_{1}

^{*}

This paper proves Riemann conjecture (RH),
*i.e*., that all the zeros in critical region of Riemann
* ξ *-function lie on symmetric line

*σ*=1/2 . Its proof is based on two important properties: the symmetry and alternative oscillation for

*=*

*ξ**u*+

*iv*. Denote

*v*≡ 0 for

*β*=0 (the symmetry). We prove that the zeros of u and v for

*β*> 0 are alternative, so

*u*(

*t*,0) is the single peak. A geometric model was proposed.

*u*(

*t*,

*β*) , if |

*u*|

*>*0 is inside

*I*

_{j }and

*u*= 0 is at its two ends. If |

*u*(

*t*,

*β*)| has only one peak on each

*I*

_{j}, which is called the single peak, else called multiple peaks (it will be proved that the multiple peaks do not exist). The important expressions of u and v for

*β*> 0 were derived. By

*u*(

*t*,

*β*) will develop toward its convex direction. Besides,

*u*(

_{t}*t*,

*β*) has opposite signs at two ends

*t*=

*t*

_{j}_{ },

*t*of

_{j+1}*I*,

_{j }*t*′ such that

*v*(

*t′*,

*β*) = 0. Therefore {|

*u*|,|

*v*|/

*β*} in

*I*

_{j}_{ }form a peak-valley structure such that

*t*∈

*I*

_{j}_{ }(

*i.e*. RH holds in

*I*

_{j}_{ }). As

*u*(

*t*,

*β*) does not have the finite condensation point (unless

*u*=

*cons*

*t*.), any finite t surely falls in some

*I*

_{j}_{ }, then

D. Hilbert [

ζ ( s ) = 1 + 1 2 s + 1 3 s + 1 4 s + ⋯ (1.1)

all have the real part 1/2, except the well-known negative integral real zeros......”

But RH has not been solved in 20^{th} century. Entering new era, S. Smale’s report [

Λ ( s ) : = π − s / 2 Γ ( s 2 ) ζ ( s ) = Λ ( 1 − s ) , (1.2)

Γ being the Gamma function. RH is the assertion that all the zeros of Λ ( s ) are on the line of symmetry for the functional equation, that is on R e ( s ) = 1 / 2 .” Where Λ ( s ) has the symmetry and alternative oscillation, which for the first time turns to ξ . Besides, J. Haglund [

Ξ ( t ) = 2 ∫ 0 ∞ Φ ( u ) cos ( u t ) d u , Φ ( u ) = 2 ∑ n = 1 ∞ ( 2 a n 2 e 9 u / 2 − 3 a n e 5 u / 2 ) e a n e 2 u , (1.3)

which is the first paper to study the equivalence of Ξ and RH. He computed Ξ and proposed a guess: if any part summation has the monotone zeros, then RH holds. He thought that the study of the Ξ -function was the right approach to RH. This is very important.

We have computed the Riemann ξ -function and other continuations of Euler ζ -function, and found that only ξ has the symmetry and alternative oscillation, which intuitively implies RH. Whereas others ζ have no the properties, and proving RH is hopeless.

We reread the original paper of Riemann (see [

1) Riemann converted Euler series (1.1) into analytical function (no symmetry)

ζ ( s ) = π s / 2 Γ − 1 ( s 2 ) { 2 s ( s − 1 ) + ∫ 1 ∞ ( x s / 2 − 1 + x − s / 2 − 1 / 2 ) ψ ( x ) d x } . (1.4)

where ψ ( x ) = ∑ n = 1 ∞ e − n 2 π x is Jacobi’s function.

2) Introduced an entire function (which is a symmetrization)

ξ ( s ) = s ( s − 1 ) 2 π s / 2 Γ ( s 2 ) ζ ( s ) . (1.5)

In critical domain ζ and ξ have the same zeros. Taking s = 1 / 2 + i t , using integration by parts twice and Jacobi equality 1 / 2 + ψ ( 1 ) + 4 ψ ′ ( 1 ) = 0 , Riemann had gotten a real function

ξ ( t ) = 4 ∫ 1 ∞ d [ x 3 / 2 ψ ′ ( x ) ] d x x − 1 / 4 cos ( t 2 ln x ) d x , Im ( ξ ) ≡ 0. (1.6)

Remark 1. In present point of view, using translating τ = s − 1 / 2 = β + i t , it should directly get Riemann’s general formula [

ξ ( s ) = ∫ 1 ∞ ( x τ / 2 + x − τ / 2 ) x − 3 / 4 f ( x ) d x , f ( x ) = 2 x 2 ψ ″ ( x ) + 3 x ψ ′ ( x ) . (1.7)

On critical line β = 0 , ξ = u + i v has the symmetry v ( t , 0 ) = 0 .

