A New Approach to the Riemann Hypothesis Summary

A new approach to the Riemann hypothesis

Jose Risomar Sousa November 28, 2020

Abstract The solution of the Riemann hypothesis has eluded the minds of some of the smartest mathematicians. This paper aims to provide some novelties and shed some light on the riddle, though it is not a proof for the problem. First we extend the formula we created m for the polylogarithm, Lik(e ), from the positive integers to the complex half-plane. This gives us an expression for the zeta function valid for <(k) > 1, and for its analytic continuation, valid for <(k) > 0. We then simplify these convoluted formulae to analyze the zeros of the zeta function on the critical strip (0 < <(k) < 1). More importantly, we oﬀer a reformulation of the Riemann hypothesis through a simpler function valid on the entire complex plane whose non-trivial zeros coincide with those of the zeta function.

Summary

1 Introduction 2

2 Partial polylogarithm sums 2 m 2.1 The analytic continuation of Ek (n)...... 3 3 The polylogarithm 4 m 3.1 The analytic continuation of Lik(e ) ...... 5

4 The analytic continuation of ζ(k) 6

5 Simplifying the problem 6 5.1 The real part equation ...... 7 5.2 The imaginary part equation ...... 8

6 Riemann hypothesis reformulation 8 6.1 Particular case, k is integer ...... 9

7 Graphics plotting 10

1 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 functions4.

The zeta function as a Dirichlet series is given by, ∞ X 1 ζ(k) = , jk j=1 and throughout here we use k for the variable instead of the usual s, to keep the same notation used in previous papers released on generalized harmonic progressions3 and related subjects2.

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 alternat- ing 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.

We start from the formula we found for the partial sums of the polylogarithm function at the positive integers, discussed in paper [5], and we create its analytic continuation by re- placing a ﬁnite sum with its general closed-form (actually an integral), which is a continuous function. We then do the same for the polylogarithm. The polylogarithm is a generalization of the zeta function, and it has the advantage of also covering 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 (C → C) to a manageable two-dimensional relation.

2 Partial polylogarithm sums

In [5] we provide the rationale for the derivation of the following formula, which are the partial sums of the polylogarithm series:

n k k X emj emn 1 X (mn)j X mk−j = − + H (n) jk 2nk 2nk j! (k − j)! j j=1 j=0 j=1 mk Z 1 mu + (1 − u)k−1 (emnu − 1) coth du 2(k − 1)! 0 2

It holds for all complex m, provided that k is a positive integer.

2 m Let’s call these partial sums Ek (n).

m 2.1 The analytic continuation of Ek (n) m As incredible as it may seem, if we just replace the ﬁnite sums in the formula of Ek (n) with closed-forms, that happens to be its analytic continuation, so let’s see how we can achieve that.

First we have:

k k n n k X mk−j X m−j X 1 X X (m q)−j H (n) = mk = mk (k − j)! j (k − j)! qj (k − j)! j=1 j=1 q=1 q=1 j=1

Let’s work with just the part that matters:

k X (m q)−j −1 + em q(m q)−k Γ(k + 1, m q) = , (k − j)! k! j=1 where Γ is the incomplete gamma function, which is given by an integral.

Let’s then rewrite the initial sum:

n k n X X (m q)−j X 1 em q Z ∞ = − + tke−t dt (k − j)! k! k!(m q)k q=1 j=1 q=1 m q

We need to transform this last integral to be able to work with it, for example:

Z ∞ Z c ∞ Z 1 Γ(k + 1, m q) = tke−t dt = m q (m q t)ke−m q t dt = (m q − log u)k e−m q du, m q 1 0 where c is the directional angle of the transformation (since m is complex), given by: ( sign <(m), if <(m) 6= 0 c = −i sign =(m), otherwise.

