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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 71049, 14 pages doi:10.1155/2007/71049 Research Article A Multiple Hilbert-Type Integral Inequality with the Best Constant Factor Baoju Sun Received 9 February 2007; Accepted 29 April 2007 Recommended by Eugene H. Dshalalow By introducing the norm x α (x ∈ R) and two parameters α, λ, we give a multiple Hilbert-type integral inequality with a best possible constant factor. Also its equivalent form is considered. Copyright © 2007 Baoju Sun. This is an open access article distributed under the Cre- ative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. 1. Introduction If p>1, 1/p+1/q = 1, f ,g ≥ 0, satisfy 0 <  ∞ 0 f p (t)dt < ∞ and 0 <  ∞ 0 g q (t)dt < ∞,then the well-known Hardy-Hilbert’s integral inequality is given by (see [1, 2])  ∞ 0 f (x)g(y) x + y dxdy < π sin(π/p)   ∞ 0 f p (t)dt  1/p   ∞ 0 g q (t)dt  1/q , (1.1) where the constant factor π/sin(π/p) is the best possible. Its equivalent form is  ∞ 0   ∞ 0 f (x) x + y dx  p dy <  π sin(π/p)  p  ∞ 0 f p (x) dx, (1.2) where the constant factor (π/sin(π/p)) p is still the best possible. Hardy-Hilbert integral inequality is important in analysis and applications. During the past few years, many researchers obtained var ious gener alizations, variants, and ex- tensions of inequality (1.1) (see [3–9] and the references cited therein). 2 Journal of Inequalities and Applications Hardy et al. [1] gave a Hilbert-type integral inequality similar to (1.1)as  ∞ 0 ln(x/y) x − y f (x)g(y)dxdy <  π sin(π/p)  2   ∞ 0 f p (x) dx  1/p   ∞ 0 g q (x) dx  1/q , (1.3) where the constant factor (π/sin(π/p)) 2 is the best possible. Recently, Yang gave a generalization of (1.3) as (see [9])  ∞ 0 ln(x/y) f (x)g(y) x λ − y λ dxdy <  π λsin(π/p)  2   ∞ 0 x (p−1)(1−λ) f p (x) dx  1/p   ∞ 0 x (q−1)(1−λ) g q (x) dx  1/q , (1.4) where the constant factor (π/λsin(π/p)) 2 is the best possible. Its equivalent form is  ∞ 0 y λ−1   ∞ 0 ln(x/y) f (x) x λ − y λ dx  p dy <  π λsin(π/p)  2p  ∞ 0 x (p−1)(1−λ) f p (x) dx, (1.5) where the constant factor (π/λsin(π/p)) 2p is the best possible. At present, because of the requirement of higher-dimensional harmonic analysis and higher-dimensional operator theory, multiple Hilbert-type integral inequalities have been studied. Hong [10] obtained the following. If a>0, n  i=1 1 p i = 1, p i > 1, r i = 1 p i n  i=1 p i , λ> 1 a  n − 1 − 1 r i  , i = 1,2, ,n, (1.6) then  ∞ α ···  ∞ α 1   n i =1  x i − α  a  λ n  i=1 f i  x i  dx 1 dx 2 ···dx n ≤ Γ n (1/α) α n−1 