Upsilon function

From HandWiki
File:Upsilon-function.png
Modular Surface of The Upsilon function

The Upsilon function, Υ(s), is a function of a complex variable s that analytically continues the sum of the infinite series

[math]\displaystyle{ \Upsilon(s) =\sum_{n=1}^\infty \frac 1 {n^2+s^2} }[/math]

for [math]\displaystyle{ s\neq ki, k\in\mathbb{Z}-\{0\} }[/math]

It was suggested in 2016 by Mohamad Emami and Alireza Jamali, analogous to the Riemann zeta function.

Definition

The original Upsilon function is defined as

[math]\displaystyle{ \Upsilon(s) =\sum_{n=1}^\infty \frac 1 {n^2+s^2} = \frac \pi {2s} \coth(\pi s) - \frac 1 {2s^2} }[/math]

For all complex numbers [math]\displaystyle{ s\in\mathbb{C}-\{ki \in\mathbb{Z}\mid k\neq 0\} }[/math].

The Upsilon function Υ(s) is a meromorphic function of a complex variable [math]\displaystyle{ s }[/math], which has isolated singularities at points [math]\displaystyle{ ki }[/math] where [math]\displaystyle{ k }[/math] is a non-zero integer. At point [math]\displaystyle{ s=0 }[/math] the function has a removable singularity; one can readily remove this singular point by assigning the value [math]\displaystyle{ \zeta(2)=\frac{\pi^2}{6} }[/math] ( see Basel problem ).

The Upsilon function is actually Generalized Upsilon function of order π when the generalized Upsilon function of order [math]\displaystyle{ \nu }[/math] is defined as

[math]\displaystyle{ \Upsilon_\nu(s)=\sum_{n=1}^\infty \frac 1 {s^2 + (\frac{n\pi}\nu)^2} = \frac \nu {2s}\coth(\nu s)-\frac{1}{2s^2} }[/math]

Singularities

Anyway, we know that at [math]\displaystyle{ s=0 }[/math] the function has a removable singularity and we can remove it as mentioned above.

Roots and Upsilon conjecture

It was conjectured by Emami and Jamali that All roots of the Upsilon function have the form of [math]\displaystyle{ s=ix }[/math] for all [math]\displaystyle{ x\in\mathbb{R}-\{0\} }[/math] satisfying the equation [math]\displaystyle{ \pi x=\tan\pi x }[/math] .

So far, it is not known whether all roots are of the above form or not, leaving it as an open problem (Upsilon conjecture).

Specific values

  • [math]\displaystyle{ \Upsilon_1(\log\sqrt{x})=\sum_{n=1}^\infty \frac 1 {(\log\sqrt{x})^2 +(n\pi)^2} = \frac 1 {2\log x} \left(\frac{x+1}{x-1} - \frac 1 {\log x}\right) }[/math]
Known as the Jamali's identity, first appeared in a paper by Alireza Jamali and Mohamad Emami.
  • [math]\displaystyle{ \Upsilon_\pi(0)=\zeta(2)=\frac{\pi^2}6. }[/math] See Basel problem.
  • [math]\displaystyle{ \Upsilon_1(\log\varphi) = \sum_{n=1}^\infty \frac 1 {(\log\varphi)^2 + (n\pi)^2} = \frac 1 {2\log\varphi} \left(\sqrt{5}-\frac 1 {\log\varphi}\right), }[/math] where φ is the Golden Ratio (Jamali's series).