The integral is not absolutely convergent. I understand its value as the limit for $t\to\infty$ of the integral you get substituting $\infty$ by t in the integral in $x$. For this case I have some computations that pointed to a value $0$. It is possible that some sign is wrong in the integrand?

In Mathematics Stack Exchange the inner log was not squared. What is the motivation that allow these differences?

@cd14 Yes I corrected the signs. Of course the limit is zero, as i said before.

@cd14 I have corrected the second part of my answer. The missing terms was not the only error in my previous version.

@cd14 Your second observation is true. I have to think about it. Numerically you appear to be true and my value correct.

@cd14 With respect your first point I checked my formula,I think it is correct. And some numerical checks with Mathematica also agree.

@Zurab Silagadze expand the integrand in powers of $\sin x/\sin z$ and then integrate term by term.

I would be surprised. All these years I have told my students that the zeta function could be defined as the only one transforming this egg shape on the left half plane $\Re(s)<0$ and taking the point $-2$ to $0$ and $0$ to $-1/2$. So this would give us a ``geometrical'' definition of $\zeta(s)$.