Γ(λ) n  i=1  Γ  1 a  1 − 1 r i  Γ  λ − 1 a  n −1 − 1 r i   ∞ α (t − α) n−1−αλ f p i i (t)dt  1/p i . (1.7) Yang and Kuang, and others obtained some multiple Hilbert-type integral inequalities (see [5, 11, 12]). The main objective of this paper is to build multiple Hilbert-type integral inequalities with best constant factor of (1.4)and(1.5). For this reason, we introduce signs as R + n =  x =  x 1 ,x 2 , ,x n  : x 1 ,x 2 , ,x n > 0  , x α =  x α 1 + x α 2 + ···+ x α n  1/α , α>0, (1.8) and we agree with x α <crepresenting {x ∈ R + n : x α <c}. Baoju Sun 3 2. Lemmas Firstwegivesomemultipleintegralformulas. Lemma 2.1 (see [13]). If p i > 0, i = 1,2, ,n, f (τ) is a measurable function, then  t 1 ,t 2 , ,t n >0;t 1 +t 2 +···+t n ≤1 f  t 1 + t 2 + ···+ t n  t p 1 −1 1 t p 2 −1 2 ···t p n −1 n dt 1 dt 2 ···dt n = Γ  p 1  Γ  p 2  ··· Γ  p n  Γ  p 1 + p 2 + ···+ p n   1 0 f (τ)τ p 1 +p 2 +···+p n −1 dτ. (2.1) Lemma 2.2. If r>0, p i > 0, i = 1,2, ,n, f (τ) is a measurable function, then  t 1 ,t 2 , ,t n >0;t 1 +t 2 +···+t n ≤r f  t 1 + t 2 + ···+ t n  t p 1 −1 1 t p 2 −1 2 ···t p n −1 n dt 1 dt 2 ···dt n = Γ  p 1  Γ  p 2  ··· Γ  p n  Γ  p 1 + p 2 + ···+ p n   r 0 f (τ)τ p 1 +p 2 +···+p n −1 dτ, (2.2)  t 1 ,t 2 , ,t n >0 f  t 1 + t 2 + ···+ t n  t p 1 −1 1 t p 2 −1 2 ···t p n −1 n dt 1 dt 2 ···dt n = Γ  p 1  Γ  p 2  ··· Γ  p n  Γ  p 1 + p 2 + ···+ p n   ∞ 0 f (τ)τ p 1 +p 2 +···+p n −1 dτ. (2.3) Proof. Setting t i /r = u i (i = 1, 2, ,n) on the left-hand side of (2.2)weobtain(2.2)from Lemma 2.1.  From (2.1)and(2.3), we have the following lemma. Lemma 2.3.  t 1 ,t 2 , ,t n >0;t 1 +t 2 +···+t n ≥1 f  t 1 + t 2 + ···+ t n  t p 1 −1 1 t p 2 −1 2 ···t p n −1 n dt 1 dt 2 ···dt n = Γ  p 1  Γ  p 2  ··· Γ  p n  Γ  p 1 + p 2 + ···+ p n   ∞ 1 f (τ)τ p 1 +p 2 +···+p n −1 dτ. (2.4) Setting t i = (x i /a i ) α i (i = 1,2, ,n)in(2.1), (2.2), (2.3), (2.4) we have the following lemma. 4 Journal of Inequalities and Applications Lemma 2.4. If p i > 0, a i > 0, α i > 0, i = 1,2, ,n, f (τ) is a measurable function, then  x 1 ,x 2 , ,x n >0;(x 1 /a 1 ) α 1 +(x 2 /a 2 ) α 2 +···+(x 1 /a n ) α n ≤1 f  x 1 a 1  α 1 +  x 2 a 2  α 2 + ···+  x 1 a n  α n  × x p 1 −1 1 x p 2 −1 2 ···x p n −1 n dx 1 dx 2 ···dx n = a p 1 1 a p 2 2 ···a p n n Γ  p 1 /α 1  Γ  p 2 /α 2  ··· Γ  p n /α n  α 1 α 2 ···α n Γ  p 1 /α 1 + p 2 /α 2 + ···+ p n /α n   1 0 f (τ)τ p 1 /α 1 +p 2 /α 2 +···+p n /α n −1 dτ,  x 1 ,x 2 , ,x n >0;(x 1 /a 1 ) α 1 +(x 2 /a 2 ) α 2 +···+(x 1 /a n ) α n ≤r f  x 1 a 1  α 1 +  x 2 a 2  α 2 + ···+  x 1 a n  α n  × x p 1 −1 1 x p 2 −1 2 ···x p n −1 n dx 1 dx 2 ···dx n = a p 1 1 a p 2 2 ···a p n n Γ  p 1 /α 1  Γ  p 2 /α 2  ··· Γ  p n /α n  α 1 α 2 ···α n Γ  p 1 /α 1 + p 2 /α 2 + ···+ p n /α n   r 0 f (τ)τ p 1 /α 1 +p 2 /α 2 +···+p n /α n −1 dτ,  x 1 ,x 2 , ,x n >0 f  x 1 a 1  α 1 +  x 2 a 2  α 2 + ···+  x 1 a n  α n  x p 1 −1 1 x p 2 −1 2 ···x p n −1 n dx 1 dx 2 ···dx n = a p 1 1 a p 2 2 ···a p n n Γ  p 1 /α 1  Γ  p 2 /α 2  ··· Γ  p n /α n  α 1 α 2 ···α n Γ  p 1 /α 1 + p 2 /α 2 + ···+ p n /α n   ∞ 0 f (τ)τ p 1 /α 1 +p 2 /α 2 +···+p n /α n −1 dτ,  x 1 ,x 2 , ,x n >0;(x 1 /a 1 ) α 1 +(x 2 /a 2 ) α 2 +···+(x 1 /a n ) α n ≥1 f  x 1 a 1  α 1 +  x 2 a 2  α 2 + ···+  x 1 a n  α n  × x p 1 −1 1 x p 2 −1 2 ···x p n −1 n dx 1 dx 2 ···dx n = a p 1 1 a p 2 2 ···a p n n Γ  p 1 /α 1  Γ  p 2 /α 2  ··· Γ  p n /α n  α 1 α 2 ···α n Γ  p 1 /α 1 + p 2 /α 2 + ···+ p n /α n   ∞ 1 f (τ)τ p 1 /α 1 +p 2 /α 2 +···+p n /α n −1 dτ. (2.5) In particular , if p>0, α>0, f (τ) is a measurable function, then  x 1 ,x 2 , ,x n >0;x α 1 +x α 2 +···+x α n ≤1 f  x α 1 + x α 2 + ···+ x α n  dx 1 dx 2 ···dx n = Γ n (1/α) α n Γ(n/α)  1 0 f (τ)τ n/α−1 dτ, (2.6)  x 1 ,x 2 , ,x n >0;x α 1 +x α 2 +···+x α n ≥1 f  x α 1 + x α 2 + ···+ x α n  dx 1 dx 2 ···dx n = Γ n (1/α) α n Γ(n/α)  ∞ 1 f (τ)τ n/α−1 dτ. (2.7) The following result holds. Baoju Sun 5 Lemma 2.5. If p>1, n ∈ Z + , α>0, λ>0,definetheweightfunctionw α,λ (x, p) as w α,λ (x, p) =  R n + ln   x α /y α  x λ α −y λ α  x α y α  n−λ/p dy. (2.8) Then w α,λ (x, p) =x n−λ α Γ n (1/α) α n−1 Γ(n/α)  π λsin(π/p)  2 . (2.9) Proof. By (2.6)and(2.7), we have w α,λ (x, p) =x n−λ/p α  R n + ln   x α /y α  x λ α −y λ α y λ/p−n α dy =x n−λ/p α  y 1 ,y 2 , ,y n >0 ln  y α 1 + y α 2 + ···+ y α n  1/α /x α   y α 1 + y α 2 + ···+ y α n  λ/α −x λ α ×  y α 1 + y α 2 + ···+ y α n  (1/α)(λ/p−n) dy 1 dy 2 ···dy n =x n−λ/p α Γ n (1/α) α n Γ(n/α)  ∞ 0 ln  t 1/α /x α  t λ/α −x λ α t (1/α)(λ/p−n) t n/α−1 dt. (2.10) Setting (t 1/α /x α ) λ = u we have w α,λ (x, p) =x n−λ α Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  ∞ 0 lnu u − 1 u 1/p−1 du. (2.11) From [1, Theorem 342] we have (1/λ 2 )  ∞ 0 (lnu/(u − 1))u 1/p−1 du = (π/λsin(π/p)) 2 . So we obtain w α,λ (x, p) =x n−λ α Γ n (1/α) α n−1 Γ(n/α)  π λsin(π/p)  2 . (2.12) Thus Lemma 2.5 is proved.  