Riemann denoted u ( t ) by ξ ( t ) in his paper. He pointed out, see [4] (pp. 301-302).

“If one denotes by α the roots of the equation ξ ( α ) = 0 , then one can express log ξ ( t ) as

∑ log ( 1 − t 2 α 2 ) + log ξ ( 0 ) , (R)

because, since the density of roots of size t grows only like log ( t / 2 π ) as t grows, this expression converges and for infinite t is only infinite like t log t ; Thus it differs from log ξ ( t ) by a function of t² which is continuous and finite for finite t and which, when divided by t², is infinitely small for infinite t. This difference is therefore a constant, the value of which can be determined by setting t = 0 .”

Because Riemann did not use α ¯ , we must regard α to be real roots, this is extremely important! But then it is misunderstood.

We now explain his analysis. On critical line u ( t ) is an even entire function and has infinite number of zeros ± t j . We consider formally the remainder of ln u ( t ) ,

∑ j = n + 1 ∞ ln ( 1 − t 2 / t j 2 ) ≈ − ∑ j = n + 1 ∞ t 2 t j 2 = − t 2 γ n ,       γ n = ∑ j = n + 1 ∞ 1 t j 2

One knows t n = O ( n / ln n ) (We have better t n = B ( n ) 2 π n / ln n , B ( n ) = 1.46 ↓ 1 , n ≥ 10 3 ). Thus

γ n = C ∑ j = n + 1 ∞ ln 2 n ( B ( n ) n ) 2 ≈ C ∫ n ∞ ln 2 x ( B ( x ) x ) 2 d x ≈ C ln 2 n B ( n ) n ,     C = ( 2 π ) − 2 . (2.1)

This series (R) converges for finite t. (Riemann said) u ( t ) has the growth ln u ( t ) = O ( t ln t ) (This expression is not suitable. We shall use ln | u ( t ) | , which admits u ( t ) = 0, see (2.6)). The series (R) differs from ln u ( t ) by a function of t², which, when divided by t², is infinitely small for infinite t. (Riemann said) “This difference is therefore a constant.” This is not rigorous, in general, it should be O ( ln t ) rather than a constant. There is a hole of the uniqueness.

Hadamard in 1893 proved product formula for general entire function, see [4] (p. 20), [5] (p. 16). Denoting τ = β + i t = i z , z = t − i β , the zeros of ξ are conjugate, one gets

ξ ( i z ) = e A + B z ∏ j ( 1 − z 2 z j 2 ) ,     z = t − i β ,     z j = t j − i β j .

which is even if β = 0 , then B = 0 . Taking z = 0 , then e A = ξ ( 0 ) .

But this Hadamard’s way from β ≠ 0 to β = 0 implies a serious contradiction. If all zeros are real, then which itself assumes RC. If conjugate complex zeros are admitted, which just denies RC. Hadamard’s theorem was referred by Von Mangoldt “The first real progression in the field in 34 years”, see [4], p. 39. We think this is a misunderstanding. Actually, Hadamard’s way is independent of proving RC and far from Riemann’s thought. Our idea is to consider u ( t ) at β = 0 as an initial value. We shall directly prove RC2 by method of analysis, so this contradiction is cast off.

We consider the functional equation

ξ ( s ) = G ( s ) ζ ( s ) ,     G ( s ) = 1 2 s ( s − 1 ) π − s / 2 Γ ( s 2 ) ,     ξ ( s ) = ξ ( 1 − s ) . (2.2)

Firstly, Γ ( s / 2 ) has asymptotic expansion

Γ ( s / 2 ) = 2 π ( s / 2 ) s / 2 − 1 / 2 e − s / 2 ( 1 + O ( 1 / s ) ) .

