aa r X i v : . [ m a t h . G M ] J a n On zeros of the Riemann zeta function
Xiaolong WuEx. Institute of Mathematics, Chinese Academy of [email protected] 20, 2021
Abstract
This paper shows that, in the critical strip, the Riemann zeta func-tion ζ ( s ) have the same set of zeros as F ( s ) := R ∞ t s − ( e t + 1) − dt ,and then discusses the behavior of F ( s ). Introduction
Define ζ ( s ) as the Riemann zeta function, and define F ( s ) := Z ∞ t s − e t + 1 dt, s ∈ C . (1)Define the open critical strip by S := { s ∈ C : 0 < R ( s ) < } (2)where R ( s ) is the real part of s.Theorem 1 shows that ζ ( s ) and F ( s ) have the same set of zeros in S . Sowe can study the zeros of F ( s ) instead of ζ ( s ), and F ( s ) looks simpler.Write s = a + ib . It is obvious that F ( s ) = Z ∞ t a − cos( b log t ) e t + 1 dt + i Z ∞ t a − sin( b log t ) e t + 1 dt =: F ( a, b )+ iF ( a, b ) . (3)1 iaolong Wu F ( a, b ) = 0 means F ( s ) = 0. This paper will work on F . We split theintegration range of F ( a, b ) in to two sub-ranges [0 , R ] and [ R, ∞ ) for somereal R = e Kπ with integer K ≥ , ≤ R < π .Define f ( t, a ) := t a − e t + 1 , g ( t ) := 1 e t + 1 . (4)Theorem 2 shows that Z R f ( t, a ) sin( b log t ) dt = − ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a . (5)Theorems 3 and 4 discuss properties of g ( n ) (0), especially, its relation withthe Bernoulli numbers.Theorem 5 shows that Z R f ( t, a ) sin( b log t ) dt > R a b ( e R + 1) − . R a b . (6)Theorems 6 to 10 study the behavior of Z ∞ R f ( t, a ) sin( b log t ) dt. In particular, Theorem 6 shows Z ∞ R f ( t, a ) sin( b log t ) dt = ∞ X k = K Z exp((2 k +1) π/b exp(2 kπ/b ) h ( t, a, b ) sin( b log t ) dt. where h ( t, a, b ) := t a − e t + 1 − t a − e aπ/b e te π/b + 1 . (7)Theorem 7 shows, for c = e π/b , Z ∞ R h ( t, a, b ) dt = Z cRR t a − e t + 1 dt. (8)Theorem 10 shows, for some range of t, a, b, h ( t, a, b ) > . (9)Write the gamma function asΓ( s ) := Z ∞ t s − e t dt. (10) iaolong Wu Main Content
Theorem 1.
For s ∈ S , ζ ( s ) = 0 if and only if F ( s ) = 0 .Proof. It is well known that, ( [Titchmarsh 1986] formula 2.4.1 ), ζ ( s ) = G ( s )Γ( s ) , (1.1)where G ( s ) := Z ∞ t s − e t − dt. (1.2)Since Γ( s ) has no zero nor pole in S , ζ ( s ) has the same set of zeros as G ( s ) in S . It is easy to see, write u = 2 t , G ( s ) − F ( s ) = Z ∞ t s − e t − dt − Z ∞ t s − e t + 1 dt = Z ∞ t s − e t − dt = Z ∞ u/ s − e u − d u
2= 2 − s Z ∞ u s − e u − du = 2 − s G ( s ) . (1.3)This means (1 − − s ) G ( s ) = F ( s ) . (1.4)Since 1 − − s has no zero nor pole in S , F ( s ) has the same set of zeros as G ( s ), hence as ζ ( s ) in S . Theorem 2.
Let b ≥ be a real, K > be an integer, and R =exp(2 Kπ/b ) such that ≤ R < π . Let f ( t, a ) and g ( t ) be defined as in (4).Then Z R f ( t, a ) sin( b log t ) dt = − ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a . (2.1) Proof.