Lemma 2.6. If λ>0, s>0, then  ∞ 1 1 x  1/x λ 0 lnu u − 1 u s−1 dudx = 2 λ ∞  n=0 1 (n + s) 3 . (2.13) Proof. Since lnu u − 1 u s−1 =−lnu ∞  n=0 u n+s−1 ,0<u<1, (2.14) 6 Journal of Inequalities and Applications then  1/x λ 0 lnu u − 1 u s−1 du = ∞  n=0  1/x λ 0 (−lnu)u n+s−1 du = ∞  n=0  λ n + s x −λ(n+s) lnx + 1 (n + s) 2 x −λ(n+s)  ,  ∞ 1 1 x  1/x λ 0 lnu u − 1 u s−1 dudx =  ∞ 1  ∞  n=0  λ n + s x −λ(n+s)−1 + 1 (n + s) 2 x −λ(n+s)−1   dx = ∞  n=0   ∞ 1 λ n + s x −λ(n+s)−1 lnxdx+  ∞ 1 1 (n + s) 2 x −λ(n+s)−1 dx  = 2 λ ∞  n=0 1 (n + s) 3 . (2.15)  We next give a key lemma in this paper. Lemma 2.7. If p>1, 1/p+1/q = 1, n ∈ Z + , α>0, λ>0, 0 <ε<qλ/2p, then A : =  x α ≥1  y α ≥1 ln   x α /y α  x λ α −y λ α x −((n−λ)(p−1)+n+ε)/p α y −((n−λ)(q−1)+n+ε)/q α dxdy ≥  Γ n (1/α) α n−1 Γ(n/α)  2  π λsin(π/p)  2 1 ε  1+o(1)  , ε −→ 0 + . (2.16) Proof. We have A =  x α ≥1 x λ/q−n−ε/p α dx ×  y α 1 +y α 2 +···+y α n ≥1 ln  y α 1 + y α 2 + ···+ y α n  1/α /x α   y α 1 + y α 2 + ···+ y α n  λ/α −x λ α ×  y α 1 + y α 2 + ···+ y α n  (1/α)(λ/p−n−ε/q) dy 1 dy 2 ···dy n =  x α ≥1 x λ/q−n−ε/p α dx Γ n (1/α) α n Γ(n/α)  ∞ 1 ln  t 1/α /x α  t λ/α −x λ α t (1/α)(λ/p−n−ε/q) t n/α−1 dt. (2.17) Setting (t 1/α /x α ) λ = u,wehave A =  x α ≥1 x −n−ε α dx Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  ∞ 1/x λ α lnu u − 1 u 1/p−1−ε/λq du =  x α ≥1 x −n−ε α dx Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  ∞ 0 lnu u − 1 u 1/p−1−ε/λq du −  x α ≥1 x −n−ε α dx Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  1/x λ α 0 lnu u − 1 u 1/p−1−ε/λq du. (2.18) Baoju Sun 7 Notice  x α ≥1 x −n−ε α dx =  x 1 ,x 2 , ,x n >0;x α 1 +x α 2 +···+x α n ≥1 (x α 1 + x α 2 + ···+ x α n ) −(n+ε)/α dx 1 dx 2 ···dx n = Γ n (1/α) α n Γ(n/α)  ∞ 1 u −(n+ε)/α u n/α−1 du = Γ n (1/α) α n Γ(n/α)  ∞ 1 u −ε/α−1 du = Γ n (1/α) α n−1 Γ(n/α) · 1 ε , 1 λ 2  ∞ 0 lnu u − 1 u 1/p−1−ε/λq du =  π λsin(π/p)  2 + o(1). (2.19) Further , from (2.7)andLemma 2.6 we have 0 ≤  x α ≥1 x −n−ε α dx Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  1/x λ α 0 lnu u − 1 u 1/p−1−ε/λq du ≤  x α ≥1 x −n α dx Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  1/x λ α 0 lnu u − 1 u 1/2p−1 du = Γ n (1/α) α n−1 Γ(n/α) 1 λ 2  ∞ 1 t −n/α   1/t λ/α 0 lnu u − 1 u 1/2p−1 du  t n/α−1 dt = Γ n (1/α) α n−1 Γ(n/α) 1 λ 2 2α λ ∞  n=0 1 (n +1/2p) 3 = 2Γ n (1/α) α n−2 λ 3 Γ(n/α) ∞  n=0 1 (n +1/2p) 3 . (2.20) Then A ≥  Γ n (1/α) α n−1 Γ(n/α)  2  π λsin(π/p)  2 1 ε  1+o(1)  . (2.21)  3. Main results Our main result is given in the following theorem. Theorem 3.1. If p>1, 1/p+1/q = 1, n ∈ Z, α>0, λ>0, f ,g ≥ 0,satisfy 0 <  R n + x (n−λ)(p−1) α f p (x) dx < ∞, 0 <  R n + y (n−λ)(q−1) α g q (y)dy <∞. (3.1) 8 Journal of Inequalities and Applications Then J : =  R n + ln   x α /y α  x λ α −y λ α f (x)g(y)dxdy < Γ n (1/α) α n−1 Γ(n/α)  π λsin(π/p)  2 ×   R n + x (n−λ)(p−1) α f p (x) dx  1/p   R n + y (n−λ)(q−1) α g q (y)dy  1/q , (3.2)  R n + y λ−n α   R n + ln   