Take logarithm and decompose the real part, c = ln ( 2 π ) ,

ln Γ ( s / 2 ) = c + ( s / 2 − 1 / 2 ) ln ( s / 2 ) − s / 2 + O ( 1 / s ) ,       s = σ + i t , = c + ( ( σ − 1 ) / 2 + i t / 2 ) { ln ( t / 2 ) + i ( π / 2 − arctan σ t ) } − ( σ + i t ) / 2 + O ( 1 / s ) ,

R e ln Γ = c + ( ( σ − 1 ) / 2 ) ln ( t / 2 ) − t π / 4 + ( t / 2 ) ( σ / t ) − σ / 2 + O ( 1 / t ) ,

Thus for σ = 1 / 2

| Γ ( s / 2 ) | = 2 π ( t 2 ) − 1 / 4 e − t π / 4 ( 1 + O ( 1 / t ) ) . (2.3)

Besides, s ( 1 − s ) / 2 = ( t 2 + 1 / 4 ) / 2 and | π − s / 2 | = π − 1 / 4 , we get

| G ( t ) | = 2 3 / 2 π 1 / 2 − 1 / 4 ( t 2 ) 7 / 4 e − t π / 4 ( 1 + O ( 1 / t ) ) . (2.4)

Secondly, there are growths of ζ ( s ) for large t, [4], p. 185, p. 200,

{ | ζ ( σ ) | ≤ C t ( 1 − σ ) / 2 ln t ,               0 ≤ σ ≤ 1, | ζ ( 1 / 2 + i t ) | ≤ C t λ , λ = 1 / 5     or     λ = 19 / 116 . (2.5)

We only need the estimate of σ = 1 / 2 , also see Remark 1.

Thus, we get an upper bound of growth (note: not ln | u | = O ( t ln t ) )

ln ( | u ( t ) | / | G ( t ) | ) ≤ 1 6 ln t + C ,     if     t ≥ T ≫ 0. (2.6)

So u ( t ) = 0 is admitted.

Finally, to prove RC2 (or RC), by contradiction. If ξ ( τ ) has conjugate complex roots t ′ ± i β ′ , β ′ > 0 , R 2 = β ′ 2 + t ′ 2 , by symmetry ξ ( τ ) = ξ ( − τ ) , then − ( t ′ ± i β ′ ) do yet. Denoting z = t − i β and using the equivalence, ξ ( τ ) ≡ u ( z ) must contain four factors

p ( z ) = ( 1 − z t ′ + i β ′ ) ( 1 − z t ′ − i β ′ ) ( 1 − z − t ′ + i β ′ ) ( 1 − z − t ′ − i β ′ ) .

Letting β = 0 , then u ( t ) must contain a real polynomial of fourth degree

p ( t ) = ( 1 − t t ′ + i β ′ ) ( 1 − t t ′ − i β ′ ) ( 1 + t t ′ − i β ′ ) ( 1 + t t ′ + i β ′ ) = ( 1 + t 2 − 2 t t ′ R 2 ) ( 1 + t 2 + 2 t t ′ R 2 ) = ( 1 − t 2 R 2 ) 2 + 4 t 2 β ′ 2 R 4 > 0 , (2.7)

and ln | u ( t ) | contains a term (as a lower bound)

ln p ( t ) ≥ 4 ln t + O ( 1 ) ,     t ≥ T ≫ 0. (2.8)

Its growth contradicts (2.6). Thus ξ ( τ ) can not have complex roots and u ( t ) does not have the factor p ( t ) . Therefore both RC and RC2 are rigorously proved. □

Originally we want to prove only RC2, fortunately, both RC and RC2 are proved.

