Let xi(s) be the Riemann Xi function, let g(s)=pi^-s/2(s-1)Gamma(s/2+1), and let sigma_a(n) denote the generalized divisor function. For s_0,a∈ C, choose sigma satisfying sigma>max(|operatornameRe(a)|-2,operatornameRe(s_0),operatornameRe(1-s_0)). Generalized Xi identity conjecture. For all s_0∈ C and all a∈ C, xi(s_0)xi(s_0-a)=frac12 pi i∑_n=1^∞int_operatornameRe(s)=sigmasigma_a(n)[fracg(s)g(s-a)s-s_0frac1n^s+fracg(s)g(s+a)s-1+s_0frac1n^s+a]ds. The identity is proved in the paper only in selected cases, notably positive odd a among 1,3,5; the conjecture extends it to arbitrary complex a and s_0.

Complex-parameter extension of the Riemann Xi exponential-sum identity

About 3 years old · traced to 2023

Let $xi(s)$ be the Riemann Xi function, let $g(s)=\pi^{-s/2}(s-1)\Gamma(s/2+1)$ , and let $\sigma_a(n)$ denote the generalized divisor function. For $s_0,a\in\mathbb C$ , choose $\sigma$ satisfying

\sigma>\max\left(|\operatorname{Re}(a)|-2,\operatorname{Re}(s_0),\operatorname{Re}(1-s_0)\right).

Generalized Xi identity conjecture. For all $s_0\in\mathbb C$ and all $a\in\mathbb C$ ,

\xi(s_0)\xi(s_0-a)=\frac{1}{2\pi i}\sum_{n=1}^{\infty}\int_{\operatorname{Re}(s)=\sigma}\sigma_a(n)\left[\frac{g(s)g(s-a)}{s-s_0}\frac{1}{n^s}+\frac{g(s)g(s+a)}{s-1+s_0}\frac{1}{n^{s+a}}\right]ds.

The identity is proved in the paper only in selected cases, notably positive odd $a$ among $1,3,5$ ; the conjecture extends it to arbitrary complex $a$ and $s_0$ .

References

Primary source

Maria Nastasescu, Nicolas Robles, Bogdan Stoica and Alexandru Zaharescu, “The Riemann zeta function and exact exponential sum identities of divisor functions”, arXiv:2311.07657 (2023).

Posted 1 month ago

Progress summary

Refreshed 1 month ago

Claimed solved

A reader-written argument claims a complete proof for every complex parameter, but the claim has not been independently verified and the original paper proves only a few special cases.

The conjecture, posed in Nastasescu, Robles, Stoica, and Zaharescu (2023), asserts the stated exponential-sum identity for every $s_0,a\in\mathbb C$ satisfying the contour condition.

Known results

The identity is proved for $a=1$ and $a=3$ .
The case $a=5$ is proved under the additional restriction $\sigma>3$ .
These cases arise from the paper's exact contour-integral identities for positive odd integers $a$ .

All are recorded in the source paper; the arbitrary-complex extension is explicitly labeled a conjecture.

Posted attempt

A reader-written argument claims a complete proof for arbitrary $a,s_0\in\mathbb C$ , by shifting contours, interchanging the divisor-function series with the integrals, and applying the functional equation of $\xi$ . The attempt has not been independently verified.

Current status (as of August 2026): the special cases $a=1,3$ and restricted $a=5$ case are settled, while the general conjecture has only an unverified complete-proof claim.

Sources

ProofThis solution needs a summarySee full solutionHide full solution

Write

g(s)=\pi^{-s/2}(s-1)\Gamma\!\left(\frac s2+1\right), \qquad \xi(s)=g(s)\zeta(s), \qquad \sigma_a(n)=\sum_{d\mid n}d^a.

The completed function $\xi$ is entire and satisfies

\xi(s)=\xi(1-s).

Fix arbitrary $a,s_0\in\mathbb C$ and an admissible contour

\sigma>\max\{|\operatorname{Re}a|-2,\operatorname{Re}s_0,1-\operatorname{Re}s_0\}.

Choose

\sigma'> \max\{1+|\operatorname{Re}a|,\operatorname{Re}s_0,1-\operatorname{Re}s_0\}.

For each fixed $n$ , the poles of the two summands in the proposed integral lie among

s=s_0,\quad s=1-s_0,\quad s=-2,-4,\ldots,\quad s=a-2,a-4,\ldots,\quad s=-a-2,-a-4,\ldots.

Consequently none lies in the closed strip between the original line and the line $\operatorname{Re}s=\sigma'$ . Stirling's estimate gives exponential decay on horizontal segments, so each individual summand integral is unchanged by shifting its contour to $\sigma'$ .

On the shifted line, both divisor Dirichlet series converge absolutely. Interchanging summation and integration and using

\sum_{n\ge1}\frac{\sigma_a(n)}{n^s} =\zeta(s)\zeta(s-a), \qquad \sum_{n\ge1}\frac{\sigma_a(n)}{n^{s+a}} =\zeta(s+a)\zeta(s),

the proposed right-hand side becomes

\frac1{2\pi i}\int_{\operatorname{Re}s=\sigma'} \left( \frac{\xi(s)\xi(s-a)}{s-s_0} + \frac{\xi(s)\xi(s+a)}{s-1+s_0} \right)\,ds.

Put

F(s)=\xi(s)\xi(s-a).

In the second integral, substitute $t=1-s$ . The functional equation gives

\xi(1-t)\xi(1-t+a)=\xi(t)\xi(t-a)=F(t),

while the denominator becomes $-(t-s_0)$ . Accounting for the vertical orientation, the expression is therefore

\frac1{2\pi i} \left( \int_{\operatorname{Re}s=\sigma'}\frac{F(s)}{s-s_0}\,ds - \int_{\operatorname{Re}s=1-\sigma'}\frac{F(s)}{s-s_0}\,ds \right).

The function $F$ is entire, has exponential decay in fixed vertical strips, and

1-\sigma'<\operatorname{Re}s_0<\sigma'.

The residue theorem consequently evaluates the difference as

\operatorname*{Res}_{s=s_0}\frac{F(s)}{s-s_0} =F(s_0) =\xi(s_0)\xi(s_0-a).

This proves the conjectured integral representation for every complex $a,s_0$ and every admissible contour.