3) Riemann said, “The number of roots of ξ ( t ) = 0 whose real parts lie between 0 andT is about = T 2 π ln T 2 π − T 2 π ” (it is proved by Mongoldt, 1905), and

pointed out that, “One finds in fact about this many real roots within these bounds and it is very likely that all of the roots are real. One would of course like to have a rigorous proof of this” (i.e. RH).

4) He guessed a multiplication formula of ξ (it is proved by Hadamard, 1893)

ξ ( t ) = ξ ( 0 ) ∏ j = 1 ∞ ( 1 − t 2 t j 2 ) . (1.8)

From these we see that Riemann had emphasized ξ , rather than ζ .

5) Siegel (1932) found a formula unpublished in Riemann’s manuscript(now called R-S formula, which was derived by ξ , and is large scale computing formula on critical line up to now), and Riemann had already computed the first several roots (due to the inspiration of R-S formula, we shall propose a new computing formula in next paper).

6) Lagarias [

property undiscussed by Riemann), which is the most essential progression since 1932, also the first equivalence to RH for ξ . Its proof requires the properties 3) and 4).

7) C. M. Chen [

Basic theorem (RC). All the zeros of Riemann ξ -function lie on symmetric line.

Therefore this paper has amended the defects and given a complete proof of RC.

Denote β = σ − 1 / 2 , τ = i t + β = s − 1 / 2 . Consider Riemann ξ -integral and its derivatives

{ ξ ( s ) = ∫ 1 ∞ ( x τ / 2 + x − τ / 2 ) x − 3 / 4 f ( x ) d x = u + i v ξ ′ ( s ) = 1 2 ∫ 1 ∞ ( x τ / 2 − x − τ / 2 ) x − 3 / 4 ln ( x ) f ( x ) d x = u β + i v β ξ ″ ( s ) = 1 4 ∫ 1 ∞ ( x τ / 2 + x − τ / 2 ) x − 3 / 4 ln 2 ( x ) f ( x ) d x = u β β + i v β β (2.1)

From decomposing expression

ξ ( s ) = ∫ 1 ∞ { ( x β / 2 + x − β / 2 ) cos ( t 2 ln x ) + i ( x β / 2 − x − β / 2 ) sin ( t 2 ln x ) } x − 3 / 4 f ( x ) d x

we have

Property 1 (symmetry). The u ( t , β ) is an even function of β and v ( t , β ) is an odd function of β . Taking β = 0 , then v ( t ,0 ) ≡ 0 , i.e.

v = 0 , u β = v t = 0 , v β β = − v t t = 0 , u β β β = − u t t β = v t t t = 0 , ⋯ (2.2)

Property 2 (alternative oscillation). All the zeros of u and v for β > 0 are alternative (proved in theorem 3), see

Definition 1 (root-interval). For any β ∈ ( 0, 1 / 2 ] , a subinterval I j = [ t j , t j + 1 ] called root-interval of u, if the real part | u ( t , β ) | > 0 inside I j , and u ( t j , β ) = 0 , u ( t j + 1 , β ) = 0 .

Definition 2 (single peak). If | u | has only one peak in each root-interval I j , called single peak, unless called multiple peaks(we shall prove no multiple peaks in theorem 3).

Using Newton-Leibnitz formula, the symmetry v ( t , 0 ) = 0 , u β ( t , 0 ) = 0 and Cauchy-Riemann conditions u β = v t , v β = − u t , we have

Lemma 1 (the expression of v). The image part v of ξ can be expressed by u t as

v ( t , β ) = v ( t , 0 ) − ∫ 0 β u t ( t , r ) d r , v ( t , 0 ) = 0 , β ∈ ( 0 , 1 / 2 ] , (2.3)

Corollary 1. | v ( t , β ) | / β is uniformly bounded with respect to β ∈ ( 0, 1 / 2 ] .

