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 $\theta$?
Best Answer
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 $\zeta(2\sigma)\log T$.