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A Reformulation of the Riemann Hypothesis

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Abstract

An elementary proof of the Riemann hypothesis was provided in a prior paper. This new paper aims to provide some novelties, unrelated to that solution. Using the analytic continuation we created for the polylogarithm, Li k (e m), we extend the zeta function from (k) > 1 to the half-complex plane, (k) > 0, by means of the Dirichlet eta function. More strikingly, we offer a reformulation of the Riemann hypothesis through a zeta's cousin, ϕ(k), a pole-free function defined on the entire complex plane whose non-trivial zeros coincide with those of the zeta function.
A Reformulation of the Riemann Hypothesis
Jose Risomar Sousa
November 28, 2020
Abstract
We present some novelties on the Riemann zeta function. Using the analytic con-
tinuation we created for the polylogarithm, Lik(em), we extend the zeta function from
<(k)>1 to the complex half-plane, <(k)>0, by means of the Dirichlet eta function.
More strikingly, we oﬀer a reformulation of the Riemann hypothesis through a zeta’s
cousin, ϕ(k), a pole-free function deﬁned on the entire complex plane whose non-trivial
zeros coincide with those of the zeta function.
1 Introduction
The Riemann Hypothesis is a long-standing problem in math, which involves the zeros of
the analytic continuation of its most famous Dirichlet series, the zeta function. This Dirichlet
series, along with its analytic continuation, constitutes a so-called L-function, whose zeros
encode information about the location of the prime numbers. Riemann provided insight into
this connection through his unnecessarily convoluted prime counting functions2.
The zeta function as a Dirichlet series is given by,
ζ(k) =
X
j=1
1
jk,
and throughout here we use kfor the variable instead of the usual s, to keep the same nota-
tion used in previous papers released on generalized harmonic numbers and progressions and
interrelated subjects.
This series only converges for <(k)>1, but it can be analytically continued to the whole
complex plane. For the purpose of analyzing the zeros of the zeta function though, we produce
its analytic continuation on the complex half-plane only, <(k)>0, by means of the alternating
zeta function, known as the Dirichlet eta function, η(k). It’s a well known fact that all the
non-trivial zeros of ζ(k) lie on the critical strip (0 <<(k)<1). The Riemann hypothesis is
then the conjecture that all such zeros have <(k)=1/2. A somewhat convincing argument
for the Riemann hypothesis was given in [4], though it’s reasonable to think it will take some
time for it to be acknowledged or dismissed.
1
We start from the formula we found for the analytic continuation of the polylogarithm
function, discussed in paper [3]. The polylogarithm is a generalization of the zeta function,
and it has the advantage of encompassing the Dirichlet eta function.
We then greatly simplify the convoluted expressions and remove the complex numbers out
of the picture, going from four-dimensional chaos (CC) to a manageable two-dimensional
relation.
2 The polylogarithm, Lik(em)
As seen in [3], the following expression for the polylogarithm holds for all complex k
with positive real part, <(k)>0, and all complex m(except msuch that <(m)>= 0 and
|=(m)|>2π– though for |=(m)|= 2πone must have <(k)>1):
Lik(em) = mk
2k!mk1(1 + log (m))
(k1)!
mk
2(k1)! Z1
0
(1 u)k1coth mu
2+21mk(mlog (1 u))k
k u2du
From this formula, we can derive two diﬀerent formulae for ζ(k), using m= 0 or m= 2πi,
but both are only valid when <(k)>1. Using m= 2πiis eﬀortless, we just need to replace
mwith 2πiin the above. Using m= 0 is not as direct, we need to take the limit of Hk(n)1as
ntends to inﬁnity:
ζ(k) = 1
k!Z1
0
(log u)k
(1 u)2du
2.1 The analytic continuation of ζ(k)
The analytic continuation of the zeta function to the complex half-plane can be achieved
using the Dirichlet eta function, as below:
ζ(k) = 1
121kη(k) = 1
121k
X
j=1
(1)j+1
jk,
which is valid in the only region that matters to the zeta function’s non-trivial zeros, the
critical strip. The exception to this mapping are the zeros of 1 21k, which are also zeros of
η(k), thus yielding an undeﬁned product.
For the purpose of studying the zeros of the zeta function, we can focus only on η(k) and
1A formula derived from the partial sums of Lik(em), as explained in [3].
2
ignore its multiplier. The below should hold whenever <(k)>0:
η(k) = (iπ)k
2k!+(iπ)k1(1 + log (iπ))
(k1)!
+(iπ)k
2(k1)! Z1
0i(1 u)k1cot πu
2+21πk(π+ilog (1 u))k
k u2du
3 Simplifying the problem
Before starting to solve equation η(k) = 0, let’s simplify it by letting h(k) = 2 k!(iπ)kη(k).