x α /y α  x λ α −y λ α f (x)dx  p dy <  π λsin(π/p)  2 Γ n (1/α) α n−1 Γ(n/α)  p  R n + x (n−λ)(p−1) α f p (x) dx. (3.3) The constant factors (Γ n (1/α)/α n−1 Γ(n/α))[π/λsin(π/p)] 2 , [(π/λsin(π/p)) 2 (Γ n (1/α)/ α n−1 Γ(n/α))] p are the best possible. Proof. By H ¨ older’s inequalit y, one has J =  R n +  ln   x α /y α  x λ α −y λ α  1/p  x α y α  (n−λ)/p+λ/ pq x (1/q−1/p)(n−λ) α f (x) ×  ln   x α /y α  x λ α −y λ α  1/q   y α x α  (n−λ)/q+λ/qp y (1/p−1/q)(n−λ) α g(y)dxdy ≤   R n + ln   x α /y α  x λ α −y λ α  x α y α  n−λ+λ/q x (p/q−1)(n−λ) α f p (x) dxdy  1/p ×   R n + ln   x α /y α  x λ α −y λ α   y α x α  n−λ+λ/p y (q/p−1)(n−λ) α g q (y)dxdy  1/q =   R n + ln   x α /y α  x λ α −y λ α   x α y α  n−λ/p x (p−2)(n−λ) α f p (x) dxdy  1/p ×   R n + ln   x α /y α  x λ α −y λ α   y α x α  n−λ/q y (q−2)(n−λ) α g q (y)dxdy  1/q =   R n + w α,λ (x, p)x (n−λ)(p−2) α f p (x) dx  1/p   R n + w α,λ (y,q)y (n−λ)(q−2) α g q (y)dy  1/q . (3.4) According to the condition of taking equality in H ¨ older’s inequality, if this inequality takes the form of an equality, then there exist constants C 1 and C 2 , such that they are not Baoju Sun 9 all zero, and C 1 ln   x α /y α  x λ α −y λ α  x α y α  n−λ/p x (p−2)(n−λ) α f p (x) = C 2 ln   x α /y α  x λ α −y λ α   y α x α  n−λ/q y (q−2)(n−λ) α g q (y), a.e. in R n + × R n + . (3.5) It follows that C 1 x n α x (p−1)(n−λ) α f p (x) = C 2 y n α y (q−1)(n−λ) α g q (y) = C (constant), a.e. in R n + × R n + , (3.6) which contradicts (3.1). Hence we have J<   R n + w α,λ (x, p)x (n−λ)(p−2) α f p (x) dx  1/p   R n + w α,λ (y,q)y (n−λ)(q−2) α g q (y)dy  1/q . (3.7) By Lemma 2.5 and since π/sin(π/p) = π/sin(π/q), we have J< Γ n (1/α) α n−1 Γ(n/α)  π λsin(π/p)  2   R n + x (n−λ)(p−1) α f p (x) dx  1/p ×   R n + y (n−λ)(q−1) α g q (y)dy  1/q . (3.8) Hence (3.2)isvalid. For 0 <a<b< ∞, let us define g a,b (y) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩  y λ−n α   R n + ln   x α /y α  x λ α −y λ α f (x)dx  p−1 , a<y α <b, 0, 0 < y α ≤ a,ory α ≥ b, g(y) =y λ−n α   R n + ln   x α /y α  x λ α −y λ α f (x)dx  p−1 , y ∈ R n + . (3.9) By (3.1), for sufficiently small a>0andsufficiently large b>0, we have 0 <  a<y α <b y (n−λ)(q−1) α g q a,b (y)dy <∞. (3.10) 10 Journal of Inequalities and Applications Hence by (3.2)wehave  a<y α <b y (n−λ)(q−1) α g q (y)dy =  a<y α <b y λ−n α   R n + ln   x α /y α  x λ α −y λ α f (x)dx  p dy =  a<y α <b y λ−n α   R n + ln   x α /y α  x λ α −y λ α f (x)dx  p−1 ×   R n + ln   x α /y α  x λ α −y λ α f (x)dx  dy =  R n + ln   x α /y α  x λ α −y λ α f (x)g a,b (y)dxdy < Γ n (1/α) α n−1 Γ(n/α)  π λsin(π/p)  2   R n + x (n−λ)(p−1) α f p (x) dx  1/p ×   R n + y (n−λ)(q−1) α g q a,b (y)dy  1/q = Γ n (1/α) α n−1 Γ(n/α)  π λsin(π/p)  2   R n + x (n−λ)(p−1) α f p (x) dx  1/p ×   a<y α <b y (n−λ)(q−1) α g q (y)dy  1/q . (3.11) It follows that  a<y α <b y (n−λ)(q−1) α g q (y)dy <  π λsin(π/p)  2 Γ n (1/α) α n−1 Γ(n/α)  p  R n + x (n−λ)(p−1) α f p (x) dx. (3.12) For a → 0 + , b → +∞,by(3.1), we have  R n + y (n−λ)(q−1) α g q (y)dy ≤  π λsin(π/p)  2 Γ n (1/α) α n−1 Γ(n/α)  p  R n + x (n−λ)(p−1) α f p (x) dx < ∞. (3.13) [...]... Hardy-Hilbert’s integral inequality with a best constant, ” Chinese Annals of Mathematics Series A, vol 21, no 4, pp 401–408, 2000 [9] B Yang, “On a generalization of a Hilbert’s type integral inequality and its applications,” Mathematica Applicata, vol 16, no 2, pp 82–86, 2003 [10] Y Hong, “All-sided generalization about Hardy-Hilbert integral inequalities,” Acta Mathematica Sinica Chinese Series, vol 44,... 619–626, 2001 [11] B Yang, A multiple Hardy-Hilbert integral inequality, ” Chinese Annals of Mathematics Series A, vol 24, no 6, pp 743–750, 2003 [12] H Yong, A multiple Hardy-Hilbert integral inequality with the best constant factor,” Journal of Inequalities in Pure and Applied Mathematics, vol 7, no 4, article 139, p 10, 2006 [13] L Hua, An Introduction to Advanced Mathematics (Remaining Sections),... Jinan, China, 2004 [6] B G Pachpatte, “On some new inequalities similar to Hilbert’s inequality, ” Journal of Mathematical Analysis and Applications, vol 226, no 1, pp 166–179, 1998 [7] B Yang and L Debnath, “On new strengthened Hardy-Hilbert’s inequality, ” International Journal of Mathematics and Mathematical Sciences, vol 21, no 2, pp 403–408, 1998 [8] B Yang, “On a general Hardy-Hilbert’s integral inequality. .. Applications [3] M Gao and B Yang, “On the extended Hilbert’s inequality, ” Proceedings of the American Mathematical Society, vol 126, no 3, pp 751–759, 1998 [4] K Jichang and L Debnath, “On new generalizations of Hilbert’s inequality and their applications,” Journal of Mathematical Analysis and Applications, vol 245, no 1, pp 248–265, 2000 [5] K Jichang, Applied Inequalities, Shandong Science and Technology... H Hardy, J E Littlewood, and G Polya, Inequalities, Cambridge University Press, Cambridge, UK, 2nd edition, 1952 [2] D S Mitrinovi´ , J E Peˇ ari´ , and A M Fink, Inequalities Involving Functions and Their Intec c c grals and Derivatives, vol 53 of Mathematics and Its Applications (East European Series), Kluwer Academic Publishers, Dordrecht, The Netherlands, 