Sohag J. Math. 5, No. 1, 9-13 (2018) 9 Sohag Journal of Mathematics An International Journal http://dx.doi.org/10.18576/sjm/050102 New Inequalities Involving the Dirichlet Beta and Euler's Gamma Functions Kwara Nantomah∗ and Mohammed Muniru Iddrisu Department of Mathematics, University for Development Studies, Navrongo Campus, P. O. Box 24, Navrongo, UE/R, Ghana. Received: 17 Mar. 2017, Revised: 18 Nov. 2017, Accepted: 23 Nov. 2017 Published online: 1 Jan. 2018 Abstract: We present some new inequalities involving the Dirichlet Beta and Euler’s Gamma functions. The concept of monotonicity of Dirichlet Beta function is also discussed. The generalized forms of the H¨older’s and Minkowski’s inequalities among other techniques are employed. Keywords: Dirichlet beta function, Gamma function, Inequality 2010 Mathematics Subject Classification: 33B15, 11M06, 33E20. 1 Introduction where G = 0.915965594177... is the Catalan constant [5]. The Dirichlet beta function (also known as the Catalan For positive integer values of n, the function β(x) may be beta function or the Dirichlet’s L-function) is defined for evaluated explicitly by x > 0by[4, p. 56] n π2n+1 β (−1) E2n ∞ x−1 (2n + 1)= n 1 β 1 t 4 + (2n)! (x)= t −t dt (1) Γ (x) 0 e + e ∞ Z where En are the Euler numbers generated by (−1)n ∑ ∞ = x 1 2 xn n=0 (2n + 1) . = x −x = ∑ En coshx e + e n=0 n! where Γ (x) is the classical Euler’s Gamma function ψ(m) defined as ∞ Also, in terms of the polygamma function (x), the Γ (x)= tx−1e−t dt function β(x) may be written as [6] 0 Z 1 1 3 and satisfying the basic relation β(n)= ψ(n−1) − ψ(n−1) . 22n(n − 1)! 4 4 Γ (x + 1)= xΓ (x). (2) The main objective of this paper is to establish some Let K(x) be defined as inequalities involving the Dirichlet Beta and Euler’s Gamma functions. We begin by recalling the following ∞ tx−1 lemmas which shall be required in order to establish our K(x)= β(x)Γ (x)= dt, x > 0. (3) et + e−t results. Z0 The Dirichlet beta function, which is closely related to the Riemann zeta function, has important applications in 2 Preliminaries Analytic Number Theory as well as other branches of mathematics. See for instance [2], [3] and the related Lemma 1(Generalized Holder’s¨ Inequality). Let β π β references therein. In particular, (1)= 4 and (2)= G, f1, f2,..., fn be functions such that the integrals exist. ∗ Corresponding author e-mail: [email protected] c 2018 NSP Natural Sciences Publishing Cor. 10 K. Nantomah, M. M. Iddrisu: New Inequalities involving the dirichlet... Then the inequality Proof.Let K(x) be defined as in (3). Then by utilizing Lemma 1, we obtain 1 α b n n b i αi ∏ fi(t) dt ≤ ∏ | fi(t)| dt (4) ∑n xi a a n ∞ ( i=1 α )−1 Z i=1 i=1 Z xi t i K ∑ = dt α t −t n 1 i=1 i ! 0 e + e holds for αi > 1 such that ∑ = 1. Z i=1 αi n xi−1 ∞ ∑ α t i=1 i Proof.See page 790-791 of [1]. = dt ∑n 1 0 t −t i=1 α Lemma 2(Generalized Minkowski’s Inequality). Let Z (e + e ) i 1 f , f ,..., f be functions such that the integrals exist. ∞ α 1 2 n n txi−1 i Then the inequality = ∏ t −t dt 0 i=1 e + e 1 Z u u 1 1 b n n b u n ∞ x −1 α u t i i ∑ fi(t) dt ≤ ∑ | fi(t)| dt (5) ≤ ∏ t −t dt a i=1 ! i=1 a 0 e + e Z Z i=1 Z n 1 holds for u ≥ 1. α = ∏(K(xi)) i Proof.See page 790-791 of [1]. i=1 Lemma 3([9]). Let f and h be continuous rapidly which gives the required result (7). decaying positive functions on [0,∞). Further, let F and H be defined as Remark.If n = 2, α1 = a, α2 = b, x1 = x and x2 = y, then, ∞ ∞ we have F(x)= f (t)tx−1 dt and H(x)= h(t)tx−1 dt. 0 0 x y 1 1 Z Z K + ≤ (K(x)) a (K(y)) b