A remark on a theorem of A. E. Ingham.

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A remark on a theorem of A. E. Ingham. K G Bhat, K Ramachandra To cite this version: K G Bhat, K Ramachandra. A remark on a theorem of A. E. Ingham.. Hardy-Ramanujan Journal, Hardy-Ramanujan Society, 2006, 29, pp.37-43. <hal-01111487> HAL Id: hal-01111487 https://hal.archives-ouvertes.fr/hal-01111487 Submitted on 2 Feb 2015 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not. The documents may come from teaching and research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

Hardy-Ramanujan Journal Vol.29 (2006) 37-43 A REMARK ON A THEOREM OF A.E.INGHAM BY K.G.BHAT and K.RAMACHANDRA ABSTRACT. The theorem of INGHAM [AEI] 1 refered to is: For all N N 0 (an absolute constant) the inequality N 3 p (N + 1) 3 is solvable in a prime p. (It may be noted that the corresponding theorem for squares is an open question even if we assume RIEMANN HYPOTHESIS). Actually INGHAM proved more, namely π(x + h) π(x) h(log x) 1 where h = x c, where c(> 5 ) is any constant. The purpose of this note is 8 to point out that even this stronger form can be proved without using the functional equation of ζ(s). 1.INTRODUCTION. The three main ingredients in the proof of ING- HAM S theorems are (A) I.M.VINOGRADOV s deep result ζ(s) 0 (s = σ + it), σ 1 K 1 (log t) 2 3 (log log t) 1 3, (1) t 100, where K 1 > 0 is an absol]ute constant. (B) Explicit formula for log p ([AEI] 2 ). (2) p x

38 A Remark on A Theorem of A. E. Ingham (C) N(σ, T ) < T ( 8 3 )(1 σ) (log T ) 100 (3) where 1 σ 1, T 1000. The precise power of log T is unimportant. Any 2 constant in place of 100 will do. N(σ, T ) denotes the number of zeros β + iγ of ζ(s) with β σ and γ T. 1) The toughest part is (A). It follows from the deep result (due to I.M.VINOGRADOV) ζ(σ + it) (t (1 σ) 3 2 log t) K 2 ( 1 2 σ 1, t 100) (4) where K 2 > 0 is an absolute constant and (1) follows from this in a relatively simple way by a method due to E.LANDAU[KR] 1. For a proof of (4) without using the functional equation see [KR,AS]. 2) Explicit formula uses the functional equation, but an alternative approach is due to [KR] 2 by the introduction of HOOLEY-HUXLEY contour. 3) The proof of (3) uses ζ( 1 2 + it) = O(t 1 6 log t), t 100, (5) where the O-constant is absolute. The main work in the present note is to sketch a proof of this without using the functional equation of ζ(s). 2. SOME REMARKS In fact we write ζ(s, α) = (n + α) s (0 < α 1, s = σ + it, σ > 1), (6) n=0 and next if X is any positive integer we have ζ(s, α) = α s Since the last term in (7) is X + α) n=1(n s + ((n + α) s n>x + X+1 n+1 n du (u + α) s ) (u + α) s du. (7) (X + 1 + α) 1 s s 1 (8)

K.G. Bhat and K. Ramachadra 39 and the rest is analytic in σ > 0, (7) gives the analytic continuation in σ > 0 of (6). We prove our main theorem which is as follows. THEOREM. We have ζ( 1 2 + it, α) α 1 2 it = O(t 1 6 log t), (t 10) (9) uniformly in the real parameter α. (Note that ζ(s, 1) = ζ(s)). 3. PROOF OF THE THEOREM We use van-der Corput s theorems (Theorems 5.9 and 5.11 of [ECT]) and after the proof of the theorem we make some comments about the Weyl-Hardy-Littlewood method of proof of (9). THEOREM 5.9. If f(x) is real and twice continuously differentiable and 0 < λ 2 f (x) hλ 2 (or 0 < λ 2 f (x) hλ 2 ) throughout the interval (a, b) and b a + 1, then a<n b e 2πif(n) = O(h(b a)λ 1 2 2 ) + O(λ 1 2 2 ). (10) THEOREM 5.11. If f(x) is real and thrice continuously differentiable and 0 < λ 3 f (x) hλ 3 ( or 0 < λ 3 f (x) hλ 3 ) throughout the interval (a,b) and b a + 1, then a<n b e 2πf(n) = O(h 1 2 (b a)λ 1 6 3 ) + O((b a) 1 2 λ 1 6 3 ). (11) We now apply these to E a n b( 2a) (n + α) it with a 10. (12) Here f(x) = t log(x + α). We have 2π f t (x) = 2π(x + α)