Therefore, if we introduce the constant inside the integral, we obtain:

n Z c ∞ X m q em q(1−t) m q tk em q(1−t) − + dt, k! k! 1 q=1 and collapsing the sum over q we ﬁnally conclude that:

k X m−j m Z c ∞ em(1−t) + n em(1−t)(n+2) − (n + 1)em(1−t)(n+1) H (n) = (tk − 1) dt (k − j)! j k! (em(1−t) − 1)2 j=1 1

3 Let’s transform this integral to get rid of the constant c:

k k! X m−j 1 Z 1 log u 1 + n un+1 − (n + 1)un H (n) = −1 + 1 − du (k − j)! j k! m (1 − u)2 j=1 0

The last piece is given by,

k X (m n)j em n Γ(k + 1, m n) 1 Z 1 = = (m n − log u)k du, j! k! k! j=0 0 which takes us to the very ﬁnal expression:

n k X emj emn 1 Z 1 log u = − m − du jk 2nk 2k! n j=1 0 k Z 1 n+1 n n m 1 k 1 + n u − (n + 1)u − + (m − log u) 2 du k! k! 0 (1 − u) mk Z 1 mu + (1 − u)k−1 (em n u − 1) coth du 2(k − 1)! 0 2

The ﬁrst term on the second line came from a simpliﬁcation of the former integral. This expression should be valid for <(k) > 0 and for any complex m (even 0, if we disregard terms with singularity).

Note that with this formula we can derive a diﬀerent formula for the generalized harmonic numbers with one degree of freedom (m can be any multiple of 2πi, though 0 makes it the simplest).

3 The polylogarithm

As seen previously5, the following expression for the polylogarithm holds for all integer k ≥ 1 and all complex m (except m such that <(m) >= 0 and |=(m)| > 2π, roughly speaking):

∞ k X emj mk mk−1 m X mk−j = − − log − + ζ(j) jk 2k! (k − 1)! 2π (k − j)! j=1 j=2 mk Z 1 mu 2π − (1 − u)k−1 coth − (1 − u) cot πu du 2(k − 1)! 0 2 m

m For the purpose of producing the analytic continuation of Lik(e ), it’s more convenient to start from the above formula and repeat some of the previous steps.

4 m 3.1 The analytic continuation of Lik(e ) For the polylogarithm, we now have:

k ∞ k X mk−j X X (m q)−j ζ(j) = mk , (k − j)! (k − j)! j=2 q=1 j=2 and one can take the limit of the partial sum:

k n X m−j Z c ∞ X em q(1−t) m ζ(j) = lim − + tk − 1 q em q(1−t) dt, (k − j)! n→∞ (k − 1)! k! j=2 1 q=1 which can be simpliﬁed collapsing the sum over q: ! Z c ∞ em(1−t) − em(1−t)(n+1) em(1−t) + n em(1−t)(n+2) − (n + 1)em(1−t)(n+1) lim − + m tk − 1 dt m(1−t) m(1−t) 2 n→∞ 1 (k − 1)! (e − 1) k!(e − 1)

This limit is not trivial. The solution may be found if one assumes that all terms with n cancel out (which is a plausible assumption if, for example, m is a positive real number) and the remaining terms constitute the limit.

It turns out this hunch is only half right, but fortunately it’s possible to ﬁgure out the term that makes up the diﬀerence through a clinical eye, leading to the below:

k X m−j 1 em Z c ∞ k em + (−k − m + m tk)emt ζ(j) = − + dt (k − j)! m(k − 1)! k! m m t 2 j=2 1 (e − e )

It’s best to make a change of variables (u = e−m(t−1)) in order to do away with c and enable the combination of the integrals:

k k X m−j 1 1 Z 1 −k u − m + m−k+1 (m − log (1 − u)) ζ(j) = − + du (k − j)! m(k − 1)! m k! u2 j=2 0

Therefore, when one puts it all together one ﬁnds that:

mk mk−1 e m Li (em) = − − log − + k 2k! (k − 1)! 2π k Z 1 k m k−1 mu 2π 2(k u + m − m(1 − log (1 − u)/m) − (1−u) coth − (1−u) cot πu+ 2 du 2(k − 1)! 0 2 m k m u

This formula is valid for all k such that <(k) > 0 and all complex m (except m such that <(m) >= 0 and |=(m)| > 2π, though for |=(m)| = 2π one must have <(k) > 1). This

5 convergence domain has been thoroughly checked but might still be subject to change.