Remark 1. In period of Riemann, no estimates (2.5), but it is possible to prove RC. As

ζ ( s ) = ∑ n = 1 ∞     n − s = ∑ n = 1 ∞   ∫ 0 1 ( n − s − ( n + x ) − s ) d x + ∫ 1 ∞     y − s d y = 1 s − 1 + ∑ n = 1 ∞     s ∫ 0 1     x − s − 1 ( 1 − x ) d x ,       by   integration   by   part

is already continued analytically to R e ( s ) > 0 (actually this is Euler’s method). Thus

| ζ ( s ) | ≤ | 1 s − 1 | + | s | ∫ 0 ∞     x − σ − 1 d x ≤ C t ,

and gets an coarse estimate (e.g. | u ( t ) | ≤ C t 3 e − t π / 4 in [5], p. 27)

ln ( | u ( t ) | / | G ( t ) | ) ≤ ln t + O ( 1 ) . (2.9)

By (2.8) one can still prove that ξ ( τ ) does not have complex roots. But nobody noted it. Therefore I feel, Riemann had already approached to prove RC. Our proof is completed to follow Riemann’s thought.

Numerical experiments 1. Using the data of the first 10⁵ zeros in Odlyzko [6], we have computed w n ( t ) = u ( 0 ) ∏ j = 1 n ( 1 − t 2 / t j 2 ) and u ( t ) in (1.2) for t ∈ [ 0,50 ] . We draw these curves by variable scale u / M , here M ( t ) = ( t / 2 + 1 ) 23 / 12 e − t π / 4 , M ( 0 ) = 1 . We see in Figure 1 that w n ( t ) approximates u ( t ) very well, of course, larger is t, then larger is its deviation.

We have for the first time computed w n and Figure 1, which make us believe the correctness of RC2, and then consider its analytic proof as before.

Consider a root-interval I j = [ t j , t j + 1 ] of u ( t , β ) , obviously t j and t j + 1 depend on β . Assume u ( t , β ) > 0 inside I j for β ∈ ( 0, 1 / 2 ] . At the left end t j , u ( t j , β ) = 0 and u t ( t j , β ) > 0 , then the slop u t ( t j , r ) > 0 , r ∈ ( 0 , β ] (which was proved in [9] ) we have

v ( t j , β ) = v ( t j , 0 ) + ∫ 0 β     u β ( t j , r ) d r = − ∫ 0 β     u t ( t j , r ) d r < 0.

At the right end t j + 1 , u ( t j + 1 , β ) = 0 and u t ( t j + 1 , β ) < 0 , similarly,

v ( t j + 1 , β ) = v ( t j + 1 , 0 ) + ∫ 0 β     u β ( t j + 1 , r ) d r = − ∫ 0 β     u t ( t j + 1 , r ) d r > 0.

Because v ( t , β ) has opposite signs at two ends of I j , there surely exists some inner point t ′ = t ′ ( β ) such that v ( t ′ , β ) = 0 . Then | v ( t , β ) | is valley and { | u | , | v | } form a peak-valley structure. There is a positive lower bound independent of t ∈ I j .

min t ∈ I j ( | u ( t , β ) | + | v ( t , β ) | ) = μ j ( β ) > 0 ,     β ∈ ( 0 , 1 / 2 ] .

So RC holds in I j . Because the zeros { t j } of analytical function u ( t , β ) do not have finite condensation point, then any finite t surely falls in some I j . RC holds for any t. □

To prove u ( t ) to be single peak, we discussed monotone growth of the argument ϕ + ψ of ξ in [9] and used Riemann’s estimate ψ ( t ) = π S ( t ) = O ( ln t ) . As ϕ ( t ) = t 2 ( ln t − ln ( 2 e π ) ) + 7 8 π + O ( t − 1 ) is of super-linear growth, when t increases to t + 1 , the increment δ ϕ = 1 2 ( ln t − ln ( 2 π ) ) + O ( t − 1 ) slowly increases,

whereas δ ψ = O ( 1 ) , then claimed that ϕ + ψ monotone increases. But here δ ψ = O ( 1 ) is not correct, which may be O ( ln t ) . We consider Riemann’s symmetrization

Z ( t ) = e i ϕ ( t ) ζ ( 1 / 2 + i t ) ,     ζ ( 1 / 2 + i t ) = U ( t ) + i V ( t ) ,

(here the decay factor | G ( s ) | ≠ 0 is reduced) and have

Z ( t ) = U cos ϕ − V sin ϕ ,     g ( t ) = U sin ϕ + V cos ϕ ≡ 0 ,     V U = − sin ϕ cos ϕ .