Since g ( t ) = ( e t + 1) − has a pole at t = πi , the radius of convergenceof series g ( t ) = ∞ X n =0 g ( n ) (0) n ! t n (2.2) iaolong Wu π . We have Z R t s − e t + 1 dt = Z R t s − ∞ X n =0 g ( n ) (0) n ! t n dt = ∞ X n =0 g ( n ) (0) n ! Z R t n + s − dt = ∞ X n =0 g ( n ) (0) n !( n + s ) R n + s = ∞ X n =0 g ( n ) (0)( n + a − ib ) n !(( n + a ) + b ) R n + a e ib log R = ∞ X n =0 g ( n ) (0)( n + a ) n !(( n + a ) + b ) R n + a − i ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a . (2.3)On the other hand, we have Z R t s − e t + 1 dt = Z R t a − bi e t + 1 dt = Z R t a − e ib log t e t + 1 dt = Z R t a − e t + 1 cos( b log t ) dt + i Z R t a − e t + 1 sin( b log t ) dt (2.4)Equate the imaginary part of (2.3) and (2.4), we get (2.1). Theorem 3.
Let g ( t ) = ( e t + 1) − as in (4), then g ( n ) (0) = 1 − n +1 n + 1 B n +1 , (3.1) where B n is the n-th Bernoulli number.Proof. The Bernoulli numbers have a generating function: te t − ∞ X n =0 B n n ! t n . (3.2)On the other hand, we have te t + 1 = ∞ X n =0 g ( n ) (0) n ! t n +1 . (3.3) iaolong Wu te t − − te t + 1 = 2 te t − ∞ X n =0 B n n ! (2 t ) n , (3.4)we have ∞ X n =0 g ( n ) (0) n ! t n +1 = te t + 1= te t − − te t − ∞ X n =0 B n n ! t n − ∞ X n =0 B n n ! (2 t ) n = ∞ X n =0 B n n ! (1 − n ) t n (3.5)Equate coefficients of t n +1 , we get g ( n ) (0) n ! = B n +1 ( n + 1)! (1 − n +1 ) (3.6) g ( n ) (0) = 1 − n +1 n + 1 B n +1 . (3.7) Theorem 4. (1) g (2 m ) (0) = 0 , ∀ m ≥ .(2) g (4 m +1) (0) < , ∀ m ≥ .(3) g (4 m − (0) > , ∀ m ≥ .(4) g (2 m − (0)(2 m − − m (1 − − m ) ζ (2 m ) 2 π m , ∀ m ≥ . (4.1) (5) − g (4 m +1) (0)(4 m + 1)! π < g (4 m − (0)(4 m − < − . g (4 m +1) (0)(4 m + 1)! π , ∀ m ≥ . (4.2) (6) π m < | g (2 m − (0) | (2 m − − − m ) ζ (2 m ) 2 π m , ∀ m ≥ . (4.3) (7) g ( t ) = 12 −
12 tanh t . (4.4) iaolong Wu Proof. (1), (2) and (3) easily follow from well-known properties of theBernoulli numbers.(4) It is well known that ζ (2 m ) = ( − m +1 (2 π ) m B m m )! , ∀ m ≥ . (4.5)Hence by Theorem 3, we have g (2 m − (0) = 1 − m m B m = 1 − m m · ( − m +1 m )! ζ (2 m )(2 π ) m = ( − m − − m )(2 m − ζ (2 m ) π m , ∀ m ≥ . (4.6)(5) By equation (4.1) we have g (4 m − (0) / (4 m − − g (4 m +1) (0) / (4 m + 1)! = (1 − − m ) ζ (4 m )2 /π m (1 − − m − ) ζ (4 m + 2)2 /π m +2 = (1 − − m ) ζ (4 m )(1 − − m − ) ζ (4 m + 2) π , ∀ m ≥ . (4.7)The last fraction in (4.7) is a decreasing function in m and approaches 1when m → ∞ . Hence, we have1 < (1 − − m ) ζ (4 m )(1 − − m − ) ζ (4 m + 2) < (1 − − ) ζ (8)(1 − − ) ζ (10) < . , ∀ m ≥ . (4.8)(6) By [Ge 2012] Theorem 1.1,2(2 m )! π m (2 m − < | B m | = ζ (2 m ) 2(2 m )!(2 π ) m . (4.9)Substitute B m = 2 mg (2 m − (0) / (1 − m ), we get2(2 m )! π m (2 m − < (cid:12)(cid:12)(cid:12)(cid:12) mg (2 m − (0)1 − m (cid:12)(cid:12)(cid:12)(cid:12) = ζ (2 m ) 2(2 m )!(2 π ) m . (4.10)2 π m < (cid:12)(cid:12)(cid:12)(cid:12) g (2 m − (0)(2 m − (cid:12)(cid:12)(cid:12)(cid:12) = (1 − − m ) ζ (2 m ) 2 π m . (4.11)(7) 1 e t + 1 −
12 = − · e t − e t + 1 = −
12 tanh t . (4.12) iaolong Wu Theorem 5.
