New Inequalities Involving the Dirichlet Beta and Euler's Gamma Functions

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

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...

Remark.We deduce from inequality (12) that the function References

β(x + 1) [1] L. M. B. C. Campos, Generalized Calculus with Applications φ(x)= x β(x) to Matter and Forces, CRC Press, Taylor and Francis Group, New York, 2014. [2] P. Cerone, Bounds for Zeta and related functions, Journal of is increasing on (0,∞). This implies Inequalities in Pure and Applied Mathematics, 6, Art. 134 (2005). β(x + 1) β(x + 1) ′ + x ≥ 0 [3] P. Cerone, On a Double Inequality for the Dirichlet Beta β(x) β(x) Function, Tamsui Oxford Journal of Mathematical Sciences, 24, 269-276 (2008). or equivalently, [4] S. R. Finch, Mathematical Constants, Cambridge University Press, 2003. β ′ [5] M. A. Idowu, Fundamental relations between the (x) ′ β(x + 1) 1 − x + xβ (x + 1) ≥ 0 Dirichlet beta function, Euler numbers, and Riemann β(x) zeta function for positive integers, Available at: https://arxiv.org/pdf/1210.5559.pdf for x > 0. ψ(k) 1 [6] S. Kölbig, The Polygamma Function (x) for x = 4 and 3 x = 4 , Journal of Computational and Applied Mathematics, Corollary 3.The inequality 75, 43-46 (1996). [7] A. Laforgia and P. Natalini, Turan type inequalities for some 4G β(x + 1) π3 special functions, Journal of Inequalities in Pure and Applied < x < (14) π β(x) 16G Mathematics, 7, 1-5 (2006). [8] E. Neuman, Inequalities involving a logarithmically convex holds for x ∈ (1,2), where G is the Catalan’s constant. function and their applications to special functions, Journal of Inequalities in Pure and Applied Mathematics, 7, Art. 16 (2006). φ β (x+1) Proof.Since (x) = x β (x) is increasing, then for [9] D. Speyer, Response to MathOverflow question x ∈ (1,2), we have φ(1) < φ(x) < φ(2) which results to No. 180716, MathOverflow, Available online at: (14). http://mathoverflow.net/questions/180716.

Theorem 5.Let α ≥ 1 and x ∈ (0,1). Then, Kwara Nantomah holds Gα [β(1 + x)Γ (1 + x)]α π α−1 ≤ ≤ PhD, MPhil, and BSc degrees β(1 + α)Γ (1 + α) β(1 + αx)Γ (1 + αx) 4 all in Mathematics. He (15) is currently a Senior Lecturer where G is the Catalan’s constant. The inequality is and the Head of Department reversed if 0 < α ≤ 1. of Mathematics, University for Development Studies, Proof.Let f (x) = β(1 + x)Γ (1 + x). Then f (x) is Ghana. His research interest differentiable and by Remark 3, it is logarithmically is in Mathematical Analysis, convex. Then by Lemma 5, the function Mathematical Inequalities, α [β (1+x)Γ (1+x)] α Special Functions related to the Gamma Function, and g(x)= β α Γ α is decreasing for ≥ 1. Hence for (1+ x) (1+ x) Convex Functions. He has published several research x ∈ (0,1), we have g(1) ≤ g(x) ≤ g(0) yielding the result articles in reputed international mathematics journals. He (15). If 0 < α ≤ 1, then g(x) is increasing and for has also served as a reviewer to a number of international x ∈ (0,1), we have g(0) ≤ g(x) ≤ g(1) which gives the journals of pure and applied mathematics. reverse inequality of (15).

4 Conclusion

In this study, we have established some inequalities involving the Dirichlet beta and Euler’s Gamma functions. We have also discussed the monotonicity of the Dirichlet beta function. The generalized forms of the H¨older’s and Minkowski’s inequalities among other analytical techniques were employed.

c 2018 NSP Natural Sciences Publishing Cor. Sohag J. Math. 5, No. 1, 9-13 (2018) / www.naturalspublishing.com/Journals.asp 13

Mohammed Muniru Iddrisu is a Senior Lecturer and a former Head of the Mathematics Department and now the Vice-Dean of the Faculty of Mathematical Sciences, University for Development Studies, Ghana. He received his BSc, MSc, and PhD degrees in Mathematics from the University of Capecoast, Ghana, Norwegian University of Science and Technology, Trondheim, Norwayand, University for Development Studies, Ghana respectively. His research interests are in Mathematical Analysis (Inequalities, Special functions and Applications), Coding Theory, Cryptography, Mathematical Statistics and Applications. He has published several research articles in reputable international Journals of mathematics. He is also a reviewer to many international journals of pure and applied mathematics. He is a member of the Ghana Mathematics Society, a member of Ghana Science Association, a member of the Management Board of the National Institute for Mathematical Sciences, Ghana and also serves on the panel for Mathematics Programmes of the National Accreditation Board, Ghana.

c 2018 NSP Natural Sciences Publishing Cor.

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