Thus the local argument π S ( t ) = arctan ( V / U ) = − ϕ in root-interval I j = [ t j , t j + 1 ] .

We see in Figure 2 (also see [6] [10] ) that S ( t ) jumps by 1 at zero t ′ j of cos ϕ , then linearly decreases in ( t ′ j , t ′ j + 1 ) with the slope ψ ′ ≈ − 1 2 ln t . Whereas ϕ has slope ϕ ′ ( t ) ≈ 1 2 ln t . It seems ϕ ′ + ψ ′ ≈ 0 , and difficult to prove ϕ ′ + ψ ′ > 0 . Besides, as g ( t ) = 0 , this research is not suitable.

This defect makes us turn to u ( t ) on critical line and re-investigate RC2.

By RC2, take logarithm ln u ( t ) and derivation to get

ln u ( t ) = ln u ( 0 ) + ∑ j = 1 ∞ ln ( 1 − t 2 t j 2 ) ,     u ′ ( t ) u ( t ) = ∑ j = 1 ∞ 2 t t 2 − t j 2 . (3.1)

Consider root-interval I n = [ t n , t n + 1 ] and decompose the summation into three parts.

u t ( t ) = 2 t u ( t ) { ∑ j ≤ n − 1 1 t 2 − t j 2 + ( 1 t 2 − t n 2 − 1 t n + 1 2 − t 2 ) − ∑ j ≥ n + 2 1 t j 2 − t 2 } , = 2 t u ( t ) g ( t ) ,         g ( t ) = g 1 ( t ) + g 2 ( t ) + g 3 ( t ) ,     t n < t < t n + 1 . (3.2)

Note that u ( t ) inside I n has same sign. We investigate each sum

{ g 1 ( t ) = ∑ j ≤ n − 1 1 t 2 − t j 2 > 0 , t > t n > t j , g 2 ( t ) = − ∑ j ≥ n + 2 1 t j 2 − t 2 < 0 , t < t n + 1 < t j . g 3 ( t ) = 1 t 2 − t n 2 − 1 t n + 1 2 − t 2 > 0 , t n < t < t n + 1 . (3.3)

For t ∈ I n , g 1 ( t ) > 0 , g 2 ( t ) < 0 are finite. If t is close to t n + 0 , then g 3 ( t ) tends to + ∞ . If t is close to t n + 1 − 0 , then g 3 ( t ) tends to − ∞ . There surely exists some inner point t * ∈ I n such that g ( t * ) = 0 . We show this point t * is unique. For this we consider their derivatives

{ g ′ 1 ( t ) = − ∑ j ≤ n − 1 2 t ( t 2 − t j 2 ) 2 < 0 , t > t n > t j , g ′ 2 ( t ) = − ∑ j ≥ n + 2 ∞ 2 t ( t j 2 − t 2 ) 2 < 0 , t < t n + 1 < t j . g ′ 3 ( t ) = − { 2 t ( t 2 − t n 2 ) 2 + 2 t ( t n + 1 2 − t 2 ) 2 } < 0 , t n < t < t n + 1 . (3.4)

Thus all g 1 , g 2 , g 3 are monotone decreasing for t ∈ I n , their sum g ( t ) does yet. Therefore this zero t * of g ( t ) is unique and then u ( t ) is single peak. This proves theorem 3. □

Using Lagarias’s positivity [11] (RC is assumed), we get monotone growth [8], p. 344,

| ξ ( β + i t ) | > | ξ ( β 0 + i t ) | ≥ 0 ,     for     β > β 0 .

This is a clear description for RC. Therefore Riemann ξ -function has mathematical beauty: The symmetry, single peak and monotone growth (i.e. the ordering).