Let
K > be the largest integer such that R = e Kπ/b ≤ . Let a ≤ . and b ≥ . Then − ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a > − R a b ( e R + 1) − . R a b . (5.1) Proof.
By maximality of K, we have Re π/b = e Kπ/b e π/b = e K +1) /b > . (5.2)Hence R > e − π/b > (cid:18) − πb (cid:19) = 2 − πb . (5.3) − ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a + R a b ( e R + 1)= − ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a + R a b ∞ X n =0 g ( n ) (0) n ! R n = R a b ∞ X n =0 g ( n ) (0) R n n ! (cid:18) − b ( n + a ) + b + 1 (cid:19) = R a b ∞ X n =0 g ( n ) (0) R n n ! (cid:18) ( n + a ) ( n + a ) + b (cid:19) = R a b (cid:18) a a + b − (1 + a ) R a ) + b ) + (3 + a ) R a ) + b ) − (5 + a ) R a ) + b ) (cid:19) + R a b ∞ X n =7 g ( n ) (0) R n n ! (cid:18) ( n + a ) ( n + a ) + b (cid:19) > R a b − . b + ∞ X m =2 c m ! , (5.4)where c m := g (4 m − (0) R m − (4 m − (cid:18) (4 m − a ) (4 m − a ) + b (cid:19) + g (4 m +1) (0) R m +1 (4 m + 1)! (cid:18) (4 m + 1 + a ) (4 m + 1 + a ) + b (cid:19) . (5.5) iaolong Wu g (4 m − (0) > , g (4 m +1) (0) and g (4 m − (0)(4 m − > − g (4 m +1) (0) π (4 m + 1)! , (5.6) − g (4 m +1) (0)(4 m + 1)! > π m +2 . (5.7)Hence c m > − g (4 m +1) (0) R m − (4 m + 1)! (cid:18) (4 m − a ) (4 m − a ) + b − (4 m + 1 + a ) R (4 m + 1 + a ) + b (cid:19) > R m − π m +2 (cid:18) (4 m + 1 + a ) (4 m + 1 + a ) + b (cid:18) π − R (cid:19)(cid:19) = 2 Rb (cid:18) Rπ (cid:19) m (cid:18) − R π (cid:19) (cid:18) (4 m + 1 + a ) (4 m + 1 + a ) b − + 1 (cid:19) (5.8)Then ∞ X m =0 c m > Rb (cid:18) − R π (cid:19) ∞ X m =2 (cid:18) Rπ (cid:19) m (cid:18) (4 m + 1 + a ) (4 m + 1 + a ) b − + 1 (cid:19) > . b ∞ X m =2 (cid:18) Rπ (cid:19) m (cid:18) b − + 1 (cid:19) > . b ∞ X m =2 (cid:18) Rπ (cid:19) m = 16 . b · ( R/π ) − ( R/pi ) > . b (5.9)Hence − ∞ X n =0 g ( n ) (0) bn !(( n + a ) + b ) R n + a + R a b ( e R + 1) > R a b (cid:18) − . b + 0 . b (cid:19) > − . R a b . (5.10) Remark . Theorem 5 chooses R ≈ iaolong Wu Theorem 6.
