For any fixed $\frac{1}{2} < \sigma < 1$, let


$$\int\_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \ dt = O(T^\theta), \qquad T \uparrow \infty. $$


It is clear that $\theta > 0$, since we have the classical asymtotic


$$\int\_0^T \frac{|\zeta(\sigma+it)|^2}{T} \ dt \sim \zeta(2\sigma), \qquad T \uparrow \infty. $$


Is there more precise information about the value of $\theta$?



Asymptotic estimate for an integral involving the squared modulus of the Riemann zeta function
Let us introduce the notation
$$M(T):=\int\_0^T|\zeta(\sigma+it)|^2\,dt.$$
Then
$$\int\_0^T \frac{|\zeta(\sigma+it)|^2}{\sqrt{1+t^2}} \,dt=\int\_0^T\frac{dM(t)}{\sqrt{1+t^2}}=\frac{M(T)}{\sqrt{1+T^2}}+\int\_0^T\frac{tM(t)}{(1+t^2)^{3/2}}\,dt$$
by writing this as a Riemann-Stieltjes integral and then integrating by parts. Using that $M(t)$ is asymptotically $\zeta(2\sigma)t$, we conclude that the left-hand side is asymptotically $\zeta(2\sigma)\log T$.