In large scale computations [6] [12] [13] [14] all zeros of u ( t ) are single, no multiple zeros are found. People believe there are only single zeros, but so far do not prove. We have to skirt round the difficulty to prove theorem 2. We shall extend theorem 2 (old) as:

Theorem 2. If u ( t ) has m-ple zeros on critical line, then { u , v } for small β > 0 will bifurcate into m alternative oscillations with single peak, and RC still holds.

Proof. Assume that there are three consecutive zeros { t j − 1 , t j , t j + 1 } of u ( t ) on critical line { β = 0 , 0 < t < ∞ } , which form two root-intervals, and t j = t * is a m-ple zero, m ≥ 1 . Denote H = min ( t j + 1 − t j , t j − t j − 1 ) and h = H / 2 . Denoting y = t − t * and the origin O ( y = 0 , β = 0 ) , and fixing a small β > 0 , we discuss a circle K δ ( t * ) with r = y 2 + β 2 ≤ δ ≤ h ( δ to be defined). Assume that the real function g ( y ) = u ( t ) has m-ple zero at O

g ( y ) = a ( t * ) y m + b ( t * ) y m + 1 + ⋯ ,     a ( t * ) ≠ 0, (3.5)

These coefficients a ( t * ) and b ( t * ) are of same order. Thus

g ( y − i β ) = a ( t * ) { ( y − i β ) m + K ( t * ) ( y − i β ) m + 1 } ,     K ≈ b ( t * ) / a ( t * ) ≠ 0 , (3.6)

Below we discuss P ( y − i β ) = ( y − i β ) m , and temporarily omit high-order remainder.

Using y − i β = r ( y / r − i β / r ) = r ( cos ϕ − i sin ϕ ) and De Moivre formula we have

( y − i β ) m = r m ( cos ϕ − i sin ϕ ) m = r m ( cos ( m ϕ ) − i sin ( m ϕ ) ) = 0 ,

and discuss r m cos ( m ϕ ) = 0 and − r m sin ( m ϕ ) = 0 respectively.

1) The zeros of cos ( m ϕ ) satisfy m ϕ j = ( 2 j − 1 ) π / 2 , i.e.

ϕ j = 2 j − 1 2 m π ,       0 < ϕ j < π ,       j = 1 , 2 , ⋯ , m ,

which are symmetric with respect to π / 2 . If m = 2 n even, then all ϕ j ≠ π / 2 . If m = 2 n − 1 odd, the middle argument ϕ n = π / 2 satisfies cos ( ϕ n ) = 0 , i.e. original point y = 0 .

Taking the roots y / r = cos ϕ j , i.e. y 2 = ( y 2 + β 2 ) cos 2 ϕ j , we have y / β = ± cos ϕ j / sin ϕ j = ± cot ϕ j and re-arrange the ordering of these m zeros as

y / β = { − cot ( ϕ 1 ) , ⋯ , − cot ( ϕ n − 1 ) , 0 ,   cot ( ϕ n − 1 ) , ⋯ , cot ( ϕ 1 ) } , if     m = 2 n − 1. y / β = { − cot ( ϕ 1 ) , ⋯ , − cot ( ϕ n ) , cot ( ϕ n ) , ⋯ , cot ( ϕ 1 ) } , if     m = 2 n , (3.7)

2) The zeros of sin ( m ψ ) satisfy m ψ k = k π , i.e.,

ψ k = k m π ,       k = 1 , 2 , ⋯ , m − 1 ,

here no k = m , as ψ m = π corresponds a trivial zero β / r = sin ( π ) = 0 (i.e. β = 0 ). Thus there are only m − 1 nontrivial zeros, whose arguments are symmetric with respect to π / 2 . When m = 2 n even, ψ n = π / 2 for k = n , i.e. sin ( ψ n ) = 1 , which corresponds the original y = 0 .