1991 14 Journal of Inequalities and Applications... contradicts the fact that k < Cn,α (λ, p), hence the constant factor in (3.2) is the best possible Since inequality (3.2) is equivalent to (3.3), the constant factor in (3.3) is also the best possible Thus the theorem is proved Remark 3.2 By using (3.3) we can obtain (3.2), hence inequality (3.2) is equivalent to (3.3) Baoju Sun 13 Corollary 3.3 If p > 1, 1/ p + 1/q = 1, n ∈ Z, α > 0, f ,g ≥ 0, satisfy... It follows that Rn + y < λ −n α ln Rn + x α/ y x λ− y α π λ sin(π/ p) 2 α λ α p f (x)dx Γn (1/α) n−1 Γ(n/α) α dy (3.15) p Rn + x (n−λ)(p−1) p f (x)dx, α hence (3.3) is valid If the constant factor Cn,α (λ, p) = (π/λ sin(π/ p))2 (Γn (1/α)/αn−1 Γ(n/α)) in (3.2) is not the best possible, then there exists a positive number k (with k < Cn,α (λ, p)), such that (3.2) is still valid if one replaces Cn,α (λ,... 4, article 139, p 10, 2006 [13] L Hua, An Introduction to Advanced Mathematics (Remaining Sections), Science Publishers, Beijing, China, 1984 Baoju Sun: Zhejiang Water Conservancy and Hydropower College, Zhejiang University, Hangzhou 310018, China Email address: sunbj@mail.zjwchc.com ... (3.21) Then ln x Rn + < x α/ y n− y α f (x)g(y)dx d y Γn (1/α) π n−1 Γ(n/α) nsin(π/ p) α ln Rn + α n α Rn + x α/ y x n− y α α n α 1/ p 2 Rn + p f (x)dx 1/q f p (x)dx dy < g q (y)d y n R+ π nsin(π/ p) 2 Γn (1/α) n−1 Γ(n/α) α , p Rn + f p (x)dx (3.22) The constant factors (Γn (1/α)/αn−1 Γ(n/α))[π/nsin(π/ p)]2 , [(π/nsin(π/ p))2 (Γn (1/α)/ αn−1 Γ(n/α))] p in (3.22) are all the best possible Corollary 3.4... ,g ≥ 0, satisfy 0< Rn + f p (x)dx < ∞, 0< Rn + g q (y)d y < ∞ (3.23) Then n i =1 x i / n n − i=1 xi ln Rn + < n i =1 y i n n i=1 yi f (x)g(y)dx d y π (n − 1)! nsin(π/ p) 1 1/ p 2 Rn + f p (x)dx 1/q g q (y)d y n R+ , (3.24) n i=1 xi / n n − i =1 x i ln Rn + Rn + < n i =1 y i n n i =1 y i π (n − 1)! nsin(π/ p) 1 p f (x)dx dy 2 p Rn + f p (x)dx, where the constant factors in (3.24) are all the best possible . still the best possible. Hardy-Hilbert integral inequality is important in analysis and applications. During the past few years, many researchers obtained var ious gener alizations, variants, and. Analysis and Applications, vol. 226, no. 1, pp. 166–179, 1998. [7] B. Yang and L. Debnath, “On new strengthened Hardy-Hilbert’s inequality, ” International Jour- nal of Mathematics and Mathematical. B. Yang, “On a generalization of a Hilbert’s type integral inequality and its applications,” Math- ematica Applicata, vol. 16, no. 2, pp. 82–86, 2003. [10] Y. Hong, “All-sided generalization about