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, 1 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 1 = − + 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 1 Z c 1 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 1 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 1 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 j=(m)j > 2π, roughly speaking): 1 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 1 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 1 X em q(1−t) m ζ(j) = lim − + tk − 1 q em q(1−t) dt, (k − j)! n!1 (k − 1)! k! j=2 1 q=1 which can be simplified collapsing the sum over q: ! Z c 1 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 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 1 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 j=(m)j > 2π, though for j=(m)j = 2π one must have <(k) > 1).