This is a reference question:


Let $\psi(x)$ be the psi-Chebyshev function. Is there any unconditional result in the literature that proves that there exists $0<a<2$ such that
$$
\int\_2^x (\psi(y)-y)^2 \mathrm dy =O(x^{a})
?$$



$\psi(x)-x$ on average
Impossible as this would imply that $\frac{\zeta'(s)}{\zeta(s)}+\frac1{s-1}$ is analytic on $\Re(s)\ge 1/2$.


Your bound
$$\int\_1^x |\psi(y)-y|^2 dy = O(x^a)$$ for some $a < 2$ implies (with Cauchy-Schwarz inequality) that $$\int\_x^{2x} (\psi(y)-y)dy = O(x^{a/2+1/2}) \implies\int\_1^2 (\psi(xt)-xt)dt = O(x^{a/2-1/2})
$$
i.e.
$$s\int\_1^\infty\int\_1^2 (\psi(tx)-tx)dt x^{-s-1}dx= \left(\frac{-\zeta'(s)}{\zeta(s)}-\frac{s}{s-1}\right)\int\_1^2 t^s dt-f(s)$$


extends analytically from $\Re(s) > 1$ to $\Re(s) > a/2-1/2$ which is absurd.


(where $f(s)=s\int\_1^2 t^s\int\_1^t x^{-s}dxdt$ is an irrelevant entire term)


