On the Product of the Gamma Function and the Riemann Zeta Function

Mathematica Aeterna, Vol. 3, 2013, no. 1, 13 - 16

Banyat Sroysang

Department of Mathematics and Statistics, Faculty of Science and Technology, Thammasat University, Pathumthani 12121 Thailand [email protected]

Abstract

In 2011, W. T. Sulaiman gave inequalities involving the product of the gamma function and the Riemann zeta function. In this paper, we generalize the inequalities.

Mathematics Subject Classiﬁcation: 26D15

Keywords: Gamma function, zeta function, inequality

1 Introduction

The gamma function Γ is deﬁned by ∞ Γ(z)= tz−1e−tdt, Z0 where Re(z) > 0. The Riemann zeta function ξ is deﬁned by

1 ∞ ts−1 ξ(s)= dt, Γ(s) et − 1 Z0 where s> 1. Now, we let h be the product of the gamma function and the Riemann zeta function, i.e., h(x)=Γ(x)ξ(x) for all x> 1. In 2011, Sulaiman [1] gave an inequality as follows.

h(1 + x + y) ≤ h1/p(1 + p(x + 1))h1/q(1 + q(y − 1)) (1) 14 Banyat Sroysang

− 1 1 for all x> 1,y> 1,p> 1 and p + q = 1. For any non-negative integer n, we denote hn be the n-th derivative of h. In 2011, Sulaiman [1] gave two inequalities as follows.

x + y h2 ≤ h (x)h (y) (2) m+n 2 2m 2n and x + y + z h3 ≤ h (x)h (y)h (z) (3) m+n+r 3 3m 3n 3r for all x,y,z > 1 and non-negative even integers n, m, r. In this paper, we present the generalizations for inequalities (1), (2) and (3).

2 Results

Theorem 2.1. Let x1, x2, ..., xn > −1, y1,y2, ..., yn > 1, p1,p2, ..., pn > 1 n and q , q , ..., q > be such that 1 1 . Then 1 2 n 1 pi + qi =1 i=1 X

n n 1/pi 1/qi h(1 + (xi + yi)) ≤ h (1 + pi(xi + 1))h (1 + qi(yi − 1)). (4) i=1 i=1 X Y

Proof. By the assumption,

n n n

h(1 + (xi + yi)) = Γ(1+ (xi + yi))ξ(1 + (xi + yi)) i=1 i=1 i=1 X ∞ Xn X tPi=1(xi+yi) dt = t 0 e − 1 Z ∞ n n − tPi=1(xi+1)tPi=1(yi 1) = dt et − 1 Z0 ∞ n txi+1 tyi−1 = dt. (et − 1)1/pi (et − 1)1/qi 0 i=1 Z Y

By the generalized H¨older inequality, On the product of the gamma function and the Riemann zeta function 15

1/pi 1/qi n n ∞ tpi(xi+1) ∞ tqi(yi−1) h(1 + (x + y )) ≤ dt dt i i et − 1 et − 1 i=1 i=1 Z0 Z0 X Yn 1/pi 1/pi = Γ (1 + pi(xi + 1))ξ (1 + pi(xi + 1)) i=1 Y 1/qi 1/qi × Γ (1 + qi(yi − 1))ξ (1 + qi(yi − 1)) n 1/pi 1/qi = h (1 + pi(xi + 1))h (1 + qi(yi − 1)). i=1 Y

We note on Theorem 2.1 that if n = 1 then we obtain the inequality (1).

Theorem 2.2. Let x1, x2, ..., xn > 1 and let k1,k2, ..., kn be non-negative n even integers and let k = i=1 ki. Then P n x n hn i ≤ h (x ). (5) k n nki i i=1 ! i=1 X Y

Proof. By the assumption,

n x n x h i = h(k) i k n n i=1 ! i=1 ! X ∞ X n xi − (log t)kt(Pi=1 n ) 1 e dt = t 0 e − 1 Z ∞ n xi−1 (log t)ktPi=1 n e dt = t 0 e − 1 Z ∞ n xi−1 (log t)ki t n = e dt (et − 1)1/n 0 i=1 Z Y ∞ n (log t)nki txi−1 1/n = e dt. et − 1 0 i=1 Z Y

By the generalized H¨older inequality, 16 Banyat Sroysang

n x n ∞ (log t)nki txi−1 1/n h i ≤ e dt k n et − 1 i=1 ! i=1 Z0 X Yn (nki) 1/n = h (xi) i=1 Yn 1/n = (hnki (xi)) i=1 Y n 1/n

= hnki (xi) . i=1 ! Y This implies the inequality (5).

We note on Theorem 2.2 that (i) if n = 2 then we obtain the inequality (2), and (ii) if n = 3 then we obtain the inequality (3).

Corollary 2.3. Let x> 1 and let k1,k2, ..., kn be non-negative even integers n and let k = i=1 ki. Then P n n hk (x) ≤ hnki (x). i=1 Y

Proof. This follows from Theorem 2.2 in case x1 = x2 = ... = xn

References

[1] W. T. Sulaiman, Turan inequalites for the Riemann zeta functions, AIP Conf. Proc., 1389 (2011), 1793–1797.

Received: December, 2012