Let a, b be reals such that < a < , b ≥ . Let K > bethe largest integer such that R = e Kπ/b ≤ . Then Z ∞ R f ( t, a ) sin( b log t ) dt = ∞ X k = K Z exp((2 k +1) π/b exp(2 kπ/b ) h ( t, a, b ) sin( b log t ) dt. (6.1) Proof.
Write Z ∞ R f ( t, a ) sin( b log t ) dt = ∞ X k = K I k , (6.2)where I k := Z exp((2( k +1) π/b )exp(2 kπ/b ) f ( t, a ) sin( b log t ) dt. (6.3)Then split the range of I k in to two sub-intervals according to the sign ofsin( b log t ). That is I k = I k, + I k, , (6.4)where I k, := Z exp(((2 k +1) π/b )exp(2 kπ/b ) f ( t, a ) sin( b log t ) dt, (6.5) I k, := Z exp((2( k +1) π/b )exp((2 k +1) π/b ) f ( t, a ) sin( b log t ) dt. (6.6)By changing variable u = te − π/b in I k, , we get I k, = Z exp((2 k +1) π/b )exp(2 kπ/b ) f ( ue π/b , a ) sin( b log( ue π/b )) d ( ue π/b )= e π/b Z exp((2 k +1) π/b )exp(2 kπ/b ) f ( ue π/b , a ) sin( b log u + π ) du = − e π/b Z exp((2 k +1) π/b )exp(2 kπ/b ) f ( ue π/b , a ) sin( b log u ) du. (6.7)Hence I k = Z exp((2 k +1) π/b )exp(2 kπ/b ) ( f ( t, a ) − e π/b f ( te π/b , a ) sin( b log t ) dt = Z exp((2 k +1) π/b )exp(2 kπ/b ) h ( t, a, b ) sin( b log t ) dt (6.8) iaolong Wu Theorem 7.
Let a, b, R be reals such that < a < , b > , R ≥ . Let c := e π/b . Then Z ∞ R h ( t, a, b ) dt = Z cRR t a − e t + 1 dt. (7.1)( c − c a − R a e cR + 1 < Z ∞ R h ( t, a, b ) dt < ( c − R a e R + 1 . (7.2) Proof.
Let 0 ≤ α < β be reals, u = ct, t = c − u . Then Z βα t a − e ct + 1 dt = Z cβcα c − a u a − e u + 1 d ( c − u ) = c − a Z cβcα u a − e u + 1 du. (7.3)Hence Z ∞ R h ( t, a, b ) dt = Z ∞ R (cid:18) t a − e t + 1 − t a − e aπ/b e te π/b + 1 (cid:19) dt = Z ∞ R t a − e t + 1 dt − c a Z ∞ R t a − e ct + 1 dt = Z ∞ R t a − e t + 1 dt − c a (cid:18) c − a Z ∞ cR t a − e t + 1 dt (cid:19) = Z ∞ R t a − e t + 1 dt − Z ∞ cR t a − e t + 1 dt = Z cRR t a − e t + 1 dt (7.4)Since t a − / ( e t + 1) is decreasing in t, we have Z cRR t a − e t + 1 dt > ( cR − R ) ( cR ) a − e cR + 1 = ( c − c a − R a e cR + 1 , (7.5) Z cRR t a − e t + 1 dt > ( cR − R ) R a − e R + 1 = ( c − R a e R + 1 . (7.6) Theorem 8.
Let k be an integer, b ≥ be a real. Define the average A := 1 e (2 k +1) π/b − e kπ/b Z exp ((2 k +1) π/b ) exp (2 kπ/b ) sin( b log t ) dt. (8.1) Then π − πb < A < π . (8.2) iaolong Wu Proof.
