by the Taylor series for the logarithm we have $$\int_{0}^{1}\sum_{\rho}\log\left(1-\frac{s}{\rho}\right)ds=-\int_{0}^{1}\sum_{\rho}\sum_{k\geq1}\left(\frac{s}{\rho}\right)^{k}\frac{1}{k}ds$$ $$=-\sum_{\rho}\frac{1}{2\rho}-\int_{0}^{1}\sum_{\rho}\sum_{k\geq2}\left(\frac{s}{\rho}\right)^{k}\frac{1}{k}ds.$$ Note that $\sum_{\rho}1/\rho$ is convergent assuming that the pair of zeros $\rho$ and $1-\rho$ are combinated. Now since $$\sum_{\rho}\int_{0}^{1}\left|\sum_{k\geq2}\left(\frac{s}{\rho}\right)^{k}\frac{1}{k}\right|ds\leq\sum_{\rho}\frac{1}{\left|\rho\right|^{2}}\int_{0}^{1}-\log\left(1-s\right)ds<\infty$$ and $$\sum_{\rho}\sum_{k\geq2}\frac{1}{\left|\rho\right|^{k}k}\int_{0}^{1}s^{k}ds\leq\sum_{\rho}\frac{1}{\left|\rho\right|^{2}}\sum_{k\geq2}\frac{1}{k\left(k+1\right)}<\infty$$ we can exchange the integral with the double series and so $$\int_{0}^{1}\sum_{\rho}\log\left(1-\frac{s}{\rho}\right)ds=-\sum_{\rho}\frac{1}{2\rho}-\sum_{\rho}\sum_{k\geq2}\frac{1}{\rho^{k}k}\int_{0}^{1}s^{k}ds$$ $$=-\sum_{\rho}\frac{1}{2\rho}-\sum_{\rho}\sum_{k\geq2}\frac{1}{\rho^{k}k\left(k+1\right)}$$ $$=-\sum_{\rho}\frac{1}{2\rho}-\sum_{\rho}\left(\log\left(1-\frac{1}{\rho}\right)\left(\rho-1\right)+1-\frac{1}{2\rho}\right)$$ then, since $\sum_{\rho}\frac{1}{\rho}=\frac{\gamma}{2}+1-\frac{\log(4\pi)}{2},$ we get

$$\int_{0}^{1}\log\left(-\zeta\left(s\right)\right)ds=-\frac{5}{2}-\frac{\gamma}{4}-\frac{13\log\left(2\right)}{12}+3\log\left(A\right)+\log\left(\pi\right)$$ $$-\sum_{\rho}\left(\log\left(1-\frac{1}{\rho}\right)\left(\rho-1\right)+1-\frac{1}{2\rho}\right).$$