Definition 3 (peak-valley structure, PVS). For any β ∈ ( 0, 1 / 2 ] , if | u | > 0 inside the root-interval I j = [ t j , t j + 1 ] , and v has opposite signs at two ends, then there surely exists some inner point t ′ ∈ I j such that v ( t ′ , β ) = 0 , and { | u | , | v | / β } form a peak-valley structure, see

Lemma 2 (the expression of u). The real part u of ξ can be expressed by u t t as

u ( t , β ) − u ( t , 0 ) = ∫ 0 β u β ( t , r ) d r = u β ( r − β ) | 0 β − ∫ 0 β u β β ( t , r ) ( r − β ) d r = 0 + d , d = − ∫ 0 β u t t ( t , r ) ( β − r ) d r . (2.4)

Corollary 2. For any β ∈ ( 0, 1 / 2 ] , in the interval of negative peak u ( t , β ) < 0 , with u t t > 0 (convex downward), then u ( t , β ) < u ( t , 0 ) ≤ 0 . In the interval of positive peak u ( t , β ) > 0 , with u t t < 0 (convex upward), then u ( t , β ) > u ( t , 0 ) ≥ 0 . So the peak u ( t , β ) will develop toward convex direction (This is the ordering of u ( t , β ) in β ).

Consider the curve segments ABC and CDE of u ( t , β ) for β ∈ ( 0, 1 / 2 ] , where C is a positive peak, A and E are negative peaks, see

Lemma 2 (the slop u_{t} of curve family). For any β ∈ ( 0, 1 / 2 ] , assume that u ( t , β ) has a positive peak C inside root-interval I j = [ t j ( β ) , t j + 1 ( β ) ] . If u ( t , β ) is from negative peak to positive one, then curve family u ( t , β ′ ) on the left line t = t j ( β ) has positive slope

u t ( t j ( β ) , β ′ ) > 0 , for any β ′ ∈ [ 0 , β ] . (2.5)

If u ( t , β ) is from positive peak to negative one, then curve family u ( t , β ′ ) on the right line t = t j + 1 ( β ) has negative slope u t ( t j + 1 ( β ) , β ′ ) < 0 for any β ′ ∈ [ 0, β ] . Discussing negative peak is similar.

Proof. For β ∈ ( 0, 1 / 2 ] , the slope u t > 0 inside ABC, u t t > 0 inside AB, u t t < 0 inside BC and u t t = 0 at B. Decreasing β ′ ∈ J = ( 0 , β ] , the corresponding curve A ′ B ′ C ′ will continuously vary and the slope u t > 0 inside A ′ B ′ C ′ . To prove u t ( t , β ′ ) > 0 for the curve family, we should avoid two peak-points t A ′ and t C ′ . Denote the upper bound p of t A ′ and the lower bound q of t C ′ by

p = max β ′ ∈ J t A ′ ( β ′ ) , q = min β ′ ∈ J t C ′ ( β ′ ) , for t ∈ I * = ( p , q ) .

Obviously u t ( t , β ′ ) > 0 , β ′ ∈ J inside common interval t ∈ I * = ( p , q ) . We prove the left line t = t j = t j ( β ) ∈ I * , so (2.5) holds.

For this, rewrite the expressions (2.4) for u ( t , β ) and u ( t , β ′ ) and their difference

u ( t , β ) − u ( t , β ′ ) = − { ∫ β ′ β u t t ( t , r ) ( β − r ) d r + ∫ 0 β ′ u t t ( t , r ) ( β − β ′ ) d r } , β > β ′ ≥ 0. (2.6)

Subdivide ABC and A ′ B ′ C ′ into three segments with different convexities. We have

1) If the zero t j ∈ ( max r ∈ J t B ′ ( r ) , t C ) , in which u t t < 0 , by (2.6), we have the ordering 0 = u ( t j , β ) > u ( t j , β ′ ) and t j < t C ′ , see

2) If the zero t j ∈ ( t A , min r ∈ J t B ′ ( r ) ] , in which u t t > 0 , by (2.6), we have the ordering 0 = u ( t j , β ) < u ( t j , β ′ ) and t j > t A ′ , see

3) If the zero t j ∈ [ min r ∈ J t B ′ ( r ) , max r ∈ J t B ′ ( r ) ] , we can not judge their ordering, but obviously t j ∈ I * .

Above three cases prove (2.5). Similarly discuss the curve CEF. □

Remark 2. We in [

We shall regard { u ( t , β ) , v ( t , β ) } as a continuously varying process from β = 0 to β = 1 / 2 . The zeros t j 0 of u ( t ,0 ) form an irregular infinite sequence

0 < t 1 0 = 14.1347 < t 2 0 = 21.0220 < ⋯ < t j − 1 0 < t j 0 < t j + 1 0 < t j + 2 0 < ⋯ → ∞ .

which are single zeros(the double zeros are admitted). For any β ∈ ( 0, 1 2 ] , the zeros t j of u ( t , β ) form an irregular infinite sequence(dependent on β )

0 < t 1 ( β ) < ⋯ < t j − 1 ( β ) < t j ( β ) < t j + 1 ( β ) < t j + 2 ( β ) < ⋯ → ∞ .

