... obtainskp=∞0 1 1+ uαλ/αu(λ 1) /2 1/ pdu = 1 α∞0 1 (1 +t)λ/αt (1/ α)[(λ +1) /2 1/ p] 1 dt= 1 αB 1 αλ +1 2− 1 p, 1 αλ +1 2− 1 q= : kpα,λ.(3 .1) 4 Journal of Inequalities and ... <kp(ε) = 1 0k (1, u)u− (1+ ε)/pdu+∞ 1 k (1, u)u− (1+ ε)/pdu= 1 0k (1, u)u− (1+ ε)/pdu+ 1 0k(v ,1) v (1+ ε)/p 1 dv= 1 0uηk (1, u)u− (1+ ε)/p−η+ u (1+ ε)/p−η 1 du≤ L 1 0u− (1+ ε)/p−η+ ... = (1/ λ)B( (1/ λ)((λ +1) /2 − 1/ p), (1/ λ)((λ +1) /2 − 1/ q)) and ∞n =1 ∞m =1 ambnmλ+ nλ< 1 λB 1 λλ +1 2− 1 p, 1 λλ +1 2− 1 qap,ωpbq,ωq;(3.5)∞n =1 n(p/2)(λ 1) ∞m =1 ammλ+...