Consider the following integral:
$$I(t,a)=\int_{-\infty}^{+\infty} \frac{\log|\zeta(it)|}{1+at²}dt$$
where $\zeta$ denotes the Riemann zeta function.
We can see this integral similar to that of Balazard Saias Yor integral
How to get value of this type of integral?
What significance this has if we could calculate its exact value in the context of prime number theory ?
For $a>0$, using the functional equation $\zeta(s)=f(s)\zeta(1-s)$ you get two integrals $$\int_{-\infty}^\infty \frac{\log |\zeta(it)|}{1+at^2}dt=\int_{-\infty}^\infty \frac{\log |f(it)|}{1+at^2}dt+\int_{-\infty}^\infty \frac{\log |\zeta(1-it)|}{1+at^2}dt$$
The second one is $$\int_{-\infty}^\infty \frac{\log |\zeta(1-it)|}{1+at^2}dt=\int_{-\infty}^\infty \frac{\log \zeta(1-it)}{1+at^2}dt=2i\pi \frac{\log \zeta(1+a^{-1/2})}{2 i a^{1/2}}$$ using the residue theorem, which is valid since $\frac{\log \zeta(1-it)}{1+at^2}$ is integrable on the real axis, it is analytic and it decreases fast enough on the upper half-plane away from a simple pole at $i a^{-1/2}$ of residue $\frac{\log \zeta(1+a^{-1/2})}{2 i a^{1/2}}$.
I don't know how to evaluate $$\int_{-\infty}^\infty \frac{\log |f(it)|}{1+at^2}dt$$
From $\pi/\sin(\pi s)=\Gamma(s)\Gamma(1-s)$ we get that $\pi/|\sin(\pi it)| = |\Gamma(it) \Gamma(1-it)|=|\Gamma(1-it) |^2/ |t|$
so that $|\Gamma(1-it)| = \frac{\pi^{1/2}}{|\sin(\pi it)|^{1/2}} |t|^{1/2}$ and $\log |f(it)| = \log (\pi^{-1/2}/2) +\log |\sin(\pi it/2)|-\frac12\log |\sin(\pi it)|+\frac12 \log|t|$. Two terms can be integrated with the residue theorem, it remains to find $\int_{-\infty}^\infty \log \left|\frac{\sin^2(\pi it/2)}{\sin(\pi it)}\right| \frac{dt}{1+at^2}$. Integration by parts and residue theorem? It should give a residue at $ia^{-1/2}$ plus something like $\sum_{n\ge 1} (-1)^n \Im(\log (1- \frac{ia^{-1/2}}{n})-\log (1+ \frac{ia^{-1/2}}{n}))$