40 A Remark on A Theorem of A. E. Ingham Thus f (x) = t 2π(x + α) 2 f (x) = and 2t 2π(x + α). 3 C 1 f (x)a 2 t 1 C 2 and C 3 f (x)a 3 t 1 C 4 (13) where C 1, C 2, C 3 and C 4 are absolute positive constants. Thus we have (n + α) it = O(t 1 2 ) + O(at 1 2 ) (14) Hence by partial summation we have (n + α) it = O(t 1 6 a 1 2 ) + O(t 1 6 a). (15) (n + α) 1 2 it = O(( t a ) 1 a 2 ) + O(( t ) 1 2 ) (16) and t (n + α) 3 2 it = O(( t a ) 3 2 ) + O(( t a ) 1 2 ). (17) Also we need (n + α) 1 2 it = O(t 1 6 ) + O(t 1 6 a 1 2 ) (18) which follows from (15). From (18) there follows (n + α) 1 2 it = O(t 1 6 log t). (19) 1 n t 2 3

K.G. Bhat and K. Ramachadra 41 From (16) there follows (n + α) 1 2 it = O(t 1 6 log t) (20) t 2 3 n t 4 3 Thus 1 n t 4 3 (n + α) 1 2 it = O(t 1 6 log t). (21) We now fix X = [t 4 3 ]. The term (8) contributes O(t 1 3 ). We note that (with s = 1 2 + it) n>x = n>x = n>x s and so its absolute value is ( n+1 (n + α) s O( a>t 4 3 n+1 n 1 This proves our main theorem. 0 n ) du (u + α) s ( (n + α) s (u + α) s) du ( u ) (n + v + α) s 1 dv du 0 ( ( t ) a ) 3 t 2 + ( a ) 1 2 = O(t 1 6 ). REMARK 1 Let X be an arbitrary positive integer 20( t + 20)(K + 1). Then by iteration of the method by which we continued ζ(s, α) in σ > 0 (incidentally the method is due to E.LANDAU (Handbuch der primzahlen) we can get the analytic continuation in σ (K + 1) (K being arbitrary constant) and also the inequality ζ(s, α) = α s + n X (n + α) s + X1 s s 1 + O(X σ ) where s = σ + it (σ arbitrary). (O constant depends on K). For this see [KR] 1. REMARK 2 A remark on Weyl-Hardy-Littlewood method is necessary here. The proof of Theorem 5.5 of [ECT] goes through to prove

42 A Remark on A Theorem of A. E. Ingham 1 n t 2 3 (n + α) 1 2 it = O(t 1 6 L) except for trival complications arising from the presence of the real parameter α. This uses the integer parameter k to be 2. However if we use the case k = 1 simple computations show that (n + α) 1 2 it = O(t 1 6 L) t 2 3 n Ct whatever the constant C 10 be. Here L is some fixed power of log t. These considerations prove the main theorem in view of Remark 1 above. We stress once again that functional equations for ζ(s) or ζ(s, α) are not necessary in the proof of INGHAM s theorems. L can be any fixed power of log t and this is enough to prove INGHAM s asymptotic formula mentioned in the abstract.

K.G. Bhat and K. Ramachadra 43 REFERENCES [AEI] 1 A.E.INGHAM, On the difference between consecutive primes,, Quart. J. Oxford(1937), 255-266. [AEI] 2 A.E.INGHAM, The distribution of prime numbers, Stechert-Hafner Service Agency, New York and London, (1964). [KR] 1 K.RAMACHANDRA, Riemann zeta-function, Publ. RAMANUJAN INSTITUTE (1979). [KR] 2 K.RAMACHANDRA, Some problems of Analytic Number Theory, Acta Arith, Vol.31(1976), 313-324. [KR,AS] K.RAMACHANDRA and A.SANKARANARAYANAN, A remark on Vinogradov s Mean Value Theorem, The J. of Analysis, 3(1995), 111-129. [ECT] E.C.TITCHMARSH, The Theory of the Riemann Zeta-function (second edition, revised and edited by D.R.HEATH-BROWN) Clarendon Press, Oxford (1986). K.G.BHAT, N I A S, IISc CAMPUS BANGALORE-560012 Email: kishor@nias.iisc.ernet.in K.RAMACHANDRA N I A S, IISc CAMPUS BANGALORE-560012 Also TIFR CENTRE P.O.BOX 1234, IISc CAMPUS BANGALORE-560012 Email: kram@math.tifrbng.res.in ADDRESS OF THE AUTHORS Accepted on 17-07-2006