Now, η(k) = 0 if and only if h(k) = 0, which implies the below equation:
1 + 2ik(1 + log (iπ))
π=Z1
0ik(1 u)k1cot πu
2+21πk(π+ilog (1 u))k
u2du
Now we need to make another transformation, with the goal of separating the real and
imaginary parts. Let’s set k=r+it, expressed in polar form, and change the integral’s
variable using the relation log (1 u) = πtan v, chosen for convenience.
k=r+it=r2+t2exp iarctan t
r
With that, taking into account the Jacobian of the transformation, our equation becomes:
π(r1) 2t(1 + log π)
π+iπ t + 2 r(1 + log π)
π=
πZπ/2
0
(sec v)2(ir2+t2tan πeπtan v
2exp π r tan v+iarctan t
rπ t tan v
+1
2csch πtan v
22
(1 exp (rlog cos v+t v +i(tlog cos vr v)))) dv
Though this expression is very complicated, it can be simpliﬁed, as we do next. Since the
parameters are real, we can separate the real and imaginary parts.
3.1 The real part equation
Below we have the equation one can derive for the real part:
π(r1) 2t(1 + log π)
π=
πZπ/2
0
(sec v)2(r2+t2tan πeπtan v
2exp (π r tan v) sin arctan t
rπ t tan v
+1
2csch πtan v
22
(1 exp (rlog cos v+t v) cos (tlog cos v+r v))) dv
3
Any positive odd integer rsatisﬁes this equation, when t= 0.
If r+itis a zero of the zeta function, so is its conjugate, rit. Hence, noting that the
ﬁrst term inside of the integral is an odd function in t, we can further simplify the above by
adding the equations for tand tup as follows:
2(r1) = π
2Zπ/2
0sec vcsch πtan v
22
2(sec v)ret v cos (tlog cos v+r v) + et v cos (tlog cos vr v) dv
Let’s call the function on the right-hand side of the equation f(r, t). After this transfor-
mation, the positive odd integers rremain zeros of f(r, 0) = 2(r1).
3.2 The imaginary part equation
Likewise, for the imaginary part we have:
π t + 2 r(1 + log π)
π=
πZπ/2
0
(sec v)2(r2+t2tan πeπtan v
2exp (π r tan v) cos arctan t
rπ t tan v
+1
2csch πtan v
22
exp (rlog cos v+t v) sin (tlog cos v+r v)) dv
Coincidentally, any positive even integer rsatisﬁes this equation when t= 0, so the two
equations (real and imaginary) are never satisﬁed simultaneously for any positive integer.
Now the ﬁrst term inside of the integral is an even function in t, so to simplify it we need
to subtract the equations for tand t, obtaining:
2t=π
2Zπ/2
0
(sec v)r+2 csch πtan v
22
et v sin (tlog cos v+r v) + et v sin (tlog cos vr v)dv
Let’s call the function on the right-hand side of the equation g(r, t). After this transforma-
tion, any rsatisﬁes g(r, 0) = 0, whereas the roots of f(r, 0) = 2(r1) are still the positive odd
integers r. This means that when t= 0, these transformations have introduced the positive
odd integers ras new zeros of the system, which weren’t there before.
4 Riemann hypothesis reformulation
If we take a linear combination of the equations for the real and imaginary parts, such as
2(r1) 2ti=f(r, t)ig(r, t), we can turn the system of equations into a simpler single
equation:
k1 = π
2Zπ/2
0sec vcsch πtan v
221cos k v
(cos v)kdv
4
Going a little further, with a simple transformation (u= tan v) we can deduce the following
theorem.
Theorem kis a non-trivial zero of the Riemann zeta function if and only if kis a non-trivial
zero of:
ϕ(k) = 1 k+π
2Z
0csch π u
221(1 + u2)k/2cos (karctan u)du
Hence the Riemann hypothesis is the statement that the zeros of ϕ(k) located on the crit-
ical strip have <(k) = 1/2.
Proof All the roots of ϕ(k) should also be roots of the zeta function, except for the positive
odd integers and the trivial zeros of the eta function (1+2πij/ log 2, for any integer j), though
this might not be true since we transformed the equations (that is, there might be other k
such that ϕ(k) = 0 but ζ(k)6= 0).
However, a little empirical research reveals the following relationship between ζ(k) and
ϕ(k):
2 Γ(k+ 1) 21k1
πkcos πk
2ζ(k) = ϕ(k), for all complex kexcept where undeﬁned. (1)
This relationship was derived from the observation that ϕ(k) = <(h(k)) for all real k.
Note this functional equation breaks down at the negative integers (k! = ±∞ but ζ(k)=0
or cos πk/2 = 0, whereas ϕ(k)6= 0) and at 1. There is a zeros trade-oﬀ between these two
functions (the negative even integers for the positive odd integers). Note also that while the
convergence domain of h(k) is <(k)>0, the domain of ϕ(k) is the whole complex plane.
If we combine equation (1) with Riemann’s functional equation, we can obtain the following
simpler equivalence, valid for all complex kexcept the zeta pole:
2(k1)(1 2k)ζ(k) = ϕ(1 k),
which in turn implies Riemann’s functional equation when combined with (1).
