Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 3 January 2018 doi:10.20944/preprints201801.0020.v1 1 Article 2 Mellin Convolution and its Extensions, Perron 3 Formula and Explicit Formulae 4 Jose Javier Garcia Moreta 5 Graduate student of Physics at the UPV/EHU (University of Basque country);In Solid State Physics;Practicantes Adan y Grijalba2 5 G;P.O 644 48920 Portugalete Vizcaya 6 (Spain);[email protected] 7 8 ABSTRACT: In this paper we use the Mellin convolution theorem, which is related to Perron's formula. Also 9 we introduce new explicit formulae for arithmetic function which generalize the explicit formulae of Weil for 10 other arithmetic functions different from the Von-Mangoldt function. 11 Keywords: Riemann-weil formula; perron formula; explicit formulae; Riemann zeros 12 MELLIN DISCRETE CONVOLUTION: 13 First of all we need to prove the following identity ci n 1 s an() f FsGsx () () (1) n1 xi2 ci an() Gs() 14 Where s is the Dirichlet generating functio of the coefficients a(n) and n1 n s 1 15 F()sdxfxx () is the Mellin transform of the function 0 16 17 The proof of (1) is easy and is obtained from the application of the Mellin convolution theorem 18 [3] applied to the integral linear operator 19 20 L g()) x dxf ( xt )() g t dt gt() an ( ) ( x n ) (2) 0 n1 21 22 The Mellin transform of the integral operator defined in (2) is a product of 2 Mellin transforms. 23 From the property of the Dirac delta functions used to define the distribution g(t) we find from 24 the Mellin convolution theorem that 25 26 dx f( xn )() a n xss1(1)1 F () s dxx g () x F () s G () s (3) 0 n1 0 27 1 28 Now if we take the change of variable x and take the inverse Mellin transforms on both x 29 sides of (3) we can prove (1) inmediatly 1 1 t>1 30 Now, if we set fHt(1) inside the test function (1)then t 0 t 1 31 we recover Perron's formula [6] for the Coefficients of the Dirichlet series x 1 ci xs 1 dx anH() 1 an () Gs () with Fs() s1 (2) nnx1 nis2 ci sx1 © 2017 by the author(s). Distributed under a Creative Commons CC BY license. Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 3 January 2018 doi:10.20944/preprints201801.0020.v1 2 of 5 32 But one of the best applications of our identity (1) is related to several Dirichlet an() Gs() 33 series (see [5]) in the form s , Where G(s) includes powers or quotients of the n1 n 34 Riemann zeta function for example 1() n '(sn ) ( ) (2sn ) ( ) s s s (3) ()snn1 ()snn1 ()snn1 ()sn | ()| (1)sn () s s (4) (2sn ) n1 ()snn1 35 The definitions of the functions inside (5) and (6) are as follows 36 1. The Möbius function, ()n 1 if the number ‘n’ is square-free (not divisible by an square) 37 with an even number of prime factors , ()n 0 if n is not squarefree and if the number 38 ‘n’ is square-free with an odd number of prime factors. 39 2. The Von Mangoldt function ()np log , in case ‘n’ is a prime or a prime power and 40 takes the value 0 otherwise ()n 41 3. The Liouville function ()n (1) ()n is the number of prime factors of the number 42 ‘n’ 43 4. |()| n is 1 if the number is square-free and 0 otherwise 1 44 5. ()nn 1 , the meaning of p | n is that the product is taken only over the pn| p 45 primes p that divide ‘n’. 46 To obtain the coefficients of the Dirichlet series we can use the Perron formula an() Ax () 1 ci xs Gs() s Ax() an () Gsds () (5) ss 1 n1 nx1 nx 2isci 47 If the function G(s) includes powers and quotients of the Riemann zeta function we can use 48 Cauchy’s theorem to obtain the explicit formulae ,see Baillie [2] by using the step function 1 1 xs 49 Hx(1) f inside (1) and evaluate the inverse Mellin transform inside (1) Fs() x 2isC 50 we find the well-known identities 2n xx Mx() () n 2 (6) nx ''(2)(2) n1 nn '(0) xx 2n ()xnx () (7) nx(0) n1 ( 2n ) xx(2 ) Lx() () n 1 (8) nx 1/2 ' Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 3 January 2018 doi:10.20944/preprints201801.0020.v1 3 of 5 2 x n 6()x 2 xn Qx() | ()|1 n (11) 2 nx ' n1 (2)nn '(2) 13xx22 ( 1) x n (21) n ()xn () 2 (12) nx6'(2)'(2) n1 nn 51 Under the assumption that all the Riemann Non-trivial zeros are simple.Also we have for the 52 Riemann zeta function and its derivatives (1)(2n nn 1)(2)! 