f (t) F(x) a b If h(t) is increasing, then so is H(x) . which implies that K(x) is logarithmically convex. Also, Lemma 4([7]). Let f and g be two nonnegative functions since every logarithmically convex function is convex, it of a real variable and m, n be real numbers such that the follows that K(x) is convex. integrals in (6) exist. Then b b Corollary 1.The inequality g(t)( f (t))m dt · g(t)( f (t))n dt Za Za 2 ′ 2 ′′ b m+n β (x) β (x) ≥ g(t)( f (t)) 2 dt (6) − ≤ ψ′(x) (8) β(x) β(x) Za Lemma 5([8]). Let f : (0,∞) → (0,∞) be a differentiable, ψ Γ ′(x) logarithmically convex function. Then the function holds for x > 0, where (x) = Γ (x) is the Digamma function. ( f (x))α g(x)= f (αx) Proof.Since K(x)= β(x)Γ (x) is logarithmically convex, is decreasing if α ≥ 1, and increasing if 0 < α ≤ 1. then (lnK(x))′′ ≥ 0 which results to (8). 3 Main Results Theorem 2.Let xi > 0, i = 1,2,...,n and u ≥ 1. Then the inequality We present the main findings of the paper in this section. 1 n u n Theorem 1.For i = 1,2,...,n, let αi > 1 such that 1 n 1 ∑ β(xi)Γ (xi) ≤ ∑ (β(xi)Γ (xi)) u (9) ∑ α = 1. Then the inequality i=1 i i=1 ! i=1 n x 1 Γ ∑ i α i 1 α ∏n β i = i i=1 ( (xi)) holds. 1 ≤ (7) α n xi ∏n Γ i β ∑ i=1 ( (xi)) i=1 αi Proof.Let K(x) bedefinedasin (3). Then by using the fact is valid for x > 0. ∑n u ∑n u i that i=1 ai ≤ ( i=1 ai) , for ai ≥ 0, u ≥ 1 in conjunction c 2018 NSP Natural Sciences Publishing Cor. Sohag J. Math. 5, No. 1, 9-13 (2018) / www.naturalspublishing.com/Journals.asp 11 with Lemma 2, we obtain Corollary 2.Let xi > 0 for i = 1,2,3,...,n. Then the inequality 1 1 u u n n n ∞ txi−1 n n ∑ K(xi) = ∑ dt ∏β(xi) ≤ β ∑ xi (11) et + e−t i=1 ! i=1 Z0 ! i=1 " i=1 !# 1 u ∞ n txi−1 is valid. = ∑ t −t dt 0 e + e Z i=1 ! Proof.Let xi > 0 for i = 1,2,3,...,n. Then since β(x) is u 1 increasing, we have ∞ n xi−1 u t u = ∑ dt t −t 1 n 0 i=1 "(e + e ) u # ! Z 0 < β(x1) ≤ β ∑ xi , 1 x −1 u i=1 ! ∞ n i u t u ≤ ∑ dt n t −t 1 β β Z0 "i=1 (e + e ) u # ! 0 < (x2) ≤ ∑ xi , i=1 ! u 1 n ∞ xi−1 u t u . ≤ ∑ dt . t t 1 i=1 0 (e + e− ) u Z " # ! n n 1 0 < β(x ) ≤ β ∑ x . = ∑ (K(x )) u n i i i=1 ! i=1 which yields the result (9). Taking products yields n β n n Theorem 3.The function (x) is monotone increasing on β β (0,∞). That is, for 0 < x ≤ y, we have ∏ (xi) ≤ ∑ xi i=1 " i=1 !# β(x) ≤ β(y). (10) as required. Proof.Let F, H, f and h be defined as Remark.In particular, if n = 2, x1 = x and x2 = y in (11), ∞ tx−1 ∞ then we obtain F(x)= dt, H(x)= tx−1e−t dt = Γ (x), et + e−t 2 Z0 Z0 β(x)β(y) ≤ [β(x + y)] . 1 f (t)= and h(t)= e−t . Theorem 4.The inequality et + e−t f t F x β β Then, ( ) = 1 is increasing and by Lemma 3, ( ) is (x + 2) (x + 1) h(t) 1+e−2t H(x) (x + 1)β ≥ x β (12) increasing as well. Thus, for 0 < x ≤ y, we have (x + 1) (x) F(x) F(y) holds for x > 0. ≤ ⇐⇒ F(x)H(y) ≤ F(y)H(x) H(x) H(y) 1 Proof.Let x > 0, g(t)= t −t , f (t)= t, m = x − 1, n = ∞ e +e which implies x + 1, a = 0 and b = . Then by Lemma 4, we have ∞ x−1 ∞ x+1 ∞ x 2 ∞ x−1 ∞ t t t t y−1 −t dt · dt ≥ dt dt · t e dt t −t t −t t −t et + e−t 0 e + e 0 e + e 0 e + e Z0 Z0 Z Z Z ∞ ty−1 ∞ ≤ dt · tx−1e−t dt which implies et + e−t Z0 Z0 β(x)Γ (x)·β(x+2)Γ (x+2) ≥ [β(x + 1)Γ (x + 1)]2 . (13) which further implies By using the functional equation (2), the relation (13) β(x)Γ (x)Γ (y) ≤ β(y)Γ (y)Γ (x). becomes Thus x 1 β x β x 2 x β x 1 2 β(x) ≤ β(y) ( + ) ( ) ( + ) ≥ ( ( + )) as required. which gives the required result. c 2018 NSP Natural Sciences Publishing Cor. 12 K. Nantomah, M. M. Iddrisu: New Inequalities involving the dirichlet..