4.1 Particular values of ϕ(k)when kis integer
From the functional equations, we can easily ﬁnd out the values of ϕ(±k) when kis a
non-negative integer:
ϕ(2k) = 222kB2k
ϕ(2k+ 1) = 0
ϕ(k)=2k12k1ζ(k+ 1)
5
And from these formulae we conclude that for large positive real k,ϕ(k)2k.
One can also create a generating function for ϕ(k), based on the following identities:
cos arctan u=1
1 + u2, and cos (karctan u) = Tk(cos arctan u),
where Tk(x) is the Chebyshev polynomial of the ﬁrst kind.
Therefore, using the generating function of Tk(x) available in the literature, for the positive
integers we have:
X
k=0 x1 + u2kTk(cos arctan u) = 1x
(1 x)2+ (x u)2,
from which it’s possible to produce the generating function of ϕ(k) (let it be q(x)):
X
k=0
xkϕ(k) = 2
1x1
(1 x)2+π x2
2(1 x)Z
0csch π u
22u2
(1 x)2+ (x u)2du
The k-th derivative of q(x) yields the value of ϕ(k), and obtaining it is not very hard (we
just need to decompose the functions in xinto a sum of fractions whose denominators have
degree 1, if the roots are simple – so we can easily generalize their k-th derivative). After we
perform all the calculations and simpliﬁcations we ﬁnd that:
ϕ(k) = q(k)(x)
k!= 1 kπ k!
2Z
0csch π u
22bk/2c
X
j=1
(1)ju2j
(2j)!(k2j)! du,
where bk/2cmeans the integer division.
In here we went from an expression that holds for all k, to an expression that only holds
for ka positive integer, the opposite of analytic continuation.
6
5 Graphics plotting
First we plot the curves obtained with the imaginary part equation, g(r, t), with values of
rstarting at 1/8 with 1/8 increments, up to 7/8, for a total of 7 curves plus the 2tline. The
points where the line crosses the curves are candidates for zeros of the Riemann zeta function
(they also need to satisfy the real part equation, f(r, t) = 2(r1)).
Let’s see what we obtain when we plot these curves with tvarying from 0 to 15. In the
graph below, higher curves have greater r, though not always, below the x-axis it’s vice-versa
– but generally the more outward the curve, the greater the r:
As we can see, it seems the line crosses the curve for r= 1/2 at its local maximum, which
must be the ﬁrst non-trivial zero (that is, its imaginary part). The line also crosses 3 other
curves (all of which have r > 1/2), but these are probably not zeros due to the real part
equation. Also, it seems there must be a line that unites the local maximum points of all the
curves, though that is just a wild guess.
One ﬁrst conclusion is that one equation seems to be enough for r= 1/2, the line seems
to only cross this curve at the zeta zeros. Another conclusion is that apparently curves with
r < 1/2 don’t even meet the ﬁrst requirement, and also apparently r= 1/2 is just right. A
third conclusion is that all curves seem to have the same inﬂection points.
7
In the below graph we plotted g(r, t) for the minimum, middle and maximum points of the
critical strip (0, 1/2 and 1), with tvarying from 0 to 26, for further comparison (0 is pink, 1
is green):
8
Now, the graph below shows plots for curves 2(r1) + f(r, t) and 2t+g(r, t) together.
The plots were created for r= 0 (red), r= 1/2 (green) and r= 1 (blue) (curves with the
same color have the same r). A point is a zero of the zeta function when both graphs cross
the x-axis at the same point (three zeta zeros are shown).
9
And ﬁnally, graphs for the diﬀerence of the two functions, 2(r1) + f(r, t) + 2tg(r, t),
were created for the same r’s as before and with the same colors as before (but now we also
have pink (r= 1/4) and cyan (r= 3/4)).
References
[1] M. Abramowitz, I. A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs,
and Mathematical Tables (9th printing ed.), New York: Dover, 1972.
[2] Risomar Sousa, Jose An Exact Formula for the Prime Counting Function, eprint
arXiv:1905.09818, 2019.
[3] Risomar Sousa, Jose The Lerch’s ΦAnalytic Continuation, eprint arXiv:2101.06835, 2021.
[4] Risomar Sousa, Jose An Elementary Proof of the Riemann Hypothesis, eprint, 2021.
10
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We demonstrate how to obtain the analytic continuation of some formulae that are only valid at the positive integers. This results in alternative expressions for the Lerch Φ and the polylogarithm functions, Φ(e m , k, b) and Li k (e m), respectively, and for their partial sums, valid in the complex half-plane. Similarly, from a formula for the Hurwitz zeta function valid at the negative integers, ζ(−k, b), we obtain its analytic continuation to the whole complex plane. Finally, the generalized harmonic numbers and progressions, H k (n) and HP k (n), respectively, are also extended to the whole complex plane with a new approach based on the Faulhaber formula.
• Risomar Sousa
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