Let Δ(x) denote the error term in the Dirichlet divisor problem, and E(T) the error term in the asymptotic formula for the mean square of $$\left|\varsigma \left(\frac{1}{2}+it\right)\right|$$ . If $$E^{*}\left(t\right)=E\left(t\right)-2\pi {\Delta }^{*}\left(t/2\pi \right)$$ with $${\Delta }^{*}\left(x\right)=-\Delta \left(x\right)+2\Delta \left(2x\right)-\frac{1}{2}\Delta \left(4x\right)$$ , then we obtain $${\int }_0^T{\left(E^{*}\left(t\right)\right)}^{4}dt{\ll }_e{T}^{16/9+\epsilon }$$ . We also show how our method of proof yields the bound $$\sum _{r=1}^R{\left({\int }_{tr-G}^{tr+G}{\left|\varsigma \left(\frac{1}{2}+it\right)\right|}^{2}dt\right)}^4{\ll }_e{T}^{2+e}{G}^{-2}+R{G}^4{T}^{\epsilon }$$ , where T 1/5+ε≤G≪T, T

Let Δ(x) denote the error term in the Dirichlet divisor problem, and E(T) the error term in the asymptotic formula for the mean square of $$\left|\zeta \left(\frac{1}{2}+it\right)\right|$$ . If E *(t)=E(t)-2πΔ*(t/2π) with $$\Delta *\left(x\right)+2\Delta \left(2x\right)-\frac{1}{2}\Delta \left(4x\right)$$ , then we obtain $${\int }_0^T{\left|E*\left(t\right)\right|}^{5}dt{\ll }_{\epsilon }{T}^{2+\epsilon }$$ and $${\int }_0^T{\left|E*\left(t\right)\right|}^{\frac{544}{75}}dt{\ll }_{\epsilon }{T}^{\frac{601}{225}+\epsilon }.$$ It is also shown how bounds for moments of | E *(t)| lead to bounds for moments of $$\left|\zeta \left(\frac{1}{2}+it\right)\right|$$ .