File:Upsilon-function.png Modular Surface of The Upsilon function The Upsilon function , Υ (s )function of a complex variable s that analytically continues the sum of the infinite series
Υ
(
s
)
=
∑
n
=
1
∞
1
n
2
+
s
2
{\displaystyle \Upsilon (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+s^{2}}}}
for
s
≠
k
i
,
k
∈
Z
−
{
0
}
{\displaystyle s\neq ki,k\in \mathbb {Z} -\{0\}}
It was suggested in 2016 by Mohamad Emami and Alireza Jamali, analogous to the Riemann zeta function .
Definition [ edit ] The original Upsilon function is defined as
Υ
(
s
)
=
∑
n
=
1
∞
1
n
2
+
s
2
=
π
2
s
coth
(
π
s
)
−
1
2
s
2
{\displaystyle \Upsilon (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{2}+s^{2}}}={\frac {\pi }{2s}}\coth(\pi s)-{\frac {1}{2s^{2}}}}
For all complex numbers
s
∈
C
−
{
k
i
∈
Z
∣
k
≠
0
}
{\displaystyle s\in \mathbb {C} -\{ki\in \mathbb {Z} \mid k\neq 0\}}
The Upsilon function Υ (s ) is a meromorphic function of a complex variable
s
{\displaystyle s}
k
i
{\displaystyle ki}
k
{\displaystyle k}
s
=
0
{\displaystyle s=0}
removable singularity ; one can readily remove this singular point by assigning the value
ζ
(
2
)
=
π
2
6
{\displaystyle \zeta (2)={\frac {\pi ^{2}}{6}}}
Basel problem ).
The Upsilon function is actually Generalized Upsilon function of order π when the generalized Upsilon function of order
ν
{\displaystyle \nu }
Υ
ν
(
s
)
=
∑
n
=
1
∞
1
s
2
+
(
n
π
ν
)
2
=
ν
2
s
coth
(
ν
s
)
−
1
2
s
2
{\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}}}}
Singularities [ edit ]
“
One can easily see that for
s
=
k
i
,
k
∈
Z
{\displaystyle s=ki,k\in \mathbb {Z} }
”
Anyway, we know that at
s
=
0
{\displaystyle s=0}
removable singularity and we can remove it as mentioned above.
Roots and Upsilon conjecture [ edit ] It was conjectured by Emami and Jamali that All roots of the Upsilon function have the form of
s
=
i
x
{\displaystyle s=ix}
x
∈
R
−
{
0
}
{\displaystyle x\in \mathbb {R} -\{0\}}
π
x
=
tan
π
x
{\displaystyle \pi x=\tan \pi x}
.
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 [ edit ]
Υ
1
(
log
x
)
=
∑
n
=
1
∞
1
(
log
x
)
2
+
(
n
π
)
2
=
1
2
log
x
(
x
+
1
x
−
1
−
1
log
x
)
{\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)}
Known as the Jamali's identity
Υ
π
(
0
)
=
ζ
(
2
)
=
π
2
6
.
{\displaystyle \Upsilon _{\pi }(0)=\zeta (2)={\frac {\pi ^{2}}{6}}.}
Basel problem .
Υ
1
(
log
φ
)
=
∑
n
=
1
∞
1
(
log
φ
)
2
+
(
n
π
)
2
=
1
2
log
φ
(
5
−
1
log
φ
)
,
{\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),}
φ is the Golden Ratio (Jamali's series
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