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GLOBALLY CONVERGENT FOR THE ZETA AND THE

SUMIT KUMAR JHA

ABSTRACT. We derive the following globally for the and the Dirichlet beta function  ‹ k  ‹ m 1 X∞ 1 2k + 1 X k ( 1) 2πin ζ(s) = (where s = 1 + ln 2 ), 2s 2 22k+1 k 1 m m− 1 s k 0 + m 0 ( + ) 6 − = = k 1 X 1 3‹(k+1) 1‹(k+1) X  k ‹ 1 m β s ∞ ( ) ( ) = 4s k 1 ! 4 4 m m− 1 s k=0 ( + ) − m=0 ( + ) using a globally convergent series for the function, and integrals representing the Riemann zeta function and the Dirichlet beta function. To the best of our knowledge, these series representations are new. Additionally, we give another proof of Hasse’s series representation for the Riemann zeta function.

1. MAIN RESULTS

Let ζ(s) represent the Riemann zeta function. H. Hasse ([1], [2]) proved that n  ‹ k 1 X∞ 1 X n ( 1) 2πin ζ(s) = (where s = 1 + ln 2 ), 1 21 s 2n+1 k k− 1 s − n 0 k 0 ( + ) 6 − = = and n k 1 X∞ 1 X n‹ 1 ζ s ( ) (where s 1). (1) ( ) = s 1 n 1 k k −1 s 1 = n 0 + k 0 ( + ) − 6 − = = We prove the following.

Theorem 1. Let ζ(s) and β(s) represent the Riemann zeta function, and the Dirichlet beta function, respectively. Then we have  ‹ k  ‹ m 1 X∞ 1 2k + 1 X k ( 1) 2πin ζ(s) = (where s = 1 + ln 2 ), (2) 2s 2 22k+1 k 1 m m− 1 s k 0 + m 0 ( + ) 6 − = = and  (k+1) (k+1) k m 1 X∞ 1 3‹ 1‹ X  k ‹ 1 β s ( ) , (3) ( ) = 4s k 1 ! 4 4 m m− 1 s k=0 ( + ) − m=0 ( + ) 2010 Mathematics Subject Classification. 11M06. Key words and phrases. Riemann zeta function, Dirichlet beta function, Global series representation. 1 2 SUMIT KUMAR JHA

n where x ( ) = (x)(x + 1) (x + n 1) denotes the rising . ··· − Proof. We begin with the following result Z ∞ n 1 Lis( x) π x − − d x = (ζ(s) ζ(s, 1 n)), (4) 0 1 + x sin nπ − −

where Lis( x) denotes the polylogarithm function, ζ(s) is the Riemann zeta function, and ζ(s, 1 n)−is the . The above integral is valid for all s C 1 , and 0 < n −< 1. This integral can be obtained from formula 3.2.1.6 in the book [3]∈. \{ } Letting n = 1/2 in equation (4) gives us Z 1/2 1 ∞ x Li x ζ s − s( ) d x (where s 1). (5) ( ) = s − = π(2 2 ) 0 1 + x 6 − The polylogarithm function, Lis(z), has the following series representation [4, 2.1] k X  z k+1 X  k ‹ 1 m+1 Li z ∞ ( ) , (6) s( ) = 1− z m m− 1 s k 0 m 0 ( + ) = − = for all s, z C with (z) < 1/2. We can now derive equation (2) using equations (5), (6), and the fact∈ that ℜ Z k 1/2 ∞ x + π 2k 1 d x p ( + )Γ k 1/2 k+2 = ( + ) 0 (1 + x) 2 (k + 1)!

π (2k + 1) 1 2k Γ (2k) = 2 − 2 (k + 1)! Γ (k) π (2k + 1) 2k (2k 1)! = 2− − (k + 1)! (k 1)!  − ‹ π 2k + 1 = . 22k+1 k + 1 We now proceed to prove equation (3). Using (4) we also have Z ∞ Lis( x) 3/4 1/4 s − (x − x − ) d x = p2π(ζ(s, 1/4) ζ(s, 3/4)) = p2π4 β(s). (7) 0 1 + x − − We can now derive equation (3) using equations (7), (6), and the fact that

Z k 1/4 k 3/4 ∞ x + x + 1 ( ) d x Γ 3/4 Γ k 5/4 Γ 1/4 Γ k 7/4 , − k+2 = ( ( ) ( + ) ( ) ( + )) 0 (1 + x) (k + 1)! −  ‹(k+1)  ‹(k+1) Γ (1/4)Γ (3/4) 1 3 = (k + 1)! 4 − 4  ‹(k+1)  ‹(k+1) p2 π 1 3 = . (k + 1)! 4 − 4 GLOBALLY CONVERGENT SERIES FOR THE RIEMANN ZETA FUNCTION AND THE DIRICHLET BETA FUNCTION 3

2. HASSE’S SERIES REPRESENTATION Letting n 0 on both sides of the equation (4) we get → Z 1 ∞ Lis 1( x) ζ(s) = − − − d x. s 1 0 (x)(1 + x) − This along with equation (6) gives us Hasse’s result (1). 

REFERENCES

1. Hasse, Helmut (1930). Ein Summierungsverfahren fÃijr die Riemannsche ζ-Reihe [A method for the Riemann ζ series]. Mathematische Zeitschrift (in Ger- man). 32 (1): 458-464. doi:10.1007/BF01194645 2. Sondow, Jonathan (1994). of Riemann’s zeta function and values at negative via Euler’s transformation of series. Proceedings of the American Mathematical Society. 120 (2): 421-424. doi:10.1090/S0002-9939-1994-1172954- 7. 3. Y. A. Brychkov, O. I. Marichev, and N. V.Savischenko, Handbook of Mellin transforms, CRC Press, Boca Raton, FL, 2019. 4. J. Guillera, J. Sondow, Double integrals and infinite products for some classical con- stants via analytic continuations of Lerch’s transcendent, J. Ramanujan J (2008) 16: 247. https://doi.org/10.1007/s11139-007-9102-0

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