It’s possible to simplify the above integral with the observation that:

Z 1 2π 2 2 log 2π − (1 − u) cot πu + du = 0 m m u m

This takes us to the ﬁnal and simplest form:

mk mk−1 (1 + log (−m)) Li (em) = − − k 2k! (k − 1)! k Z 1 −k k m k−1 mu 2(1 − m (m − log (1 − u)) − (1 − u) coth + 2 du 2(k − 1)! 0 2 k u 4 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:

∞ 1 1 X (−1)j ζ(k) = η(k) = , 21−k − 1 21−k − 1 jk j=1 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 21−k − 1, 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 ignore its multiplier. The below should hold whenever <(k) > 0:

(iπ)k (iπ)k−1 (1 + log (−iπ)) η(k) = − − 2k! (k − 1)! k Z 1 −k k (iπ) k−1 πu 2(1 − π (π + i log (1 − u)) − −i(1 − u) cot + 2 du 2(k − 1)! 0 2 k u 5 Simplifying the problem

First oﬀ, let’s make η(k) = 0 and simplify the equation:

Z 1 −k k 2i k (1 + log (−iπ)) k−1 πu 2(1 − π (π + i log (1 − u)) − 1 + = −i k(1 − u) cot + 2 du π 0 2 u

Let the integral minus the left-hand side be the function s(k), so that solving η(k) = 0 is equivalent to solving s(k) = 0. We now need to make a transformation, with the intent of

6 separating the real and imaginary parts. Let’s set k = r + i t, expressed in polar form, and change the integral’s variable using the relation log (1 − u) = π tan v, chosen for convenience.

√ t k = r + i t = r2 + t2 exp i arctan r

With that, taking into account the Jacobian of the transformation, our equation becomes:

π(r − 1) − 2 t(1 + log π) π t + 2 r(1 + log π) + i = π π Z π/2 √ πe−π tan v t π (sec v)2(−i r2 + t2 tan exp −π r tan v + i arctan − π t tan v 0 2 r 1 π tan v 2 + csch (1 − exp (−r log cos v + t v + i(−t log cos v − r v)))) dv 2 2

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.

5.1 The real part equation Below we have the equation one can derive for the real part:

π(r − 1) − 2 t(1 + log π) = π Z π/2 √ πe−π tan v t π (sec v)2( r2 + t2 tan exp (−π r tan v) sin arctan − π t tan v 0 2 r 1 π tan v 2 + csch (1 − exp (−r log cos v + t v) cos (t log cos v + r v))) dv 2 2

Any positive odd integer r satisﬁes this equation, when t = 0.

If r + i t is a zero of the zeta function, so is its conjugate, r − i t. 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 t and −t up as follows:

π Z π/2 π tan v 2 2(r−1) = sec v csch 2 − (sec v)r et v cos (t log cos v + r v) + e−t v cos (t log cos v − r v) dv 2 0 2

Let’s call the function on the right-hand side of the equation f(r, t). After this transformation, the positive odd integers r remain zeros of f(r, 0) = 2(r − 1).

7 5.2 The imaginary part equation Likewise, for the imaginary part we have:

π t + 2 r(1 + log π) = π Z π/2 √ πe−π tan v t π (sec v)2(− r2 + t2 tan exp (−π r tan v) cos arctan − π t tan v 0 2 r 1 π tan v 2 + csch exp (−r log cos v + t v) sin (t log cos v + r v)) dv 2 2

Coincidentally, any positive even integer r satisﬁes 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 t and −t, obtaining:

π Z π/2 π tan v 2 2 t = (sec v)r+2 csch et v sin (t log cos v + r v) + e−t v sin (t log cos v − r v) dv 2 0 2

Let’s call the function on the right-hand side of the equation g(r, t). After this transformation, any r satisﬁes g(r, 0) = 0, whereas the roots of f(r, 0) = 2(r −1) are still the positive odd integers r. This means that when t = 0, these transformations have introduced the positive odd integers r as new zeros of the system, which weren’t there before.

