If (k,ℓ)(k,ℓ) is an exponent pair such that k+ℓ<1k+ℓ<1, then we have ∫1T∣ζ(1/2+it)∣4∣ζ(σ+it)∣2dt≪ϵT1+ϵ(σ>min(56,max(ℓ−k,5k+ℓ4k+1)),∫1T∣ζ(1/2+it)∣4∣ζ(σ+it)∣2dt≪ϵT1+ϵ(σ>min(65,max(ℓ−k,4k+15k+ℓ)), while if (k,ℓ)(k,ℓ) is an exponent pair such that 3k+ℓ<13k+ℓ<1, then we have ∫1T∣ζ(1/2+it)∣4∣ζ(σ+it)∣4dt≪ϵT1+ϵ(σ>11k+ℓ+18k+2).∫1T∣ζ(1/2+it)∣4∣ζ(σ+it)∣4dt≪ϵT1+ϵ(σ>8k+211k+ℓ+1).
生成讲解博客的中文版本
生成AI智能校验的中文版本