The initial I j 0 = [ t j 0 , t j + 1 0 ] continuously becomes to root-interval I j = [ t j , t j + 1 ] . We prove

Basic theorem. All zeros of Riemann ξ -function lie on critical line.

Its proof consists of three theorems as follows.

Theorem 1. If u ( t ,0 ) is single peak, then peak-valley structure and RH are valid.

Proof. Consider a root-interval I 0 = [ t j 0 , t j + 1 0 ] of u ( t ,0 ) . When increasing β > 0 , by corollary 2, the peak u ( t , β ) will develop toward convex direction. Assuming u ( t , β ) > 0 inside I j = [ t j , t j + 1 ] for β ∈ ( 0, 1 / 2 ] , we analyze two cases as follows.

At the left end t j , u ( t j , β ) = 0 and u t ( t j , β ) > 0 , by Lemma 2 the slop u t ( t j , r ) > 0 , r ∈ ( 0 , β ] (author’s remark: in the proof [

{ v ( t j , β ) / β = − 1 β ∫ 0 β u t ( t j , r ) d r < 0 , lim β → + 0 v ( t j , β ) / β = − u t ( t j , 0 ) < 0. (3.1)

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

{ v ( t j + 1 , β ) / β = − 1 β ∫ 0 β u t ( t j + 1 , r ) d r > 0 , lim β → + 0 v ( t j + 1 , β ) / β = − u t ( t j + 1 , 0 ) > 0. (3.2)

They are valid and numerically stable for β ∈ ( 0, 1 / 2 ] .

Because v ( t , β ) has opposite signs at two ends of I j = [ t j , t j + 1 ] , there surely exists some inner point t ′ j = t ′ j ( β ) such that v ( t ′ , β ) = 0 . Then | v ( t , β ) | / β is valley, and { | u | , | v | / β } form a peak-valley structure, see

ϕ ( t , β ) = | u ( t , β ) | + | v ( t , β ) | / β , β ∈ ( 0, 1 / 2 ] , t ∈ I j = [ t j , t j + 1 ] .

which certainly has a positive lower bound independent of t ∈ I j

min t ∈ I j ϕ ( t , β ) = μ j ( β ) > 0 , β ∈ ( 0, 1 / 2 ] . (3.3)

So RH holds in I j . This is a refine local geometric analysis.

As each root-interval I j will repeat the peak-valley structure, we get a positive irregular infinite sequence

μ 1 ( β ) , μ 2 ( β ) , ⋯ , μ j ( β ) , μ j + 1 ( β ) , ⋯ (3.4)

Because all the zeros { t j } of analytical function u ( t , β ) do not have the finite condensation point (unless u ≡ c o n s t . ), then any finite t surely falls in some I j . RH holds for any t. □

Theorem 2. If two roots of u ( t ,0 ) are very close to each other (including double root), then the peak-valley structure for β > 0 and RH still hold.

Proof. Let u ( t ,0 ) have root-interval I j 0 = [ t j 0 , t j + 1 0 ] , u ( t ,0 ) ≥ 0 inside I j and u ( t ,0 ) < 0 outside I j , see

u ( t , β ) = u ( t , 0 ) + d , d = − ∫ 0 β u t t ( t , r ) ( β − r ) d r > 0 , t ∈ I ,

i.e. u ( t , β ) translates u ( t ,0 ) by a positive distance d upward. So u ( t , β ) has a larger root-interval I j = [ t j ( β ) , t j + 1 ( β ) ] ⊃ I j 0 . The u ( t , β ) has a positive peak inside the new interval I j . Besides v ( t , β ) has opposite signs at two ends of I j and surely v ( t , β ) = 0 at some inner point, then | v | / β is valley. Therefore { | u | , | v | / β } form a peak-valley structure and ‖ ξ ‖ > 0 in I j . □

Remark 3. In large scale computation [

that u ( t ,0 ) is single peak, no multiple one.

Theorem 3. u ( t ,0 ) is single peak.