6 Riemann hypothesis reformulation

If we take a linear combination of the equations for the real and imaginary parts, such as 2(r − 1) − 2 t i = f(r, t) − i g(r, t), we can turn the system of equations into a simpler single equation: π Z π/2 π tan v 2 cos k v k − 1 = sec v csch 1 − k dv 2 0 2 (cos v)

Going a little further, with a simple transformation (u = tan v) we can deduce the following theorem.

Theorem k is a non-trivial zero of the Riemann zeta function if and only if k is a non-trivial zero of: π Z ∞ π u2 h(k) = 1 − k + csch 1 − (1 + u2)k/2 cos (k arctan u) du 2 0 2

Hence the Riemann hypothesis is the statement that the zeros of h(k) located on the critical strip have <(k) = 1/2.

8 Proof All the roots of h(k) should also be roots of the zeta function, except for the positive odd integers and the trivial zeros of the eta function (1+2πi j/ log 2, for any integer j), though this might not be a certainty since we transformed the equations (that is, there might be other k such that h(k) = 0 but ζ(k) 6= 0).

However, a little empirical research reveals the following relationship between ζ(k) and h(k): 2 k! 21−k − 1 πk − cos ζ(k) = h(k), for all complex k except where undeﬁned. πk 2

This relationship was derived from the observation that h(k) = <(s(k)) for all real k > 0.

Note this functional equation breaks down at the negative integers (k! = ±∞ but ζ(k) = 0 or cos πk/2 = 0, whereas h(k) 6= 0) and at 1. It seems 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 s(k) is <(k) > 0, the domain of h(k) is the whole complex plane.

Per the equation, k = 1 could be a non-trivial zero of the zeta function if it weren’t its pole, so we need to disregard any such zeros that are not on the critical strip.

6.1 Particular case, k is integer It’s possible to simplify h(k) when k is a positive integer, based on the following facts: 1 cos arctan u = √ , and cos (k arctan u) = Tk(cos arctan u), 1 + u2 where Tk(x) is the Chebyshev polynomial of the ﬁrst kind.

Therefore, using the generating function of Tk(x), available in the literature, we can infer that: ∞ X √ k 1 − x x 1 + u2 T (cos arctan u) = , k (1 − x)2 + (x u)2 k=0 from which it’s possible to produce the generating function of h(k), let’s say it’s q(x):

∞ X 2 1 π x2 Z ∞ π u2 u2 xk h(k) = − + csch du 1 − x (1 − x)2 2(1 − x) 2 (1 − x)2 + (x u)2 k=0 0

The k-th derivative of q(x) yields the value of h(k), and obtaining it is not very hard (we just need to decompose the functions in x into 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:

bk/2c q(k)(x) π k! Z ∞ π u2 X (−1)j u2j h(k) = = 1 − k − csch du, k! 2 2 (2j)!(k − 2j)! 0 j=1

9 where bk/2c means the integer division.

In here we went from an expression that holds for all k, to an expression that only holds for k a positive integer, the opposite of analytic continuation.

7 Graphics plotting

First we plot the curves obtained with the imaginary part equation, g(r, t), with values of r starting at 1/8 with 1/8 increments, up to 7/8, for a total of 7 curves plus the 2 t line. 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(r − 1)).

Let’s see what we obtain when we plot these curves with t varying 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.

10 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.

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 t varying from 0 to 26, for further comparison (0 is pink, 1 is green):

11 Now, the graph below shows plots for curves −2(r − 1) + 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).

12 And ﬁnally, graphs for the diﬀerence of the two functions, −2(r − 1) + f(r, t) + 2t − g(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 Generalized Harmonic Numbers, eprint arXiv:1810.07877, 2018 2018arXiv181007877R.

[3] Risomar Sousa, Jose Generalized Harmonic Progression, eprint arXiv:1811.11305, 2018 2018arXiv181111305R.

[4] Risomar Sousa, Jose An Exact Formula for the Prime Counting Function, eprint arXiv:1905.09818, 2019 2019arXiv190509818R.

[5] Risomar Sousa, Jose Lerch’s Φ and the Polylogarithm at the Positive Integers, eprint arXiv:2006.08406, 2020 2020arXiv200608406R.