1 1 '( 2n ) '(0) log 2 (0) (13) 21nn 2 2 2 2 53 The reader will remember the relation between Perron's formula and our discrete convolution , 54 using the work of Baillie [2] we will give different explicit formulae, to do so we use Cauchy's theorem 55 on complex integration and we also evaluate the closed mellin inverse transform inside (1) 56 1 s 57 Using the residue theorem F()sGsx () where 'C' is a closed circuit including all the 2i C 58 poles of the Dirichlet series G(s) , if we apply the Residue theorem inside (1) the contribution to the 59 closed integral (Mellin transform) come from the Poles 60 61 The pole of the Riemann Zeta function at s 1 62 The Riemann non-trivial zeros on the critical strip 0Re()1s 63 The Trivial zeros of the Riemann zeta (2)n 0 64 Poles of the Mellin transform (if any) 65 66 This is what we have done insidethe expression (8-12) but we can generalize these results to 67 different test functions to compute more sums. 68 69 Then we can obtain some interesting and useful new formulae , which we think they have never 70 been published (at least all of them) n 1 ()nf xF (1) xF ( ) F (2) n 2n (14) nn11x x nFFn() (2)1 ()nf x 2n (15) nn11x '( ) '( 2nx ) nx 1(2)() F ()nf F x n 1 x 1 2'() (16) 2 2 nFFnn6(1)()(2)(21) ()nf F (2) x2 x 22 n (9) n1 x '( )n1 xn '( 2 ) Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 3 January 2018 doi:10.20944/preprints201801.0020.v1 4 of 5 F nFnn6()()2 22 (18) |()|nf Fx (1) x 2 n n1 x 2'()n1 xn 2'(2) s 71 If the Mellin transform has poles inside the closed circuit 'C' FsGsx() () , then this poles C 72 will contribute with a remainder term due to the Residue theorem [1] in this case we have the extra 73 term s k1 rx() Re sFsGsx () () with Fk() dxfxx () (19) k sk 0 74 This is what happens in Perron's formula , due to the step function Hx(1) in this case there 1 75 is a pole at s 0 since Fs() this is why in formulae (8-12) there is a constant term. s 76 77 These formulae (14-18) may be regarded as a generalization of the Riemann-Weil formula 78 ()nirih 1 ' 1 (0) gn(log ) dr h log h k (20) n1 n 44222 k0 79 But we have used the Mellin integral transform , instead of the Fourier integral transform to 80 relate some new sums over primes and over Riemann Zeros 81 An easier derivation of our explicit formulae 82 There is an easier derivation for our explicit formulae, in general after Perron's formula is 83 applied we find the following identity dnr 2 aPQxhxnn () cx2 (21) nx n1 84 For some real constants PQdc,,,2n , r and a function h() ,which includes the Riemann zeta 85 function and its first derivative. 86 Taking the distributional derivative for an step function of the form an nx d dnr11 21 aaxndQxhxcnrxnn() () 2 n (2) (22) dx nx n11 n t 87 So if we apply a certain test function with a parameter 'x' f and its Mellin transform x 88 defined by Preprints (www.preprints.org) | NOT PEER-REVIEWED | Posted: 3 January 2018 doi:10.20944/preprints201801.0020.v1 5 of 5 t ss11 ss dt f t x dt f t t x F() s (23) 00x 89 Then we find the desired explicit formula n afnn dQFd() h () F () c2 (2 nrF )(2 nr ) (24) nn11x d 90 to evaluate the derivative of the Step function an we can use the identities (8-12) these dx nx 91 are defined and derived in Baillie [2] assuming that ALL the Non-trivial zeros of the Riemann Zeta 92 function are simple 93 94 A similar method can be applied to derive the Poisson summation formula, let be the Floor 95 functionx , then we have a formula valid on the whole real line 11 sin 2 nx xx (25) 2 n1 n 96 Taking the distributional derivative of (32) and using the Euler's formula for the cosine function d eeix ix x ()12cos(2)xn nxcos(x ) (26) dx nn 1 2 97 Now if we use a test function inside (32) we have the Poisson summation formula 2inx fn() fx () e (27) nn 98 REFERENCES: 99 [1] Apostol Tom “Introduction to Analytic Number theory” ED: Springuer-Verlag,(1976) 100 [2] Baillie R.