Taking the roots β / r = sin ψ k , i.e. β 2 = ( y 2 + β 2 ) sin 2 ψ k , we have y / β = ± cos ( ψ k ) / sin ( ψ k ) = ± cot ( ψ k ) and re-arrange the ordering of these m − 1 roots as

y / β = { − cot ( ψ 1 ) , ⋯ , − cot ( ψ n − 1 ) , cot ( ψ n − 1 ) , ⋯ , cot ( ψ 1 ) } , if     m = 2 n − 1 , y / β = { − cot ( ψ 1 ) , ⋯ , − cot ( ψ n − 1 ) , 0 , cot ( ψ n − 1 ) , ⋯ , cot ( ψ 1 ) } , if     m = 2 n . (3.8)

Comparing (3.7) and (3,8) we see that for fixing β > 0 , the maximum of these roots is | y | = β cot ( ϕ 1 ) = β cot ( π / 2 m ) , thus the radius of circle K δ ( t * ) satisfies r = ( y 2 + β 2 ) 1 / 2 = β / sin ( π / 2 m ) ≤ h , i.e. we should confine β ≤ h sin ( π / 2 m ) < h π / 2 m = H h π / 4 m = d .

For g ( y ) = a ( t * ) ( y − i β ) m = u 1 + i v 1 , we have the following conclusions:

1) The real part u 1 ( y , β ) has m zeros y / β = cot ϕ j , and the imaginary part v 1 ( y , β ) has m − 1 zeros y / β = cot ψ k . Due to

ϕ j = 2 j − 1 2 m π < ψ j = j m π ,   j = 1 , 2 , ⋯ , n − 1 ,

obviously cot ϕ j > cot ψ j . Hence all zeros of (3.7) and (3.8) are alternatively arranged.

2) At zeros ψ j = j π / m of v 1 , the peak values u 1 = a ( t * ) r m cos ( j π ) = a ( t * ) r m ( − 1 ) j alternatively change their signs. At zeros ϕ j = ( j − 1 / 2 ) π / m of u 1 , the peak values v 1 = a ( t * ) r m sin ( ( j − 1 / 2 ) π ) = a ( t * ) r m ( − 1 ) j − 1 also alternatively change their signs.

3) Because v 1 ≠ 0 at zero of u 1 and u 1 ≠ 0 at zero of v 1 , they all are single peak. Thus in the m root-intervals of u 1 , all { | u 1 | , | v 1 | } form local peak-valley structures, and RH locally holds.

Finally, in K δ ( t * ) with β ∈ ( 0, δ ] suitably small we discuss a general case

g ( y − i β ) = a ( t * ) { ( y − i β ) m + R m } ,     R m = O ( r m + 1 ) ,     r ≤ δ , (3.9)

The actual zeros of ξ ( τ ) are small perturbations of these zeros mentioned above, which do not change these m peak-valley structures in K δ ( t * ) . When β increases, they will continue to develop toward locally convex direction by

theorem 1, so RC holds. Hence theorem 2 is proved. □

Numerical experiments 2. With the suggestion of Dr.XM Jiao, we have computed ( ξ ) m = u + i v , m = 2 , 3 at the second zero t 2 = 21.0220 of ξ ( s ) . We see in Figure 3 for β = 0.05 , 0.1 , u ( t , β ) indeed bifurcate into m curves of single peak, and { u , v } are alternative oscillation. The peak of u ( t , β ) develops toward its convex direction.

Remark 2. The author of this paper should sincerely thank Dr. Xiangmin Jiao (Stone Brook University, US). He, on 22 April in 2021, sent e-mail to me to discuss the highest super-convergence (see [15] ), we know for the first time. I told him I’m studying RC, and brings special interests and discuss together. He has carefully verified my papers and proposed valuable comments. He suggested the example ( ξ ( s ) ) m and sent papers [10] [16] to me. If no support from him, I very hard, at least in a shorter time, propose the newest proof.