Proof. We know that analytic function f ( s ) = | f ( s ) | ( cos ψ + i sin ψ ) is alternative oscillation with single peak, if ψ monotonously trends to the infinity when t increases. Riemann introduced

ξ ( s ) = 1 2 s ( s − 1 ) G ( s ) ζ ( s ) = u + i v , G ( s ) = π − s / 2 Γ ( s / 2 ) ,

where G ( s ) = | G ( s ) | e i ϕ , its amplitude

ϕ ( t ) = t 2 ln t 2 e π + ( β − 1 2 ) π 4 + O ( t − 1 ) , (3.5)

is super-linearly increasing with t, so G ( s ) is alternative high-frequency oscillation. We recall [

critical rectangle t ≤ T is about N ( T ) = T 2 π ln T 2 e π + O ( ln T ) , its main part is ϕ ( T ) / π , but ζ has only O ( ln T ) . If take an increment Δ t = 1 for larger t, we have Δ ϕ = 1 2 { ln t + 1 − ln ( 2 e π ) } + O ( t − 1 ) > 0 and Δ ψ = O ( 1 ) > 0 , then the amplitude ϕ + ψ monotonously tends to the infinity. As s ( s − 1 ) / 2 is not oscillation function, it is proved that G ( s ) ζ ( s ) = | G ( s ) ζ ( s ) | e i ( ϕ + ψ ) and then ξ are alternative oscillation with single peak for β > 0 . Below prove that its limit value u ( t ,0 ) still is single peak.

By contradiction, assume that u ( t ,0 ) inside root-interval I j 0 = [ t j 0 , t j + 1 0 ] has three positive peaks A, B and C, see

goes against the alternative oscillation and then u ( t ,0 ) is single peak (Author’s remark: in multiple peak case, [

Finally summarizing three theorems above, our basic theorem (RH) is proved.

It is interesting that the proof of RH looks like to solve the Cauchy problem of Cauchy-Riemann system u β = v t , v β = − u t with analytic initial values u ( t , 0 ) , v ( t , 0 ) = 0 on line β = 0 . Follow Riemann, by τ = β + i t = i z , z = t − i β , its solution is ξ ( τ ) = u ( z , 0 ) .

Assume that RH holds, Lagarias (1999) proved the following wonderful result:

Lagarias theorem. RH is equivalent to the positivity R e ( ξ ′ ξ ) > 0 for any β > 0 .

Denoting τ = β + i t and ξ ( τ ) = u + i v , the positivity can be expressed in the form

R e ( ξ ′ ξ ) = R e ( ξ ′ ξ ¯ | ξ | 2 ) = ψ ( t ) | ξ | 2 > 0 , ψ ( t ) = u u β + v v β > 0 , u β = v t , v β = − u t . (4.1)

This result is sharp (can not be improved), which has three important corollaries.

Corollary 3. The peak-valley structure and RH are equivalent.

Proof. Above RH is proved by peak-valley structure. Now assume that RH holds, then the positivity also is valid, we shall prove peak-valley structure. Let I j = [ t j , t j + 1 ] be root-interval of u ( t , β ) , u > 0 , u t t < 0 inside I j , and u t ( t j , β ) > 0 , u t ( t j + 1 , β ) < 0 .

At the left end t j , u = 0 , u t > 0 , then ψ = u v t − v u t = − v u t > 0 leads to v < 0 .

At the right end t j + 1 , u = 0 , u t < 0 , then ψ = u v t − v u t = − v u t > 0 leads to v > 0 .

Then | v | is valley in I j and { | u | , | v | } form peak-valley structure. □

Therefore the peak-valley structure is the exact geometric description to RH.

Corollary 4. If RH holds, then u ( t ,0 ) is single peak.

Proof. If u ( t ,0 ) has three peaks, see

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

At this point t ′ , ψ ( t ) = u v t − v u t = u v t < 0 is contradiction to the positivity. □

So we see that the correctness of our geometric proof of RH can be derived by Lagarias’ theorem.

Corollary 5 (Monotone). If RH holds, then | ξ ( t , β ) | > | ξ ( t , β 0 ) | ≥ 0 for β > β 0 ≥ 0 .

Proof. Using (4.1), the positive integral

| ξ ( t , β ) | 2 − | ξ ( t , β 0 ) | 2 = 2 ∫ β 0 β ( u u β + v v β ) d β > 0 , for β > β 0 ≥ 0 ,

means the monotone. So | ξ ( t , β ) | > 0 for β > 0 is a concise statement of RH. □

Remark 4. We recall that Aristotle, an ancient Greek philosopher and mathematician, thought: “order and symmetry are important elements of beauty”. Therefore we can say, the symmetry and monotone of ξ are mathematical beauty of Riemann conjecture.

The author expresses sincere gratitude to the reviewer for his careful remark, valuable and constructive comments. Besides, I should thank Prof. Zhengtin Hou and Prof. Xinwen Jiang for their precious opinion in the discussion.

The author declares no conflicts of interest regarding the publication of this paper.

Chen, C.M. (2021) Geometric Proof of Riemann Conjecture. Advances in Pure Mathematics, 11, 334-345. https://doi.org/10.4236/apm.2021.114021