Substitute u = b log t, t = e u/b , we get Z exp ((2 k +1) π/b ) exp (2 kπ/b ) sin( b log t ) dt = 1 b Z (2 k +1) π/b kπ/b e u/b sin udu = e u/b b ( b − + 1) (cid:20) b sin u − cos u (cid:21) (2 k +1) π kπ = e (2 k +1) π/b + e kπ/b b ( b − + 1) . (8.3)So the average is A = e (2 k +1) π/b + e kπ/b b ( b − + 1)( e (2 k +1) π/b − e kπ/b ) = 1 + e − π/b b ( b − + 1)(1 − e − π/b ) . (8.4)Since1 + e − π/b − bπ (1 − e − π/b ) = 1 + ∞ X n =0 ( − π ) n n ! b n + 2 bπ ∞ X n =1 ( − π ) n n ! b n = 2 + ∞ X n =1 ( − π ) n n ! b n − ∞ X n =1 ( − n +1 π n ( n + 1)! b n = ∞ X n =1 ( − π ) n b n (cid:18) n ! − n + 1)! (cid:19) = ∞ X n =1 ( − π ) n b n · n − n + 1)!= π b − · · · , (8.5)we have 2 π < e − π/b b (1 − e − π/b ) < π + π b b ( π/b − ( π/b ) / π + π b (1 − π/ (2 b )) ≤ π + π b (1 − π/ < π + π b (5 /
6) = 2 π + π b , ∀ b ≥ . (8.6) iaolong Wu A < b − + 1 (cid:18) π + π b (cid:19) < (cid:18) − b + 1 b (cid:19) (cid:18) π + π b (cid:19) = 2 π + π b − πb − π b + 2 πb + π b = 2 π − . b + 0 . b + 0 . b < π , (8.7) A > b − + 1 (cid:18) π (cid:19) > π (cid:18) − b (cid:19) . (8.8) Theorem 9.
Let k be an integer, b ≥ be a real. Write t = exp(2 kπ/b ) , t = exp((2 k + 1) π/b ) . Let h ( t ) be a continuous function and M = min t ≤ t ≤ t h ( t ) , M = max t ≤ t ≤ t h ( t ) . (9.1) Then (cid:18) π − πb (cid:19) Z t t M dt < Z t t h ( t ) sin( b log t ) dt < π Z t t M dt. (9.2) Proof.
Since sin( b log t ) ≥ t , t ], we have, by Theorem 8, Z t t h ( t ) sin( b log t ) dt ≥ Z t t M sin( b log t ) dt> M (cid:18) π − πb (cid:19) ( t − t )= (cid:18) π − πb (cid:19) Z t t M dt. (9.3)The other inequality can be proved similarly. Theorem 10.
Let t, a, b be reals such that t ≥ , < a ≤ e/ ( e + 1) =0 . · · · and b > . We have h ( t, a, b ) > . (10.1) iaolong Wu Proof.
We need only to prove that1 e t + 1 − e aπ/b e te π/b + 1 > . (10.2)1 e t + 1 − e aπ/b e te π/b + 1 = e te π/b + 1 − e t + aπ/b − e aπ/b ( e t + 1)( e te π/b + 1) > e t (1+ π/b ) + 1 − e t + aπ/b − e aπ/b ( e t + 1)( e te π/b + 1)= e t ( e tπ/b − e aπ/b ) − ( e aπ/b − e t + 1)( e te π/b + 1) > e t ( tπ/b − aπ/b ) − aπ/b ( e t + 1)( e te π/b + 1)= πb · te t − ae t − a ( e t + 1)( e te π/b + 1) (10.3)Since te t − ae t − a > t ≥ , a ≤ e/ ( e + 1) = 0 . · · · , (10.2) holds. References [Ge 2012] Hua-feng Ge.
New Sharp Bounds for the Bernoulli Numbers andRefinement of Becker-Stark Inequalities . Journal of Applied MathematicsVol 2012, (2012)[Titchmarsh 1986] E.C. Titchmarsh and D.R. Heath Brown.
The Theory ofthe Riemann Zeta-function . Oxford University, 2nd edition (1986)
Appendix g ( n ) (0) for n ≤ g (0) (0) = 12 , g (1) (0) = − , g (3) (0) = 18 ,g (5) (0) = − g (7) (0) = 1716 , g (9) (0) = − ,g (11) (0) = 6918 , g (13) (0) = − , g (15) (0) = 92956932(0) = 92956932